MTH303 — Real Analysis, samjha ke (LaTeX + Themes)

Chapter 5

Differentiation & MVT

Differentiability at a Point

Definition

Let \( f: A \to \mathbb{R} \). We say \( f \) is differentiable at \( c \in A \) if the limit of the secant slopes exists: \[ f'(c) = \lim_{x \to c} \frac{f(x) - f(c)}{x - c} = \lim_{h \to 0} \frac{f(c+h) - f(c)}{h} \]

Hinglish: Derivative ka matlab hai curve pe do points (secant) lo, aur unhe paas laate jao. Agar left side se aane wali secant aur right side se aane wali secant ek hi tangent line banayein, toh function differentiable hai. Agar graph mein sharp "corner" hai (jaise \( |x| \)), toh tangents clash karenge aur derivative exist nahi karega!

Theorem — Differentiability \( \Rightarrow \) Continuity

If \( f \) is differentiable at \( c \), then \( f \) is continuous at \( c \). The converse is FALSE!

Hinglish: Differentiable → Continuous (hamesha sach). Lekin ulta sach nahi! Example: \( f(x) = |x| \) at \( x = 0 \): continuous hai ✅ (graph mein koi break nahi) lekin differentiable nahi ❌ (sharp corner hai). Toh "smooth" (differentiable) hona "connected" (continuous) hone se zyada strong condition hai.

Rolle's Theorem (MVT ka Special Case)

Theorem — Rolle's Theorem

If \( f \) is continuous on \( [a,b] \), differentiable on \( (a,b) \), and \( f(a) = f(b) \), then there exists at least one point \( c \in (a,b) \) where \( f'(c) = 0 \).

Hinglish: Agar function start aur end pe same height pe hai (f(a) = f(b)), toh beech mein kahin na kahin ek flat point (slope = 0) zaroor hoga! Example: Ek ball ko oopar phenko — wo same height pe wapas aayegi. Toh beech mein kisi point pe ball ki speed exactly 0 hogi (jab wo apni maximum height pe hogi). Mathematical Example: \( f(x) = x^2 - 4x + 3 \) on \( [1, 3] \). \( f(1) = 0, f(3) = 0 \). \( f'(x) = 2x - 4 = 0 \implies x = 2 \). At \( c = 2 \), slope is 0. ✅

The Mean Value Theorem (MVT)

Theorem

If \( f \) is continuous on \( [a,b] \) and differentiable on \( (a,b) \), then there exists at least one point \( c \in (a,b) \) such that: \[ f'(c) = \frac{f(b) - f(a)}{b - a} \]

Hinglish: MVT geometric terms mein kehta hai: interval ke start aur end points ko jodne wali line (secant) ke parallel, graph pe kahin na kahin ek tangent zaroor milegi. Physical terms mein: Agar tumhari average speed 60 km/h thi, toh safar mein kisi ek point pe tumhari exact speedometer speed exactly 60 km/h rahi hogi.

Chapter 6

Quiz 1 Solutions (Visualized)

Complete solutions for SampleQuiz1_MTH303_2026.pdf — har question color boxes mein: Question / Math / Hinglish / Source.

Gold = Question Teal = Math steps Coral = Exam-style answer Violet = Detailed Hinglish Green = Source / theorem

Key Definitions for Quiz Solutions

Theorem — Heine-Borel (The Compactness Criterion)

A subset \( K \subseteq \mathbb{R} \) is compact if and only if it is closed and bounded.

Hinglish: ℝ mein compactness check karna bohot easy hai — bas do sawaal pucho: (1) Kya set bounded hai? (Kya ek dabba mein fit hota hai?) (2) Kya set closed hai? (Kya apne limit points ko contain karta hai?) Agar dono yes, toh compact hai! Examples: \( [0, 1] \) compact ✅. \( (0, 1) \) compact nahi ❌ (closed nahi). \( [0, \infty) \) compact nahi ❌ (bounded nahi). \( \{1, 2, 3\} \) compact ✅ (finite set = closed + bounded).

Definition — Countable vs Uncountable Sets

A set is countable if its elements can be listed in a sequence (put in 1-to-1 correspondence with \( \mathbb{N} \)). A set is uncountable if no such listing exists.

Hinglish: Countable = elements ko ek-ek karke gin sakte ho (1st, 2nd, 3rd...). Uncountable = ginne ki koshish karo lekin koi list complete nahi hogi. Examples:

\( \mathbb{N} = \{1, 2, 3, \dots\} \) — countable ✅ (wo toh counting numbers hi hain!)
\( \mathbb{Z} = \{\dots, -2, -1, 0, 1, 2, \dots\} \) — countable ✅ (list: 0, 1, -1, 2, -2, ...)
\( \mathbb{Q} \) — countable ✅ (surprising! Cantor's diagonal argument se list ban sakti hai)
\( \mathbb{R} \) — uncountable ❌ (Cantor ne prove kiya ki koi bhi list mein ek real number zaroor chhoot jayega)
\( (0, 1) \) — uncountable ❌ (har open interval mein utne hi elements hain jitne poore \( \mathbb{R} \) mein!)

Definition — Clopen Set (Open + Closed)

A set that is simultaneously open and closed is called clopen. In \( \mathbb{R} \), the only clopen sets are \( \emptyset \) and \( \mathbb{R} \) itself (because \( \mathbb{R} \) is connected).

Hinglish: Clopen = open bhi aur closed bhi. Sunne mein ajeeb lagta hai, lekin \( \emptyset \) (khaali set) aur \( \mathbb{R} \) (poori real line) dono clopen hain. ℝ mein yahi do iklaute clopen sets hain — yahi fact quiz mein A1 mein aata hai!

Section A: True or False

Question · A1 · Clopen subsets of ℝ

True or False: If \( A \subseteq \mathbb{R} \) is both open and closed in \( \mathbb{R} \), then \( A = \emptyset \) or \( A = \mathbb{R} \).

English Mathematical Steps — full marks checklist

Recall: A space \( X \) is connected if it cannot be written as \( U \cup V \) with ( U,V ) nonempty, disjoint, and both open in \( X \). Equivalently, the only clopen subsets of a connected space are \( \emptyset \) and \( X \).
Fact: \( \mathbb{R} \) (with the standard topology) is connected. (Any two points can be joined by the interval between them; an open split would break that.)
Conclusion: Therefore the only subsets of \( \mathbb{R} \) that are simultaneously open and closed are \( \emptyset \) and \( \mathbb{R} \). The statement is TRUE.

Exam-style answer (copy into answer sheet)

Given / Statement. We are asked to decide whether the following is true or false: If \( A \subseteq \mathbb{R} \) is both open and closed in \( \mathbb{R} \) (with the standard topology), then \( A = \emptyset \) or \( A = \mathbb{R} \).

To prove / To find. Determine the truth value and justify using the definition of connectedness and the connectedness of \( \mathbb{R} \).

Step 1 (Definition). A topological space \( X \) is connected if it cannot be written as \( X = U \cup V \) where \( U \) and \( V \) are nonempty, disjoint, and both open in \( X \). An equivalent formulation: the only subsets of \( X \) that are simultaneously open and closed (clopen) are \( \emptyset \) and \( X \) itself.

Step 2 (Known fact). The real line \( \mathbb{R} \) with the usual topology is connected. (Intuitively, one cannot split \( \mathbb{R} \) into two nonempty disjoint open pieces; any such split would disconnect the interval between points in different pieces.)

Step 3 (Apply to clopen sets). Suppose \( A \subseteq \mathbb{R} \) is clopen. Then \( A \) and \( \mathbb{R} \setminus A \) are both open and disjoint, and their union is \( \mathbb{R} \). If both were nonempty, \( \mathbb{R} \) would be a union of two nonempty disjoint open sets, contradicting connectedness. Hence at least one of \( A \) or \( \mathbb{R} \setminus A \) is empty, i.e. \( A = \emptyset \) or \( A = \mathbb{R} \).

Final answer (box this). The statement is TRUE: the only clopen subsets of \( \mathbb{R} \) are \( \emptyset \) and \( \mathbb{R} \).

Method. Present the solution as a proof or counterexample with numbered steps. Each step must state the definition or theorem used and why the next line follows — no “obvious” jumps.

Marks tip. Underline the final boxed answer. For TRUE/FALSE, include a one-line justification name (e.g. connectedness, counterexample, EVT). Missing quantifiers or confused open/closed/compact costs marks.

Hinglish — detailed (examiner-safe, beginner-clear)

Pehle question samjho: Bolo: koi set \( A \) real line mein open bhi hai aur closed bhi — kya wo sirf khaali ya poori line ho sakta hai? Answer sheet pe pehle likho kya verify karna hai: clopen sets ki poori list.

Step 1 — Connected definition: Connected ka matlab: space ko do alag nonempty open tukdon mein tod nahi sakte. Clopen set wahi hota hai jo open bhi closed bhi — agar dono \( A \) aur uska complement open hon, toh do tukde ban gaye. Isliye connected space mein clopen sirf \( \emptyset \) aur poora space.

Step 2 — \( \mathbb{R} \) connected: Real line connected hai (course fact). Exam mein ek line likh do: “\( \mathbb{R} \) is connected in the standard topology.” Iske bina conclusion weak lagega.

Step 3 — Conclusion likhna: Agar \( A \) clopen hai aur beech ka koi proper subset hota, dono sides open + disjoint mil jate — contradiction. Isliye sirf extreme cases. Final line: TRUE, aur examples mat bhoolo: \( \emptyset \), \( \mathbb{R} \) dono clopen hain.

Marks tip: Definition of connected + clopen characterization + “\( \mathbb{R} \) is connected” — teenon likhe bina full marks risky.

Exam mindset: Copy-paste feel mat do — har step pe “definition → apply → conclude” likho. English math symbols same rakh sakte ho; explanation Hinglish mein full sentences mein.

Common cutters: Definition skip, “obviously” bina reason, ∀/∃ galat, compact vs closed confuse, image vs pre-image confuse — in sab se bacho.

Final line (marks save): End pe clearly likho jo construct/prove maanga tha — ek alag “Hence …” sentence. Counterexample ho toh pehle example define, phir verify, phir conclude.

Exam mindset: Copy-paste feel nahi dena; har line pe “definition → apply → conclude” dikhana. Neeche wala write-up answer sheet pe utaar sakte ho (Hinglish soch + English math symbols).

Exam Step 1 (Hinglish): Recall: A space \( X \) is connected if it cannot be written as \( U \cup V \) with ( U,V ) nonempty, disjoint, and both open in \( X \). Equivalently, the only clopen subsets of a connected space are \( \emptyset \) and \( X \).
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — Recall: A space \( X \) is connected if it cannot be written as \( U \cup V \) with ( U,V ) nonempty, disjoint, and both… ko full sentence mein expand karke likho.

Exam Step 2 (Hinglish): Fact: \( \mathbb{R} \) (with the standard topology) is connected. (Any two points can be joined by the interval between them; an open split would break that.)
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — Fact: \( \mathbb{R} \) (with the standard topology) is connected. (Any two points can be joined by the interval between … ko full sentence mein expand karke likho.

Exam Step 3 (Hinglish): Conclusion: Therefore the only subsets of \( \mathbb{R} \) that are simultaneously open and closed are \( \emptyset \) and \( \mathbb{R} \). The statement is TRUE.
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — Conclusion: Therefore the only subsets of \( \mathbb{R} \) that are simultaneously open and closed are \( \emptyset \) a… ko full sentence mein expand karke likho.

Final line (marks save): End pe clearly likho TRUE/FALSE ya jo construct/prove maanga tha — ek alag sentence: “Hence …”. Agar counterexample hai toh pehle example fully define karo, phir check karo, phir conclude.

Common cutters: (1) definition skip (2) “obviously” without reason (3) wrong quantifiers ∀/∃ (4) compact vs closed confuse (5) image vs pre-image confuse. Inse bacho.

Source / Idea kahan se aayi

Uses: definition of connected space and the theorem “\( \mathbb{R} \) is connected”. Course topic: Topology of \( \mathbb{R} \) / connectedness.

A2.

Question · A2 · Arbitrary intersections of open sets

True or False: The intersection of an arbitrary family of open subsets of \( \mathbb{R} \) is open.

English Mathematical Steps — full marks checklist

What is true: Finite intersections of open sets are open. Arbitrary (possibly infinite) unions of open sets are open.
Counterexample: Let \( A_n = \big(-\frac{1}{n}, \frac{1}{n}\big) \) for \( n \in \mathbb{N} \). Each ( A_n ) is open.
Compute the intersection: \( \bigcap_{n=1}^{\infty} A_n = \{0\} \). Why? Any \( x \neq 0 \) has ( |x| > 0 ), so choose ( n > 1/|x| ); then \( x \notin A_n \). Only \( 0 \) lies in every ( A_n ).
Conclusion: \( \{0\} \) is not open in \( \mathbb{R} \) (no open interval around \( 0 \) stays inside \( \{0\} \)). Hence the statement is FALSE.

Exam-style answer (copy into answer sheet)

Given / Statement. True or False: The intersection of an arbitrary (possibly infinite) family of open subsets of \( \mathbb{R} \) is open.

To prove / To find. Disprove or prove; if false, give a counterexample with full verification.

Step 1 (What topology guarantees). In any topological space, finite intersections of open sets are open, and arbitrary unions of open sets are open. The statement here is about arbitrary intersections — this is not an axiom of general topology.

Step 2 (Counterexample setup). Define \( A_n = \big(-\frac{1}{n}, \frac{1}{n}\big) \) for each \( n \in \mathbb{N} \). Each \( A_n \) is open in \( \mathbb{R} \) (open interval).

Step 3 (Compute the intersection). Claim: \( \bigcap_{n=1}^{\infty} A_n = \{0\} \). If \( x \neq 0 \), then \( |x| > 0 \); choose \( n > 1/|x| \). Then \( 1/n < |x| \), so \( x \notin (-1/n, 1/n) = A_n \). Thus \( x \) is not in the intersection. Clearly \( 0 \in A_n \) for all \( n \). Hence the intersection is exactly \( \{0\} \).

Step 4 (Show \( \{0\} \) is not open). If \( \{0\} \) were open, there would exist \( \varepsilon > 0 \) with \( (-\varepsilon, \varepsilon) \subseteq \{0\} \), impossible since \( \varepsilon/2 \neq 0 \) lies in the interval but not in \( \{0\} \). So \( \{0\} \) is not open.

Final answer. The statement is FALSE: an arbitrary intersection of open sets in \( \mathbb{R} \) need not be open.

Method. Present the solution as a proof or counterexample with numbered steps. Each step must state the definition or theorem used and why the next line follows — no “obvious” jumps.

Hinglish — detailed (examiner-safe, beginner-clear)

Question: Infinite open sets ka intersection hamesha open — sahi ya galat? Yaad rakho finite intersection open hota hai; infinite pe trap hai.

Step 2: Intervals \( (-1/n, 1/n) \) har n pe open hain aur zero ke around shrink hote hain — standard counterexample.

Step 3: Intersection sirf \( 0 \) bachta hai kyunki koi nonzero x eventually interval se bahar ho jata hai. Ye calculation likhna zaroori hai.

Step 4: Single point open nahi — koi bhi interval us point ke around aur numbers bhi contain karta hai.

Final: FALSE bold karke counterexample ka naam likh do.

Exam mindset: Copy-paste feel mat do — har step pe “definition → apply → conclude” likho. English math symbols same rakh sakte ho; explanation Hinglish mein full sentences mein.

Common cutters: Definition skip, “obviously” bina reason, ∀/∃ galat, compact vs closed confuse, image vs pre-image confuse — in sab se bacho.

Final line (marks save): End pe clearly likho jo construct/prove maanga tha — ek alag “Hence …” sentence. Counterexample ho toh pehle example define, phir verify, phir conclude.

Exam mindset: Copy-paste feel nahi dena; har line pe “definition → apply → conclude” dikhana. Neeche wala write-up answer sheet pe utaar sakte ho (Hinglish soch + English math symbols).

Exam Step 1 (Hinglish): What is true: Finite intersections of open sets are open. Arbitrary (possibly infinite) unions of open sets are open.
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — What is true: Finite intersections of open sets are open. Arbitrary (possibly infinite) unions of open sets are open. ko full sentence mein expand karke likho.

Exam Step 2 (Hinglish): Counterexample: Let \( A_n = \big(-\frac{1}{n}, \frac{1}{n}\big) \) for \( n \in \mathbb{N} \). Each ( A_n ) is open.
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — Counterexample: Let \( A_n = \big(-\frac{1}{n}, \frac{1}{n}\big) \) for \( n \in \mathbb{N} \). Each ( A_n ) is open. ko full sentence mein expand karke likho.

Exam Step 3 (Hinglish): Compute the intersection: \( \bigcap_{n=1}^{\infty} A_n = \{0\} \). Why? Any \( x \neq 0 \) has ( |x| > 0 ), so choose ( n > 1/|x| ); then \( x \notin A_n \). Only \( 0 \) lies in every ( A_n ).
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — Compute the intersection: \( \bigcap_{n=1}^{\infty} A_n = \{0\} \). Why? Any \( x \neq 0 \) has ( |x| > 0 ), so choos… ko full sentence mein expand karke likho.

Exam Step 4 (Hinglish): Conclusion: \( \{0\} \) is not open in \( \mathbb{R} \) (no open interval around \( 0 \) stays inside \( \{0\} \)). Hence the statement is FALSE.
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — Conclusion: \( \{0\} \) is not open in \( \mathbb{R} \) (no open interval around \( 0 \) stays inside \( \{0\} \)). Henc… ko full sentence mein expand karke likho.

Common cutters: (1) definition skip (2) “obviously” without reason (3) wrong quantifiers ∀/∃ (4) compact vs closed confuse (5) image vs pre-image confuse. Inse bacho.

Source / Idea kahan se aayi

Uses: definition of open set; standard axioms of topology (finite ∩ of opens is open; infinite ∩ need not be). Course topic: Topology of \( \mathbb{R} \).

Interactive — A2

Question · A3 · Continuous images of closed sets

True or False: If \( f:\mathbb{R}\to\mathbb{R} \) is continuous and \( A\subset\mathbb{R} \) is closed, then \( f(A) \) is closed.

English Mathematical Steps — full marks checklist

Claim to test: Continuous maps send closed sets to closed sets. This is false without extra hypotheses (e.g. compactness).
Counterexample: Take \( f(x)=e^{x} \) (continuous on \( \mathbb{R} \)) and \( A=\mathbb{R} \) (closed in \( \mathbb{R} \)).
Image: \( f(A)=f(\mathbb{R})=(0,\infty) \). The set \( (0,\infty) \) does not contain its limit point \( 0 \), so it is not closed.
Conclusion: Statement is FALSE. (By contrast: continuous image of a compact set is compact, hence closed and bounded.)

Exam-style answer (copy into answer sheet)

Given / Statement. True or False: If \( f:\mathbb{R}\to\mathbb{R} \) is continuous and \( A\subset\mathbb{R} \) is closed, then \( f(A) \) is closed in \( \mathbb{R} \).

To prove / To find. Test the claim; if false, exhibit continuous \( f \) and closed \( A \) with non-closed image.

Step 1. Continuous functions do not always send closed sets to closed sets (unlike pre-images of closed sets). A standard repair: if \( A \) is compact, then \( f(A) \) is compact, hence closed and bounded in \( \mathbb{R} \).

Step 2 (Counterexample). Let \( f(x) = e^{x} \), which is continuous on \( \mathbb{R} \). Let \( A = \mathbb{R} \), which is closed in \( \mathbb{R} \).

Step 3 (Image). Then \( f(A) = f(\mathbb{R}) = (0, \infty) \). The point \( 0 \) is a limit point of \( (0,\infty) \) (e.g. \( 1/n \to 0 \)) but \( 0 \notin (0,\infty) \), so \( (0,\infty) \) is not closed.

Final answer. The statement is FALSE.

Method. Present the solution as a proof or counterexample with numbered steps. Each step must state the definition or theorem used and why the next line follows — no “obvious” jumps.

Hinglish — detailed (examiner-safe, beginner-clear)

Idea: Continuous + closed domain se image closed — ye automatically nahi hota. Compact case alag theorem hai.

Example: \( e^x \) poori line (closed) ko positive open ray pe bhejti hai — \( 0 \) missing limit point.

Exam line: Counterexample define → image compute → limit point check → FALSE.

Exam mindset: Copy-paste feel mat do — har step pe “definition → apply → conclude” likho. English math symbols same rakh sakte ho; explanation Hinglish mein full sentences mein.

Common cutters: Definition skip, “obviously” bina reason, ∀/∃ galat, compact vs closed confuse, image vs pre-image confuse — in sab se bacho.

Final line (marks save): End pe clearly likho jo construct/prove maanga tha — ek alag “Hence …” sentence. Counterexample ho toh pehle example define, phir verify, phir conclude.

Exam mindset: Copy-paste feel nahi dena; har line pe “definition → apply → conclude” dikhana. Neeche wala write-up answer sheet pe utaar sakte ho (Hinglish soch + English math symbols).

Exam Step 1 (Hinglish): Claim to test: Continuous maps send closed sets to closed sets. This is false without extra hypotheses (e.g. compactness).
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — Claim to test: Continuous maps send closed sets to closed sets. This is false without extra hypotheses (e.g. compactness… ko full sentence mein expand karke likho.

Exam Step 2 (Hinglish): Counterexample: Take \( f(x)=e^{x} \) (continuous on \( \mathbb{R} \)) and \( A=\mathbb{R} \) (closed in \( \mathbb{R} \)).
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — Counterexample: Take \( f(x)=e^{x} \) (continuous on \( \mathbb{R} \)) and \( A=\mathbb{R} \) (closed in \( \mathbb{R} \… ko full sentence mein expand karke likho.

Exam Step 3 (Hinglish): Image: \( f(A)=f(\mathbb{R})=(0,\infty) \). The set \( (0,\infty) \) does not contain its limit point \( 0 \), so it is not closed.
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — Image: \( f(A)=f(\mathbb{R})=(0,\infty) \). The set \( (0,\infty) \) does not contain its limit point \( 0 \), so it is … ko full sentence mein expand karke likho.

Exam Step 4 (Hinglish): Conclusion: Statement is FALSE. (By contrast: continuous image of a compact set is compact, hence closed and bounded.)
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — Conclusion: Statement is FALSE. (By contrast: continuous image of a compact set is compact, hence closed and bounded.) ko full sentence mein expand karke likho.

Common cutters: (1) definition skip (2) “obviously” without reason (3) wrong quantifiers ∀/∃ (4) compact vs closed confuse (5) image vs pre-image confuse. Inse bacho.

Source / Idea kahan se aayi

Uses: counterexample method; contrast with “continuous image of compact = compact” (Heine–Borel / EVT family). Course: Continuity + closed/compact sets.

Question · A4 · Is sup A always a limit point?

True or False: If a nonempty set \( A\subseteq\mathbb{R} \) is bounded above, then \( \sup A \) is a limit point of \( A \).

English Mathematical Steps — full marks checklist

Definitions: \( s=\sup A \) is the least upper bound. A point \( p \) is a limit point of \( A \) if every neighborhood of \( p \) contains a point of \( A \) different from \( p \).
Counterexample: Let \( A=\{1\} \). Then \( A \) is nonempty and bounded above, and \( \sup A=1 \).
Check limit point: Take neighborhood \( (1-\tfrac12,1+\tfrac12) \). The only point of \( A \) inside is \( 1 \) itself. So \( 1 \) is an isolated point, not a limit point.
Conclusion: Statement is FALSE. (If \( A \) has no maximum, then \( \sup A \) is a limit point — but that extra assumption is needed.)

Exam-style answer (copy into answer sheet)

Given / Statement. True or False: If a nonempty set \( A\subseteq\mathbb{R} \) is bounded above, then \( \sup A \) is a limit point of \( A \).

To prove / To find. Decide truth; recall definitions of supremum and limit point.

Step 1 (Definitions). \( s = \sup A \) is the least upper bound. A point \( p \) is a limit point (accumulation point) of \( A \) if every neighborhood of \( p \) contains some point of \( A \) different from \( p \).

Step 2 (Counterexample). Take \( A = \{1\} \). Then \( A \neq \emptyset \), bounded above, and \( \sup A = 1 \).

Step 3. Neighborhood \( (1 - \tfrac12, 1 + \tfrac12) \) contains only the point \( 1 \) from \( A \); no other point of \( A \) is arbitrarily close. So \( 1 \) is an isolated point, not a limit point.

Final answer. The statement is FALSE. (Remark: if \( A \) has no maximum element, then \( \sup A \) is a limit point — extra hypothesis.)

Method. Present the solution as a proof or counterexample with numbered steps. Each step must state the definition or theorem used and why the next line follows — no “obvious” jumps.

Hinglish — detailed (examiner-safe, beginner-clear)

Trap: Supremum hamesha limit point nahi — singleton pe sup wahi point hai, isolated.

Definitions: Limit point = har neighborhood mein dusra point of A.

Final: FALSE; agar max nahi hai tab sup often limit point — ye remark marks ke liye accha.

Exam mindset: Copy-paste feel mat do — har step pe “definition → apply → conclude” likho. English math symbols same rakh sakte ho; explanation Hinglish mein full sentences mein.

Common cutters: Definition skip, “obviously” bina reason, ∀/∃ galat, compact vs closed confuse, image vs pre-image confuse — in sab se bacho.

Final line (marks save): End pe clearly likho jo construct/prove maanga tha — ek alag “Hence …” sentence. Counterexample ho toh pehle example define, phir verify, phir conclude.

Exam mindset: Copy-paste feel nahi dena; har line pe “definition → apply → conclude” dikhana. Neeche wala write-up answer sheet pe utaar sakte ho (Hinglish soch + English math symbols).

Exam Step 1 (Hinglish): Definitions: \( s=\sup A \) is the least upper bound. A point \( p \) is a limit point of \( A \) if every neighborhood of \( p \) contains a point of \( A \) different from \( p \).
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — Definitions: \( s=\sup A \) is the least upper bound. A point \( p \) is a limit point of \( A \) if every neighborhood … ko full sentence mein expand karke likho.

Exam Step 2 (Hinglish): Counterexample: Let \( A=\{1\} \). Then \( A \) is nonempty and bounded above, and \( \sup A=1 \).
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — Counterexample: Let \( A=\{1\} \). Then \( A \) is nonempty and bounded above, and \( \sup A=1 \). ko full sentence mein expand karke likho.

Exam Step 3 (Hinglish): Check limit point: Take neighborhood \( (1-\tfrac12,1+\tfrac12) \). The only point of \( A \) inside is \( 1 \) itself. So \( 1 \) is an isolated point, not a limit point.
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — Check limit point: Take neighborhood \( (1-\tfrac12,1+\tfrac12) \). The only point of \( A \) inside is \( 1 \) itself. … ko full sentence mein expand karke likho.

