Derivatives & Quiz

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.

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 \( f'(c) = 0 \).

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: \[ f'(c) = \frac{f(b) - f(a)}{b - a} \]

Let \( A=\{(-1)^{n}(1+\frac{1}{n}) : n \in \mathbb{N}\} \). Find \( \sup A \), \( \inf A \), limit points, closure \( \overline{A} \), and determine if it is compact.

Analysis: If \( n \) is even, terms are positive: \( 3/2, 5/4, 7/6 \dots \to 1 \). If \( n \) is odd, terms are negative: \( -2, -4/3, -6/5 \dots \to -1 \).

• sup A: \( 3/2 \) (Maximum value at \( n=2 \)).
• inf A: \( -2 \) (Minimum value at \( n=1 \)).
• Limit points: \( \{1, -1\} \) (Sequences converge here as \( n \to \infty \)).
• Closure \( \overline{A} \): \( A \cup \{1, -1\} \).
• Compact?: No. It is bounded but not closed (limit points \( 1, -1 \notin A \)).

Hinglish: Jaise jaise \( n \) ki value badhegi, even numbers 1 ki taraf aur odd numbers -1 ki taraf jayenge, yahi limit points hain. Compact nahi hai kyunki limit points set ke andar nahi hain, toh set closed nahi hai.

(i) Infinite compact subset of \( \mathbb{R} \) with exactly one limit point:
\( A = \{0\} \cup \{\frac{1}{n} : n \in \mathbb{N}\} = \{0, 1, 1/2, 1/3 \dots \} \). It is bounded (in \( [0,1] \)) and closed (contains its only limit point \( 0 \)). Hence, compact.

(ii) A subset of \( \mathbb{R} \) neither open nor closed:
Half-open interval \( A = [0, 1) \). Not open (contains boundary \( 0 \)). Not closed (missing limit point \( 1 \)).

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: Define \( h(x) = f(x) - g(x) \). Since the difference of two continuous functions is continuous, \( h \) is continuous. \( A = \{x : h(x) = 0\} = h^{-1}(\{0\}) \). Since \( \{0\} \) is a closed set in \( \mathbb{R} \), and the continuous pre-image of a closed set is always closed, \( A \) must be closed.

Hinglish: Ek naya function maante hain: \( h(x) = f(x) - g(x) \). Set \( A \) wahi set hai jahan \( h(x) = 0 \) hoga. Ek rule hai ki kisi continuous function mein agar output (image) closed hai, toh uska input (pre-image) bhi closed hota hai.

(ii) Is \( \{x : f(x)/g(x) = 1\} \) closed?
No. For \( f(x)/g(x) \) to be defined, \( g(x) \neq 0 \). Example: \( f(x)=x \), \( g(x)=x \). Both continuous. \( x/x = 1 \) for all \( x \neq 0 \). The set is \( \mathbb{R} \setminus \{0\} \), which is open, not closed!

Let \( K \subseteq \mathbb{R} \) be a nonempty compact set and \( x_0 \in \mathbb{R} \). Prove \( \exists p \in K \) such that \( |x_0 - p| \le |x_0 - y| \) for all \( y \in K \).

Proof: Define a distance function \( d(y) = |x_0 - y| \) for \( y \in K \). The absolute value function is continuous. By the Extreme Value Theorem, a continuous function evaluated on a compact set is bounded and must attain its absolute minimum. Thus, \( d(y) \) achieves its minimum at some point \( p \in K \). This means \( d(p) \le d(y) \), i.e., \( |x_0 - p| \le |x_0 - y| \).

Hinglish: Distance function \( d(y) = |x_0 - y| \) continuous hota hai. 'Extreme Value Theorem' kehti hai ki kisi compact set par continuous function apni minimum value zaroor chhoota hai. Toh \( K \) mein ek sabse kareebi point \( p \) zaroor hoga jahan distance sabse kam ho.

Statement: Let \( U \subseteq \mathbb{R} \) be a bounded open set. Prove that \( U \) cannot contain \( \sup U \).