Exam Step 4 (Hinglish): Conclusion: Statement is FALSE. (If \( A \) has no maximum, then \( \sup A \) is a limit point — but that extra assumption is needed.)
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — Conclusion: Statement is FALSE. (If \( A \) has no maximum, then \( \sup A \) is a limit point — but that extra assumpti… ko full sentence mein expand karke likho.

Common cutters: (1) definition skip (2) “obviously” without reason (3) wrong quantifiers ∀/∃ (4) compact vs closed confuse (5) image vs pre-image confuse. Inse bacho.

Source / Idea kahan se aayi

Uses: definitions of supremum, limit point, isolated point. Course: Completeness / topology of \( \mathbb{R} \).

Question · A5 · Must every point of a closed set be a limit point?

True or False: Every point of a closed set \( A \) is a limit point of \( A \).

English Mathematical Steps — full marks checklist

True fact about closed sets: \( A \) is closed iff \( A \) contains all of its limit points (i.e. \( A' \subseteq A \)).
What the statement claims (wrongly): every point of \( A \) is itself a limit point (i.e. \( A \subseteq A' \)). That would mean \( A \) has no isolated points.
Counterexample: \( A=\{1,2\} \) is finite, hence closed in \( \mathbb{R} \). Both \( 1 \) and \( 2 \) are isolated (small enough neighborhoods contain no other points of \( A \)).
Conclusion: Statement is FALSE. Closed ⇒ “no missing limit points”, not “no isolated points”.

Exam-style answer (copy into answer sheet)

Given / Statement. True or False: Every point of a closed set \( A \subseteq \mathbb{R} \) is a limit point of \( A \).

To prove / To find. Clarify closed vs “every point is accumulation point”.

Step 1. \( A \) is closed iff \( A \) contains all its limit points: \( A' \subseteq A \). This does not say every point of \( A \) is in \( A' \).

Step 2 (Counterexample). Let \( A = \{1, 2\} \). Finite sets in \( \mathbb{R} \) are closed. Points \( 1 \) and \( 2 \) are isolated (small intervals contain no other points of \( A \)).

Final answer. The statement is FALSE. Closed means no missing limit points, not “no isolated points”.

Method. Present the solution as a proof or counterexample with numbered steps. Each step must state the definition or theorem used and why the next line follows — no “obvious” jumps.

Method. We solve this in the exam format: state definitions used, justify each implication, and end with a clear final line.

Step 1. True fact about closed sets: \( A \) is closed iff \( A \) contains all of its limit points (i.e. \( A' \subseteq A \)). (Write this justification fully on the answer sheet — do not skip definitions or “why”.)

Step 2. What the statement claims (wrongly): every point of \( A \) is itself a limit point (i.e. \( A \subseteq A' \)). That would mean \( A \) has no isolated points. (Write this justification fully on the answer sheet — do not skip definitions or “why”.)

Step 3. Counterexample: \( A=\{1,2\} \) is finite, hence closed in \( \mathbb{R} \). Both \( 1 \) and \( 2 \) are isolated (small enough neighborhoods contain no other points of \( A \)). (Write this justification fully on the answer sheet — do not skip definitions or “why”.)

Step 4. Conclusion: Statement is FALSE. Closed ⇒ “no missing limit points”, not “no isolated points”. (Write this justification fully on the answer sheet — do not skip definitions or “why”.)

Final answer (box this). Conclusion: Statement is FALSE. Closed ⇒ “no missing limit points”, not “no isolated points”.

Marks tip. Examiner cuts marks for missing definitions, unjustified jumps, and unclear TRUE/FALSE or final claim. Keep every “because” explicit.

Hinglish — detailed (examiner-safe, beginner-clear)

Confusion clear karo: Closed = saare limit points andar; ye sab points limit points hon iska matlab nahi.

Example: \( \{1,2\} \) closed hai lekin dono isolated.

Final: FALSE.
Exam mindset: Copy-paste feel mat do — har step pe “definition → apply → conclude” likho. English math symbols same rakh sakte ho; explanation Hinglish mein full sentences mein.
Common cutters: Definition skip, “obviously” bina reason, ∀/∃ galat, compact vs closed confuse, image vs pre-image confuse — in sab se bacho.
Final line (marks save): End pe clearly likho jo construct/prove maanga tha — ek alag “Hence …” sentence. Counterexample ho toh pehle example define, phir verify, phir conclude.

Exam mindset: Copy-paste feel nahi dena; har line pe “definition → apply → conclude” dikhana. Neeche wala write-up answer sheet pe utaar sakte ho (Hinglish soch + English math symbols).

Exam Step 1 (Hinglish): True fact about closed sets: \( A \) is closed iff \( A \) contains all of its limit points (i.e. \( A' \subseteq A \)).
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — True fact about closed sets: \( A \) is closed iff \( A \) contains all of its limit points (i.e. \( A' \subseteq A \)). ko full sentence mein expand karke likho.

Exam Step 2 (Hinglish): What the statement claims (wrongly): every point of \( A \) is itself a limit point (i.e. \( A \subseteq A' \)). That would mean \( A \) has no isolated points.
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — What the statement claims (wrongly): every point of \( A \) is itself a limit point (i.e. \( A \subseteq A' \)). That wo… ko full sentence mein expand karke likho.

Exam Step 3 (Hinglish): Counterexample: \( A=\{1,2\} \) is finite, hence closed in \( \mathbb{R} \). Both \( 1 \) and \( 2 \) are isolated (small enough neighborhoods contain no other points of \( A \)).
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — Counterexample: \( A=\{1,2\} \) is finite, hence closed in \( \mathbb{R} \). Both \( 1 \) and \( 2 \) are isolated (smal… ko full sentence mein expand karke likho.

Exam Step 4 (Hinglish): Conclusion: Statement is FALSE. Closed ⇒ “no missing limit points”, not “no isolated points”.
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — Conclusion: Statement is FALSE. Closed ⇒ “no missing limit points”, not “no isolated points”. ko full sentence mein expand karke likho.

Final line (marks save): End pe clearly likho TRUE/FALSE ya jo construct/prove maanga tha — ek alag sentence: “Hence …”. Agar counterexample hai toh pehle example fully define karo, phir check karo, phir conclude.

Common cutters: (1) definition skip (2) “obviously” without reason (3) wrong quantifiers ∀/∃ (4) compact vs closed confuse (5) image vs pre-image confuse. Inse bacho.

Source / Idea kahan se aayi

Uses: characterization of closed sets via limit points; finite sets are closed. Course: Topology of \( \mathbb{R} \).

Section B: Examples & Calculations

Question · B1 · Sequence Analysis

Let \( A=\{(-1)^{n}(1+\frac{1}{n}) : n \in \mathbb{N}\} \). Find \( \sup A \), \( \inf A \), the set of limit points, the closure \( \overline{A} \), and decide whether \( A \) is compact.

English Mathematical Steps — full marks checklist

Write the first few terms carefully. For even \( n=2k \): \( a_{2k}=+(1+\frac{1}{2k}) \), so the positive subsequence is \( \frac{3}{2},\frac{5}{4},\frac{7}{6},\ldots \) which decreases toward \( 1 \).

Odd terms. For odd \( n=2k-1 \): \( a_{2k-1}=-(1+\frac{1}{2k-1}) \), so \( -2,-\frac{4}{3},-\frac{6}{5},\ldots \) increases toward \( -1 \).

Supremum / maximum. The largest value among all terms is the first positive term \( a_2=\frac{3}{2} \). Hence \( \sup A=\max A=\frac{3}{2} \).

Infimum / minimum. The smallest value is the first odd term \( a_1=-2 \). Hence \( \inf A=\min A=-2 \).

Limit points. Every neighborhood of \( 1 \) contains infinitely many even terms, and every neighborhood of \( -1 \) contains infinitely many odd terms. No other real number is approached by a subsequence. So the derived set is \( A'=\{-1,1\} \).

Important check: Is \( 1\in A \)? Solve \( (-1)^n(1+1/n)=1 \). For even \( n \), \( 1+1/n=1 \) forces \( 1/n=0 \), impossible. Similarly \( -1\notin A \). So the limit points are not in \( A \).

Closure. \( \overline{A}=A\cup A'=A\cup\{-1,1\} \).

Compactness (Heine–Borel). \( A \) is bounded (lies in \( [-2,3/2] \)) but not closed (missing \( \pm 1 \)). Therefore \( A \) is not compact.

Exam-style answer (copy into answer sheet)

Given. Let \( A = \{ (-1)^{n}(1 + \frac{1}{n}) : n \in \mathbb{N} \} \).

To find. Find \( \sup A \), \( \inf A \), the set \( A' \) of limit points, the closure \( \overline{A} \), and decide whether \( A \) is compact.

Step 1 (Terms). For even \( n = 2k \): \( a_{2k} = +(1 + \frac{1}{2k}) \), decreasing toward \( 1 \). For odd \( n = 2k-1 \): \( a_{2k-1} = -(1 + \frac{1}{2k-1}) \), increasing toward \( -1 \).

Step 2 (Extrema). Largest term: \( a_2 = 3/2 \). Smallest: \( a_1 = -2 \). Hence \( \sup A = \max A = 3/2 \), \( \inf A = \min A = -2 \).

Step 3 (Limit points). Even subsequence \( \to 1 \), odd subsequence \( \to -1 \). No other accumulation (all terms lie in \( [-2, 3/2] \) and cluster only at \( \pm 1 \)). So \( A' = \{-1, 1\} \).

Step 4 (Membership check). \( 1 \notin A \) (even \( n \) would need \( 1/n = 0 \)); similarly \( -1 \notin A \).

Step 5 (Closure and compactness). \( \overline{A} = A \cup A' = A \cup \{-1, 1\} \). \( A \) is bounded but not closed (missing \( \pm 1 \)), so by Heine–Borel \( A \) is not compact.

Final answer. \( \sup A = 3/2 \), \( \inf A = -2 \), \( A' = \{-1,1\} \), \( \overline{A} = A \cup \{-1,1\} \), \( A \) is not compact.
Method. Present the solution as a proof or counterexample with numbered steps. Each step must state the definition or theorem used and why the next line follows — no “obvious” jumps.
Marks tip. Underline the final boxed answer. For TRUE/FALSE, include a one-line justification name (e.g. connectedness, counterexample, EVT). Missing quantifiers or confused open/closed/compact costs marks.

Hinglish — detailed (examiner-safe, beginner-clear)

Sequence pattern: Even n → positive terms \( 1 + 1/n \to 1 \); odd n → negative \( -(1+1/n) \to -1 \).

Sup/Inf: Pehla even max \( 3/2 \), pehla odd min \( -2 \) — explicitly terms likho.

Limit points: \( \pm 1 \) par infinite terms crowd; verify \( \pm 1 \) set mein nahi.

Compact: Bounded hai, closed nahi → not compact. Heine–Borel cite karo.
Exam mindset: Copy-paste feel mat do — har step pe “definition → apply → conclude” likho. English math symbols same rakh sakte ho; explanation Hinglish mein full sentences mein.
Common cutters: Definition skip, “obviously” bina reason, ∀/∃ galat, compact vs closed confuse, image vs pre-image confuse — in sab se bacho.
Final line (marks save): End pe clearly likho jo construct/prove maanga tha — ek alag “Hence …” sentence. Counterexample ho toh pehle example define, phir verify, phir conclude.

Exam mindset: Copy-paste feel nahi dena; har line pe “definition → apply → conclude” dikhana. Neeche wala write-up answer sheet pe utaar sakte ho (Hinglish soch + English math symbols).

Exam Step 1 (Hinglish): Write the first few terms carefully. For even \( n=2k \): \( a_{2k}=+(1+\frac{1}{2k}) \), so the positive subsequence is \( \frac{3}{2},\frac{5}{4},\frac{7}{6},\ldots \) which decreases toward \( 1 \).
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — Write the first few terms carefully. For even \( n=2k \): \( a_{2k}=+(1+\frac{1}{2k}) \), so the positive subsequence is… ko full sentence mein expand karke likho.

Exam Step 2 (Hinglish): Odd terms. For odd \( n=2k-1 \): \( a_{2k-1}=-(1+\frac{1}{2k-1}) \), so \( -2,-\frac{4}{3},-\frac{6}{5},\ldots \) increases toward \( -1 \).
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — Odd terms. For odd \( n=2k-1 \): \( a_{2k-1}=-(1+\frac{1}{2k-1}) \), so \( -2,-\frac{4}{3},-\frac{6}{5},\ldots \) increa… ko full sentence mein expand karke likho.

Exam Step 3 (Hinglish): Supremum / maximum. The largest value among all terms is the first positive term \( a_2=\frac{3}{2} \). Hence \( \sup A=\max A=\frac{3}{2} \).
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — Supremum / maximum. The largest value among all terms is the first positive term \( a_2=\frac{3}{2} \). Hence \( \sup A=… ko full sentence mein expand karke likho.

Exam Step 4 (Hinglish): Infimum / minimum. The smallest value is the first odd term \( a_1=-2 \). Hence \( \inf A=\min A=-2 \).
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — Infimum / minimum. The smallest value is the first odd term \( a_1=-2 \). Hence \( \inf A=\min A=-2 \). ko full sentence mein expand karke likho.

Exam Step 5 (Hinglish): Limit points. Every neighborhood of \( 1 \) contains infinitely many even terms, and every neighborhood of \( -1 \) contains infinitely many odd terms. No other real number is approached by a subsequence. So the derived set is \( A'=\{-1,1\} \).
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — Limit points. Every neighborhood of \( 1 \) contains infinitely many even terms, and every neighborhood of \( -1 \) cont… ko full sentence mein expand karke likho.

Exam Step 6 (Hinglish): Important check: Is \( 1\in A \)? Solve \( (-1)^n(1+1/n)=1 \). For even \( n \), \( 1+1/n=1 \) forces \( 1/n=0 \), impossible. Similarly \( -1\notin A \). So the limit points are not in \( A \).
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — Important check: Is \( 1\in A \)? Solve \( (-1)^n(1+1/n)=1 \). For even \( n \), \( 1+1/n=1 \) forces \( 1/n=0 \), impos… ko full sentence mein expand karke likho.

Exam Step 7 (Hinglish): Closure. \( \overline{A}=A\cup A'=A\cup\{-1,1\} \).
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — Closure. \( \overline{A}=A\cup A'=A\cup\{-1,1\} \). ko full sentence mein expand karke likho.

Exam Step 8 (Hinglish): Compactness (Heine–Borel). \( A \) is bounded (lies in \( [-2,3/2] \)) but not closed (missing \( \pm 1 \)). Therefore \( A \) is not compact.
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — Compactness (Heine–Borel). \( A \) is bounded (lies in \( [-2,3/2] \)) but not closed (missing \( \pm 1 \)). Therefore \… ko full sentence mein expand karke likho.

Final line (marks save): End pe clearly likho TRUE/FALSE ya jo construct/prove maanga tha — ek alag sentence: “Hence …”. Agar counterexample hai toh pehle example fully define karo, phir check karo, phir conclude.

Common cutters: (1) definition skip (2) “obviously” without reason (3) wrong quantifiers ∀/∃ (4) compact vs closed confuse (5) image vs pre-image confuse. Inse bacho.

Source / Idea kahan se aayi

Uses: definitions of \( \sup/\inf \), limit point, closure; Heine–Borel (compact ⇔ closed and bounded in \( \mathbb{R} \)). Course: Sequences + topology of \( \mathbb{R} \).

Interactive — B1 Visualization: Oscillating Sequence Convergence

Max terms to show (N):

Question · B2 · Explicit Examples

(i) Give an infinite compact subset of \( \mathbb{R} \) with exactly one limit point.
(ii) Give a subset of \( \mathbb{R} \) that is neither open nor closed.

English Mathematical Steps — full marks checklist

(i) Candidate: Let \( A=\{0\}\cup\{\frac{1}{n}:n\in\mathbb{N}\}=\{0,1,\tfrac12,\tfrac13,\ldots\} \).

(i) Limit points: The only accumulation point is \( 0 \) (terms \( 1/n\to 0 \)). Each \( 1/n \) is isolated: choose radius smaller than the gap to neighbours.

(i) Closed + bounded: \( A\subset[0,1] \) (bounded). Closed because it contains its only limit point \( 0 \). Heine–Borel ⇒ compact, infinite, exactly one limit point.

(ii) Candidate: Take the half-open interval \( A=[0,1) \).

(ii) Not open: The endpoint \( 0\in A \), but every neighborhood of \( 0 \) sticks out to the left of \( 0 \), leaving \( A \).

(ii) Not closed: \( 1 \) is a limit point of \( A \) (approach from the left) but \( 1\notin A \).

(ii) Conclusion: \( [0,1) \) is neither open nor closed.

Exam-style answer (copy into answer sheet)

Given. (i) Give an infinite compact subset of \( \mathbb{R} \) with exactly one limit point.
(ii) Give a subset of \( \mathbb{R} \) that is neither open nor closed.

Step 1 (i) — Example. \( A = \{0\} \cup \{ \frac{1}{n} : n \in \mathbb{N} \} \).

Step 2 (i) — One limit point. \( 1/n \to 0 \), so \( 0 \) is a limit point. Each \( 1/n \) is isolated for large enough gap; no other accumulation.

Step 3 (i) — Compact. \( A \subset [0,1] \) bounded; closed because \( A' = \{0\} \subseteq A \). Heine–Borel ⇒ compact, infinite.

Step 4 (ii) — Example. \( B = [0, 1) \).

Step 5 (ii) — Not open. \( 0 \in B \) but every neighborhood of \( 0 \) contains negative numbers not in \( B \).

Step 6 (ii) — Not closed. \( 1 \) is limit from left but \( 1 \notin B \).

Final answer. (i) \( \{0\} \cup \{1/n\} \). (ii) \( [0,1) \).
Method. Present the solution as a proof or counterexample with numbered steps. Each step must state the definition or theorem used and why the next line follows — no “obvious” jumps.
Marks tip. Underline the final boxed answer. For TRUE/FALSE, include a one-line justification name (e.g. connectedness, counterexample, EVT). Missing quantifiers or confused open/closed/compact costs marks.

Hinglish — detailed (examiner-safe, beginner-clear)

(i): \( 1/n \) sequence + limit 0 — ek hi accumulation; bounded closed → compact.

(ii): Half-open interval classic — left endpoint open issue, right endpoint missing for closed.
Exam mindset: Copy-paste feel mat do — har step pe “definition → apply → conclude” likho. English math symbols same rakh sakte ho; explanation Hinglish mein full sentences mein.
Common cutters: Definition skip, “obviously” bina reason, ∀/∃ galat, compact vs closed confuse, image vs pre-image confuse — in sab se bacho.
Final line (marks save): End pe clearly likho jo construct/prove maanga tha — ek alag “Hence …” sentence. Counterexample ho toh pehle example define, phir verify, phir conclude.

Exam mindset: Copy-paste feel nahi dena; har line pe “definition → apply → conclude” dikhana. Neeche wala write-up answer sheet pe utaar sakte ho (Hinglish soch + English math symbols).

Exam Step 1 (Hinglish): (i) Candidate: Let \( A=\{0\}\cup\{\frac{1}{n}:n\in\mathbb{N}\}=\{0,1,\tfrac12,\tfrac13,\ldots\} \).
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — (i) Candidate: Let \( A=\{0\}\cup\{\frac{1}{n}:n\in\mathbb{N}\}=\{0,1,\tfrac12,\tfrac13,\ldots\} \). ko full sentence mein expand karke likho.

Exam Step 2 (Hinglish): (i) Limit points: The only accumulation point is \( 0 \) (terms \( 1/n\to 0 \)). Each \( 1/n \) is isolated: choose radius smaller than the gap to neighbours.
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — (i) Limit points: The only accumulation point is \( 0 \) (terms \( 1/n\to 0 \)). Each \( 1/n \) is isolated: choose radi… ko full sentence mein expand karke likho.

Exam Step 3 (Hinglish): (i) Closed + bounded: \( A\subset[0,1] \) (bounded). Closed because it contains its only limit point \( 0 \). Heine–Borel ⇒ compact, infinite, exactly one limit point.
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — (i) Closed + bounded: \( A\subset[0,1] \) (bounded). Closed because it contains its only limit point \( 0 \). Heine–Bore… ko full sentence mein expand karke likho.

Exam Step 4 (Hinglish): (ii) Candidate: Take the half-open interval \( A=[0,1) \).
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — (ii) Candidate: Take the half-open interval \( A=[0,1) \). ko full sentence mein expand karke likho.

Exam Step 5 (Hinglish): (ii) Not open: The endpoint \( 0\in A \), but every neighborhood of \( 0 \) sticks out to the left of \( 0 \), leaving \( A \).
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — (ii) Not open: The endpoint \( 0\in A \), but every neighborhood of \( 0 \) sticks out to the left of \( 0 \), leaving \… ko full sentence mein expand karke likho.

Exam Step 6 (Hinglish): (ii) Not closed: \( 1 \) is a limit point of \( A \) (approach from the left) but \( 1\notin A \).
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — (ii) Not closed: \( 1 \) is a limit point of \( A \) (approach from the left) but \( 1\notin A \). ko full sentence mein expand karke likho.

Exam Step 7 (Hinglish): (ii) Conclusion: \( [0,1) \) is neither open nor closed.
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — (ii) Conclusion: \( [0,1) \) is neither open nor closed. ko full sentence mein expand karke likho.

Final line (marks save): End pe clearly likho TRUE/FALSE ya jo construct/prove maanga tha — ek alag sentence: “Hence …”. Agar counterexample hai toh pehle example fully define karo, phir check karo, phir conclude.

Common cutters: (1) definition skip (2) “obviously” without reason (3) wrong quantifiers ∀/∃ (4) compact vs closed confuse (5) image vs pre-image confuse. Inse bacho.

Source / Idea kahan se aayi

Uses: limit point / isolated point; Heine–Borel; definitions of open and closed sets. Course: Compactness + topology of \( \mathbb{R} \).

Section C: Proofs

Question · C1 · Equality set of continuous functions

Let \( f,g:\mathbb{R}\to\mathbb{R} \) be continuous. Let \( A=\{x\in\mathbb{R}: f(x)=g(x)\} \).
(i) Prove \( A \) is closed.
(ii) Is \( \{x: f(x)/g(x)=1\} \) necessarily closed?

English Mathematical Steps — full marks checklist

(i) Build an auxiliary function. Define \( h(x)=f(x)-g(x) \). Difference of continuous functions ⇒ \( h \) is continuous on \( \mathbb{R} \).

(i) Rewrite the set. \( A=\{x:h(x)=0\}=h^{-1}(\{0\}) \).

(i) Use the closed-preimage theorem. Singleton \( \{0\} \) is closed in \( \mathbb{R} \). Continuous pre-image of a closed set is closed. Therefore \( A \) is closed.

(ii) Domain issue. The quotient \( f/g \) is only defined where \( g(x)\neq 0 \). So the set in (ii) is not automatically comparable to \( A \).

(ii) Counterexample. Take \( f(x)=x \) and \( g(x)=x \). Both continuous. Then \( f(x)/g(x)=1 \) for all \( x\neq 0 \), so the set equals \( \mathbb{R}\setminus\{0\} \), which is open, not closed.

(ii) Conclusion. The answer is No — that set need not be closed.

Exam-style answer (copy into answer sheet)

Given. Continuous \( f,g:\mathbb{R}\to\mathbb{R} \), \( A = \{ x : f(x) = g(x) \} \).

To prove / To find. (i) Prove \( A \) is closed. (ii) Is \( \{ x : f(x)/g(x) = 1 \} \) necessarily closed?

Step 1 (i). Define \( h(x) = f(x) - g(x) \), continuous. Then \( A = h^{-1}(\{0\}) \).

Step 2 (i). \( \{0\} \) is closed in \( \mathbb{R} \). Continuous pre-image of closed set is closed. Hence \( A \) is closed.

Step 3 (ii). Quotient only defined where \( g(x) \neq 0 \). Not the same set as \( A \) in general.

Step 4 (ii) — Counterexample. \( f(x) = g(x) = x \). Then \( f/g = 1 \) for \( x \neq 0 \), set is \( \mathbb{R} \setminus \{0\} \), which is open, not closed.

Final answer. (i) \( A \) is closed. (ii) No — not necessarily closed.
Method. Present the solution as a proof or counterexample with numbered steps. Each step must state the definition or theorem used and why the next line follows — no “obvious” jumps.
Marks tip. Underline the final boxed answer. For TRUE/FALSE, include a one-line justification name (e.g. connectedness, counterexample, EVT). Missing quantifiers or confused open/closed/compact costs marks.

Method. We solve this in the exam format: state definitions used, justify each implication, and end with a clear final line.

Step 1. (i) Build an auxiliary function. Define \( h(x)=f(x)-g(x) \). Difference of continuous functions ⇒ \( h \) is continuous on \( \mathbb{R} \). (Write this justification fully on the answer sheet — do not skip definitions or “why”.)

Step 2. (i) Rewrite the set. \( A=\{x:h(x)=0\}=h^{-1}(\{0\}) \). (Write this justification fully on the answer sheet — do not skip definitions or “why”.)

Step 3. (i) Use the closed-preimage theorem. Singleton \( \{0\} \) is closed in \( \mathbb{R} \). Continuous pre-image of a closed set is closed. Therefore \( A \) is closed. (Write this justification fully on the answer sheet — do not skip definitions or “why”.)

Step 4. (ii) Domain issue. The quotient \( f/g \) is only defined where \( g(x)\neq 0 \). So the set in (ii) is not automatically comparable to \( A \). (Write this justification fully on the answer sheet — do not skip definitions or “why”.)

Step 5. (ii) Counterexample. Take \( f(x)=x \) and \( g(x)=x \). Both continuous. Then \( f(x)/g(x)=1 \) for all \( x\neq 0 \), so the set equals \( \mathbb{R}\setminus\{0\} \), which is open, not closed. (Write this justification fully on the answer sheet — do not skip definitions or “why”.)

Step 6. (ii) Conclusion. The answer is No — that set need not be closed. (Write this justification fully on the answer sheet — do not skip definitions or “why”.)

Final answer (box this). (ii) Conclusion. The answer is No — that set need not be closed.

Marks tip. Examiner cuts marks for missing definitions, unjustified jumps, and unclear TRUE/FALSE or final claim. Keep every “because” explicit.

Hinglish — detailed (examiner-safe, beginner-clear)

(i): Difference function zero set — standard closed preimage argument.

(ii): Division domain hole at zero — \( \mathbb{R}\setminus\{0\} \) open counterexample.
Exam mindset: Copy-paste feel mat do — har step pe “definition → apply → conclude” likho. English math symbols same rakh sakte ho; explanation Hinglish mein full sentences mein.
Common cutters: Definition skip, “obviously” bina reason, ∀/∃ galat, compact vs closed confuse, image vs pre-image confuse — in sab se bacho.
Final line (marks save): End pe clearly likho jo construct/prove maanga tha — ek alag “Hence …” sentence. Counterexample ho toh pehle example define, phir verify, phir conclude.

Exam mindset: Copy-paste feel nahi dena; har line pe “definition → apply → conclude” dikhana. Neeche wala write-up answer sheet pe utaar sakte ho (Hinglish soch + English math symbols).

Exam Step 1 (Hinglish): (i) Build an auxiliary function. Define \( h(x)=f(x)-g(x) \). Difference of continuous functions ⇒ \( h \) is continuous on \( \mathbb{R} \).
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — (i) Build an auxiliary function. Define \( h(x)=f(x)-g(x) \). Difference of continuous functions ⇒ \( h \) is continuous… ko full sentence mein expand karke likho.

Exam Step 2 (Hinglish): (i) Rewrite the set. \( A=\{x:h(x)=0\}=h^{-1}(\{0\}) \).
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — (i) Rewrite the set. \( A=\{x:h(x)=0\}=h^{-1}(\{0\}) \). ko full sentence mein expand karke likho.

Exam Step 3 (Hinglish): (i) Use the closed-preimage theorem. Singleton \( \{0\} \) is closed in \( \mathbb{R} \). Continuous pre-image of a closed set is closed. Therefore \( A \) is closed.
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — (i) Use the closed-preimage theorem. Singleton \( \{0\} \) is closed in \( \mathbb{R} \). Continuous pre-image of a clos… ko full sentence mein expand karke likho.

Exam Step 4 (Hinglish): (ii) Domain issue. The quotient \( f/g \) is only defined where \( g(x)\neq 0 \). So the set in (ii) is not automatically comparable to \( A \).
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — (ii) Domain issue. The quotient \( f/g \) is only defined where \( g(x)\neq 0 \). So the set in (ii) is not automaticall… ko full sentence mein expand karke likho.

Exam Step 5 (Hinglish): (ii) Counterexample. Take \( f(x)=x \) and \( g(x)=x \). Both continuous. Then \( f(x)/g(x)=1 \) for all \( x\neq 0 \), so the set equals \( \mathbb{R}\setminus\{0\} \), which is open, not closed.
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — (ii) Counterexample. Take \( f(x)=x \) and \( g(x)=x \). Both continuous. Then \( f(x)/g(x)=1 \) for all \( x\neq 0 \), … ko full sentence mein expand karke likho.

Exam Step 6 (Hinglish): (ii) Conclusion. The answer is No — that set need not be closed.
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — (ii) Conclusion. The answer is No — that set need not be closed. ko full sentence mein expand karke likho.

Final line (marks save): End pe clearly likho TRUE/FALSE ya jo construct/prove maanga tha — ek alag sentence: “Hence …”. Agar counterexample hai toh pehle example fully define karo, phir check karo, phir conclude.

Common cutters: (1) definition skip (2) “obviously” without reason (3) wrong quantifiers ∀/∃ (4) compact vs closed confuse (5) image vs pre-image confuse. Inse bacho.

Source / Idea kahan se aayi

Uses: algebra of continuous functions; continuous pre-image of closed sets is closed. Course: Continuity + closed sets.

Question · C2 · Closest point in a compact set

Let \( K\subseteq\mathbb{R} \) be nonempty and compact, and fix \( x_0\in\mathbb{R} \). Prove there exists \( p\in K \) such that \( |x_0-p|\le|x_0-y| \) for every \( y\in K \) (a nearest point).

English Mathematical Steps — full marks checklist

Define the distance function on \( K \). Let \( d:K\to\mathbb{R} \) by \( d(y)=|x_0-y| \). Absolute value is continuous, so \( d \) is continuous on \( K \).

Apply Extreme Value Theorem. Compact domain + continuous real function ⇒ \( d \) attains its minimum: there exists \( p\in K \) with \( d(p)=\min_{y\in K}d(y) \).

Translate back. \( d(p)\le d(y) \) for all \( y\in K \) means exactly \( |x_0-p|\le|x_0-y| \) for all \( y\in K \).

Remark. Compactness is essential: on an open interval the infimum of distance may not be attained.

Exam-style answer (copy into answer sheet)

Given. Nonempty compact \( K \subseteq \mathbb{R} \), fixed \( x_0 \in \mathbb{R} \).

To prove. There exists \( p \in K \) such that \( |x_0 - p| \le |x_0 - y| \) for all \( y \in K \) (nearest point in \( K \)).

Step 1. Define \( d(y) = |x_0 - y| \) for \( y \in K \). Continuous on \( K \).

Step 2. \( K \) compact, \( d \) continuous ⇒ by Extreme Value Theorem \( d \) attains minimum at some \( p \in K \).

Step 3. Then \( |x_0 - p| = d(p) \le d(y) = |x_0 - y| \) for all \( y \in K \).

Final answer. Such \( p \) exists by EVT on compact \( K \).
Method. Present the solution as a proof or counterexample with numbered steps. Each step must state the definition or theorem used and why the next line follows — no “obvious” jumps.
Marks tip. Underline the final boxed answer. For TRUE/FALSE, include a one-line justification name (e.g. connectedness, counterexample, EVT). Missing quantifiers or confused open/closed/compact costs marks.

Method. We solve this in the exam format: state definitions used, justify each implication, and end with a clear final line.

Step 1. Define the distance function on \( K \). Let \( d:K\to\mathbb{R} \) by \( d(y)=|x_0-y| \). Absolute value is continuous, so \( d \) is continuous on \( K \). (Write this justification fully on the answer sheet — do not skip definitions or “why”.)

Step 2. Apply Extreme Value Theorem. Compact domain + continuous real function ⇒ \( d \) attains its minimum: there exists \( p\in K \) with \( d(p)=\min_{y\in K}d(y) \). (Write this justification fully on the answer sheet — do not skip definitions or “why”.)

Step 3. Translate back. \( d(p)\le d(y) \) for all \( y\in K \) means exactly \( |x_0-p|\le|x_0-y| \) for all \( y\in K \). (Write this justification fully on the answer sheet — do not skip definitions or “why”.)

Step 4. Remark. Compactness is essential: on an open interval the infimum of distance may not be attained. (Write this justification fully on the answer sheet — do not skip definitions or “why”.)

Final answer (box this). Remark. Compactness is essential: on an open interval the infimum of distance may not be attained.

Marks tip. Examiner cuts marks for missing definitions, unjustified jumps, and unclear TRUE/FALSE or final claim. Keep every “because” explicit.

Hinglish — detailed (examiner-safe, beginner-clear)

Strategy: Distance function continuous + compact domain → minimum attain hota hai.

Exam: EVT ka naam likho; compactness kyun zaroori — remark optional.
Exam mindset: Copy-paste feel mat do — har step pe “definition → apply → conclude” likho. English math symbols same rakh sakte ho; explanation Hinglish mein full sentences mein.
Common cutters: Definition skip, “obviously” bina reason, ∀/∃ galat, compact vs closed confuse, image vs pre-image confuse — in sab se bacho.
Final line (marks save): End pe clearly likho jo construct/prove maanga tha — ek alag “Hence …” sentence. Counterexample ho toh pehle example define, phir verify, phir conclude.

Exam mindset: Copy-paste feel nahi dena; har line pe “definition → apply → conclude” dikhana. Neeche wala write-up answer sheet pe utaar sakte ho (Hinglish soch + English math symbols).

Exam Step 1 (Hinglish): Define the distance function on \( K \). Let \( d:K\to\mathbb{R} \) by \( d(y)=|x_0-y| \). Absolute value is continuous, so \( d \) is continuous on \( K \).
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — Define the distance function on \( K \). Let \( d:K\to\mathbb{R} \) by \( d(y)=|x_0-y| \). Absolute value is continuous,… ko full sentence mein expand karke likho.

Exam Step 2 (Hinglish): Apply Extreme Value Theorem. Compact domain + continuous real function ⇒ \( d \) attains its minimum: there exists \( p\in K \) with \( d(p)=\min_{y\in K}d(y) \).
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — Apply Extreme Value Theorem. Compact domain + continuous real function ⇒ \( d \) attains its minimum: there exists \( p\… ko full sentence mein expand karke likho.

Exam Step 3 (Hinglish): Translate back. \( d(p)\le d(y) \) for all \( y\in K \) means exactly \( |x_0-p|\le|x_0-y| \) for all \( y\in K \).
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — Translate back. \( d(p)\le d(y) \) for all \( y\in K \) means exactly \( |x_0-p|\le|x_0-y| \) for all \( y\in K \). ko full sentence mein expand karke likho.

Exam Step 4 (Hinglish): Remark. Compactness is essential: on an open interval the infimum of distance may not be attained.
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — Remark. Compactness is essential: on an open interval the infimum of distance may not be attained. ko full sentence mein expand karke likho.

Final line (marks save): End pe clearly likho TRUE/FALSE ya jo construct/prove maanga tha — ek alag sentence: “Hence …”. Agar counterexample hai toh pehle example fully define karo, phir check karo, phir conclude.

Common cutters: (1) definition skip (2) “obviously” without reason (3) wrong quantifiers ∀/∃ (4) compact vs closed confuse (5) image vs pre-image confuse. Inse bacho.

Source / Idea kahan se aayi

Uses: Extreme Value Theorem (continuous function on compact set attains min/max). Course: Compactness + continuity.

Interactive — C2 Visualization: Distance to a Compact Set

Drag \( x_0 \) to see the closest point \( p \) in the compact set \( K \).

Section D: Additional Exam Proofs

Term Definitions (Q1)

Bounded Set (Seemit Set): A set is bounded if it fits entirely within some finite interval; it has both an upper limit and a lower limit.

Open Set (Khula Set): A set where for every point inside it, you can draw a small interval (neighborhood) around that point which is also completely inside the set.

Supremum (\( \sup U \)): The least upper bound of a set. It is the smallest number that is greater than or equal to every element in the set.

Question · D1 · Supremum of an open set

Let \( U\subseteq\mathbb{R} \) be a nonempty bounded open set. Prove that \( \sup U\notin U \).

English Mathematical Steps — full marks checklist

Setup. Let \( s=\sup U \) (exists by completeness / bounded above). We must show \( s\notin U \).

Contradiction assumption. Suppose \( s\in U \).

Use openness. Openness at \( s \) gives \( \varepsilon>0 \) with \( (s-\varepsilon,s+\varepsilon)\subseteq U \).

Produce a bigger element. Then \( s+\varepsilon/2\in U \) and \( s+\varepsilon/2>s \).

Contradiction. No element of \( U \) can exceed the upper bound \( s \). Therefore the assumption is false: \( s\notin U \).

Moral. A bounded open set in \( \mathbb{R} \) never contains its least upper bound (same argument for \( \inf \)).

Exam-style answer (copy into answer sheet)

Given. \( U \subseteq \mathbb{R} \) nonempty, bounded, open.

To prove. \( \sup U \notin U \).

Step 1. Let \( s = \sup U \) (exists by completeness). Suppose for contradiction \( s \in U \).

Step 2. Openness: \( \exists \varepsilon > 0 \) with \( (s - \varepsilon, s + \varepsilon) \subseteq U \).

Step 3. Then \( s + \varepsilon/2 \in U \) and \( s + \varepsilon/2 > s \), contradicting that \( s \) is an upper bound for \( U \).

Final answer. Therefore \( \sup U \notin U \).
Method. Present the solution as a proof or counterexample with numbered steps. Each step must state the definition or theorem used and why the next line follows — no “obvious” jumps.
Marks tip. Underline the final boxed answer. For TRUE/FALSE, include a one-line justification name (e.g. connectedness, counterexample, EVT). Missing quantifiers or confused open/closed/compact costs marks.

Method. We solve this in the exam format: state definitions used, justify each implication, and end with a clear final line.

Step 1. Setup. Let \( s=\sup U \) (exists by completeness / bounded above). We must show \( s\notin U \). (Write this justification fully on the answer sheet — do not skip definitions or “why”.)

Step 2. Contradiction assumption. Suppose \( s\in U \). (Write this justification fully on the answer sheet — do not skip definitions or “why”.)

Step 3. Use openness. Openness at \( s \) gives \( \varepsilon>0 \) with \( (s-\varepsilon,s+\varepsilon)\subseteq U \). (Write this justification fully on the answer sheet — do not skip definitions or “why”.)

Step 4. Produce a bigger element. Then \( s+\varepsilon/2\in U \) and \( s+\varepsilon/2>s \). (Write this justification fully on the answer sheet — do not skip definitions or “why”.)

Step 5. Contradiction. No element of \( U \) can exceed the upper bound \( s \). Therefore the assumption is false: \( s\notin U \). (Write this justification fully on the answer sheet — do not skip definitions or “why”.)

Step 6. Moral. A bounded open set in \( \mathbb{R} \) never contains its least upper bound (same argument for \( \inf \)). (Write this justification fully on the answer sheet — do not skip definitions or “why”.)

Final answer (box this). Moral. A bounded open set in \( \mathbb{R} \) never contains its least upper bound (same argument for \( \inf \)).

Marks tip. Examiner cuts marks for missing definitions, unjustified jumps, and unclear TRUE/FALSE or final claim. Keep every “because” explicit.

Hinglish — detailed (examiner-safe, beginner-clear)

Proof idea: Sup upper bound hai — agar andar hota toh thoda upar wala point bhi open interval se andar, contradiction.

Likho: Contradiction structure clear: assume in U → openness → element > s.
Exam mindset: Copy-paste feel mat do — har step pe “definition → apply → conclude” likho. English math symbols same rakh sakte ho; explanation Hinglish mein full sentences mein.
Common cutters: Definition skip, “obviously” bina reason, ∀/∃ galat, compact vs closed confuse, image vs pre-image confuse — in sab se bacho.
Final line (marks save): End pe clearly likho jo construct/prove maanga tha — ek alag “Hence …” sentence. Counterexample ho toh pehle example define, phir verify, phir conclude.

Exam mindset: Copy-paste feel nahi dena; har line pe “definition → apply → conclude” dikhana. Neeche wala write-up answer sheet pe utaar sakte ho (Hinglish soch + English math symbols).

Exam Step 1 (Hinglish): Setup. Let \( s=\sup U \) (exists by completeness / bounded above). We must show \( s\notin U \).
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — Setup. Let \( s=\sup U \) (exists by completeness / bounded above). We must show \( s\notin U \). ko full sentence mein expand karke likho.

Exam Step 2 (Hinglish): Contradiction assumption. Suppose \( s\in U \).
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — Contradiction assumption. Suppose \( s\in U \). ko full sentence mein expand karke likho.

Exam Step 3 (Hinglish): Use openness. Openness at \( s \) gives \( \varepsilon>0 \) with \( (s-\varepsilon,s+\varepsilon)\subseteq U \).
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — Use openness. Openness at \( s \) gives \( \varepsilon>0 \) with \( (s-\varepsilon,s+\varepsilon)\subseteq U \). ko full sentence mein expand karke likho.

Exam Step 4 (Hinglish): Produce a bigger element. Then \( s+\varepsilon/2\in U \) and \( s+\varepsilon/2>s \).
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — Produce a bigger element. Then \( s+\varepsilon/2\in U \) and \( s+\varepsilon/2>s \). ko full sentence mein expand karke likho.

Exam Step 5 (Hinglish): Contradiction. No element of \( U \) can exceed the upper bound \( s \). Therefore the assumption is false: \( s\notin U \).
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — Contradiction. No element of \( U \) can exceed the upper bound \( s \). Therefore the assumption is false: \( s\notin U… ko full sentence mein expand karke likho.

Exam Step 6 (Hinglish): Moral. A bounded open set in \( \mathbb{R} \) never contains its least upper bound (same argument for \( \inf \)).
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — Moral. A bounded open set in \( \mathbb{R} \) never contains its least upper bound (same argument for \( \inf \)). ko full sentence mein expand karke likho.

Final line (marks save): End pe clearly likho TRUE/FALSE ya jo construct/prove maanga tha — ek alag sentence: “Hence …”. Agar counterexample hai toh pehle example fully define karo, phir check karo, phir conclude.

Common cutters: (1) definition skip (2) “obviously” without reason (3) wrong quantifiers ∀/∃ (4) compact vs closed confuse (5) image vs pre-image confuse. Inse bacho.

Source / Idea kahan se aayi

Uses: definition of supremum; definition of open set; proof by contradiction. Course: Completeness + open sets in \( \mathbb{R} \).

Interactive — D1: The Contradiction of \( s \in U \)

Term Definitions (Q2 Continuity)

Continuous Function (Nirantar Function): A function that has no abrupt jumps or breaks. Small changes in input \( x \) lead to small changes in output \( f(x) \).

\( \varepsilon-\delta \) Definition: A formal way to prove continuity. For every \( \varepsilon > 0 \) (error tolerance), there exists a \( \delta > 0 \) (allowable variance) such that if \( |x - c| < \delta \), then \( |f(x) - f(c)| < \varepsilon \).

Sequential Characterization: A function \( f \) is continuous at \( c \) if, for every sequence \( (x_n) \) converging to \( c \), the sequence of outputs \( f(x_n) \) converges to \( f(c) \).

Question · D2(a) · Continuity of √x on (0,∞)

Determine whether \( f:(0,\infty)\to\mathbb{R} \), \( f(x)=\sqrt{x} \) is continuous. Give a full \( \varepsilon \)–\( \delta \) proof.

English Mathematical Steps — full marks checklist

Claim. Yes — \( f \) is continuous at every \( c\in(0,\infty) \).

Fix \( c>0 \) and \( \varepsilon>0 \). We must find \( \delta>0 \) so that \( |x-c|<\delta \) ⇒ \( |\sqrt{x}-\sqrt{c}|<\varepsilon \).

Algebra identity. Rationalize: \( |\sqrt{x}-\sqrt{c}|=\dfrac{|x-c|}{\sqrt{x}+\sqrt{c}} \).

Lower-bound the denominator. For \( x>0 \), \( \sqrt{x}+\sqrt{c}>\sqrt{c} \), hence \( |\sqrt{x}-\sqrt{c}|<\dfrac{|x-c|}{\sqrt{c}} \).

Choose δ. Take \( \delta=\varepsilon\sqrt{c} \). Then \( |x-c|<\delta \) ⇒ \( |\sqrt{x}-\sqrt{c}|<\varepsilon \).

Conclusion. Continuous on \( (0,\infty) \).

Exam-style answer (copy into answer sheet)

Given. \( f:(0,\infty) \to \mathbb{R} \), \( f(x) = \sqrt{x} \).

To prove. \( f \) is continuous at every \( c \in (0,\infty) \) via \( \varepsilon \)–\( \delta \).

Step 1. Fix \( c > 0 \), \( \varepsilon > 0 \). Need \( \delta > 0 \) with \( |x - c| < \delta \Rightarrow |\sqrt{x} - \sqrt{c}| < \varepsilon \).

Step 2. Identity: \( |\sqrt{x} - \sqrt{c}| = \frac{|x - c|}{\sqrt{x} + \sqrt{c}} \le \frac{|x - c|}{\sqrt{c}} \) for \( x > 0 \).

Step 3. Choose \( \delta = \varepsilon \sqrt{c} \). If \( |x - c| < \delta \), then \( |\sqrt{x} - \sqrt{c}| < \varepsilon \).

Final answer. \( f \) is continuous on \( (0,\infty) \).
Method. Present the solution as a proof or counterexample with numbered steps. Each step must state the definition or theorem used and why the next line follows — no “obvious” jumps.
Marks tip. Underline the final boxed answer. For TRUE/FALSE, include a one-line justification name (e.g. connectedness, counterexample, EVT). Missing quantifiers or confused open/closed/compact costs marks.

Method. We solve this in the exam format: state definitions used, justify each implication, and end with a clear final line.

Step 1. Claim. Yes — \( f \) is continuous at every \( c\in(0,\infty) \). (Write this justification fully on the answer sheet — do not skip definitions or “why”.)

Step 2. Fix \( c>0 \) and \( \varepsilon>0 \). We must find \( \delta>0 \) so that \( |x-c|<\delta \) ⇒ \( |\sqrt{x}-\sqrt{c}|<\varepsilon \). (Write this justification fully on the answer sheet — do not skip definitions or “why”.)

Step 3. Algebra identity. Rationalize: \( |\sqrt{x}-\sqrt{c}|=\dfrac{|x-c|}{\sqrt{x}+\sqrt{c}} \). (Write this justification fully on the answer sheet — do not skip definitions or “why”.)

Step 4. Lower-bound the denominator. For \( x>0 \), \( \sqrt{x}+\sqrt{c}>\sqrt{c} \), hence \( |\sqrt{x}-\sqrt{c}|<\dfrac{|x-c|}{\sqrt{c}} \). (Write this justification fully on the answer sheet — do not skip definitions or “why”.)

Step 5. Choose δ. Take \( \delta=\varepsilon\sqrt{c} \). Then \( |x-c|<\delta \) ⇒ \( |\sqrt{x}-\sqrt{c}|<\varepsilon \). (Write this justification fully on the answer sheet — do not skip definitions or “why”.)

Step 6. Conclusion. Continuous on \( (0,\infty) \). (Write this justification fully on the answer sheet — do not skip definitions or “why”.)

Final answer (box this). Conclusion. Continuous on \( (0,\infty) \).

Marks tip. Examiner cuts marks for missing definitions, unjustified jumps, and unclear TRUE/FALSE or final claim. Keep every “because” explicit.

Hinglish — detailed (examiner-safe, beginner-clear)

ε–δ: Rationalize identity yaad karo; denominator \( \sqrt{c} \) se bound.

δ choice: \( \varepsilon\sqrt{c} \) explicitly likhna — examiner check karta hai.
Exam mindset: Copy-paste feel mat do — har step pe “definition → apply → conclude” likho. English math symbols same rakh sakte ho; explanation Hinglish mein full sentences mein.
Common cutters: Definition skip, “obviously” bina reason, ∀/∃ galat, compact vs closed confuse, image vs pre-image confuse — in sab se bacho.
Final line (marks save): End pe clearly likho jo construct/prove maanga tha — ek alag “Hence …” sentence. Counterexample ho toh pehle example define, phir verify, phir conclude.

Exam mindset: Copy-paste feel nahi dena; har line pe “definition → apply → conclude” dikhana. Neeche wala write-up answer sheet pe utaar sakte ho (Hinglish soch + English math symbols).

Exam Step 1 (Hinglish): Claim. Yes — \( f \) is continuous at every \( c\in(0,\infty) \).
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — Claim. Yes — \( f \) is continuous at every \( c\in(0,\infty) \). ko full sentence mein expand karke likho.

Exam Step 2 (Hinglish): Fix \( c>0 \) and \( \varepsilon>0 \). We must find \( \delta>0 \) so that \( |x-c|<\delta \) ⇒ \( |\sqrt{x}-\sqrt{c}|<\varepsilon \).
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — Fix \( c>0 \) and \( \varepsilon>0 \). We must find \( \delta>0 \) so that \( |x-c|<\delta \) ⇒ \( |\sqrt{x}-\sqrt{c}|<\… ko full sentence mein expand karke likho.

Exam Step 3 (Hinglish): Algebra identity. Rationalize: \( |\sqrt{x}-\sqrt{c}|=\dfrac{|x-c|}{\sqrt{x}+\sqrt{c}} \).
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — Algebra identity. Rationalize: \( |\sqrt{x}-\sqrt{c}|=\dfrac{|x-c|}{\sqrt{x}+\sqrt{c}} \). ko full sentence mein expand karke likho.

Exam Step 4 (Hinglish): Lower-bound the denominator. For \( x>0 \), \( \sqrt{x}+\sqrt{c}>\sqrt{c} \), hence \( |\sqrt{x}-\sqrt{c}|<\dfrac{|x-c|}{\sqrt{c}} \).
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — Lower-bound the denominator. For \( x>0 \), \( \sqrt{x}+\sqrt{c}>\sqrt{c} \), hence \( |\sqrt{x}-\sqrt{c}|<\dfrac{|x-c|}… ko full sentence mein expand karke likho.

Exam Step 5 (Hinglish): Choose δ. Take \( \delta=\varepsilon\sqrt{c} \). Then \( |x-c|<\delta \) ⇒ \( |\sqrt{x}-\sqrt{c}|<\varepsilon \).
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — Choose δ. Take \( \delta=\varepsilon\sqrt{c} \). Then \( |x-c|<\delta \) ⇒ \( |\sqrt{x}-\sqrt{c}|<\varepsilon \). ko full sentence mein expand karke likho.

Exam Step 6 (Hinglish): Conclusion. Continuous on \( (0,\infty) \).
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — Conclusion. Continuous on \( (0,\infty) \). ko full sentence mein expand karke likho.

Final line (marks save): End pe clearly likho TRUE/FALSE ya jo construct/prove maanga tha — ek alag sentence: “Hence …”. Agar counterexample hai toh pehle example fully define karo, phir check karo, phir conclude.

Common cutters: (1) definition skip (2) “obviously” without reason (3) wrong quantifiers ∀/∃ (4) compact vs closed confuse (5) image vs pre-image confuse. Inse bacho.

Source / Idea kahan se aayi

Uses: \( \varepsilon \)–\( \delta \) definition; rationalizing identity. Course: Continuity.

Question · D2(b) · Continuity of x² on ℝ

Determine whether \( f:\mathbb{R}\to\mathbb{R} \), \( f(x)=x^2 \) is continuous. Give a full \( \varepsilon \)–\( \delta \) proof.

English Mathematical Steps — full marks checklist

Claim. Yes — continuous at every \( c\in\mathbb{R} \).

Factor. \( |x^2-c^2|=|x-c|\,|x+c| \).

Local bound. Restrict to \( |x-c|<1 \). Then \( |x|<1+|c| \), so \( |x+c|<2|c|+1 \).

Combine. \( |x^2-c^2|<|x-c|(2|c|+1) \). Want this \( <\varepsilon \) ⇒ need \( |x-c|<\varepsilon/(2|c|+1) \).

Choose δ. \( \delta=\min\bigl(1,\tfrac{\varepsilon}{2|c|+1}\bigr) \).

Conclusion. Continuous on \( \mathbb{R} \) (but not uniformly continuous on all of \( \mathbb{R} \)).

Exam-style answer (copy into answer sheet)

Given. \( f:\mathbb{R}\to\mathbb{R} \), \( f(x) = x^2 \).

To prove. Continuous at every \( c \in \mathbb{R} \) (\( \varepsilon \)–\( \delta \)).

Step 1. \( |x^2 - c^2| = |x - c| \, |x + c| \).

Step 2. If \( |x - c| < 1 \), then \( |x| < 1 + |c| \), so \( |x + c| \le |x| + |c| < 2|c| + 1 \).

Step 3. Take \( \delta = \min\bigl(1, \frac{\varepsilon}{2|c| + 1}\bigr) \). Then \( |x - c| < \delta \Rightarrow |x^2 - c^2| < \varepsilon \).