English Proof: Let us assume for the sake of contradiction that the supremum of \( U \), let's call it \( s = \sup U \), is actually contained inside the set \( U \) (i.e., \( s \in U \)).
Since we are given that \( U \) is an open set, every point in \( U \) must have an open interval around it that is completely contained within \( U \). Therefore, since \( s \in U \), there must exist some small positive number \( \varepsilon > 0 \) such that the interval \( (s - \varepsilon, s + \varepsilon) \) is entirely a subset of \( U \).
This implies that the number \( s + \frac{\varepsilon}{2} \), which is strictly greater than \( s \), is also an element of the set \( U \).
However, \( s \) was defined as the supremum (upper bound) of \( U \), meaning no element in \( U \) can be strictly greater than \( s \). We have found an element \( s + \frac{\varepsilon}{2} \in U \) that is greater than \( s \), which is a direct contradiction.
Thus, our initial assumption must be wrong. The set \( U \) cannot contain its supremum.

Hinglish Proof: Maan lijiye thodi der ke liye ulta sochte hain (contradiction ke liye): hum maante hain ki set \( U \) ka supremum, jise hum \( s \) keh dete hain (\( s = \sup U \)), set \( U \) ke andar hi maujood hai (\( s \in U \)).
Kyunki question mein diya hai ki \( U \) ek "open set" hai, iska matlab set ke har point ke aas-paas ek chhota sa open interval banaya ja sakta hai jo poora ka poora set \( U \) ke andar hi hoga. Ab kyunki \( s \in U \), toh zaroor ek chhota sa positive number \( \varepsilon > 0 \) hoga jisse interval \( (s - \varepsilon, s + \varepsilon) \) poori tarah set \( U \) ke andar aayega.
Iska matlab yeh hua ki number \( s + \frac{\varepsilon}{2} \), jo ki \( s \) se bada hai, wo bhi set \( U \) ka hi hissa hai.
Lekin wait, humne shuru mein \( s \) ko supremum mana tha (yaani set ka sabse bada upper bound), jiska matlab hai set \( U \) ka koi bhi element \( s \) se bada nahi ho sakta. Par hume toh \( s + \frac{\varepsilon}{2} \) mil gaya jo set mein hai aur \( s \) se bada hai! Yeh ek virodhabhas (contradiction) hai. Isliye, ek bounded open set \( U \) kabhi bhi apne supremum ko apne andar nahi rakh sakta.

Statement: Determine whether \( f:(0,\infty) \to \mathbb{R} \), \( f(x) = \sqrt{x} \) is continuous.

Answer: Yes, it is continuous.

English Proof (\( \varepsilon-\delta \)): Let \( c \in (0, \infty) \). We need to show that for any given \( \varepsilon > 0 \), there exists a \( \delta > 0 \) such that if \( |x - c| < \delta \), then \( |\sqrt{x} - \sqrt{c}| < \varepsilon \).
Consider the expression: \( |\sqrt{x} - \sqrt{c}| = \left| \frac{(\sqrt{x} - \sqrt{c})(\sqrt{x} + \sqrt{c})}{\sqrt{x} + \sqrt{c}} \right| = \frac{|x - c|}{\sqrt{x} + \sqrt{c}} \).
Since \( x \in (0, \infty) \) and \( c > 0 \), we know that \( \sqrt{x} + \sqrt{c} > \sqrt{c} \).
Therefore, \( \frac{|x - c|}{\sqrt{x} + \sqrt{c}} < \frac{|x - c|}{\sqrt{c}} \).
To make this entire term less than \( \varepsilon \), we need \( \frac{|x - c|}{\sqrt{c}} < \varepsilon \), which implies \( |x - c| < \varepsilon\sqrt{c} \).
So, we can choose \( \boldsymbol{\delta = \varepsilon\sqrt{c}} \). If \( |x - c| < \delta \), then \( |\sqrt{x} - \sqrt{c}| < \varepsilon \). The function is continuous.

Hinglish Proof: Maan lijiye domain \( (0, \infty) \) mein ek point \( c \) hai. Hume \( \varepsilon-\delta \) proof se dikhana hai ki har \( \varepsilon > 0 \) ke liye ek \( \delta > 0 \) hai.
Pehle hum \( |\sqrt{x} - \sqrt{c}| \) ko simplify karte hain usko rationalize karke: \( |\sqrt{x} - \sqrt{c}| = \frac{|x - c|}{\sqrt{x} + \sqrt{c}} \).
Kyunki \( x \) aur \( c \) dono positive hain, denominator hamesha \( \sqrt{c} \) se bada hoga. Toh hum likh sakte hain: \( \frac{|x - c|}{\sqrt{x} + \sqrt{c}} < \frac{|x - c|}{\sqrt{c}} \).
Hume answer \( \varepsilon \) se chota chahiye, isliye hum likhenge \( \frac{|x - c|}{\sqrt{c}} < \varepsilon \), jisse mila \( |x - c| < \varepsilon\sqrt{c} \). Yahan hum apni \( \delta \) ki value \( \varepsilon\sqrt{c} \) set kar sakte hain.