Final answer. \( f \) is continuous on \( \mathbb{R} \).
Method. Present the solution as a proof or counterexample with numbered steps. Each step must state the definition or theorem used and why the next line follows — no “obvious” jumps.
Marks tip. Underline the final boxed answer. For TRUE/FALSE, include a one-line justification name (e.g. connectedness, counterexample, EVT). Missing quantifiers or confused open/closed/compact costs marks.

Method. We solve this in the exam format: state definitions used, justify each implication, and end with a clear final line.

Step 1. Claim. Yes — continuous at every \( c\in\mathbb{R} \). (Write this justification fully on the answer sheet — do not skip definitions or “why”.)

Step 2. Factor. \( |x^2-c^2|=|x-c|\,|x+c| \). (Write this justification fully on the answer sheet — do not skip definitions or “why”.)

Step 3. Local bound. Restrict to \( |x-c|<1 \). Then \( |x|<1+|c| \), so \( |x+c|<2|c|+1 \). (Write this justification fully on the answer sheet — do not skip definitions or “why”.)

Step 4. Combine. \( |x^2-c^2|<|x-c|(2|c|+1) \). Want this \( <\varepsilon \) ⇒ need \( |x-c|<\varepsilon/(2|c|+1) \). (Write this justification fully on the answer sheet — do not skip definitions or “why”.)

Step 5. Choose δ. \( \delta=\min\bigl(1,\tfrac{\varepsilon}{2|c|+1}\bigr) \). (Write this justification fully on the answer sheet — do not skip definitions or “why”.)

Step 6. Conclusion. Continuous on \( \mathbb{R} \) (but not uniformly continuous on all of \( \mathbb{R} \)). (Write this justification fully on the answer sheet — do not skip definitions or “why”.)

Final answer (box this). Conclusion. Continuous on \( \mathbb{R} \) (but not uniformly continuous on all of \( \mathbb{R} \)).

Marks tip. Examiner cuts marks for missing definitions, unjustified jumps, and unclear TRUE/FALSE or final claim. Keep every “because” explicit.

Hinglish — detailed (examiner-safe, beginner-clear)

Trick: Pehle \( |x-c|<1 \) se \( |x+c| \) bound — local + global mix.

δ: Min of 1 and \( \varepsilon/(2|c|+1) \) dono conditions likho.
Exam mindset: Copy-paste feel mat do — har step pe “definition → apply → conclude” likho. English math symbols same rakh sakte ho; explanation Hinglish mein full sentences mein.
Common cutters: Definition skip, “obviously” bina reason, ∀/∃ galat, compact vs closed confuse, image vs pre-image confuse — in sab se bacho.
Final line (marks save): End pe clearly likho jo construct/prove maanga tha — ek alag “Hence …” sentence. Counterexample ho toh pehle example define, phir verify, phir conclude.

Exam mindset: Copy-paste feel nahi dena; har line pe “definition → apply → conclude” dikhana. Neeche wala write-up answer sheet pe utaar sakte ho (Hinglish soch + English math symbols).

Exam Step 1 (Hinglish): Claim. Yes — continuous at every \( c\in\mathbb{R} \).
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — Claim. Yes — continuous at every \( c\in\mathbb{R} \). ko full sentence mein expand karke likho.

Exam Step 2 (Hinglish): Factor. \( |x^2-c^2|=|x-c|\,|x+c| \).
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — Factor. \( |x^2-c^2|=|x-c|\,|x+c| \). ko full sentence mein expand karke likho.

Exam Step 3 (Hinglish): Local bound. Restrict to \( |x-c|<1 \). Then \( |x|<1+|c| \), so \( |x+c|<2|c|+1 \).
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — Local bound. Restrict to \( |x-c|<1 \). Then \( |x|<1+|c| \), so \( |x+c|<2|c|+1 \). ko full sentence mein expand karke likho.

Exam Step 4 (Hinglish): Combine. \( |x^2-c^2|<|x-c|(2|c|+1) \). Want this \( <\varepsilon \) ⇒ need \( |x-c|<\varepsilon/(2|c|+1) \).
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — Combine. \( |x^2-c^2|<|x-c|(2|c|+1) \). Want this \( <\varepsilon \) ⇒ need \( |x-c|<\varepsilon/(2|c|+1) \). ko full sentence mein expand karke likho.

Exam Step 5 (Hinglish): Choose δ. \( \delta=\min\bigl(1,\tfrac{\varepsilon}{2|c|+1}\bigr) \).
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — Choose δ. \( \delta=\min\bigl(1,\tfrac{\varepsilon}{2|c|+1}\bigr) \). ko full sentence mein expand karke likho.

Exam Step 6 (Hinglish): Conclusion. Continuous on \( \mathbb{R} \) (but not uniformly continuous on all of \( \mathbb{R} \)).
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — Conclusion. Continuous on \( \mathbb{R} \) (but not uniformly continuous on all of \( \mathbb{R} \)). ko full sentence mein expand karke likho.

Final line (marks save): End pe clearly likho TRUE/FALSE ya jo construct/prove maanga tha — ek alag sentence: “Hence …”. Agar counterexample hai toh pehle example fully define karo, phir check karo, phir conclude.

Common cutters: (1) definition skip (2) “obviously” without reason (3) wrong quantifiers ∀/∃ (4) compact vs closed confuse (5) image vs pre-image confuse. Inse bacho.

Source / Idea kahan se aayi

Uses: \( \varepsilon \)–\( \delta \) definition; triangle inequality. Course: Continuity.

Question · D2(c) · Dirichlet variant discontinuous everywhere

Let \( f:[0,1]\to\mathbb{R} \) with \( f(x)=0 \) for rational \( x \) and \( f(x)=1 \) for irrational \( x \). Is \( f \) continuous anywhere?

English Mathematical Steps — full marks checklist

Claim. \( f \) is discontinuous at every \( c\in[0,1] \).

Case \( c \) rational. Then \( f(c)=0 \). Density of irrationals ⇒ sequence of irrationals \( x_n\to c \) with \( f(x_n)=1\to 1\neq 0=f(c) \).

Case \( c \) irrational. Then \( f(c)=1 \). Density of rationals ⇒ rationals \( q_n\to c \) with \( f(q_n)=0\to 0\neq 1=f(c) \).

Sequential test fails everywhere ⇒ nowhere continuous.

Exam-style answer (copy into answer sheet)

Given. \( f:[0,1]\to\mathbb{R} \): \( f(x)=0 \) if \( x \in \mathbb{Q} \), \( f(x)=1 \) if \( x \notin \mathbb{Q} \).

To prove / To find. Is \( f \) continuous at any point of \( [0,1] \)?

Step 1 (c rational). \( f(c) = 0 \). Irrationals dense ⇒ sequence \( x_n \to c \), \( x_n \notin \mathbb{Q} \), \( f(x_n) = 1 \not\to 0 \).

Step 2 (c irrational). \( f(c) = 1 \). Rationals dense ⇒ \( q_n \to c \), \( f(q_n) = 0 \not\to 1 \).

Final answer. \( f \) is discontinuous at every point of \( [0,1] \).
Method. Present the solution as a proof or counterexample with numbered steps. Each step must state the definition or theorem used and why the next line follows — no “obvious” jumps.
Marks tip. Underline the final boxed answer. For TRUE/FALSE, include a one-line justification name (e.g. connectedness, counterexample, EVT). Missing quantifiers or confused open/closed/compact costs marks.

Method. We solve this in the exam format: state definitions used, justify each implication, and end with a clear final line.

Step 1. Claim. \( f \) is discontinuous at every \( c\in[0,1] \). (Write this justification fully on the answer sheet — do not skip definitions or “why”.)

Step 2. Case \( c \) rational. Then \( f(c)=0 \). Density of irrationals ⇒ sequence of irrationals \( x_n\to c \) with \( f(x_n)=1\to 1\neq 0=f(c) \). (Write this justification fully on the answer sheet — do not skip definitions or “why”.)

Step 3. Case \( c \) irrational. Then \( f(c)=1 \). Density of rationals ⇒ rationals \( q_n\to c \) with \( f(q_n)=0\to 0\neq 1=f(c) \). (Write this justification fully on the answer sheet — do not skip definitions or “why”.)

Step 4. Sequential test fails everywhere ⇒ nowhere continuous. (Write this justification fully on the answer sheet — do not skip definitions or “why”.)

Final answer (box this). Sequential test fails everywhere ⇒ nowhere continuous.

Marks tip. Examiner cuts marks for missing definitions, unjustified jumps, and unclear TRUE/FALSE or final claim. Keep every “because” explicit.

Hinglish — detailed (examiner-safe, beginner-clear)

Dirichlet type: Rational/irrational density se sequence values alag limit — sequential continuity fail.

Dono cases: Rational c aur irrational c alag-alag likhna full marks ke liye.
Exam mindset: Copy-paste feel mat do — har step pe “definition → apply → conclude” likho. English math symbols same rakh sakte ho; explanation Hinglish mein full sentences mein.
Common cutters: Definition skip, “obviously” bina reason, ∀/∃ galat, compact vs closed confuse, image vs pre-image confuse — in sab se bacho.
Final line (marks save): End pe clearly likho jo construct/prove maanga tha — ek alag “Hence …” sentence. Counterexample ho toh pehle example define, phir verify, phir conclude.

Exam mindset: Copy-paste feel nahi dena; har line pe “definition → apply → conclude” dikhana. Neeche wala write-up answer sheet pe utaar sakte ho (Hinglish soch + English math symbols).

Exam Step 1 (Hinglish): Claim. \( f \) is discontinuous at every \( c\in[0,1] \).
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — Claim. \( f \) is discontinuous at every \( c\in[0,1] \). ko full sentence mein expand karke likho.

Exam Step 2 (Hinglish): Case \( c \) rational. Then \( f(c)=0 \). Density of irrationals ⇒ sequence of irrationals \( x_n\to c \) with \( f(x_n)=1\to 1\neq 0=f(c) \).
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — Case \( c \) rational. Then \( f(c)=0 \). Density of irrationals ⇒ sequence of irrationals \( x_n\to c \) with \( f(x_n)… ko full sentence mein expand karke likho.

Exam Step 3 (Hinglish): Case \( c \) irrational. Then \( f(c)=1 \). Density of rationals ⇒ rationals \( q_n\to c \) with \( f(q_n)=0\to 0\neq 1=f(c) \).
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — Case \( c \) irrational. Then \( f(c)=1 \). Density of rationals ⇒ rationals \( q_n\to c \) with \( f(q_n)=0\to 0\neq 1=… ko full sentence mein expand karke likho.

Exam Step 4 (Hinglish): Sequential test fails everywhere ⇒ nowhere continuous.
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — Sequential test fails everywhere ⇒ nowhere continuous. ko full sentence mein expand karke likho.

Final line (marks save): End pe clearly likho TRUE/FALSE ya jo construct/prove maanga tha — ek alag sentence: “Hence …”. Agar counterexample hai toh pehle example fully define karo, phir check karo, phir conclude.

Common cutters: (1) definition skip (2) “obviously” without reason (3) wrong quantifiers ∀/∃ (4) compact vs closed confuse (5) image vs pre-image confuse. Inse bacho.

Source / Idea kahan se aayi

Uses: sequential characterization of continuity; density of \( \mathbb{Q} \) and \( \mathbb{R}\setminus\mathbb{Q} \). Course: Continuity.

Interactive — D2(c): Why Dirichlet fails everywhere

Target Point \( c \):

Question · D2(d) · Distance to a closed set

Let \( F\subseteq\mathbb{R} \) be nonempty and closed. Define \( f(x)=\inf\{|x-a|:a\in F\} \). Is \( f \) continuous?

English Mathematical Steps — full marks checklist

Claim. Yes — in fact \( |f(x)-f(y)|\le|x-y| \) (Lipschitz, hence uniformly continuous).

Triangle step. For any \( a\in F \): \( |x-a|\le|x-y|+|y-a| \).

Take inf over \( a \). \( f(x)\le|x-y|+f(y) \) ⇒ \( f(x)-f(y)\le|x-y| \).

Swap \( x,y \). Similarly \( f(y)-f(x)\le|x-y| \). Combine: \( |f(x)-f(y)|\le|x-y| \).

ε–δ. Given \( \varepsilon>0 \), take \( \delta=\varepsilon \). Then \( |x-y|<\delta \) ⇒ \( |f(x)-f(y)|<\varepsilon \).

Note. Closedness of \( F \) is used when one wants \( f(x)=0\Leftrightarrow x\in F \); continuity of \( f \) holds for any nonempty \( F \).

Exam-style answer (copy into answer sheet)

Given. Nonempty closed \( F \subseteq \mathbb{R} \), \( f(x) = \inf\{|x - a| : a \in F\} \) (distance to \( F \)).

To prove. \( f \) is continuous (in fact Lipschitz).

Step 1. For any \( a \in F \): \( |x - a| \le |x - y| + |y - a| \).

Step 2. Inf over \( a \): \( f(x) \le |x - y| + f(y) \), so \( f(x) - f(y) \le |x - y| \).

Step 3. Swap roles: \( |f(x) - f(y)| \le |x - y| \). Given \( \varepsilon > 0 \), \( \delta = \varepsilon \) works.

Final answer. \( f \) is continuous on \( \mathbb{R} \).
Method. Present the solution as a proof or counterexample with numbered steps. Each step must state the definition or theorem used and why the next line follows — no “obvious” jumps.
Marks tip. Underline the final boxed answer. For TRUE/FALSE, include a one-line justification name (e.g. connectedness, counterexample, EVT). Missing quantifiers or confused open/closed/compact costs marks.

Method. We solve this in the exam format: state definitions used, justify each implication, and end with a clear final line.

Step 1. Claim. Yes — in fact \( |f(x)-f(y)|\le|x-y| \) (Lipschitz, hence uniformly continuous). (Write this justification fully on the answer sheet — do not skip definitions or “why”.)

Step 2. Triangle step. For any \( a\in F \): \( |x-a|\le|x-y|+|y-a| \). (Write this justification fully on the answer sheet — do not skip definitions or “why”.)

Step 3. Take inf over \( a \). \( f(x)\le|x-y|+f(y) \) ⇒ \( f(x)-f(y)\le|x-y| \). (Write this justification fully on the answer sheet — do not skip definitions or “why”.)

Step 4. Swap \( x,y \). Similarly \( f(y)-f(x)\le|x-y| \). Combine: \( |f(x)-f(y)|\le|x-y| \). (Write this justification fully on the answer sheet — do not skip definitions or “why”.)

Step 5. ε–δ. Given \( \varepsilon>0 \), take \( \delta=\varepsilon \). Then \( |x-y|<\delta \) ⇒ \( |f(x)-f(y)|<\varepsilon \). (Write this justification fully on the answer sheet — do not skip definitions or “why”.)

Step 6. Note. Closedness of \( F \) is used when one wants \( f(x)=0\Leftrightarrow x\in F \); continuity of \( f \) holds for any nonempty \( F \). (Write this justification fully on the answer sheet — do not skip definitions or “why”.)

Final answer (box this). Note. Closedness of \( F \) is used when one wants \( f(x)=0\Leftrightarrow x\in F \); continuity of \( f \) holds for any nonempty \( F \).

Marks tip. Examiner cuts marks for missing definitions, unjustified jumps, and unclear TRUE/FALSE or final claim. Keep every “because” explicit.

Hinglish — detailed (examiner-safe, beginner-clear)

Distance function: Triangle inequality + inf — standard 1-Lipschitz proof.

Note: Closed F tab useful jab \( f(x)=0 \iff x\in F \); continuity har nonempty F pe.
Exam mindset: Copy-paste feel mat do — har step pe “definition → apply → conclude” likho. English math symbols same rakh sakte ho; explanation Hinglish mein full sentences mein.
Common cutters: Definition skip, “obviously” bina reason, ∀/∃ galat, compact vs closed confuse, image vs pre-image confuse — in sab se bacho.
Final line (marks save): End pe clearly likho jo construct/prove maanga tha — ek alag “Hence …” sentence. Counterexample ho toh pehle example define, phir verify, phir conclude.

Exam mindset: Copy-paste feel nahi dena; har line pe “definition → apply → conclude” dikhana. Neeche wala write-up answer sheet pe utaar sakte ho (Hinglish soch + English math symbols).

Exam Step 1 (Hinglish): Claim. Yes — in fact \( |f(x)-f(y)|\le|x-y| \) (Lipschitz, hence uniformly continuous).
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — Claim. Yes — in fact \( |f(x)-f(y)|\le|x-y| \) (Lipschitz, hence uniformly continuous). ko full sentence mein expand karke likho.

Exam Step 2 (Hinglish): Triangle step. For any \( a\in F \): \( |x-a|\le|x-y|+|y-a| \).
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — Triangle step. For any \( a\in F \): \( |x-a|\le|x-y|+|y-a| \). ko full sentence mein expand karke likho.

Exam Step 3 (Hinglish): Take inf over \( a \). \( f(x)\le|x-y|+f(y) \) ⇒ \( f(x)-f(y)\le|x-y| \).
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — Take inf over \( a \). \( f(x)\le|x-y|+f(y) \) ⇒ \( f(x)-f(y)\le|x-y| \). ko full sentence mein expand karke likho.

Exam Step 4 (Hinglish): Swap \( x,y \). Similarly \( f(y)-f(x)\le|x-y| \). Combine: \( |f(x)-f(y)|\le|x-y| \).
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — Swap \( x,y \). Similarly \( f(y)-f(x)\le|x-y| \). Combine: \( |f(x)-f(y)|\le|x-y| \). ko full sentence mein expand karke likho.

Exam Step 5 (Hinglish): ε–δ. Given \( \varepsilon>0 \), take \( \delta=\varepsilon \). Then \( |x-y|<\delta \) ⇒ \( |f(x)-f(y)|<\varepsilon \).
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — ε–δ. Given \( \varepsilon>0 \), take \( \delta=\varepsilon \). Then \( |x-y|<\delta \) ⇒ \( |f(x)-f(y)|<\varepsilon \). ko full sentence mein expand karke likho.

Exam Step 6 (Hinglish): Note. Closedness of \( F \) is used when one wants \( f(x)=0\Leftrightarrow x\in F \); continuity of \( f \) holds for any nonempty \( F \).
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — Note. Closedness of \( F \) is used when one wants \( f(x)=0\Leftrightarrow x\in F \); continuity of \( f \) holds for a… ko full sentence mein expand karke likho.

Final line (marks save): End pe clearly likho TRUE/FALSE ya jo construct/prove maanga tha — ek alag sentence: “Hence …”. Agar counterexample hai toh pehle example fully define karo, phir check karo, phir conclude.

Common cutters: (1) definition skip (2) “obviously” without reason (3) wrong quantifiers ∀/∃ (4) compact vs closed confuse (5) image vs pre-image confuse. Inse bacho.

Source / Idea kahan se aayi

Uses: triangle inequality; \( \varepsilon \)–\( \delta \) definition; Lipschitz continuity. Course: Continuity / metric ideas.

Interactive — D2(d): Graphing the Distance Function

Move \( x \) to trace \( f(x) = \inf|x-a| \):

Term Definitions (Q3 Gluing Lemma)

Well-defined: A function is well-defined if each input corresponds to exactly one unambiguous output. (Ek function tab well-defined hota hai jab ek input daalne par sirf ek hi pakka result aaye, confusion na ho).

Open set characterization of continuity: A function \( f \) is continuous if and only if the pre-image \( f^{-1}(V) \) of every open set \( V \) in the codomain is an open set in the domain. (Agar output side ka koi open set lein, toh input side par uske corresponding points ka set bhi open hona chahiye).

Union (\( \cup \)): The combination of all elements from multiple sets. (Alag-alag sets ke saare elements ko mila kar ek bada set banana).

Question · D3 · Gluing / Pasting Lemma

Let \( \{U_\alpha\} \) be open sets in \( \mathbb{R} \) and \( f_\alpha:U_\alpha\to\mathbb{R} \) continuous. Assume \( f_\alpha=f_\beta \) on overlaps \( U_\alpha\cap U_\beta \). Define \( f \) on \( U=\bigcup U_\alpha \) by \( f(x)=f_\alpha(x) \) when \( x\in U_\alpha \). Show \( f \) is well-defined and continuous.

English Mathematical Steps — full marks checklist

Well-defined. If \( x\in U_\alpha\cap U_\beta \), agreement says \( f_\alpha(x)=f_\beta(x) \). Assigned value does not depend on the chart.

Open pre-images. For open \( V\subseteq\mathbb{R} \), \( f^{-1}(V)=\bigcup_\alpha f_\alpha^{-1}(V) \).

Each piece open in \( \mathbb{R} \). Continuity of \( f_\alpha \) + openness of \( U_\alpha \) ⇒ each \( f_\alpha^{-1}(V) \) open in \( \mathbb{R} \).

Union of opens is open ⇒ \( f^{-1}(V) \) open ⇒ \( f \) continuous.

Exam-style answer (copy into answer sheet)

Given. Open cover \( \{U_\alpha\} \), continuous \( f_\alpha: U_\alpha \to \mathbb{R} \), agree on overlaps. \( f \) on \( U = \bigcup U_\alpha \) by \( f(x) = f_\alpha(x) \) if \( x \in U_\alpha \).

To prove. \( f \) is well-defined and continuous (Pasting Lemma).

Step 1. Well-defined: overlap agreement \( f_\alpha = f_\beta \) on \( U_\alpha \cap U_\beta \).

Step 2. For open \( V \subseteq \mathbb{R} \): \( f^{-1}(V) = \bigcup_\alpha f_\alpha^{-1}(V) \), each preimage open in \( \mathbb{R} \).

Final answer. Union of open sets is open ⇒ \( f \) continuous.
Method. Present the solution as a proof or counterexample with numbered steps. Each step must state the definition or theorem used and why the next line follows — no “obvious” jumps.
Marks tip. Underline the final boxed answer. For TRUE/FALSE, include a one-line justification name (e.g. connectedness, counterexample, EVT). Missing quantifiers or confused open/closed/compact costs marks.

Method. We solve this in the exam format: state definitions used, justify each implication, and end with a clear final line.

Step 1. Well-defined. If \( x\in U_\alpha\cap U_\beta \), agreement says \( f_\alpha(x)=f_\beta(x) \). Assigned value does not depend on the chart. (Write this justification fully on the answer sheet — do not skip definitions or “why”.)

Step 2. Open pre-images. For open \( V\subseteq\mathbb{R} \), \( f^{-1}(V)=\bigcup_\alpha f_\alpha^{-1}(V) \). (Write this justification fully on the answer sheet — do not skip definitions or “why”.)

Step 3. Each piece open in \( \mathbb{R} \). Continuity of \( f_\alpha \) + openness of \( U_\alpha \) ⇒ each \( f_\alpha^{-1}(V) \) open in \( \mathbb{R} \). (Write this justification fully on the answer sheet — do not skip definitions or “why”.)

Step 4. Union of opens is open ⇒ \( f^{-1}(V) \) open ⇒ \( f \) continuous. (Write this justification fully on the answer sheet — do not skip definitions or “why”.)

Final answer (box this). Union of opens is open ⇒ \( f^{-1}(V) \) open ⇒ \( f \) continuous.

Marks tip. Examiner cuts marks for missing definitions, unjustified jumps, and unclear TRUE/FALSE or final claim. Keep every “because” explicit.

Hinglish — detailed (examiner-safe, beginner-clear)

Step 1 — Well-defined: Agar x do charts mein hai, overlap pe \( f_\alpha = f_\beta \) — isliye value unique. Ye likhna mat bhoolo warna “function” hi doubt mein aa jata hai.

Step 2 — Preimage: Open \( V \) ka preimage har piece mein \( f_\alpha^{-1}(V) \) — continuous \( f_\alpha \) se ye open hai.

Step 3 — Union: Sab pieces ka union open ⇒ global \( f \) continuous. Pasting Lemma ka core yahi hai.

Pasting: Overlap pe same value → function unique; continuity ke liye preimage open union.
Exam mindset: Copy-paste feel mat do — har step pe “definition → apply → conclude” likho. English math symbols same rakh sakte ho; explanation Hinglish mein full sentences mein.
Common cutters: Definition skip, “obviously” bina reason, ∀/∃ galat, compact vs closed confuse, image vs pre-image confuse — in sab se bacho.
Final line (marks save): End pe clearly likho jo construct/prove maanga tha — ek alag “Hence …” sentence. Counterexample ho toh pehle example define, phir verify, phir conclude.

Exam mindset: Copy-paste feel nahi dena; har line pe “definition → apply → conclude” dikhana. Neeche wala write-up answer sheet pe utaar sakte ho (Hinglish soch + English math symbols).

Exam Step 1 (Hinglish): Well-defined. If \( x\in U_\alpha\cap U_\beta \), agreement says \( f_\alpha(x)=f_\beta(x) \). Assigned value does not depend on the chart.
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — Well-defined. If \( x\in U_\alpha\cap U_\beta \), agreement says \( f_\alpha(x)=f_\beta(x) \). Assigned value does not d… ko full sentence mein expand karke likho.

Exam Step 2 (Hinglish): Open pre-images. For open \( V\subseteq\mathbb{R} \), \( f^{-1}(V)=\bigcup_\alpha f_\alpha^{-1}(V) \).
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — Open pre-images. For open \( V\subseteq\mathbb{R} \), \( f^{-1}(V)=\bigcup_\alpha f_\alpha^{-1}(V) \). ko full sentence mein expand karke likho.

Exam Step 3 (Hinglish): Each piece open in \( \mathbb{R} \). Continuity of \( f_\alpha \) + openness of \( U_\alpha \) ⇒ each \( f_\alpha^{-1}(V) \) open in \( \mathbb{R} \).
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — Each piece open in \( \mathbb{R} \). Continuity of \( f_\alpha \) + openness of \( U_\alpha \) ⇒ each \( f_\alpha^{-1}(V… ko full sentence mein expand karke likho.

Exam Step 4 (Hinglish): Union of opens is open ⇒ \( f^{-1}(V) \) open ⇒ \( f \) continuous.
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — Union of opens is open ⇒ \( f^{-1}(V) \) open ⇒ \( f \) continuous. ko full sentence mein expand karke likho.

Final line (marks save): End pe clearly likho TRUE/FALSE ya jo construct/prove maanga tha — ek alag sentence: “Hence …”. Agar counterexample hai toh pehle example fully define karo, phir check karo, phir conclude.

Common cutters: (1) definition skip (2) “obviously” without reason (3) wrong quantifiers ∀/∃ (4) compact vs closed confuse (5) image vs pre-image confuse. Inse bacho.

Source / Idea kahan se aayi

Uses: open-set characterization of continuity; arbitrary unions of open sets. Course: Continuity (gluing lemma).

Term Definitions (Q4 Composition)

Composition of functions (\( f \circ g \)): Applying one function to the results of another. Mathematically, \( (f \circ g)(x) = f(g(x)) \). (Ek function ke output ko dusre function ke input ki tarah istemal karna).

Discontinuous Function: A function that is not continuous, meaning its graph has jumps, breaks, or holes. (Ek function jiske graph mein break ho, jo lagatar na chale).