Statement: Determine whether \( f:\mathbb{R} \to \mathbb{R} \), \( f(x) = x^2 \) is continuous.

Answer: Yes, it is continuous.

English Proof (\( \varepsilon-\delta \)): Let \( c \in \mathbb{R} \). For a given \( \varepsilon > 0 \), we want \( \delta > 0 \) such that if \( |x - c| < \delta \), then \( |x^2 - c^2| < \varepsilon \).
We factor the expression: \( |x^2 - c^2| = |x - c||x + c| \).
We need to put an upper bound on \( |x + c| \). Let's restrict \( \delta \le 1 \). If we assume \( |x - c| < 1 \), then by the triangle inequality, \( |x| = |x - c + c| \le |x - c| + |c| < 1 + |c| \).
Now, let's bound \( |x + c| \): \( |x + c| \le |x| + |c| < (1 + |c|) + |c| = 2|c| + 1 \).
Substituting this back: \( |x^2 - c^2| = |x - c||x + c| < |x - c|(2|c| + 1) \).
We want this to be less than \( \varepsilon \), so we need \( |x - c| < \frac{\varepsilon}{2|c| + 1} \).
To satisfy both restrictions, we choose \( \boldsymbol{\delta = \min(1, \frac{\varepsilon}{2|c| + 1})} \). Thus, the function is continuous.

Hinglish Proof: Koi bhi point \( c \in \mathbb{R} \) lijiye. Hume \( |x^2 - c^2| \) ko \( \varepsilon \) se chota dikhana hai.
Factorize karte hain: \( |x^2 - c^2| = |x - c||x + c| \). Yahan \( |x + c| \) ki value ko control karne ke liye hum \( \delta \) ki ek limit set karte hain, maan lijiye \( \delta \le 1 \). Agar \( |x - c| < 1 \) hai, toh \( |x + c| \) zyada se zyada kitna bada ho sakta hai? \( |x + c| \le |x| + |c| < 1 + |c| + |c| = 2|c| + 1 \).
Ab isko wapas equation mein dalte hain: \( |x^2 - c^2| < |x - c|(2|c| + 1) \). Hume final result \( \varepsilon \) se chota chahiye, toh \( |x - c| < \frac{\varepsilon}{2|c| + 1} \) chahiye hoga.
Toh humari final \( \delta \) ki value hogi 1 aur \( \frac{\varepsilon}{2|c| + 1} \) mein se jo choti ho. Is tarah yeh continuous prove hota hai.

Statement: Determine whether \( f:[0,1] \to \mathbb{R} \) defined by \( f(x) = 0 \) for \( x \in \mathbb{Q} \) and \( f(x) = 1 \) for \( x \in \mathbb{R} \setminus \mathbb{Q} \) is continuous.

Answer: No, it is not continuous anywhere.

English Proof (Sequential Characterization): Let \( c \) be any point in \( [0, 1] \). There are two cases:
Case 1: \( c \in \mathbb{Q} \). Then \( f(c) = 0 \). Because the irrational numbers are dense in \( \mathbb{R} \), we can find a sequence of irrational numbers \( (x_n) \) that converges to \( c \). Since each \( x_n \) is irrational, \( f(x_n) = 1 \) for all \( n \). Therefore, \( \lim f(x_n) = 1 \). But \( f(c) = 0 \). Since \( \lim f(x_n) \neq f(c) \), the function is not continuous at \( c \).
Case 2: \( c \in \mathbb{R} \setminus \mathbb{Q} \). Then \( f(c) = 1 \). Because the rational numbers are dense in \( \mathbb{R} \), we can find a sequence of rational numbers \( (q_n) \) that converges to \( c \). Since each \( q_n \) is rational, \( f(q_n) = 0 \) for all \( n \). Therefore, \( \lim f(q_n) = 0 \). But \( f(c) = 1 \). Since \( \lim f(q_n) \neq f(c) \), the function is not continuous at \( c \).
The function is discontinuous at every point.