Constant Function: A function whose output value is the same for every input value. (Aisa function jiska answer hamesha ek hi aaye, chahe aap \( x \) ki value kuch bhi daalo).

Question · D4 · Discontinuous ∘ discontinuous can be continuous

Find discontinuous \( f,g \) such that \( f\circ g \) is continuous.

English Mathematical Steps — full marks checklist

Jump \( g \). \( g(x)=1 \) if \( x\ge 0 \), \( g(x)=-1 \) if \( x<0 \) — discontinuous at \( 0 \).

Spike \( f \). \( f(\pm 1)=0 \) and \( f=5 \) elsewhere — discontinuous at \( \pm 1 \).

Compose. \( g(x)\in\{\pm 1\} \) always ⇒ \( f(g(x))=0 \) constantly ⇒ continuous.

Moral. Continuous composition does not force continuous factors.

Exam-style answer (copy into answer sheet)

Given. Find discontinuous \( f, g:\mathbb{R}\to\mathbb{R} \) with continuous \( f \circ g \).

Step 1. \( g(x) = 1 \) if \( x \ge 0 \), \( g(x) = -1 \) if \( x < 0 \) (jump at 0).

Step 2. \( f(t) = 0 \) for \( t = \pm 1 \), else \( f(t) = 5 \) (discontinuous at \( \pm 1 \)).

Step 3. \( g(x) \in \{\pm 1\} \) always, so \( (f \circ g)(x) = 0 \) constant ⇒ continuous.

Final answer. Such \( f, g \) exist; composition can be continuous without both factors continuous.
Method. Present the solution as a proof or counterexample with numbered steps. Each step must state the definition or theorem used and why the next line follows — no “obvious” jumps.
Marks tip. Underline the final boxed answer. For TRUE/FALSE, include a one-line justification name (e.g. connectedness, counterexample, EVT). Missing quantifiers or confused open/closed/compact costs marks.

Method. We solve this in the exam format: state definitions used, justify each implication, and end with a clear final line.

Step 1. Jump \( g \). \( g(x)=1 \) if \( x\ge 0 \), \( g(x)=-1 \) if \( x<0 \) — discontinuous at \( 0 \). (Write this justification fully on the answer sheet — do not skip definitions or “why”.)

Step 2. Spike \( f \). \( f(\pm 1)=0 \) and \( f=5 \) elsewhere — discontinuous at \( \pm 1 \). (Write this justification fully on the answer sheet — do not skip definitions or “why”.)

Step 3. Compose. \( g(x)\in\{\pm 1\} \) always ⇒ \( f(g(x))=0 \) constantly ⇒ continuous. (Write this justification fully on the answer sheet — do not skip definitions or “why”.)

Step 4. Moral. Continuous composition does not force continuous factors. (Write this justification fully on the answer sheet — do not skip definitions or “why”.)

Final answer (box this). Moral. Continuous composition does not force continuous factors.

Marks tip. Examiner cuts marks for missing definitions, unjustified jumps, and unclear TRUE/FALSE or final claim. Keep every “because” explicit.

Hinglish — detailed (examiner-safe, beginner-clear)

Idea: Inner function sirf do values pe stuck; outer un values pe same output de.

Verify: Dono discontinuous points explicitly mention karo.
Exam mindset: Copy-paste feel mat do — har step pe “definition → apply → conclude” likho. English math symbols same rakh sakte ho; explanation Hinglish mein full sentences mein.
Common cutters: Definition skip, “obviously” bina reason, ∀/∃ galat, compact vs closed confuse, image vs pre-image confuse — in sab se bacho.
Final line (marks save): End pe clearly likho jo construct/prove maanga tha — ek alag “Hence …” sentence. Counterexample ho toh pehle example define, phir verify, phir conclude.

Exam mindset: Copy-paste feel nahi dena; har line pe “definition → apply → conclude” dikhana. Neeche wala write-up answer sheet pe utaar sakte ho (Hinglish soch + English math symbols).

Exam Step 1 (Hinglish): Jump \( g \). \( g(x)=1 \) if \( x\ge 0 \), \( g(x)=-1 \) if \( x<0 \) — discontinuous at \( 0 \).
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — Jump \( g \). \( g(x)=1 \) if \( x\ge 0 \), \( g(x)=-1 \) if \( x<0 \) — discontinuous at \( 0 \). ko full sentence mein expand karke likho.

Exam Step 2 (Hinglish): Spike \( f \). \( f(\pm 1)=0 \) and \( f=5 \) elsewhere — discontinuous at \( \pm 1 \).
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — Spike \( f \). \( f(\pm 1)=0 \) and \( f=5 \) elsewhere — discontinuous at \( \pm 1 \). ko full sentence mein expand karke likho.

Exam Step 3 (Hinglish): Compose. \( g(x)\in\{\pm 1\} \) always ⇒ \( f(g(x))=0 \) constantly ⇒ continuous.
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — Compose. \( g(x)\in\{\pm 1\} \) always ⇒ \( f(g(x))=0 \) constantly ⇒ continuous. ko full sentence mein expand karke likho.

Exam Step 4 (Hinglish): Moral. Continuous composition does not force continuous factors.
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — Moral. Continuous composition does not force continuous factors. ko full sentence mein expand karke likho.

Final line (marks save): End pe clearly likho TRUE/FALSE ya jo construct/prove maanga tha — ek alag sentence: “Hence …”. Agar counterexample hai toh pehle example fully define karo, phir check karo, phir conclude.

Common cutters: (1) definition skip (2) “obviously” without reason (3) wrong quantifiers ∀/∃ (4) compact vs closed confuse (5) image vs pre-image confuse. Inse bacho.

Source / Idea kahan se aayi

Uses: continuity definitions; composition. Course: Continuity counterexamples.

Interactive — D4: Discontinuous \( f \) and \( g \) creating a continuous \( f \circ g \)

Term Definitions (Q5 Image of Sets)

Pre-image (\( f^{-1}(K) \)): The set of all inputs in the domain that map to a specific set of outputs \( K \) in the codomain. (Wo saare inputs jinko function mein daalne par answer set \( K \) ke andar se aaye).

Compact Set: A set that is both closed and bounded (in \( \mathbb{R} \)). (Aisa set jo band ho aur jiska size limited ho, infinite tak na phaila ho).

Question · D5(a) · Continuous image of closed need not be closed

Can closed \( A\subseteq\mathbb{R} \) and continuous \( f \) have non-closed \( f(A) \)?

English Mathematical Steps — full marks checklist

Yes. Take \( A=\mathbb{N} \) (closed, discrete) and \( f(x)=\frac{1}{x^2+1} \).

Image \( \{\frac{1}{n^2+1}\} \) has limit point \( 0\notin f(A) \) ⇒ not closed.

Contrast: continuous image of compact sets is compact (hence closed).

Exam-style answer (copy into answer sheet)

Given. Can closed \( A \subseteq \mathbb{R} \) and continuous \( f \) have non-closed \( f(A) \)?

Step 1. Yes: \( A = \mathbb{N} \) (closed, discrete), \( f(x) = \frac{1}{x^2 + 1} \).

Step 2. \( f(A) = \{ \frac{1}{n^2 + 1} : n \in \mathbb{N} \} \) accumulates at \( 0 \notin f(A) \).

Final answer. Yes — image need not be closed without compactness of domain.
Method. Present the solution as a proof or counterexample with numbered steps. Each step must state the definition or theorem used and why the next line follows — no “obvious” jumps.
Marks tip. Underline the final boxed answer. For TRUE/FALSE, include a one-line justification name (e.g. connectedness, counterexample, EVT). Missing quantifiers or confused open/closed/compact costs marks.

Method. We solve this in the exam format: state definitions used, justify each implication, and end with a clear final line.

Step 1. Yes. Take \( A=\mathbb{N} \) (closed, discrete) and \( f(x)=\frac{1}{x^2+1} \). (Write this justification fully on the answer sheet — do not skip definitions or “why”.)

Step 2. Image \( \{\frac{1}{n^2+1}\} \) has limit point \( 0\notin f(A) \) ⇒ not closed. (Write this justification fully on the answer sheet — do not skip definitions or “why”.)

Step 3. Contrast: continuous image of compact sets is compact (hence closed). (Write this justification fully on the answer sheet — do not skip definitions or “why”.)

Final answer (box this). Contrast: continuous image of compact sets is compact (hence closed).

Marks tip. Examiner cuts marks for missing definitions, unjustified jumps, and unclear TRUE/FALSE or final claim. Keep every “because” explicit.

Hinglish — detailed (examiner-safe, beginner-clear)

Example: Natural numbers closed; reciprocal-square image limit 0 miss.

Contrast: Compact A pe image compact/closed.
Exam mindset: Copy-paste feel mat do — har step pe “definition → apply → conclude” likho. English math symbols same rakh sakte ho; explanation Hinglish mein full sentences mein.
Common cutters: Definition skip, “obviously” bina reason, ∀/∃ galat, compact vs closed confuse, image vs pre-image confuse — in sab se bacho.
Final line (marks save): End pe clearly likho jo construct/prove maanga tha — ek alag “Hence …” sentence. Counterexample ho toh pehle example define, phir verify, phir conclude.

Exam mindset: Copy-paste feel nahi dena; har line pe “definition → apply → conclude” dikhana. Neeche wala write-up answer sheet pe utaar sakte ho (Hinglish soch + English math symbols).

Exam Step 1 (Hinglish): Yes. Take \( A=\mathbb{N} \) (closed, discrete) and \( f(x)=\frac{1}{x^2+1} \).
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — Yes. Take \( A=\mathbb{N} \) (closed, discrete) and \( f(x)=\frac{1}{x^2+1} \). ko full sentence mein expand karke likho.

Exam Step 2 (Hinglish): Image \( \{\frac{1}{n^2+1}\} \) has limit point \( 0\notin f(A) \) ⇒ not closed.
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — Image \( \{\frac{1}{n^2+1}\} \) has limit point \( 0\notin f(A) \) ⇒ not closed. ko full sentence mein expand karke likho.

Exam Step 3 (Hinglish): Contrast: continuous image of compact sets is compact (hence closed).
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — Contrast: continuous image of compact sets is compact (hence closed). ko full sentence mein expand karke likho.

Final line (marks save): End pe clearly likho TRUE/FALSE ya jo construct/prove maanga tha — ek alag sentence: “Hence …”. Agar counterexample hai toh pehle example fully define karo, phir check karo, phir conclude.

Common cutters: (1) definition skip (2) “obviously” without reason (3) wrong quantifiers ∀/∃ (4) compact vs closed confuse (5) image vs pre-image confuse. Inse bacho.

Source / Idea kahan se aayi

Uses: closed ⇔ contains limit points. Course: Continuity + closed/compact.

Interactive — D5(a): Projection of \( A = \mathbb{N} \) through \( f(x) = 1/(x^2+1) \)

Number of points to evaluate (N):

Question · D5(b) · Pre-image of compact need not be compact

Can compact \( K \) and continuous \( f \) have non-compact \( f^{-1}(K) \)?

English Mathematical Steps — full marks checklist

Yes. Constant \( f\equiv 0 \), \( K=\{0\} \) compact.

Pre-image \( =\mathbb{R} \) unbounded ⇒ not compact (Heine–Borel).

Moral: continuous pre-images preserve closedness, not compactness.

Exam-style answer (copy into answer sheet)

Given. Can compact \( K \) and continuous \( f \) have non-compact \( f^{-1}(K) \)?

Step 1. \( f \equiv 0 \), \( K = \{0\} \) compact.

Step 2. \( f^{-1}(K) = \mathbb{R} \), unbounded ⇒ not compact in \( \mathbb{R} \).

Final answer. Yes. Continuous pre-images need not preserve compactness (unlike closed sets).
Method. Present the solution as a proof or counterexample with numbered steps. Each step must state the definition or theorem used and why the next line follows — no “obvious” jumps.
Marks tip. Underline the final boxed answer. For TRUE/FALSE, include a one-line justification name (e.g. connectedness, counterexample, EVT). Missing quantifiers or confused open/closed/compact costs marks.

Method. We solve this in the exam format: state definitions used, justify each implication, and end with a clear final line.

Step 1. Yes. Constant \( f\equiv 0 \), \( K=\{0\} \) compact. (Write this justification fully on the answer sheet — do not skip definitions or “why”.)

Step 2. Pre-image \( =\mathbb{R} \) unbounded ⇒ not compact (Heine–Borel). (Write this justification fully on the answer sheet — do not skip definitions or “why”.)

Step 3. Moral: continuous pre-images preserve closedness, not compactness. (Write this justification fully on the answer sheet — do not skip definitions or “why”.)

Final answer (box this). Moral: continuous pre-images preserve closedness, not compactness.

Marks tip. Examiner cuts marks for missing definitions, unjustified jumps, and unclear TRUE/FALSE or final claim. Keep every “because” explicit.

Hinglish — detailed (examiner-safe, beginner-clear)

Constant map: Poori line preimage — Heine–Borel fail.

Moral: Forward image compact good for compact domain; inverse alag story.
Exam mindset: Copy-paste feel mat do — har step pe “definition → apply → conclude” likho. English math symbols same rakh sakte ho; explanation Hinglish mein full sentences mein.
Common cutters: Definition skip, “obviously” bina reason, ∀/∃ galat, compact vs closed confuse, image vs pre-image confuse — in sab se bacho.
Final line (marks save): End pe clearly likho jo construct/prove maanga tha — ek alag “Hence …” sentence. Counterexample ho toh pehle example define, phir verify, phir conclude.

Exam mindset: Copy-paste feel nahi dena; har line pe “definition → apply → conclude” dikhana. Neeche wala write-up answer sheet pe utaar sakte ho (Hinglish soch + English math symbols).

Exam Step 1 (Hinglish): Yes. Constant \( f\equiv 0 \), \( K=\{0\} \) compact.
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — Yes. Constant \( f\equiv 0 \), \( K=\{0\} \) compact. ko full sentence mein expand karke likho.

Exam Step 2 (Hinglish): Pre-image \( =\mathbb{R} \) unbounded ⇒ not compact (Heine–Borel).
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — Pre-image \( =\mathbb{R} \) unbounded ⇒ not compact (Heine–Borel). ko full sentence mein expand karke likho.

Exam Step 3 (Hinglish): Moral: continuous pre-images preserve closedness, not compactness.
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — Moral: continuous pre-images preserve closedness, not compactness. ko full sentence mein expand karke likho.

Final line (marks save): End pe clearly likho TRUE/FALSE ya jo construct/prove maanga tha — ek alag sentence: “Hence …”. Agar counterexample hai toh pehle example fully define karo, phir check karo, phir conclude.

Common cutters: (1) definition skip (2) “obviously” without reason (3) wrong quantifiers ∀/∃ (4) compact vs closed confuse (5) image vs pre-image confuse. Inse bacho.

Source / Idea kahan se aayi

Uses: Heine–Borel. Course: Compactness + continuity.

Term Definitions (Q6 Disjoint Intervals)

Density of Rationals (Parimey sankhyaon ka ghanatva): Between any two distinct real numbers, there exists at least one rational number. (Koi bhi do alag-alag real numbers ke beech mein humesha ek rational number \( p/q \) maujood hota hai).

Countable Set (Ginne yogya set): A set whose elements can be matched one-to-one with the natural numbers (\( 1, 2, 3... \)).

Disjoint Intervals (Alag-alag intervals): Intervals that do not overlap or share any common points.

Question · D6 · No uncountable disjoint open intervals

Show there cannot be uncountably many pairwise disjoint nonempty open intervals in \( \mathbb{R} \).

English Mathematical Steps — full marks checklist

Pick rationals. Each open interval contains a rational (density).

Injective map from intervals to \( \mathbb{Q} \) (disjoint ⇒ distinct rationals).

\( \mathbb{Q} \) countable ⇒ at most countably many such intervals.

Exam-style answer (copy into answer sheet)

Given. Show there cannot be uncountably many pairwise disjoint nonempty open intervals in \( \mathbb{R} \).

Step 1. Each open interval contains a rational (density of \( \mathbb{Q} \)).

Step 2. Pick one rational per interval; disjoint intervals ⇒ distinct rationals ⇒ injective map to \( \mathbb{Q} \).

Step 3. \( \mathbb{Q} \) countable ⇒ at most countably many such intervals.

Final answer. Uncountably many pairwise disjoint nonempty open intervals in \( \mathbb{R} \) cannot exist.
Method. Present the solution as a proof or counterexample with numbered steps. Each step must state the definition or theorem used and why the next line follows — no “obvious” jumps.
Marks tip. Underline the final boxed answer. For TRUE/FALSE, include a one-line justification name (e.g. connectedness, counterexample, EVT). Missing quantifiers or confused open/closed/compact costs marks.

Method. We solve this in the exam format: state definitions used, justify each implication, and end with a clear final line.

Step 1. Pick rationals. Each open interval contains a rational (density). (Write this justification fully on the answer sheet — do not skip definitions or “why”.)

Step 2. Injective map from intervals to \( \mathbb{Q} \) (disjoint ⇒ distinct rationals). (Write this justification fully on the answer sheet — do not skip definitions or “why”.)

Step 3. \( \mathbb{Q} \) countable ⇒ at most countably many such intervals. (Write this justification fully on the answer sheet — do not skip definitions or “why”.)

Final answer (box this). \( \mathbb{Q} \) countable ⇒ at most countably many such intervals.

Marks tip. Examiner cuts marks for missing definitions, unjustified jumps, and unclear TRUE/FALSE or final claim. Keep every “because” explicit.

Hinglish — detailed (examiner-safe, beginner-clear)

Counting trick: Har interval mein ek rational — disjointness se injection.

Conclusion: Uncountable family impossible.
Exam mindset: Copy-paste feel mat do — har step pe “definition → apply → conclude” likho. English math symbols same rakh sakte ho; explanation Hinglish mein full sentences mein.
Common cutters: Definition skip, “obviously” bina reason, ∀/∃ galat, compact vs closed confuse, image vs pre-image confuse — in sab se bacho.
Final line (marks save): End pe clearly likho jo construct/prove maanga tha — ek alag “Hence …” sentence. Counterexample ho toh pehle example define, phir verify, phir conclude.

Exam mindset: Copy-paste feel nahi dena; har line pe “definition → apply → conclude” dikhana. Neeche wala write-up answer sheet pe utaar sakte ho (Hinglish soch + English math symbols).

Exam Step 1 (Hinglish): Pick rationals. Each open interval contains a rational (density).
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — Pick rationals. Each open interval contains a rational (density). ko full sentence mein expand karke likho.

Exam Step 2 (Hinglish): Injective map from intervals to \( \mathbb{Q} \) (disjoint ⇒ distinct rationals).
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — Injective map from intervals to \( \mathbb{Q} \) (disjoint ⇒ distinct rationals). ko full sentence mein expand karke likho.

Exam Step 3 (Hinglish): \( \mathbb{Q} \) countable ⇒ at most countably many such intervals.
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — \( \mathbb{Q} \) countable ⇒ at most countably many such intervals. ko full sentence mein expand karke likho.

Final line (marks save): End pe clearly likho TRUE/FALSE ya jo construct/prove maanga tha — ek alag sentence: “Hence …”. Agar counterexample hai toh pehle example fully define karo, phir check karo, phir conclude.

Common cutters: (1) definition skip (2) “obviously” without reason (3) wrong quantifiers ∀/∃ (4) compact vs closed confuse (5) image vs pre-image confuse. Inse bacho.

Source / Idea kahan se aayi

Uses: density + countability of \( \mathbb{Q} \). Course: Countability / topology of \( \mathbb{R} \).

Term Definitions (Q7 Open Sets Structure)

Equivalence Relation: A relation that is Reflexive (\( x \sim x \)), Symmetric (if \( x \sim y \), then \( y \sim x \)), and Transitive (if \( x \sim y \) and \( y \sim z \), then \( x \sim z \)).

Equivalence Class (\( [x] \)): The set of all elements that are related to \( x \). (Wo saare points ka group jo \( x \) se jude hue hain).

Question · D7 · Open sets = countable unions of disjoint open intervals

For open \( U\subseteq\mathbb{R} \), relate points that share an open interval inside \( U \). Show classes are disjoint open intervals and \( U \) is their countable union.

English Mathematical Steps — full marks checklist

Equivalence. Reflexive (openness), symmetric (rewrite interval), transitive (glue overlapping intervals).

Classes = open intervals (union of intervals through a point).

Disjoint classes (equivalence-class property).

Countability by D6 ⇒ \( U \) countable union of those intervals.

Exam-style answer (copy into answer sheet)

Given. Let \(U\subseteq\mathbb R\) be open. For \(x,y\in U\), define \(x\sim y\) when there is an open interval \(I\subseteq U\) containing both \(x\) and \(y\).

To prove. The equivalence classes are pairwise disjoint open intervals, and \(U\) is the union of at most countably many such intervals.

Step 1 — reflexivity. If \(x\in U\), openness gives \(\varepsilon>0\) with \((x-\varepsilon,x+\varepsilon)\subseteq U\). This interval contains \(x\), so \(x\sim x\).

Step 2 — symmetry and transitivity. Symmetry follows because an interval containing \(x,y\) also contains \(y,x\). If \(x\sim y\) through \(I\) and \(y\sim z\) through \(J\), then \(y\in I\cap J\). Two intersecting intervals have an interval as their union, so \(I\cup J\subseteq U\) contains \(x,z\). Hence \(x\sim z\).

Step 3 — each class is open. Let \(y\in[x]\). Choose an open interval \(I\subseteq U\) containing \(x,y\). Because \(I\) is open, some neighborhood of \(y\) lies in \(I\), and every point of that neighborhood is also related to \(x\). Thus \([x]\) is open.

Step 4 — each class is an interval. If \(y,z\in[x]\) and \(y
Step 5 — decomposition and countability. Equivalence classes are equal or disjoint and cover \(U\). Every nonempty open interval contains a rational. Assigning one rational to each class is injective because distinct classes are disjoint. Since \(\mathbb Q\) is countable, there are at most countably many classes.

Final answer. Every open \(U\subseteq\mathbb R\) is a finite or countable union of pairwise disjoint open interval components.

Hinglish — detailed (examiner-safe, beginner-clear)

Question ka matlab: \(U\) ke do points ko same group mein rakho jab \(U\) ke andar ka ek open interval un dono ko connect kare. Prove karna hai ki ye groups exactly disjoint open intervals hain.

Step 1 — relation check: Reflexive openness se milta hai; symmetric definition se. Transitive ke liye \(x,y\) wala interval aur \(y,z\) wala interval common point \(y\) par overlap karte hain. Unka union bhi interval hai, so \(x\sim z\).

Step 2 — class open kyun: Agar \(y\in[x]\), connecting interval ke andar \(y\) ka chhota neighborhood bhi hai. Us neighborhood ka har point \(x\) se related hai, hence class open.

Step 3 — class interval kyun: Class mein \(y
Step 4 — countability: Classes partition \(U\), so equal ya disjoint hain. Har class ko ek rational ticket do. Distinct classes ke tickets distinct honge aur \(\mathbb Q\) countable hai; hence classes at most countable.

Final line: \(U\) pairwise disjoint open intervals ka finite ya countable union hai. Common trap: Reflexive/transitive proof aur rays \((a,\infty)\), \((-\infty,b)\) ko skip mat karo.

Exam mindset: Copy-paste feel nahi dena; har line pe “definition → apply → conclude” dikhana. Neeche wala write-up answer sheet pe utaar sakte ho (Hinglish soch + English math symbols).

Exam Step 1 (Hinglish): Equivalence. Reflexive (openness), symmetric (rewrite interval), transitive (glue overlapping intervals).
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — Equivalence. Reflexive (openness), symmetric (rewrite interval), transitive (glue overlapping intervals). ko full sentence mein expand karke likho.

Exam Step 2 (Hinglish): Classes = open intervals (union of intervals through a point).
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — Classes = open intervals (union of intervals through a point). ko full sentence mein expand karke likho.

Exam Step 3 (Hinglish): Disjoint classes (equivalence-class property).
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — Disjoint classes (equivalence-class property). ko full sentence mein expand karke likho.

Exam Step 4 (Hinglish): Countability by D6 ⇒ \( U \) countable union of those intervals.
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — Countability by D6 ⇒ \( U \) countable union of those intervals. ko full sentence mein expand karke likho.

Final line (marks save): End pe clearly likho TRUE/FALSE ya jo construct/prove maanga tha — ek alag sentence: “Hence …”. Agar counterexample hai toh pehle example fully define karo, phir check karo, phir conclude.

Common cutters: (1) definition skip (2) “obviously” without reason (3) wrong quantifiers ∀/∃ (4) compact vs closed confuse (5) image vs pre-image confuse. Inse bacho.

Source / Idea kahan se aayi

Uses: equivalence relations; D6. Course: Topology of \( \mathbb{R} \).

Interactive — Q6/Q7: Mapping Intervals to Rationals

Click the button to see why uncountably many disjoint intervals cannot exist.

Term Definitions (Q8 Unbounded Continuous Functions)

Unbounded Function (Aseemit Function): A function whose outputs grow to positive or negative infinity; it cannot be trapped between two finite horizontal lines.

Question · D8 · Unbounded continuous function on every open set

On every \( (a,b) \) build continuous unbounded \( f \). Using D7+D3, do the same for every nonempty open \( U \).

English Mathematical Steps — full marks checklist

On \( (a,b) \): \( f(x)=\frac{1}{x-a} \) continuous and \( \to+\infty \) as \( x\to a^+ \).

Decompose \( U \) into countable disjoint open intervals (D7).

Paste the same explosion on each piece via D3 (overlaps empty).

Result: continuous unbounded on all of \( U \).

Exam-style answer (copy into answer sheet)

Given. Let \(U\subseteq\mathbb R\) be nonempty and open.

To construct. A continuous function \(f:U\to\mathbb R\) which is unbounded.

Step 1 — bounded interval. On \(I=(a,b)\), set \(h_I(x)=1/(x-a)\). It is continuous because its denominator is nonzero on \(I\). Moreover \(h_I(x)\to+\infty\) as \(x\to a^+\), so it is unbounded.

Step 2 — handle every interval type. If \(I=(a,\infty)\), use \(h_I(x)=1/(x-a)\). If \(I=(-\infty,b)\), use \(h_I(x)=1/(b-x)\). If \(I=\mathbb R\), use \(h_I(x)=x\). Each formula is continuous and unbounded on its stated domain.

Step 3 — decompose \(U\). By D7, \(U=\bigcup_\alpha I_\alpha\), where the \(I_\alpha\) are pairwise disjoint nonempty open interval components. This includes bounded intervals, rays, and possibly \(\mathbb R\).

Step 4 — define and paste. For \(x\in I_\alpha\), define \(f(x)=h_{I_\alpha}(x)\) using the appropriate formula from Steps 1–2. The definition is unambiguous because components are disjoint. Every local function is continuous, and the domains \(I_\alpha\) are open; therefore D3, the open gluing lemma, implies that \(f\) is continuous on \(U\).

Step 5 — verify unboundedness. Since \(U\ne\emptyset\), at least one component \(I_\alpha\) exists. The restriction \(f|_{I_\alpha}=h_{I_\alpha}\) is unbounded, so \(f\) is unbounded on all of \(U\).

Final answer. Every nonempty open subset of \(\mathbb R\) admits a continuous unbounded real-valued function.

Hinglish — detailed (examiner-safe, beginner-clear)

Question ka matlab: Kisi bhi nonempty open set \(U\) par continuous function banana hai jiske outputs kisi finite bound ke andar na rahen.

Step 1 — interval recipe: \((a,b)\) ya \((a,\infty)\) par \(1/(x-a)\) boundary \(a\) ke paas explode karta hai. Left ray \((-\infty,b)\) par \(1/(b-x)\), aur poori \(\mathbb R\) par \(x\) use karo.

Step 2 — D7: Open set ko disjoint open interval components \(I_\alpha\) mein split karo. Har component par uske type ke according above formula define karo.

Step 3 — D3: Components open aur disjoint hain, so overlap conflict nahi. Har local formula continuous hai; open gluing lemma se combined \(f\) poore \(U\) par continuous hai.

Step 4 — unboundedness: \(U\) nonempty hai, hence at least one component hai. Function us component par unbounded hai, therefore \(U\) par bhi unbounded hai.

Final line: Every nonempty open \(U\subseteq\mathbb R\) supports a continuous unbounded function. Common trap: Sirf \(1/(x-a)\) mat likho; \(\mathbb R\) aur left-ray cases mein finite left endpoint hota hi nahi.

Exam mindset: Copy-paste feel nahi dena; har line pe “definition → apply → conclude” dikhana. Neeche wala write-up answer sheet pe utaar sakte ho (Hinglish soch + English math symbols).

Exam Step 1 (Hinglish): On \( (a,b) \): \( f(x)=\frac{1}{x-a} \) continuous and \( \to+\infty \) as \( x\to a^+ \).
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — On \( (a,b) \): \( f(x)=\frac{1}{x-a} \) continuous and \( \to+\infty \) as \( x\to a^+ \). ko full sentence mein expand karke likho.

Exam Step 2 (Hinglish): Decompose \( U \) into countable disjoint open intervals (D7).
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — Decompose \( U \) into countable disjoint open intervals (D7). ko full sentence mein expand karke likho.

Exam Step 3 (Hinglish): Paste the same explosion on each piece via D3 (overlaps empty).
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — Paste the same explosion on each piece via D3 (overlaps empty). ko full sentence mein expand karke likho.

Exam Step 4 (Hinglish): Result: continuous unbounded on all of \( U \).
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — Result: continuous unbounded on all of \( U \). ko full sentence mein expand karke likho.

Final line (marks save): End pe clearly likho TRUE/FALSE ya jo construct/prove maanga tha — ek alag sentence: “Hence …”. Agar counterexample hai toh pehle example fully define karo, phir check karo, phir conclude.

Common cutters: (1) definition skip (2) “obviously” without reason (3) wrong quantifiers ∀/∃ (4) compact vs closed confuse (5) image vs pre-image confuse. Inse bacho.

Source / Idea kahan se aayi

Uses: D7, D3 pasting lemma. Course: Continuity on open sets.

Interactive — Q8: Unbounded Continuous Function on U

This demonstrates \( f(x) = \frac{0.5}{x - a_\alpha} \) plotted across three disjoint open intervals.

Term Definitions (Q9 Continuity Set)

Set of Continuity Points \( U \): All \( x \) where \( f \) is continuous. (Wo saari \( x \) values jahan function smooth hai, toota nahi.)

Thomae's Function (Popcorn): \( f(p/q)=1/q \) in lowest terms, \( f=0 \) on irrationals — classic counterexample tool.

Question · D9 · Continuity set open/closed/neither

Let \( f:\mathbb{R}\to\mathbb{R} \). Let \( U=\{x: f\text{ is continuous at }x\} \). Must \( U \) be open? closed? Or can it be neither?

English Mathematical Steps — full marks checklist

Goal: Show that \( U \) need not be open and need not be closed — all three behaviours are possible.

Case Open: Define \( f(x)=0 \) for \( x\le 0 \) and \( f(x)=1 \) for \( x>0 \). Then \( f \) fails continuity only at \( 0 \), so \( U=\mathbb{R}\setminus\{0\} \), which is open.

Case Closed: Define \( f(x)=x \) on \( \mathbb{Q} \) and \( f(x)=0 \) on irrationals. Then ( |f(x)|\le|x| \), so by squeeze \( f \) is continuous only at \( 0 \). Thus \( U=\{0\} \), which is closed.

Case Neither: Thomae's function \( f(p/q)=1/q \), \( f=0 \) on irrationals, is continuous exactly on \( \mathbb{R}\setminus\mathbb{Q} \). Irrationals are neither open nor closed in \( \mathbb{R} \).

Conclusion: \( U \) is not forced to be open or closed.

Exam-style answer (copy into answer sheet)

Given / Statement. Let \( f:\mathbb{R}\to\mathbb{R} \). Let \( U=\{x: f\text{ is continuous at }x\} \). Must \( U \) be open? closed? Or can it be neither?

To prove / To find. Show all three behaviours are possible; \( U \) is not forced to be open or closed.

Step 1 (Goal). Exhibit examples where \( U \) is open, where \( U \) is closed, and where \( U \) is neither open nor closed in \( \mathbb{R} \).

Step 2 (Case open). Define \( f(x)=0 \) for \( x\le 0 \) and \( f(x)=1 \) for \( x>0 \). Then \( f \) fails continuity only at \( 0 \) (e.g. \( 1/n\to 0 \) but \( f(1/n)=1\not\to f(0)=0 \)). Hence \( U=\mathbb{R}\setminus\{0\} \), which is open in \( \mathbb{R} \).

Step 3 (Case closed). Define \( f(x)=x \) on \( \mathbb{Q} \) and \( f(x)=0 \) on irrationals. Then \( |f(x)|\le |x| \), so by squeeze \( f \) is continuous only at \( 0 \). Thus \( U=\{0\} \), which is closed.

Step 4 (Case neither). Thomae's function: \( f(p/q)=1/q \) in lowest terms, \( f=0 \) on irrationals. It is continuous exactly on \( \mathbb{R}\setminus\mathbb{Q} \). Irrationals are neither open (every interval contains rationals) nor closed (rationals are limit points) in \( \mathbb{R} \).

Final answer (box this). \( U \) is not forced to be open or closed; it can be open, closed, or neither.
Method. Present the solution as a proof or counterexample with numbered steps. Each step must state the definition or theorem used and why the next line follows — no “obvious” jumps.
Marks tip. Underline the final boxed answer. For TRUE/FALSE, include a one-line justification name (e.g. connectedness, counterexample, EVT). Missing quantifiers or confused open/closed/compact costs marks.

Hinglish — detailed (examiner-safe, beginner-clear)

Pehle question samjho: \( U \) wahi points jahan \( f \) continuous hai — kya ye set hamesha open ya closed hota hai? Teen alag examples se teen possibilities dikhani hain.

Step 2 — Open \( U \): Step function zero ke left/right alag constant — sirf \( 0 \) par jump; continuity set \( \mathbb{R}\setminus\{0\} \) open interval union jaisa.

Step 3 — Closed \( U \): Rational/irrational wala weird \( f \) — squeeze se sirf \( 0 \) par continuous; \( U=\{0\} \) closed singleton.

Step 4 — Neither: Thomae (popcorn) function irrationals par continuous; irrationals na open na closed — ye classic.

Final: Koi forced category nahi — open, closed, dono mein se kuch nahi — sab possible.
Exam mindset: Copy-paste feel mat do — har step pe “definition → apply → conclude” likho. English math symbols same rakh sakte ho; explanation Hinglish mein full sentences mein.
Common cutters: Definition skip, “obviously” bina reason, ∀/∃ galat, compact vs closed confuse, image vs pre-image confuse — in sab se bacho.
Final line (marks save): End pe clearly likho jo construct/prove maanga tha — ek alag “Hence …” sentence. Counterexample ho toh pehle example define, phir verify, phir conclude.

Exam mindset: Copy-paste feel nahi dena; har line pe “definition → apply → conclude” dikhana. Neeche wala write-up answer sheet pe utaar sakte ho (Hinglish soch + English math symbols).

Exam Step 1 (Hinglish): Goal: Show that \( U \) need not be open and need not be closed — all three behaviours are possible.
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — Goal: Show that \( U \) need not be open and need not be closed — all three behaviours are possible. ko full sentence mein expand karke likho.

Exam Step 2 (Hinglish): Case Open: Define \( f(x)=0 \) for \( x\le 0 \) and \( f(x)=1 \) for \( x>0 \). Then \( f \) fails continuity only at \( 0 \), so \( U=\mathbb{R}\setminus\{0\} \), which is open.
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — Case Open: Define \( f(x)=0 \) for \( x\le 0 \) and \( f(x)=1 \) for \( x>0 \). Then \( f \) fails continuity only at \(… ko full sentence mein expand karke likho.

Exam Step 3 (Hinglish): Case Closed: Define \( f(x)=x \) on \( \mathbb{Q} \) and \( f(x)=0 \) on irrationals. Then ( |f(x)|\le|x| \), so by squeeze \( f \) is continuous only at \( 0 \). Thus \( U=\{0\} \), which is closed.
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — Case Closed: Define \( f(x)=x \) on \( \mathbb{Q} \) and \( f(x)=0 \) on irrationals. Then ( |f(x)|\le|x| \), so by sque… ko full sentence mein expand karke likho.

Exam Step 4 (Hinglish): Case Neither: Thomae's function \( f(p/q)=1/q \), \( f=0 \) on irrationals, is continuous exactly on \( \mathbb{R}\setminus\mathbb{Q} \). Irrationals are neither open nor closed in \( \mathbb{R} \).
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — Case Neither: Thomae's function \( f(p/q)=1/q \), \( f=0 \) on irrationals, is continuous exactly on \( \mathbb{R}\setmi… ko full sentence mein expand karke likho.

Exam Step 5 (Hinglish): Conclusion: \( U \) is not forced to be open or closed.
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — Conclusion: \( U \) is not forced to be open or closed. ko full sentence mein expand karke likho.

Final line (marks save): End pe clearly likho TRUE/FALSE ya jo construct/prove maanga tha — ek alag sentence: “Hence …”. Agar counterexample hai toh pehle example fully define karo, phir check karo, phir conclude.

Common cutters: (1) definition skip (2) “obviously” without reason (3) wrong quantifiers ∀/∃ (4) compact vs closed confuse (5) image vs pre-image confuse. Inse bacho.

Source / Idea kahan se aayi

Uses: ε–δ continuity; squeeze theorem; Thomae's function theorem; open/closed definitions. Course: Continuity + topology of \( \mathbb{R} \).

Interactive — Q9: Continuity set \( U \) (open / closed / neither)

Example

Term Definitions (Q10 Bijective Inverse)

Bijective: Injective (one-to-one) and surjective (onto).

Strictly monotonic: Strictly increasing everywhere, or strictly decreasing everywhere.

Inverse \( f^{-1} \): Undoes \( f \).

Question · D10 · Continuous bijection inverse

Can there exist a continuous bijection \( f:\mathbb{R}\to\mathbb{R} \) such that ( f^{-1} ) is discontinuous?

English Mathematical Steps — full marks checklist

Answer: No.

Step 1: Continuous + injective on an interval ⇒ strictly monotonic. (If it rose and then fell, a horizontal line would hit the graph twice, contradicting injectivity.)

Step 2: A continuous strictly monotonic function on an interval has a continuous inverse on its image (standard theorem from IVT).

Step 3: Surjectivity ⇒ image \( =\mathbb{R} \), so \( f^{-1}:\mathbb{R}\to\mathbb{R} \) is continuous.

Conclusion: Such a discontinuous inverse cannot exist under these hypotheses.

Exam-style answer (copy into answer sheet)

Given / Statement. Can there exist a continuous bijection \( f:\mathbb{R}\to\mathbb{R} \) such that \( f^{-1} \) is discontinuous?

To prove / To find. Answer no; justify via strict monotonicity and continuity of inverse on an interval.

Step 1 (Answer). No such example exists under these hypotheses.

Step 2. Continuous + injective on an interval \( \Rightarrow \) strictly monotonic. If the graph rose then fell, IVT would give a horizontal line hitting the graph twice, contradicting injectivity.

Step 3. A continuous strictly monotonic function on an interval has a continuous inverse on its image (standard theorem from IVT / order preservation).

Step 4. Surjectivity \( \Rightarrow \) image \( =\mathbb{R} \), so \( f^{-1}:\mathbb{R}\to\mathbb{R} \) is continuous everywhere.

Final answer (box this). A continuous bijection \( f:\mathbb{R}\to\mathbb{R} \) cannot have a discontinuous inverse; \( f^{-1} \) is continuous.
Method. Present the solution as a proof or counterexample with numbered steps. Each step must state the definition or theorem used and why the next line follows — no “obvious” jumps.
Marks tip. Underline the final boxed answer. For TRUE/FALSE, include a one-line justification name (e.g. connectedness, counterexample, EVT). Missing quantifiers or confused open/closed/compact costs marks.

Hinglish — detailed (examiner-safe, beginner-clear)

Idea: Continuous + one-to-one graph U-turn nahi le sakta — strictly up ya strictly down. Aise map ka ulta bhi continuous hota hai; onto hone se inverse poori line par defined.

Step 2: Injectivity + continuity se monotonicity — exam mein “IVT se do pre-images” wala one-liner likh do.

Step 3–4: Monotone continuous inverse theorem cite karo; surjective image \( \mathbb{R} \) ⇒ inverse domain bhi \( \mathbb{R} \).

Final: No — discontinuous inverse impossible.
Exam mindset: Copy-paste feel mat do — har step pe “definition → apply → conclude” likho. English math symbols same rakh sakte ho; explanation Hinglish mein full sentences mein.
Common cutters: Definition skip, “obviously” bina reason, ∀/∃ galat, compact vs closed confuse, image vs pre-image confuse — in sab se bacho.
Final line (marks save): End pe clearly likho jo construct/prove maanga tha — ek alag “Hence …” sentence. Counterexample ho toh pehle example define, phir verify, phir conclude.

Exam mindset: Copy-paste feel nahi dena; har line pe “definition → apply → conclude” dikhana. Neeche wala write-up answer sheet pe utaar sakte ho (Hinglish soch + English math symbols).

Exam Step 1 (Hinglish): Answer: No.
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — Answer: No. ko full sentence mein expand karke likho.

Exam Step 2 (Hinglish): Step 1: Continuous + injective on an interval ⇒ strictly monotonic. (If it rose and then fell, a horizontal line would hit the graph twice, contradicting injectivity.)
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — Step 1: Continuous + injective on an interval ⇒ strictly monotonic. (If it rose and then fell, a horizontal line would h… ko full sentence mein expand karke likho.

Exam Step 3 (Hinglish): Step 2: A continuous strictly monotonic function on an interval has a continuous inverse on its image (standard theorem from IVT).
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — Step 2: A continuous strictly monotonic function on an interval has a continuous inverse on its image (standard theorem … ko full sentence mein expand karke likho.

Exam Step 4 (Hinglish): Step 3: Surjectivity ⇒ image \( =\mathbb{R} \), so \( f^{-1}:\mathbb{R}\to\mathbb{R} \) is continuous.
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — Step 3: Surjectivity ⇒ image \( =\mathbb{R} \), so \( f^{-1}:\mathbb{R}\to\mathbb{R} \) is continuous. ko full sentence mein expand karke likho.

Exam Step 5 (Hinglish): Conclusion: Such a discontinuous inverse cannot exist under these hypotheses.
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — Conclusion: Such a discontinuous inverse cannot exist under these hypotheses. ko full sentence mein expand karke likho.

Final line (marks save): End pe clearly likho TRUE/FALSE ya jo construct/prove maanga tha — ek alag sentence: “Hence …”. Agar counterexample hai toh pehle example fully define karo, phir check karo, phir conclude.

Common cutters: (1) definition skip (2) “obviously” without reason (3) wrong quantifiers ∀/∃ (4) compact vs closed confuse (5) image vs pre-image confuse. Inse bacho.

Source / Idea kahan se aayi

Uses: continuous injective ⇒ strictly monotone; inverse of continuous strictly monotone map on an interval is continuous. Course: Continuity / IVT.

Interactive — Q10: Why continuous bijections have continuous inverses

View

Term Definitions (Q11 Relative / Subspace Topology)

Open in \( X \): \( A=U\cap X \) for some open \( U\subseteq\mathbb{R} \).

Closed in \( X \): \( A=C\cap X \) for some closed \( C\subseteq\mathbb{R} \).

Clopen: Both open and closed in \( X \).

Question · D11 · Relative open/closed in a larger set

For each set below, decide whether it is open, closed, both, or neither relative to the larger set \( X \).

English Mathematical Steps — full marks checklist

Relative definitions: \( A \subseteq X \) is open in \( X \) iff \( A=U\cap X \) for some open \( U\subseteq\mathbb{R} \). Closed in \( X \) iff \( A=C\cap X \) for some closed \( C\subseteq\mathbb{R} \).

(a) \( A=\{1/n\}\subseteq X=(0,1) \) (effectively \( \{1/2,1/3,\ldots\} \)). Not open (isolated points). Closed in \( X \) because the only \( \mathbb{R} \)-limit point \( 0 \) lies outside \( X \). Closed only.

(b) \( A=\{x\in\mathbb{Q}:x>\sqrt{2}\}\subseteq X=\mathbb{Q} \). Then \( A=(\sqrt{2},\infty)\cap\mathbb{Q} \) (open in \( \mathbb{Q} \)) and also ( A=[\sqrt{2},\infty)\cap\mathbb{Q} ) (closed in \( \mathbb{Q} \)) since \( \sqrt{2}\notin\mathbb{Q} \). Both (clopen).

(c) \( A=\{2\}\subseteq X=[0,1]\cup\{2\} \). \( A=(1.5,2.5)\cap X \) ⇒ open in \( X \); \( A=\{2\}\cap X \) with \( \{2\} \) closed ⇒ closed in \( X \). Both.

(d) Any \( A\subseteq X=\mathbb{Z} \): each integer is isolated by a small interval, so every subset is open in \( \mathbb{Z} \); also closed (no limit points). Both (discrete topology).

Exam-style answer (copy into answer sheet)

Given / Statement. For each set below, decide whether it is open, closed, both, or neither relative to the larger set \( X \).

Step 1 (Definitions). \( A\subseteq X \) is open in \( X \) iff \( A=U\cap X \) for some open \( U\subseteq\mathbb{R} \). Closed in \( X \) iff \( A=C\cap X \) for some closed \( C\subseteq\mathbb{R} \).

Step 2 (a). \( A=\{1/n\}\subseteq X=(0,1) \) (effectively \( \{1/2,1/3,\ldots\} \)). Not open (isolated points need neighbourhoods with extra points of \( X \)). Closed in \( X \) because the only \( \mathbb{R} \)-limit point \( 0 \) lies outside \( X \). Closed only.

Step 3 (b). \( A=\{x\in\mathbb{Q}:x>\sqrt{2}\}\subseteq X=\mathbb{Q} \). Then \( A=(\sqrt{2},\infty)\cap\mathbb{Q} \) (open in \( \mathbb{Q} \)) and \( A=[\sqrt{2},\infty)\cap\mathbb{Q} \) (closed in \( \mathbb{Q} \)) since \( \sqrt{2}\notin\mathbb{Q} \). Both (clopen).

Step 4 (c). \( A=\{2\}\subseteq X=[0,1]\cup\{2\} \). \( A=(1.5,2.5)\cap X \) ⇒ open in \( X \); \( \{2\} \) closed in \( \mathbb{R} \) ⇒ closed in \( X \). Both.

Step 5 (d). Any \( A\subseteq X=\mathbb{Z} \): each integer is isolated by \( (n-1/2,n+1/2)\cap\mathbb{Z}=\{n\} \), so every subset is open; complements likewise open ⇒ every subset closed. Both (discrete topology).

Final answer (box this). (a) closed only; (b) clopen; (c) clopen; (d) every subset clopen.
Method. Present the solution as a proof or counterexample with numbered steps. Each step must state the definition or theorem used and why the next line follows — no “obvious” jumps.
Marks tip. Underline the final boxed answer. For TRUE/FALSE, include a one-line justification name (e.g. connectedness, counterexample, EVT). Missing quantifiers or confused open/closed/compact costs marks.

Hinglish — detailed (examiner-safe, beginner-clear)

Relative topology: Open/closed ab \( X \) ke andar measure hota hai — \( A=X\cap O \) ya \( A=X\cap F \) likhna zaroori.

(a): \( 1/n \) isolated; limit \( 0 \) universe \( (0,1) \) ke bahar — isliye relative closed only.

(b): \( \sqrt{2} \) rationals mein nahi — open ray aur closed ray dono se same set; clopen.

(c): \( \{2\} \) alag point — chhota interval ∩ \( X \) sirf 2; singleton closed bhi.

(d): Integers discrete — har subset clopen.
Exam mindset: Copy-paste feel mat do — har step pe “definition → apply → conclude” likho. English math symbols same rakh sakte ho; explanation Hinglish mein full sentences mein.
Common cutters: Definition skip, “obviously” bina reason, ∀/∃ galat, compact vs closed confuse, image vs pre-image confuse — in sab se bacho.
Final line (marks save): End pe clearly likho jo construct/prove maanga tha — ek alag “Hence …” sentence. Counterexample ho toh pehle example define, phir verify, phir conclude.

Exam mindset: Copy-paste feel nahi dena; har line pe “definition → apply → conclude” dikhana. Neeche wala write-up answer sheet pe utaar sakte ho (Hinglish soch + English math symbols).

Exam Step 1 (Hinglish): Relative definitions: \( A \subseteq X \) is open in \( X \) iff \( A=U\cap X \) for some open \( U\subseteq\mathbb{R} \). Closed in \( X \) iff \( A=C\cap X \) for some closed \( C\subseteq\mathbb{R} \).
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — Relative definitions: \( A \subseteq X \) is open in \( X \) iff \( A=U\cap X \) for some open \( U\subseteq\mathbb{R} \… ko full sentence mein expand karke likho.

Exam Step 2 (Hinglish): (a) \( A=\{1/n\}\subseteq X=(0,1) \) (effectively \( \{1/2,1/3,\ldots\} \)). Not open (isolated points). Closed in \( X \) because the only \( \mathbb{R} \)-limit point \( 0 \) lies outside \( X \). Closed only.
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — (a) \( A=\{1/n\}\subseteq X=(0,1) \) (effectively \( \{1/2,1/3,\ldots\} \)). Not open (isolated points). Closed in \( X … ko full sentence mein expand karke likho.

Exam Step 3 (Hinglish): (b) \( A=\{x\in\mathbb{Q}:x>\sqrt{2}\}\subseteq X=\mathbb{Q} \). Then \( A=(\sqrt{2},\infty)\cap\mathbb{Q} \) (open in \( \mathbb{Q} \)) and also ( A=[\sqrt{2},\infty)\cap\mathbb{Q} ) (closed in \( \mathbb{Q} \)) since \( \sqrt{2}\notin\mathbb{Q} \). Both (clopen).
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — (b) \( A=\{x\in\mathbb{Q}:x>\sqrt{2}\}\subseteq X=\mathbb{Q} \). Then \( A=(\sqrt{2},\infty)\cap\mathbb{Q} \) (open i… ko full sentence mein expand karke likho.

Exam Step 4 (Hinglish): (c) \( A=\{2\}\subseteq X=[0,1]\cup\{2\} \). \( A=(1.5,2.5)\cap X \) ⇒ open in \( X \); \( A=\{2\}\cap X \) with \( \{2\} \) closed ⇒ closed in \( X \). Both.
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — (c) \( A=\{2\}\subseteq X=[0,1]\cup\{2\} \). \( A=(1.5,2.5)\cap X \) ⇒ open in \( X \); \( A=\{2\}\cap X \) with \( \{2\… ko full sentence mein expand karke likho.

Exam Step 5 (Hinglish): (d) Any \( A\subseteq X=\mathbb{Z} \): each integer is isolated by a small interval, so every subset is open in \( \mathbb{Z} \); also closed (no limit points). Both (discrete topology).
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — (d) Any \( A\subseteq X=\mathbb{Z} \): each integer is isolated by a small interval, so every subset is open in \( \math… ko full sentence mein expand karke likho.

Final line (marks save): End pe clearly likho TRUE/FALSE ya jo construct/prove maanga tha — ek alag sentence: “Hence …”. Agar counterexample hai toh pehle example fully define karo, phir check karo, phir conclude.

Common cutters: (1) definition skip (2) “obviously” without reason (3) wrong quantifiers ∀/∃ (4) compact vs closed confuse (5) image vs pre-image confuse. Inse bacho.

Source / Idea kahan se aayi

Uses: subspace (relative) topology definitions. Course: Topology of \( \mathbb{R} \) / relative open-closed.

Interactive — Q11: Relative open/closed pictures

Part

Term Definitions (Q12 Restriction)

Restriction \( f\vert_B \): Same rule as \( f \), but domain shrunk to \( B\subseteq A \).

Subset: Every element of \( B \) already lives in \( A \).

Question · D12 · Restriction remains continuous

Let \( f:A\to\mathbb{R} \) be continuous and \( B\subseteq A \). Show that the restriction \( f\vert_B:B\to\mathbb{R} \) is continuous.

English Mathematical Steps — full marks checklist

Fix an arbitrary point \( c\in B \). Since \( B\subseteq A \), also \( c\in A \).

Given continuity of \( f \) at \( c \): for every \( \varepsilon>0 \) there exists \( \delta>0 \) such that if \( x\in A \) and \( |x-c|<\delta \), then \( |f(x)-f(c)|<\varepsilon \).

Restrict to \( B \): take any \( x\in B \) with \( |x-c|<\delta \). Then \( x\in A \) automatically, so the same inequality holds.

Identify values: \( f\vert_B(x)=f(x) \) and \( f\vert_B(c)=f(c) \). Hence the ε–δ condition holds for \( f\vert_B \) at \( c \).

Since \( c\in B \) was arbitrary, \( f\vert_B \) is continuous on \( B \).

Exam-style answer (copy into answer sheet)

Given / Statement. Let \( f:A\to\mathbb{R} \) be continuous and \( B\subseteq A \). Show that the restriction \( f|_B:B\to\mathbb{R} \) is continuous.

Step 1. Fix an arbitrary point \( c\in B \). Since \( B\subseteq A \), also \( c\in A \).