Hinglish Proof: Is function ki continuity check karne ke liye hum 'sequences' ka use karenge. Maan lijiye \( c \) koi bhi point hai.
Pehla Case: Agar \( c \) ek rational number hai. Yahan \( f(c) = 0 \) hoga. Hume pata hai ki irrationals real line par har jagah faile hote hain (dense). Toh hum ek aisi sequence bana sakte hain irrationals ki \( (x_n) \) jo dreere dreere \( c \) ke paas ja rahi ho. Kyunki sequence ke saare number irrational hain, output un sab par 1 hoga. Par exact \( c \) par output 0 hai. Sequence 1 ki taraf ja rahi hai, par destination point par value 0 hai. Continuity break ho gayi.
Dusra Case: Agar \( c \) irrational hai, toh \( f(c) = 1 \). Same logic se hum rationals ki ek sequence le sakte hain jo \( c \) ki taraf badh rahi ho. Un sab rationals par output 0 aayega. Yahan bhi sequence 0 ki taraf ja rahi hai par destination \( c \) par value 1 hai. Yeh function kahin bhi continuous nahi hai.

Statement: Let \( F \subseteq \mathbb{R} \) be a non-empty closed set, and let \( f:\mathbb{R} \to \mathbb{R} \) be \( f(x) = \inf\{|x-a| : a \in F\} \). Determine if \( f \) is continuous.

Answer: Yes, it is continuous (in fact, uniformly continuous).

English Proof: The function \( f(x) \) represents the shortest distance from a point \( x \) to the set \( F \). We will prove this using the triangle inequality.
For any two points \( x, y \in \mathbb{R} \) and any point \( a \in F \):
\( |x - a| = |(x - y) + (y - a)| \le |x - y| + |y - a| \) (by Triangle Inequality).
Taking the infimum over all \( a \in F \) on both sides:
\( \inf_{a \in F} |x - a| \le |x - y| + \inf_{a \in F} |y - a| \)
\( f(x) \le |x - y| + f(y) \implies f(x) - f(y) \le |x - y| \).
By swapping \( x \) and \( y \), we can symmetrically prove \( f(y) - f(x) \le |x - y| \). Combining these gives \( |f(x) - f(y)| \le |x - y| \).
Now, applying the \( \varepsilon-\delta \) definition: for any given \( \varepsilon > 0 \), we can simply choose \( \boldsymbol{\delta = \varepsilon} \). If \( |x - y| < \delta \), then \( |f(x) - f(y)| \le |x - y| < \delta = \varepsilon \). Therefore, the function is continuous.

Hinglish Proof: Yeh function \( f(x) \) asal mein kisi bhi point \( x \) ki set \( F \) se sabse chhoti doori batata hai. Isko prove karne ke liye hum 'Triangle Inequality' ka use karenge.
Agar hum koi bhi do points \( x, y \) lein, aur set \( F \) mein ek point \( a \) lein: \( |x - a| \le |x - y| + |y - a| \).
Agar hum is equation mein \( a \) ki sabse chhoti distance (infimum) nikalen, toh equation banegi: \( f(x) \le |x - y| + f(y) \) jisko hum aise likh sakte hain: \( f(x) - f(y) \le |x - y| \).
Agar hum \( x \) aur \( y \) ki jagah badal dein toh aayega \( f(y) - f(x) \le |x - y| \). In dono ko milakar milta hai: \( |f(x) - f(y)| \le |x - y| \).
\( \varepsilon-\delta \) proof ke liye, har \( \varepsilon > 0 \) ke liye hum seedha \( \delta = \varepsilon \) chun sakte hain. Kyunki jab \( |x - y| < \delta \) hoga, toh \( |f(x) - f(y)| \le |x - y| < \delta = \varepsilon \). Yeh function bilkul continuous hai.

Statement: Let \( U_{\alpha} \subset \mathbb{R} \) be open sets, with \( f_{\alpha} : U_{\alpha} \to \mathbb{R} \) continuous functions. Assuming that \( f_{\alpha}(x) = f_{\beta}(x) \) for all \( x \in U_{\alpha} \cap U_{\beta} \), show that the function \( f \) defined on \( U = \bigcup_{\alpha} U_{\alpha} \) by \( f(x) = f_{\alpha}(x) \) for \( x \in U_{\alpha} \) is well-defined and continuous.