Step 2. Given continuity of \( f \) at \( c \): for every \( \varepsilon>0 \) there exists \( \delta>0 \) such that if \( x\in A \) and \( |x-c|<\delta \), then \( |f(x)-f(c)|<\varepsilon \).

Step 3. Restrict to \( B \): take any \( x\in B \) with \( |x-c|<\delta \). Then \( x\in A \) automatically, so the same inequality holds.

Step 4. \( (f|_B)(x)=f(x) \) and \( (f|_B)(c)=f(c) \). Hence the \( \varepsilon \)–\( \delta \) condition holds for \( f|_B \) at \( c \).

Final answer (box this). Since \( c\in B \) was arbitrary, \( f|_B \) is continuous on \( B \).
Method. Present the solution as a proof or counterexample with numbered steps. Each step must state the definition or theorem used and why the next line follows — no “obvious” jumps.
Marks tip. Underline the final boxed answer. For TRUE/FALSE, include a one-line justification name (e.g. connectedness, counterexample, EVT). Missing quantifiers or confused open/closed/compact costs marks.

Method. We solve this in the exam format: state definitions used, justify each implication, and end with a clear final line.

Step 1. Fix an arbitrary point \( c\in B \). Since \( B\subseteq A \), also \( c\in A \). (Write this justification fully on the answer sheet — do not skip definitions or “why”.)

Step 2. Given continuity of \( f \) at \( c \): for every \( \varepsilon>0 \) there exists \( \delta>0 \) such that if \( x\in A \) and \( |x-c|<\delta \), then \( |f(x)-f(c)|<\varepsilon \). (Write this justification fully on the answer sheet — do not skip definitions or “why”.)

Step 3. Restrict to \( B \): take any \( x\in B \) with \( |x-c|<\delta \). Then \( x\in A \) automatically, so the same inequality holds. (Write this justification fully on the answer sheet — do not skip definitions or “why”.)

Step 4. Identify values: \( f\vert_B(x)=f(x) \) and \( f\vert_B(c)=f(c) \). Hence the ε–δ condition holds for \( f\vert_B \) at \( c \). (Write this justification fully on the answer sheet — do not skip definitions or “why”.)

Step 5. Since \( c\in B \) was arbitrary, \( f\vert_B \) is continuous on \( B \). (Write this justification fully on the answer sheet — do not skip definitions or “why”.)

Final answer (box this). Since \( c\in B \) was arbitrary, \( f\vert_B \) is continuous on \( B \).

Marks tip. Examiner cuts marks for missing definitions, unjustified jumps, and unclear TRUE/FALSE or final claim. Keep every “because” explicit.

Hinglish — detailed (examiner-safe, beginner-clear)

Picture: Poori road \( A \) pe smooth function — chhota tukda \( B \) pe bhi same formula; naya \( \delta \) invent karne ki zaroorat nahi.

Proof flow: \( c\in B\subseteq A \) → \( f \) ka \( \delta \) milta hai → \( x\in B \) bhi \( A \) mein → restriction values same → continuity at \( c \).

Final: Arbitrary \( c \) ⇒ \( f|_B \) continuous on \( B \).
Exam mindset: Copy-paste feel mat do — har step pe “definition → apply → conclude” likho. English math symbols same rakh sakte ho; explanation Hinglish mein full sentences mein.
Common cutters: Definition skip, “obviously” bina reason, ∀/∃ galat, compact vs closed confuse, image vs pre-image confuse — in sab se bacho.
Final line (marks save): End pe clearly likho jo construct/prove maanga tha — ek alag “Hence …” sentence. Counterexample ho toh pehle example define, phir verify, phir conclude.

Exam mindset: Copy-paste feel nahi dena; har line pe “definition → apply → conclude” dikhana. Neeche wala write-up answer sheet pe utaar sakte ho (Hinglish soch + English math symbols).

Exam Step 1 (Hinglish): Fix an arbitrary point \( c\in B \). Since \( B\subseteq A \), also \( c\in A \).
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — Fix an arbitrary point \( c\in B \). Since \( B\subseteq A \), also \( c\in A \). ko full sentence mein expand karke likho.

Exam Step 2 (Hinglish): Given continuity of \( f \) at \( c \): for every \( \varepsilon>0 \) there exists \( \delta>0 \) such that if \( x\in A \) and \( |x-c|<\delta \), then \( |f(x)-f(c)|<\varepsilon \).
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — Given continuity of \( f \) at \( c \): for every \( \varepsilon>0 \) there exists \( \delta>0 \) such that if \( … ko full sentence mein expand karke likho.

Exam Step 3 (Hinglish): Restrict to \( B \): take any \( x\in B \) with \( |x-c|<\delta \). Then \( x\in A \) automatically, so the same inequality holds.
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — Restrict to \( B \): take any \( x\in B \) with \( |x-c|<\delta \). Then \( x\in A \) automatically, so the same ineq… ko full sentence mein expand karke likho.

Exam Step 4 (Hinglish): Identify values: \( f\vert_B(x)=f(x) \) and \( f\vert_B(c)=f(c) \). Hence the ε–δ condition holds for \( f\vert_B \) at \( c \).
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — Identify values: \( f\vert_B(x)=f(x) \) and \( f\vert_B(c)=f(c) \). Hence the ε–δ condition holds for \( f\vert_B \) at … ko full sentence mein expand karke likho.

Exam Step 5 (Hinglish): Since \( c\in B \) was arbitrary, \( f\vert_B \) is continuous on \( B \).
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — Since \( c\in B \) was arbitrary, \( f\vert_B \) is continuous on \( B \). ko full sentence mein expand karke likho.

Final line (marks save): End pe clearly likho TRUE/FALSE ya jo construct/prove maanga tha — ek alag sentence: “Hence …”. Agar counterexample hai toh pehle example fully define karo, phir check karo, phir conclude.

Common cutters: (1) definition skip (2) “obviously” without reason (3) wrong quantifiers ∀/∃ (4) compact vs closed confuse (5) image vs pre-image confuse. Inse bacho.

Source / Idea kahan se aayi

Uses: ε–δ definition of continuity on a domain; restriction of a function. Course: Continuity (basic theorems). Related to Q14(b).

Interactive — Q12: Same \( \delta \) works on the smaller domain

ε band:

Term Definitions (Q13 Relative Compactness)

Open cover / finite subcover: Open sets covering \( K \); finitely many still enough.

Open relative to \( A \): \( U=V\cap A \) for some open \( V\subseteq\mathbb{R} \).

Compact relative to \( A \): Every cover of \( K \) by sets open in \( A \) has a finite subcover.

Question · D13 · Compact relative to A iff compact in ℝ

Let \( K\subset A\subset\mathbb{R} \). Show \( K \) is compact relative to \( A \) iff \( K \) is compact in \( \mathbb{R} \) (no extra conditions on \( A \)).

English Mathematical Steps — full marks checklist

Definition: Compact relative to \( A \) means every cover of \( K \) by sets open in \( A \) has a finite subcover.

(⇒) Assume relative compactness. Let \( \{V_\alpha\} \) be an open cover of \( K \) in \( \mathbb{R} \). Set \( U_\alpha=V_\alpha\cap A \) (open in \( A \)). These still cover \( K \). Take a finite subcover of ( U \)'s; since each \( U\subseteq V \), the corresponding ( V \)'s cover \( K \). So \( K \) is compact in \( \mathbb{R} \).

(⇐) Assume compactness in \( \mathbb{R} \). Let \( \{U_\alpha\} \) cover \( K \) with \( U_\alpha=V_\alpha\cap A \), \( V_\alpha \) open in \( \mathbb{R} \). Then \( \{V_\alpha\} \) covers \( K \) in \( \mathbb{R} \); take finite subcover; intersect with \( A \) (using \( K\subseteq A \)) to get a finite relative subcover.

Moral: Compactness of \( K \) is intrinsic — the ambient \( A \) does not change it.

Exam-style answer (copy into answer sheet)

Given / Statement. Let \( K\subset A\subset\mathbb{R} \). Show \( K \) is compact relative to \( A \) iff \( K \) is compact in \( \mathbb{R} \) (no extra conditions on \( A \)).

Step 1 (Definition). Compact relative to \( A \) means every cover of \( K \) by sets open in \( A \) has a finite subcover.

Step 2 (⇒). Assume relative compactness. Let \( \{V_\alpha\} \) be an open cover of \( K \) in \( \mathbb{R} \). Set \( U_\alpha=V_\alpha\cap A \) (open in \( A \)). These still cover \( K \). Take a finite subcover of the \( U \)'s; since each \( U\subseteq V \), the corresponding \( V \)'s cover \( K \). So \( K \) is compact in \( \mathbb{R} \).

Step 3 (⇐). Assume compactness in \( \mathbb{R} \). Let \( \{U_\alpha\} \) cover \( K \) with \( U_\alpha=V_\alpha\cap A \), \( V_\alpha \) open in \( \mathbb{R} \). Then \( \{V_\alpha\} \) covers \( K \) in \( \mathbb{R} \); take finite subcover; intersect with \( A \) (using \( K\subseteq A \)) to get a finite relative subcover.

Final answer (box this). \( K \) is compact relative to \( A \) iff \( K \) is compact in \( \mathbb{R} \); compactness of \( K \) is intrinsic to \( K \), not the ambient \( A \).
Method. Present the solution as a proof or counterexample with numbered steps. Each step must state the definition or theorem used and why the next line follows — no “obvious” jumps.
Marks tip. Underline the final boxed answer. For TRUE/FALSE, include a one-line justification name (e.g. connectedness, counterexample, EVT). Missing quantifiers or confused open/closed/compact costs marks.

Hinglish — detailed (examiner-safe, beginner-clear)

Idea: Relative-open = \( V\cap A \). Cover convert karke dono directions same finite subcover idea use karte hain.

⇒: Real cover → intersect with \( A \) → relative cover → finite \( U \)'s → bigger \( V \)'s finite.

⇐: Relative cover → lift to \( V_\alpha \) in \( \mathbb{R} \) → compact in \( \mathbb{R} \) → intersect back.

Final: \( A \) open/closed hona zaroori nahi — iff statement pure \( K \) ki property hai.
Exam mindset: Copy-paste feel mat do — har step pe “definition → apply → conclude” likho. English math symbols same rakh sakte ho; explanation Hinglish mein full sentences mein.
Common cutters: Definition skip, “obviously” bina reason, ∀/∃ galat, compact vs closed confuse, image vs pre-image confuse — in sab se bacho.
Final line (marks save): End pe clearly likho jo construct/prove maanga tha — ek alag “Hence …” sentence. Counterexample ho toh pehle example define, phir verify, phir conclude.

Exam mindset: Copy-paste feel nahi dena; har line pe “definition → apply → conclude” dikhana. Neeche wala write-up answer sheet pe utaar sakte ho (Hinglish soch + English math symbols).

Exam Step 1 (Hinglish): Definition: Compact relative to \( A \) means every cover of \( K \) by sets open in \( A \) has a finite subcover.
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — Definition: Compact relative to \( A \) means every cover of \( K \) by sets open in \( A \) has a finite subcover. ko full sentence mein expand karke likho.

Exam Step 2 (Hinglish): (⇒) Assume relative compactness. Let \( \{V_\alpha\} \) be an open cover of \( K \) in \( \mathbb{R} \). Set \( U_\alpha=V_\alpha\cap A \) (open in \( A \)). These still cover \( K \). Take a finite subcover of ( U \)'s; since each \( U\subseteq V \), the corresponding ( V \)'s cover \( K \). So \( K \) is compact in \( \mathbb{R} \).
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — (⇒) Assume relative compactness. Let \( \{V_\alpha\} \) be an open cover of \( K \) in \( \mathbb{R} \). Set \( U_\alpha… ko full sentence mein expand karke likho.

Exam Step 3 (Hinglish): (⇐) Assume compactness in \( \mathbb{R} \). Let \( \{U_\alpha\} \) cover \( K \) with \( U_\alpha=V_\alpha\cap A \), \( V_\alpha \) open in \( \mathbb{R} \). Then \( \{V_\alpha\} \) covers \( K \) in \( \mathbb{R} \); take finite subcover; intersect with \( A \) (using \( K\subseteq A \)) to get a finite relative subcover.
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — (⇐) Assume compactness in \( \mathbb{R} \). Let \( \{U_\alpha\} \) cover \( K \) with \( U_\alpha=V_\alpha\cap A \), \( … ko full sentence mein expand karke likho.

Exam Step 4 (Hinglish): Moral: Compactness of \( K \) is intrinsic — the ambient \( A \) does not change it.
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — Moral: Compactness of \( K \) is intrinsic — the ambient \( A \) does not change it. ko full sentence mein expand karke likho.

Final line (marks save): End pe clearly likho TRUE/FALSE ya jo construct/prove maanga tha — ek alag sentence: “Hence …”. Agar counterexample hai toh pehle example fully define karo, phir check karo, phir conclude.

Common cutters: (1) definition skip (2) “obviously” without reason (3) wrong quantifiers ∀/∃ (4) compact vs closed confuse (5) image vs pre-image confuse. Inse bacho.

Source / Idea kahan se aayi

Uses: relative openness \( U=V\cap A \); definition of compactness via open covers. Course: Compactness / Heine–Borel chapter.

Interactive — Q13: Relative covers ↔ absolute covers

Direction

Term Definitions (Q14 Heine–Borel)

Heine–Borel: In \( \mathbb{R} \), compact \( \Leftrightarrow \) closed and bounded.

Limit point outside \( A \): Used to build \( 1/(x-a) \) when \( A \) is not closed.

Question · D14 · Continuous unbounded function on non-compact A

(a) If \( a\notin A \) is a limit point of \( A \), show ( f(x)=1/(x-a) ) is continuous on \( A \). (b) If \( A \) unbounded, show \( f(x)=x \) is continuous on \( A \). (c) Conclude every non-compact \( A\subseteq\mathbb{R} \) admits a continuous unbounded function.

English Mathematical Steps — full marks checklist

(a) On \( A \), denominator \( x-a\neq 0 \). Quotient of continuous functions with nonzero denominator ⇒ continuous on \( A \).

(b) Identity is continuous on \( \mathbb{R} \); by Q12 its restriction to \( A \) is continuous. Unboundedness of \( A \) affects range size, not continuity.

(c) Heine–Borel: non-compact ⇒ not closed or not bounded.

If not bounded: use \( f(x)=x \) from (b) — continuous and unbounded on \( A \).

If bounded but not closed: some limit point \( a\notin A \); use ( f(x)=1/(x-a) ) from (a) — continuous on \( A \), unbounded near \( a \).

Exam-style answer (copy into answer sheet)

Given / Statement. (a) If \( a\notin A \) is a limit point of \( A \), show \( f(x)=1/(x-a) \) is continuous on \( A \). (b) If \( A \) unbounded, show \( f(x)=x \) is continuous on \( A \). (c) Conclude every non-compact \( A\subseteq\mathbb{R} \) admits a continuous unbounded function.

Step 1 (a). On \( A \), denominator \( x-a\neq 0 \). Quotient of continuous functions with nonzero denominator ⇒ continuous on \( A \). Unbounded: given \( M>0 \), limit-point property gives \( x\in A \) with \( 0<|x-a|<1/M \), so \( |f(x)|>M \).

Step 2 (b). Identity is continuous on \( \mathbb{R} \); by restriction (Q12) its restriction to \( A \) is continuous. Unboundedness of \( A \) gives \( |x|>M \) in \( A \), so \( |f(x)|>M \).

Step 3 (c). Heine–Borel: non-compact \( \Rightarrow \) not closed or not bounded. If not bounded, use \( f(x)=x \) from (b). If bounded but not closed, pick \( a\in\overline{A}\setminus A \) (limit point outside); use \( f(x)=1/(x-a) \) from (a).

Final answer (box this). Every non-compact \( A\subseteq\mathbb{R} \) has a continuous unbounded real-valued function on \( A \).
Method. Present the solution as a proof or counterexample with numbered steps. Each step must state the definition or theorem used and why the next line follows — no “obvious” jumps.
Marks tip. Underline the final boxed answer. For TRUE/FALSE, include a one-line justification name (e.g. connectedness, counterexample, EVT). Missing quantifiers or confused open/closed/compact costs marks.

Hinglish — detailed (examiner-safe, beginner-clear)

Recipe: Non-compact = bounded fail ya closed fail. Unbounded domain → \( f(x)=x \). Missing limit point → \( 1/(x-a) \) pole ke paas explode, domain par continuous.

(a): Denominator kabhi zero nahi on \( A \) — quotient rule; limit point se arbitrarily large values.

(c): Heine–Borel cite karke do cases cover karo — dono parts (a)(b) link karna marks ke liye important.
Exam mindset: Copy-paste feel mat do — har step pe “definition → apply → conclude” likho. English math symbols same rakh sakte ho; explanation Hinglish mein full sentences mein.
Common cutters: Definition skip, “obviously” bina reason, ∀/∃ galat, compact vs closed confuse, image vs pre-image confuse — in sab se bacho.
Final line (marks save): End pe clearly likho jo construct/prove maanga tha — ek alag “Hence …” sentence. Counterexample ho toh pehle example define, phir verify, phir conclude.

Exam mindset: Copy-paste feel nahi dena; har line pe “definition → apply → conclude” dikhana. Neeche wala write-up answer sheet pe utaar sakte ho (Hinglish soch + English math symbols).

Exam Step 1 (Hinglish): (a) On \( A \), denominator \( x-a\neq 0 \). Quotient of continuous functions with nonzero denominator ⇒ continuous on \( A \).
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — (a) On \( A \), denominator \( x-a\neq 0 \). Quotient of continuous functions with nonzero denominator ⇒ continuous on \… ko full sentence mein expand karke likho.

Exam Step 2 (Hinglish): (b) Identity is continuous on \( \mathbb{R} \); by Q12 its restriction to \( A \) is continuous. Unboundedness of \( A \) affects range size, not continuity.
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — (b) Identity is continuous on \( \mathbb{R} \); by Q12 its restriction to \( A \) is continuous. Unboundedness of \( A \… ko full sentence mein expand karke likho.

Exam Step 3 (Hinglish): (c) Heine–Borel: non-compact ⇒ not closed or not bounded.
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — (c) Heine–Borel: non-compact ⇒ not closed or not bounded. ko full sentence mein expand karke likho.

Exam Step 4 (Hinglish): If not bounded: use \( f(x)=x \) from (b) — continuous and unbounded on \( A \).
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — If not bounded: use \( f(x)=x \) from (b) — continuous and unbounded on \( A \). ko full sentence mein expand karke likho.

Exam Step 5 (Hinglish): If bounded but not closed: some limit point \( a\notin A \); use ( f(x)=1/(x-a) ) from (a) — continuous on \( A \), unbounded near \( a \).
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — If bounded but not closed: some limit point \( a\notin A \); use ( f(x)=1/(x-a) ) from (a) — continuous on \( A \), unbo… ko full sentence mein expand karke likho.

Final line (marks save): End pe clearly likho TRUE/FALSE ya jo construct/prove maanga tha — ek alag sentence: “Hence …”. Agar counterexample hai toh pehle example fully define karo, phir check karo, phir conclude.

Common cutters: (1) definition skip (2) “obviously” without reason (3) wrong quantifiers ∀/∃ (4) compact vs closed confuse (5) image vs pre-image confuse. Inse bacho.

Source / Idea kahan se aayi

Uses: Heine–Borel theorem; Q12 restriction; rational functions continuous off poles. Course: Compactness + continuity.

Interactive — Q14: Which continuous unbounded function?

Case

Term Definitions (Q15 Disconnection)

Disconnected: \( A = U \cup V \) with \( U,V \) nonempty, disjoint, and open in \( A \).

Proper subset: Strictly smaller than \( A \).

Clopen: Both open and closed (in \( A \)).

Question · D15 · Disconnected iff nonempty proper clopen subset

Show \( A \) is disconnected iff there exists a nonempty proper subset of \( A \) that is both open and closed in \( A \).

English Mathematical Steps — full marks checklist

(⇒) If disconnected, \( A=U\cup V \) with ( U,V ) nonempty, disjoint, open in \( A \). Then \( U=A\setminus V \). Complement of open \( V \) is closed ⇒ \( U \) closed; already open ⇒ clopen. Proper because \( V\neq\emptyset \).

(⇐) If \( U\subset A \) nonempty proper clopen, set ( V=A\setminus U \). Then \( V\neq\emptyset \), open (complement of closed), and \( U \) open, disjoint, union \( A \) ⇒ disconnected.

Exam-style answer (copy into answer sheet)

Given / Statement. Show \( A \) is disconnected iff there exists a nonempty proper subset of \( A \) that is both open and closed in \( A \).

Step 1 (⇒). If disconnected, \( A=U\cup V \) with \( U,V \) nonempty, disjoint, open in \( A \). Then \( U=A\setminus V \). Complement of open \( V \) is closed ⇒ \( U \) closed; already open ⇒ clopen. Proper because \( V\neq\emptyset \).

Step 2 (⇐). If \( U\subset A \) nonempty proper clopen, set \( V=A\setminus U \). Then \( V\neq\emptyset \), open (complement of closed \( U \)), disjoint from \( U \), and \( A=U\cup V \) ⇒ disconnected.

Final answer (box this). Disconnectedness ⇔ existence of a nonempty proper clopen subset (not \( \emptyset \) or \( A \) alone).
Method. Present the solution as a proof or counterexample with numbered steps. Each step must state the definition or theorem used and why the next line follows — no “obvious” jumps.
Marks tip. Underline the final boxed answer. For TRUE/FALSE, include a one-line justification name (e.g. connectedness, counterexample, EVT). Missing quantifiers or confused open/closed/compact costs marks.

Method. We solve this in the exam format: state definitions used, justify each implication, and end with a clear final line.

Step 1. (⇒) If disconnected, \( A=U\cup V \) with ( U,V ) nonempty, disjoint, open in \( A \). Then \( U=A\setminus V \). Complement of open \( V \) is closed ⇒ \( U \) closed; already open ⇒ clopen. Proper because \( V\neq\emptyset \). (Write this justification fully on the answer sheet — do not skip definitions or “why”.)

Step 2. (⇐) If \( U\subset A \) nonempty proper clopen, set ( V=A\setminus U \). Then \( V\neq\emptyset \), open (complement of closed), and \( U \) open, disjoint, union \( A \) ⇒ disconnected. (Write this justification fully on the answer sheet — do not skip definitions or “why”.)

Final answer (box this). (⇐) If \( U\subset A \) nonempty proper clopen, set ( V=A\setminus U \). Then \( V\neq\emptyset \), open (complement of closed), and \( U \) open, disjoint, union \( A \) ⇒ disconnected.

Marks tip. Examiner cuts marks for missing definitions, unjustified jumps, and unclear TRUE/FALSE or final claim. Keep every “because” explicit.

Hinglish — detailed (examiner-safe, beginner-clear)

One line: Disconnect = do open pieces; ek piece ka complement dusra piece → closed bhi → proper clopen. Ulta: clopen + complement = separation.

⇒: \( U \) open tha, \( V \) open ⇒ \( U \) closed; nonempty proper check likho.

⇐: \( V=A\setminus U \) nonempty, open, disjoint — definition of disconnected satisfied.
Exam mindset: Copy-paste feel mat do — har step pe “definition → apply → conclude” likho. English math symbols same rakh sakte ho; explanation Hinglish mein full sentences mein.
Common cutters: Definition skip, “obviously” bina reason, ∀/∃ galat, compact vs closed confuse, image vs pre-image confuse — in sab se bacho.
Final line (marks save): End pe clearly likho jo construct/prove maanga tha — ek alag “Hence …” sentence. Counterexample ho toh pehle example define, phir verify, phir conclude.

Exam mindset: Copy-paste feel nahi dena; har line pe “definition → apply → conclude” dikhana. Neeche wala write-up answer sheet pe utaar sakte ho (Hinglish soch + English math symbols).

Exam Step 1 (Hinglish): (⇒) If disconnected, \( A=U\cup V \) with ( U,V ) nonempty, disjoint, open in \( A \). Then \( U=A\setminus V \). Complement of open \( V \) is closed ⇒ \( U \) closed; already open ⇒ clopen. Proper because \( V\neq\emptyset \).
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — (⇒) If disconnected, \( A=U\cup V \) with ( U,V ) nonempty, disjoint, open in \( A \). Then \( U=A\setminus V \). Comple… ko full sentence mein expand karke likho.

Exam Step 2 (Hinglish): (⇐) If \( U\subset A \) nonempty proper clopen, set ( V=A\setminus U \). Then \( V\neq\emptyset \), open (complement of closed), and \( U \) open, disjoint, union \( A \) ⇒ disconnected.
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — (⇐) If \( U\subset A \) nonempty proper clopen, set ( V=A\setminus U \). Then \( V\neq\emptyset \), open (complement of … ko full sentence mein expand karke likho.

Final line (marks save): End pe clearly likho TRUE/FALSE ya jo construct/prove maanga tha — ek alag sentence: “Hence …”. Agar counterexample hai toh pehle example fully define karo, phir check karo, phir conclude.

Common cutters: (1) definition skip (2) “obviously” without reason (3) wrong quantifiers ∀/∃ (4) compact vs closed confuse (5) image vs pre-image confuse. Inse bacho.

Source / Idea kahan se aayi

Uses: definition of disconnected set in relative topology; open/closed complements. Course: Connectedness.

Interactive — Q15: Two pieces ↔ one clopen piece

View

Term Definitions (Q16 Even-Degree Polynomials)

Even degree + \( a_n\gt 0 \): both ends of the graph go to \( +\infty \).

Triangle inequality: \( |A+B|\ge|A|-|B| \) and \( |\sum|\le\sum|\cdot| \).

EVT: continuous on compact \( [-R,R] \) attains a minimum.

Question · D16 · Even-degree polynomial global minimum

Let \( p \) be even degree with leading coefficient ( a_n>0 ). (a) Triangle bound for ( |x|>1 ). (b) Unbounded above. (c) EVT ⇒ global minimum.

English Mathematical Steps — full marks checklist

(a) Write ( p=a_n x^n+S ). Then \( p\ge a_n x^n-|S| \). Bound ( |S|\le\sum|a_k||x|^k \). For ( |x|>1 ) and \( k\le n-1 \), \( |x|^k\le|x|^{n-1} \), so \( |S|\le C|x|^{n-1} \). Even \( n \) + ( a_n>0 ) ⇒ ( a_n x^n=a_n|x|^n ).

(b) Factor \( p\ge|x|^{n-1}(a_n|x|-C) \). For given \( M \), set ( R=\max(1,(M+C)/a_n) ). Then ( |x|>R ) ⇒ ( p(x)>M ).

(c) Fix ( x_0 ), set ( M=p(x_0) ), enlarge \( R \) so ( |x_0|<R ). On compact ( [-R,R] ), EVT gives minimizer \( x_{\min} \). Outside, \( p>p(x_0)\ge p(x_{\min}) \). Hence global min.