English Proof (Well-defined): Let \( x \) be an arbitrary point in \( U \). By definition of union, \( x \) must belong to at least one open set, say \( U_\alpha \). If \( x \) happens to also belong to another open set \( U_\beta \), the problem explicitly states that \( f_\alpha(x) = f_\beta(x) \). Therefore, assigning \( f(x) \) the value of \( f_\alpha(x) \) yields a single, consistent result regardless of which set covering \( x \) we choose. Thus, \( f \) is well-defined.

Hinglish Proof (Well-defined): Maan lijiye \( x \) set \( U \) ka ek point hai. \( U \) banaya gaya hai chote sets ko milakar, toh \( x \) kam se kam kisi ek \( U_\alpha \) mein toh hoga hi. Agar \( x \) galti se kisi dusre set \( U_\beta \) mein bhi aata hai, toh question ne pehle hi bata diya hai ki \( f_\alpha(x) \) aur \( f_\beta(x) \) ka result ekdum barabar hoga. Iska matlab hai ki hum chahe kisi bhi set ke through value nikalein, answer ek hi aayega. Isliye function well-defined hai.

English Proof (Continuity): Let \( V \) be any open set in \( \mathbb{R} \). We need to show that the pre-image \( f^{-1}(V) \) is open in \( U \). By definition, \( f^{-1}(V) = \{x \in U : f(x) \in V\} \). We can rewrite this set by looking at how \( f \) behaves on each piece \( U_\alpha \):
\( f^{-1}(V) = \bigcup_\alpha \{x \in U_\alpha : f_\alpha(x) \in V\} = \bigcup_\alpha f_\alpha^{-1}(V) \).
Since \( f_\alpha \) is a continuous function, \( f_\alpha^{-1}(V) \) is an open set in \( U_\alpha \). Because \( U_\alpha \) itself is an open set in \( \mathbb{R} \), \( f_\alpha^{-1}(V) \) is also an open set in \( \mathbb{R} \). Since the arbitrary union of open sets is always open, the union \( \bigcup_\alpha f_\alpha^{-1}(V) \) is open. Therefore, \( f^{-1}(V) \) is open, which proves \( f \) is continuous.

Hinglish Proof (Continuity): Continuity check karne ka ek rule hai ki agar output (\( V \)) ek open set hai, toh uska input \( f^{-1}(V) \) bhi open hona chahiye. Hum input set ko alag-alag tukdon mein tod sakte hain: \( f^{-1}(V) = \bigcup_\alpha f_\alpha^{-1}(V) \). Hume pata hai ki \( f_\alpha \) continuous hai, toh uske liye output ko input tak trace back karenge toh result \( f_\alpha^{-1}(V) \) open hi milega. Ek rule yeh bhi hai ki kitne bhi 'open sets' ko aapas mein mila lo (union kar lo), banne wala naya set hamesha open hi rehta hai. Toh \( \bigcup_\alpha f_\alpha^{-1}(V) \) bhi open hoga. Isse sabit hota hai ki poora function \( f \) continuous hai.

Statement: Find two discontinuous functions \( f, g \) such that \( f \circ g \) is continuous.

English Example & Justification: Let us define \( g:\mathbb{R} \to \mathbb{R} \) as a step function: \( g(x) = \begin{cases} 1 & \text{if } x \ge 0 \\ -1 & \text{if } x < 0 \end{cases} \)
The function \( g \) is clearly discontinuous at \( x = 0 \) because there is a jump from \(-1\) to \( 1 \).

Now let us define \( f:\mathbb{R} \to \mathbb{R} \) as another piecewise function: \( f(x) = \begin{cases} 0 & \text{if } x = 1 \text{ or } x = -1 \\ 5 & \text{otherwise} \end{cases} \)
The function \( f \) is discontinuous at \( x = 1 \) and \( x = -1 \) due to the points being displaced from the rest of the graph.

Now, let's find the composition \( h(x) = (f \circ g)(x) = f(g(x)) \). For any value of \( x \), \( g(x) \) will always output either \( 1 \) or \( -1 \). When we feed this result into \( f \), we are always evaluating either \( f(1) \) or \( f(-1) \). According to our definition of \( f \), both \( f(1) \) and \( f(-1) \) equal \( 0 \). Therefore, \( (f \circ g)(x) = 0 \) for all \( x \in \mathbb{R} \). This is a constant function, which is continuous everywhere.

Hinglish Justification: Maan lijiye hum ek function banate hain \( g(x) \). Agar \( x \) ki value \( 0 \) ya usse badi hai, toh output \( 1 \) aayega, aur agar \( x \) zero se chota hai, toh output \( -1 \) aayega. Yeh function \( x=0 \) par achanak jump karta hai, isliye yeh discontinuous hai. Ab dusra function \( f(x) \) banate hain. Agar hum isme \( 1 \) ya \( -1 \) daalte hain toh answer \( 0 \) aayega. In do values ke alawa kuch bhi dalenge toh answer \( 5 \) aayega. Yeh function \( x=1 \) aur \( x=-1 \) par toota hua hai. Ab in dono ko jodte hain: \( f(g(x)) \). Aap \( x \) ki koi bhi value sochiye, \( g(x) \) humesha usko ya toh \( 1 \) bana dega ya \( -1 \). Ab jab hum yeh value aage \( f \) mein dalenge, toh hum asal mein hamesha \( f(1) \) ya \( f(-1) \) hi nikal rahe honge, jo humesha \( 0 \) hota hai. Toh overall result \( (f \circ g)(x) = 0 \) ban gaya. Yeh ek flat, straight line hai jo bilkul continuous hai!

Statement: Can there exist a closed set \( A \subseteq \mathbb{R} \) and a continuous function \( f:\mathbb{R} \to \mathbb{R} \) such that \( f(A) \) is not a closed subset of \( \mathbb{R} \)?

Answer: Yes.

English Proof / Example: Let the closed set be the set of natural numbers \( A = \{1, 2, 3, 4, ...\} \subset \mathbb{R} \). (Any discrete set of points with no accumulation points in \( \mathbb{R} \) is closed). Let our continuous function be \( f(x) = \frac{1}{x^2 + 1} \). This function is continuous for all real numbers. Let us find the image of \( A \): \( f(A) = \{\frac{1}{2}, \frac{1}{5}, \frac{1}{10}, \frac{1}{17}, ...\} \). As \( n \) grows larger, \( \frac{1}{n^2 + 1} \) gets closer and closer to \( 0 \). Therefore, \( 0 \) is a limit point of the set \( f(A) \). However, the number \( 0 \) is not actually in the set \( f(A) \) because there is no finite natural number \( x \) that makes \( \frac{1}{x^2 + 1} = 0 \). Since the set \( f(A) \) does not contain its limit point \( 0 \), it is not a closed set.

Hinglish Proof: Haan, aisa ho sakta hai. Maan lijiye hamara closed set saare natural numbers ka set hai: \( A = \{1, 2, 3, ...\} \). (Yeh closed isliye hai kyunki iske points door-door hain aur iska koi limit point real line par nahi banta). Ab hum ek continuous function lete hain: \( f(x) = \frac{1}{x^2 + 1} \). Jab hum set \( A \) ki values isme dalte hain, toh output set banta hai \( f(A) = \{\frac{1}{2}, \frac{1}{5}, \frac{1}{10}, ...\} \). Jaise jaise hum aage badhenge, yeh numbers chhote hokar \( 0 \) ki taraf jayenge. Iska matlab hai \( 0 \) is output set ka limit point hai. Lekin, \( 0 \) khud is set ke andar maujood nahi hai. Kyunki set \( f(A) \) mein apna limit point nahi hai, isliye yeh output set closed nahi hai.

Statement: Can there exist a compact set \( K \) and a continuous function such that \( f^{-1}(K) \) is not a compact set?

Answer: Yes.

English Proof / Example: Consider the constant function \( f(x) = 0 \) for all \( x \in \mathbb{R} \). A constant function is continuous everywhere. Let the set \( K = \{0\} \). Because \( K \) contains only a single finite point, it is closed and bounded, making it a compact set. Now, let's find the pre-image \( f^{-1}(K) \). This means finding all \( x \) values such that \( f(x) = 0 \). Since \( f(x) \) is \( 0 \) for every real number, the pre-image is the entire real line: \( f^{-1}(K) = \mathbb{R} \). The set \( \mathbb{R} \) is closed, but it is not bounded (it stretches from negative infinity to positive infinity). Therefore, by the Heine-Borel theorem, \( \mathbb{R} \) is not a compact set.