Exam-style answer (copy into answer sheet)

Given / Statement. Let \( p \) be even degree with leading coefficient \( a_n>0 \). (a) Triangle bound for \( |x|>1 \). (b) Unbounded above. (c) EVT ⇒ global minimum.

Step 1 (a). Write \( p=a_n x^n+S \). Then \( p\ge a_n x^n-|S| \). Bound \( |S|\le\sum|a_k||x|^k \). For \( |x|>1 \) and \( k\le n-1 \), \( |x|^k\le|x|^{n-1} \), so \( |S|\le C|x|^{n-1} \). Even \( n \) + \( a_n>0 \) ⇒ \( a_n x^n=a_n|x|^n \).

Step 2 (b). Factor \( p\ge|x|^{n-1}(a_n|x|-C) \). For given \( M \), set \( R=\max(1,(M+C)/a_n) \) (enlarge as needed). Then \( |x|>R \Rightarrow p(x)>M \).

Step 3 (c). Fix \( x_0 \), set \( M_0=p(x_0) \), enlarge \( R \) so \( |x_0|R\Rightarrow p(x)>M_0 \). On compact \( [-R,R] \), EVT gives minimizer \( x_{\min} \). Outside, \( p>M_0\ge p(x_{\min}) \). Hence global min at \( x_{\min} \).

Final answer (box this). Even degree with \( a_n>0 \) ⇒ \( p\to+\infty \) at both ends, unbounded above, and attains a global minimum on \( \mathbb{R} \).
Method. Present the solution as a proof or counterexample with numbered steps. Each step must state the definition or theorem used and why the next line follows — no “obvious” jumps.
Marks tip. Underline the final boxed answer. For TRUE/FALSE, include a one-line justification name (e.g. connectedness, counterexample, EVT). Missing quantifiers or confused open/closed/compact costs marks.

Hinglish — detailed (examiner-safe, beginner-clear)

Picture: Even + positive leading ⇒ dono ends \( +\infty \). (a) triangle inequality se leading term dominate; (b) large \( |x| \) pe \( p>M \); (c) bahar badi values, andar compact box pe EVT minimum — wahi global min.

(a): \( C=\sum_{k1 \) pe lower degrees \( |x|^{n-1} \) se bound.

(c): \( R \) choose karke bahar \( p(x_0) \) se upar, andar EVT — teen steps alag likhna.
Exam mindset: Copy-paste feel mat do — har step pe “definition → apply → conclude” likho. English math symbols same rakh sakte ho; explanation Hinglish mein full sentences mein.
Common cutters: Definition skip, “obviously” bina reason, ∀/∃ galat, compact vs closed confuse, image vs pre-image confuse — in sab se bacho.
Final line (marks save): End pe clearly likho jo construct/prove maanga tha — ek alag “Hence …” sentence. Counterexample ho toh pehle example define, phir verify, phir conclude.

Exam mindset: Copy-paste feel nahi dena; har line pe “definition → apply → conclude” dikhana. Neeche wala write-up answer sheet pe utaar sakte ho (Hinglish soch + English math symbols).

Exam Step 1 (Hinglish): (a) Write ( p=a_n x^n+S ). Then \( p\ge a_n x^n-|S| \). Bound ( |S|\le\sum|a_k||x|^k \). For ( |x|>1 ) and \( k\le n-1 \), \( |x|^k\le|x|^{n-1} \), so \( |S|\le C|x|^{n-1} \). Even \( n \) + ( a_n>0 ) ⇒ ( a_n x^n=a_n|x|^n ).
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — (a) Write ( p=a_n x^n+S ). Then \( p\ge a_n x^n-|S| \). Bound ( |S|\le\sum|a_k||x|^k \). For ( |x|>1 ) and \( k\le n-… ko full sentence mein expand karke likho.

Exam Step 2 (Hinglish): (b) Factor \( p\ge|x|^{n-1}(a_n|x|-C) \). For given \( M \), set ( R=\max(1,(M+C)/a_n) ). Then ( |x|>R ) ⇒ ( p(x)>M ).
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — (b) Factor \( p\ge|x|^{n-1}(a_n|x|-C) \). For given \( M \), set ( R=\max(1,(M+C)/a_n) ). Then ( |x|>R ) ⇒ ( p(x)>… ko full sentence mein expand karke likho.

Exam Step 3 (Hinglish): (c) Fix ( x_0 ), set ( M=p(x_0) ), enlarge \( R \) so ( |x_0|<R ). On compact ( [-R,R] ), EVT gives minimizer \( x_{\min} \). Outside, \( p>p(x_0)\ge p(x_{\min}) \). Hence global min.
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — (c) Fix ( x_0 ), set ( M=p(x_0) ), enlarge \( R \) so ( |x_0|<R ). On compact ( [-R,R] ), EVT gives minimizer \( x_{\… ko full sentence mein expand karke likho.

Final line (marks save): End pe clearly likho TRUE/FALSE ya jo construct/prove maanga tha — ek alag sentence: “Hence …”. Agar counterexample hai toh pehle example fully define karo, phir check karo, phir conclude.

Common cutters: (1) definition skip (2) “obviously” without reason (3) wrong quantifiers ∀/∃ (4) compact vs closed confuse (5) image vs pre-image confuse. Inse bacho.

Source / Idea kahan se aayi

Uses: triangle inequality; Extreme Value Theorem; even-degree end behaviour. Course: Continuity on compact sets / EVT.

Interactive — Q16: Even degree → global minimum

Part M:

Term Definitions (Q17 Odd-Degree Root)

Odd degree: ends of the graph go to opposite infinities.

IVT: continuous path from negative to positive must hit 0.

Question · D17 · Odd-degree polynomial has a real root

Use IVT to show every odd-degree polynomial has a real root.

English Mathematical Steps — full marks checklist

WLOG ( a_n>0 ) (else replace \( p \) by ( -p )).

End behaviour: odd degree ⇒ \( p(x)\to+\infty \) as \( x\to+\infty \) and \( p(x)\to-\infty \) as \( x\to-\infty \).

Pick signs: choose ( a<0 ) with ( p(a)<0 ) and ( b>0 ) with ( p(b)>0 ).

IVT: \( p \) continuous on ( [a,b] ) and \( 0 \) lies between ( p(a) ) and ( p(b) ) ⇒ some \( c\in(a,b) \) with ( p(c)=0 ).

Exam-style answer (copy into answer sheet)

Given / Statement. Use IVT to show every odd-degree polynomial has a real root.

Step 1. WLOG \( a_n>0 \) (else replace \( p \) by \( -p \); zeros unchanged).

Step 2. End behaviour: odd degree ⇒ \( p(x)\to+\infty \) as \( x\to+\infty \) and \( p(x)\to-\infty \) as \( x\to-\infty \).

Step 3. Choose \( a<0 \) with \( p(a)<0 \) and \( b>0 \) with \( p(b)>0 \) (large enough \( |a|,|b| \)).

Step 4. IVT: \( p \) continuous on \( [a,b] \) and \( 0 \) lies between \( p(a) \) and \( p(b) \) ⇒ some \( c\in(a,b) \) with \( p(c)=0 \).

Final answer (box this). Every odd-degree real polynomial has at least one real zero.
Method. Present the solution as a proof or counterexample with numbered steps. Each step must state the definition or theorem used and why the next line follows — no “obvious” jumps.
Marks tip. Underline the final boxed answer. For TRUE/FALSE, include a one-line justification name (e.g. connectedness, counterexample, EVT). Missing quantifiers or confused open/closed/compact costs marks.

Method. We solve this in the exam format: state definitions used, justify each implication, and end with a clear final line.

Step 1. WLOG ( a_n>0 ) (else replace \( p \) by ( -p )). (Write this justification fully on the answer sheet — do not skip definitions or “why”.)

Step 2. End behaviour: odd degree ⇒ \( p(x)\to+\infty \) as \( x\to+\infty \) and \( p(x)\to-\infty \) as \( x\to-\infty \). (Write this justification fully on the answer sheet — do not skip definitions or “why”.)

Step 3. Pick signs: choose ( a<0 ) with ( p(a)<0 ) and ( b>0 ) with ( p(b)>0 ). (Write this justification fully on the answer sheet — do not skip definitions or “why”.)

Step 4. IVT: \( p \) continuous on ( [a,b] ) and \( 0 \) lies between ( p(a) ) and ( p(b) ) ⇒ some \( c\in(a,b) \) with ( p(c)=0 ). (Write this justification fully on the answer sheet — do not skip definitions or “why”.)

Final answer (box this). IVT: \( p \) continuous on ( [a,b] ) and \( 0 \) lies between ( p(a) ) and ( p(b) ) ⇒ some \( c\in(a,b) \) with ( p(c)=0 ).

Marks tip. Examiner cuts marks for missing definitions, unjustified jumps, and unclear TRUE/FALSE or final claim. Keep every “because” explicit.

Hinglish — detailed (examiner-safe, beginner-clear)

IVT scene: Odd degree ⇒ ek taraf \( +\infty \), dusri \( -\infty \). Continuous path negative se positive — zero cross karna padega.

WLOG: Leading coeff sign flip se roots same — examiner ko dikhao sign assume kiya.

Step 3–4: Explicit \( a,b \) with opposite signs; IVT hypothesis checklist likho.
Exam mindset: Copy-paste feel mat do — har step pe “definition → apply → conclude” likho. English math symbols same rakh sakte ho; explanation Hinglish mein full sentences mein.
Common cutters: Definition skip, “obviously” bina reason, ∀/∃ galat, compact vs closed confuse, image vs pre-image confuse — in sab se bacho.
Final line (marks save): End pe clearly likho jo construct/prove maanga tha — ek alag “Hence …” sentence. Counterexample ho toh pehle example define, phir verify, phir conclude.

Exam mindset: Copy-paste feel nahi dena; har line pe “definition → apply → conclude” dikhana. Neeche wala write-up answer sheet pe utaar sakte ho (Hinglish soch + English math symbols).

Exam Step 1 (Hinglish): WLOG ( a_n>0 ) (else replace \( p \) by ( -p )).
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — WLOG ( a_n>0 ) (else replace \( p \) by ( -p )). ko full sentence mein expand karke likho.

Exam Step 2 (Hinglish): End behaviour: odd degree ⇒ \( p(x)\to+\infty \) as \( x\to+\infty \) and \( p(x)\to-\infty \) as \( x\to-\infty \).
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — End behaviour: odd degree ⇒ \( p(x)\to+\infty \) as \( x\to+\infty \) and \( p(x)\to-\infty \) as \( x\to-\infty \). ko full sentence mein expand karke likho.

Exam Step 3 (Hinglish): Pick signs: choose ( a<0 ) with ( p(a)<0 ) and ( b>0 ) with ( p(b)>0 ).
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — Pick signs: choose ( a<0 ) with ( p(a)<0 ) and ( b>0 ) with ( p(b)>0 ). ko full sentence mein expand karke likho.

Exam Step 4 (Hinglish): IVT: \( p \) continuous on ( [a,b] ) and \( 0 \) lies between ( p(a) ) and ( p(b) ) ⇒ some \( c\in(a,b) \) with ( p(c)=0 ).
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — IVT: \( p \) continuous on ( [a,b] ) and \( 0 \) lies between ( p(a) ) and ( p(b) ) ⇒ some \( c\in(a,b) \) with ( p(c)=0… ko full sentence mein expand karke likho.

Final line (marks save): End pe clearly likho TRUE/FALSE ya jo construct/prove maanga tha — ek alag sentence: “Hence …”. Agar counterexample hai toh pehle example fully define karo, phir check karo, phir conclude.

Common cutters: (1) definition skip (2) “obviously” without reason (3) wrong quantifiers ∀/∃ (4) compact vs closed confuse (5) image vs pre-image confuse. Inse bacho.

Source / Idea kahan se aayi

Uses: Intermediate Value Theorem; end behaviour of odd-degree polynomials. Course: Continuity / IVT.

Interactive — Q17: Odd degree crosses the axis

Example

Term Definitions (Q18 Uniform Continuity)

Pointwise continuity: for each \( a \), a \( \delta=\delta(\varepsilon,a) \) may depend on \( a \).

Uniform continuity: one \( \delta=\delta(\varepsilon) \) works for all points in the domain.

Question · D18 · ε–δ continuity and uniform continuity

For each function, give an explicit \( \delta(\varepsilon,a) \) (or show UC with \( \delta(\varepsilon) \) only) and decide uniform continuity.

English Mathematical Steps — full marks checklist

(a) Step across √2 gap: continuous with \( \delta=|a-\sqrt{2}| \) (cannot cross gap). Not UC — \( \delta\to 0 \) as \( a\to\sqrt{2} \).

(b) ( x^2 ) on \( \mathbb{R} \): ( \delta=\min(1,\varepsilon/(1+2|a|)) ). Continuous; not UC (depends on \( a \)).

(c) \( \sqrt{x} \) on ( [0,\infty) ): \( |\sqrt{x}-\sqrt{a}|\le\sqrt{|x-a|} \) ⇒ \( \delta=\varepsilon^2 \) independent of \( a \) ⇒ UC.

(d) ( x^3 ) on \( \mathbb{R} \): ( \delta=\min(1,\varepsilon/(3|a|^2+3|a|+1)) ). Continuous; not UC.

(e) \( 1/\sqrt{x} \) on \( (0,1) \): \( \delta=\min(a/2, c a^{3/2}\varepsilon) \). Continuous; not UC as \( a\to 0^+ \).

Exam-style answer (copy into answer sheet)

Given / Statement. For each function, give an explicit \( \delta(\varepsilon,a) \) (or show UC with \( \delta(\varepsilon) \) only) and decide uniform continuity.

Step 1 (a). Step across \( \sqrt{2} \) gap on \( \mathbb{R}\setminus\{\sqrt{2}\} \): continuous with \( \delta=|a-\sqrt{2}| \) (stay on same side). Not UC — \( \delta\to 0 \) as \( a\to\sqrt{2} \); sequences \( \sqrt{2}\pm 1/n \) have distance \( \to 0 \) but output gap \( 1 \).

Step 2 (b). \( x^2 \) on \( \mathbb{R} \): \( \delta=\min(1,\varepsilon/(1+2|a|)) \) after \( |x-a|<1 \) bound. Continuous; not UC (depends on \( a \); e.g. \( n \) vs \( n+1/n \)).

Step 3 (c). \( \sqrt{x} \) on \( [0,\infty) \): \( |\sqrt{x}-\sqrt{a}|\le\sqrt{|x-a|} \) ⇒ \( \delta=\varepsilon^2 \) independent of \( a \) ⇒ UC.

Step 4 (d). \( x^3 \) on \( \mathbb{R} \): \( \delta=\min(1,\varepsilon/(3|a|^2+3|a|+1)) \). Continuous; not UC.

Step 5 (e). \( 1/\sqrt{x} \) on \( (0,1) \): \( \delta=\min(a/2,\varepsilon a^{3/2}/\sqrt{2}) \). Continuous; not UC as \( a\to 0^+ \).

Final answer (box this). All five are continuous on their domains; only \( \sqrt{x} \) on \( [0,\infty) \) is uniformly continuous.
Method. Present the solution as a proof or counterexample with numbered steps. Each step must state the definition or theorem used and why the next line follows — no “obvious” jumps.
Marks tip. Underline the final boxed answer. For TRUE/FALSE, include a one-line justification name (e.g. connectedness, counterexample, EVT). Missing quantifiers or confused open/closed/compact costs marks.

Hinglish — detailed (examiner-safe, beginner-clear)

UC rule: Ek hi \( \delta(\varepsilon) \) poori domain — agar \( \delta \) point \( a \) pe depend kare aur kisi edge pe \( \to 0 \), UC fail.

(a): Gap cross nahi kar sakte — local constant; \( \sqrt{2} \) ke paas delta chhota hota jata hai.

(c): Sirf yahan \( \delta=\varepsilon^2 \) — \( a \) free — UC pass.

(b)(d)(e): Pointwise delta likho + sequence counterexample UC ke against.
Exam mindset: Copy-paste feel mat do — har step pe “definition → apply → conclude” likho. English math symbols same rakh sakte ho; explanation Hinglish mein full sentences mein.
Common cutters: Definition skip, “obviously” bina reason, ∀/∃ galat, compact vs closed confuse, image vs pre-image confuse — in sab se bacho.
Final line (marks save): End pe clearly likho jo construct/prove maanga tha — ek alag “Hence …” sentence. Counterexample ho toh pehle example define, phir verify, phir conclude.

Exam mindset: Copy-paste feel nahi dena; har line pe “definition → apply → conclude” dikhana. Neeche wala write-up answer sheet pe utaar sakte ho (Hinglish soch + English math symbols).

Exam Step 1 (Hinglish): (a) Step across √2 gap: continuous with \( \delta=|a-\sqrt{2}| \) (cannot cross gap). Not UC — \( \delta\to 0 \) as \( a\to\sqrt{2} \).
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — (a) Step across √2 gap: continuous with \( \delta=|a-\sqrt{2}| \) (cannot cross gap). Not UC — \( \delta\to 0 \) as \( a… ko full sentence mein expand karke likho.

Exam Step 2 (Hinglish): (b) ( x^2 ) on \( \mathbb{R} \): ( \delta=\min(1,\varepsilon/(1+2|a|)) ). Continuous; not UC (depends on \( a \)).
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — (b) ( x^2 ) on \( \mathbb{R} \): ( \delta=\min(1,\varepsilon/(1+2|a|)) ). Continuous; not UC (depends on \( a \)). ko full sentence mein expand karke likho.

Exam Step 3 (Hinglish): (c) \( \sqrt{x} \) on ( [0,\infty) ): \( |\sqrt{x}-\sqrt{a}|\le\sqrt{|x-a|} \) ⇒ \( \delta=\varepsilon^2 \) independent of \( a \) ⇒ UC.
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — (c) \( \sqrt{x} \) on ( [0,\infty) ): \( |\sqrt{x}-\sqrt{a}|\le\sqrt{|x-a|} \) ⇒ \( \delta=\varepsilon^2 \) independent … ko full sentence mein expand karke likho.

Exam Step 4 (Hinglish): (d) ( x^3 ) on \( \mathbb{R} \): ( \delta=\min(1,\varepsilon/(3|a|^2+3|a|+1)) ). Continuous; not UC.
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — (d) ( x^3 ) on \( \mathbb{R} \): ( \delta=\min(1,\varepsilon/(3|a|^2+3|a|+1)) ). Continuous; not UC. ko full sentence mein expand karke likho.

Exam Step 5 (Hinglish): (e) \( 1/\sqrt{x} \) on \( (0,1) \): \( \delta=\min(a/2, c a^{3/2}\varepsilon) \). Continuous; not UC as \( a\to 0^+ \).
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — (e) \( 1/\sqrt{x} \) on \( (0,1) \): \( \delta=\min(a/2, c a^{3/2}\varepsilon) \). Continuous; not UC as \( a\to 0^+ \). ko full sentence mein expand karke likho.

Final line (marks save): End pe clearly likho TRUE/FALSE ya jo construct/prove maanga tha — ek alag sentence: “Hence …”. Agar counterexample hai toh pehle example fully define karo, phir check karo, phir conclude.

Common cutters: (1) definition skip (2) “obviously” without reason (3) wrong quantifiers ∀/∃ (4) compact vs closed confuse (5) image vs pre-image confuse. Inse bacho.

Source / Idea kahan se aayi

Uses: ε–δ definition; uniform continuity definition; standard algebraic estimates. Course: Continuity vs uniform continuity.

Interactive — Q18: Which are uniformly continuous?

Case ε:

Term Definitions (Q19 Extension)

Continuous extension: continuous \( g:\overline{A}\to\mathbb{R} \) with \( g|_A=f \).

Cauchy sequence: terms get arbitrarily close to each other.

Completeness: every real Cauchy sequence converges in \( \mathbb{R} \).

Question · D19 · UC functions extend continuously to the closure

If \( f:A\to\mathbb{R} \) is uniformly continuous, define \( g \) on \( \overline{A} \) by limits along sequences in \( A \). Justify well-definedness, continuity, uniqueness, and give a non-UC counterexample that does not extend.

English Mathematical Steps — full marks checklist

(a) \( x_n\to x \) in \( A \) is Cauchy; UC ⇒ \( f(x_n) \) Cauchy; completeness ⇒ limit exists. Interleaving two sequences to \( x \) shows the limit is unique ⇒ \( g \) well-defined.

(b) Use UC of \( f \) with \( \varepsilon/3 \). For ( |x-y| ) small, approximate by \( x_n,y_n\in A \) and apply triangle inequality three ways ⇒ \( g \) continuous (actually UC) on \( \overline{A} \).

(c) Any continuous extension \( h \) satisfies \( h(x)=\lim h(x_n)=\lim f(x_n)=g(x) \) ⇒ uniqueness.

(d) \( f(x)=1/x \) on \( (0,1) \) is continuous but not UC; \( \lim_{x\to 0^+}1/x=+\infty\notin\mathbb{R} \) ⇒ no continuous real extension to ( [0,1] ). Similarly \( \sin(1/x) \) (limit missing).

Exam-style answer (copy into answer sheet)

Given / Statement. If \( f:A\to\mathbb{R} \) is uniformly continuous, define \( g \) on \( \overline{A} \) by limits along sequences in \( A \). Justify well-definedness, continuity, uniqueness, and give a non-UC counterexample that does not extend.

Step 1 (a). For \( x\in\overline{A} \), pick \( x_n\to x \) in \( A \), set \( g(x)=\lim f(x_n) \). \( (x_n) \) Cauchy; UC ⇒ \( (f(x_n)) \) Cauchy; completeness of \( \mathbb{R} \) ⇒ limit exists. Interleaving two sequences to \( x \) shows the limit is unique ⇒ \( g \) well-defined.

Step 2 (b). Use UC of \( f \) with \( \varepsilon/3 \). For \( |x-y| \) small, approximate by \( x_n,y_n\in A \) and triangle inequality ⇒ \( g \) continuous (actually UC) on \( \overline{A} \).

Step 3 (c). Any continuous extension \( h \) satisfies \( h(x)=\lim h(x_n)=\lim f(x_n)=g(x) \) ⇒ uniqueness.

Step 4 (d). \( f(x)=1/x \) on \( (0,1) \) is continuous but not UC; \( \lim_{x\to 0^+}1/x=+\infty\notin\mathbb{R} \) ⇒ no continuous real extension to \( [0,1] \). Similarly \( \sin(1/x) \) (limit missing).

Final answer (box this). UC \( f \) extends uniquely and continuously to \( \overline{A} \); mere continuity without UC does not guarantee a real extension.
Method. Present the solution as a proof or counterexample with numbered steps. Each step must state the definition or theorem used and why the next line follows — no “obvious” jumps.
Marks tip. Underline the final boxed answer. For TRUE/FALSE, include a one-line justification name (e.g. connectedness, counterexample, EVT). Missing quantifiers or confused open/closed/compact costs marks.

Hinglish — detailed (examiner-safe, beginner-clear)

Extension story: UC Cauchy inputs ko Cauchy outputs deta hai — boundary pe limit exist; interleaving se well-defined; continuous extension unique.

(a): \( x_n\to x \) Cauchy → UC → \( f(x_n) \) Cauchy → \( \mathbb{R} \) complete; do sequences interleave karke same limit.

(d): \( 1/x \) UC nahi — \( 1/n\to 0 \) par outputs \( n\to\infty \); real continuous value at 0 impossible.

Final: UC sufficient for closure extension; counterexample bina UC dikhana zaroori.
Exam mindset: Copy-paste feel mat do — har step pe “definition → apply → conclude” likho. English math symbols same rakh sakte ho; explanation Hinglish mein full sentences mein.
Common cutters: Definition skip, “obviously” bina reason, ∀/∃ galat, compact vs closed confuse, image vs pre-image confuse — in sab se bacho.
Final line (marks save): End pe clearly likho jo construct/prove maanga tha — ek alag “Hence …” sentence. Counterexample ho toh pehle example define, phir verify, phir conclude.

Exam mindset: Copy-paste feel nahi dena; har line pe “definition → apply → conclude” dikhana. Neeche wala write-up answer sheet pe utaar sakte ho (Hinglish soch + English math symbols).

Exam Step 1 (Hinglish): (a) \( x_n\to x \) in \( A \) is Cauchy; UC ⇒ \( f(x_n) \) Cauchy; completeness ⇒ limit exists. Interleaving two sequences to \( x \) shows the limit is unique ⇒ \( g \) well-defined.
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — (a) \( x_n\to x \) in \( A \) is Cauchy; UC ⇒ \( f(x_n) \) Cauchy; completeness ⇒ limit exists. Interleaving two sequenc… ko full sentence mein expand karke likho.

Exam Step 2 (Hinglish): (b) Use UC of \( f \) with \( \varepsilon/3 \). For ( |x-y| ) small, approximate by \( x_n,y_n\in A \) and apply triangle inequality three ways ⇒ \( g \) continuous (actually UC) on \( \overline{A} \).
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — (b) Use UC of \( f \) with \( \varepsilon/3 \). For ( |x-y| ) small, approximate by \( x_n,y_n\in A \) and apply triangl… ko full sentence mein expand karke likho.

Exam Step 3 (Hinglish): (c) Any continuous extension \( h \) satisfies \( h(x)=\lim h(x_n)=\lim f(x_n)=g(x) \) ⇒ uniqueness.
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — (c) Any continuous extension \( h \) satisfies \( h(x)=\lim h(x_n)=\lim f(x_n)=g(x) \) ⇒ uniqueness. ko full sentence mein expand karke likho.

Exam Step 4 (Hinglish): (d) \( f(x)=1/x \) on \( (0,1) \) is continuous but not UC; \( \lim_{x\to 0^+}1/x=+\infty\notin\mathbb{R} \) ⇒ no continuous real extension to ( [0,1] ). Similarly \( \sin(1/x) \) (limit missing).
Kyun likhen: Examiner ko dikhana hai ki aap sirf answer nahi, reason bhi jaante ho. Is step mein clearly likho: kya assume kiya, kaunsa definition/theorem use hua, aur next step kaise follow hota hai. Short one-liner mat chhodna — (d) \( f(x)=1/x \) on \( (0,1) \) is continuous but not UC; \( \lim_{x\to 0^+}1/x=+\infty\notin\mathbb{R} \) ⇒ no contin… ko full sentence mein expand karke likho.

Final line (marks save): End pe clearly likho TRUE/FALSE ya jo construct/prove maanga tha — ek alag sentence: “Hence …”. Agar counterexample hai toh pehle example fully define karo, phir check karo, phir conclude.

Common cutters: (1) definition skip (2) “obviously” without reason (3) wrong quantifiers ∀/∃ (4) compact vs closed confuse (5) image vs pre-image confuse. Inse bacho.

Source / Idea kahan se aayi

Uses: UC preserves Cauchy sequences; completeness of \( \mathbb{R} \); sequential continuity. Course: Uniform continuity / metric completion ideas.

Interactive — Q19: Extension succeeds vs fails

Example