Hinglish Proof: Haan, yeh bhi bilkul possible hai. Ek sabse simple function sochiye: \( f(x) = 0 \). Yeh function duniya ki har \( x \) value par sirf \( 0 \) answer deta hai aur yeh completely continuous hai. Ab hum ek compact set \( K = \{0\} \) lete hain. Kyunki isme sirf ek hi element hai, yeh bounded bhi hai aur closed bhi, isliye yeh compact hai. Ab hum pre-image nikalte hain yaani ulta sochte hain: aise kaunse \( x \) hain jinko function mein dalne par answer \( 0 \) aaye? Kyunki har value par answer 0 hi aata hai, iska jawab hai saare real numbers, yaani poori \( \mathbb{R} \) line. \( f^{-1}(K) = \mathbb{R} \). Ab jo yeh set \( \mathbb{R} \) aaya hai, yeh infinite tak phaila hua hai, toh yeh bounded nahi hai. Bounded na hone ki wajah se yeh pre-image compact nahi ban sakti.

Statement: Using the density of the rationals, and the fact that \( \mathbb{Q} \) is a countable set, show that there cannot be uncountably many disjoint open intervals in \( \mathbb{R} \).

English Proof: Let us assume we have an arbitrary collection of disjoint open intervals in \( \mathbb{R} \). Let's denote this collection as \( \{I_\alpha\}_{\alpha \in A} \), where \( A \) is some index set.
Because each interval \( I_\alpha \) is open and non-empty, it must have a left endpoint \( a \) and a right endpoint \( b \) such that \( a < b \), making the interval \( (a, b) \).
According to the density of rational numbers (\( \mathbb{Q} \)), between any two real numbers \( a \) and \( b \), there must exist at least one rational number \( q \). Therefore, we can pick exactly one rational number \( q_\alpha \in \mathbb{Q} \) such that \( q_\alpha \in I_\alpha \) for every interval in our collection.
Because the intervals are given to be strictly disjoint, they do not overlap. Consequently, no two different intervals can contain the same rational number. This establishes a one-to-one (injective) mapping from the collection of intervals to the set of rational numbers \( \mathbb{Q} \).
Since the set of rational numbers \( \mathbb{Q} \) is a countable set, any subset of \( \mathbb{Q} \) must also be at most countable. Because we mapped each interval to a distinct rational number, the total number of disjoint open intervals cannot exceed the total number of rational numbers. Therefore, the index set \( A \) is at most countable, proving that uncountably many disjoint open intervals cannot exist in \( \mathbb{R} \).

Hinglish Proof: Maan lijiye hamare paas \( \mathbb{R} \) par disjoint open intervals (aise intervals jinme kuch bhi common nahi hai) ka ek bada collection hai.
Har interval "open" hai, toh uske andar koi na koi do points \( a \) aur \( b \) zaroor honge jahan \( a < b \) hoga.
Rationals ki "density property" kehti hai ki kisi bhi do real numbers ke beech kam se kam ek rational number (\( p/q \) form) zaroor hota hai. Iska matlab har ek interval ke andar hume ek rational number toh mil hi jayega. Hum har interval ke liye ek aisa rational number (\( q_\alpha \)) chun lete hain.
Kyunki ye sabhi intervals "disjoint" hain (bilkul alag-alag hain), toh kisi bhi do intervals mein aane wala rational number same nahi ho sakta. Is tarah humne har interval ko ek unique rational number pakda diya.
Kyunki total rational numbers (\( \mathbb{Q} \)) "countable" hote hain (unhe natural numbers ki tarah gina ja sakta hai), toh intervals ki ginti bhi un assigned rational numbers ki ginti se zyada nahi ho sakti. Isliye, \( \mathbb{R} \) line par uncountable (anginat, jinko gina na ja sake) disjoint open intervals ka hona mathematical roop se namumkin hai.

Statement: Let \( U \subseteq \mathbb{R} \) be an open set. For points \( x, y \in U \), define a relation by \( x \sim y \) if there exists \( \delta_1, \delta_2 > 0 \) such that \( y \in (x - \delta_1, x + \delta_2) \subseteq U \). Show it's an equivalence relation, classes are disjoint open intervals, and conclude every open set in \( \mathbb{R} \) is a union of countably many disjoint open intervals.

(a) Equivalence Relation:
Reflexive: Let \( x \in U \). Since \( U \) is open, there exists an open interval around \( x \), say \( (x - \varepsilon, x + \varepsilon) \subseteq U \). If we choose \( \delta_1 = \varepsilon \) and \( \delta_2 = \varepsilon \), then \( x \in (x - \delta_1, x + \delta_2) \subseteq U \). Thus, \( x \sim x \).
Symmetric: Assume \( x \sim y \). This means \( y \in (x - \delta_1, x + \delta_2) \subseteq U \). Let \( a = x - \delta_1 \) and \( b = x + \delta_2 \). This means both \( x \) and \( y \) belong to the open interval \( (a, b) \subseteq U \). Because \( (a, b) \) is an open interval containing \( y \), we can write it in terms of \( y \): \( (a, b) = (y - (y - a), y + (b - y)) \). Let \( \delta_3 = y - a \) and \( \delta_4 = b - y \). Both are positive. Then \( x \in (a,b) = (y - \delta_3, y + \delta_4) \subseteq U \). Thus, \( y \sim x \).
Transitive: Assume \( x \sim y \) and \( y \sim z \). This means \( x \) and \( y \) share some open interval \( I_1 \subseteq U \), and \( y \) and \( z \) share some open interval \( I_2 \subseteq U \). Because \( y \) is in both intervals, their union \( I_1 \cup I_2 \) forms a single, larger connected open interval that is fully contained in \( U \). This combined interval contains both \( x \) and \( z \). Thus, we can find an interval around \( x \) containing \( z \) inside \( U \), meaning \( x \sim z \).

(b) \( [x] \) is an open interval:
By definition, \( [x] \) is the set of all points \( y \) related to \( x \). This is exactly the union of all open intervals \( I \subseteq U \) that contain the point \( x \). The union of a collection of open intervals that all intersect at a common point (here, \( x \)) is always a single connected open interval. Therefore, \( [x] \) is an open interval.

(c) \( [x] \) and \( [y] \) are disjoint:
This is a fundamental property of equivalence classes. Suppose for a contradiction that \( [x] \) and \( [y] \) are not disjoint, meaning there is some point \( z \) that belongs to both. This would mean \( z \sim x \) and \( z \sim y \). By symmetry and transitivity, \( x \sim z \) and \( z \sim y \) implies \( x \sim y \). But the premise states \( x \) and \( y \) are not related. Thus, our assumption is false, and they must be strictly disjoint.

(d) & (e) Countably many distinct classes forming \( U \):
From part (b) and (c), the distinct equivalence classes \( [x] \) form a collection of disjoint open intervals in \( \mathbb{R} \). By the proof established in Question 6, any collection of disjoint open intervals in \( \mathbb{R} \) must be at most countable.
Because every point \( x \in U \) is in its own equivalence class \( [x] \) (reflexivity), the entire set \( U \) is simply the union of all distinct equivalence classes. Thus, \( U = \bigcup [x] \). Since each \( [x] \) is a disjoint open interval, and there are only countably many of them, \( U \) is the union of countably many disjoint open intervals.

Hinglish Summary: Set \( U \) ka har ek element kisi na kisi \( [x] \) class ke andar hoga. Prove hua ki yeh classes asal mein ek dusre se alag (disjoint) open intervals hain. Toh agar hum in saare alag-alag \( [x] \) intervals ko mila dein, toh poora set \( U \) ban jayega. Aur Q6 ne bataya ki aisi alag-alag intervals anant (uncountable) nahi ho sakti, sirf countable ho sakti hain. Isse sabit hota hai ki Real line par koi bhi open set asal mein countable disjoint open intervals ke milne se hi banta hai.

Statement: On every open interval \( (a, b) \), construct a continuous function \( f \) which is not bounded. Use Q7 and Q3 to conclude that on every open set \( U \), there is a continuous function \( f \) which is not bounded.

Part 1: Constructing \( f \) on \( (a, b) \):
Let the open interval be \( (a, b) \). We can construct a simple rational function: \( f(x) = \frac{1}{x - a} \).
Since the denominator \( (x - a) \) is not zero for any \( x \in (a, b) \), this function is continuous on the entire interval \( (a, b) \).
However, as \( x \) approaches the left boundary \( a \) from the right (\( x \to a^+ \)), the denominator becomes infinitesimally small, causing the function's value to approach positive infinity (\( f(x) \to \infty \)). Thus, the function is continuous but strictly unbounded on the open interval \( (a, b) \).

Part 2: Extending to an open set \( U \):
By the conclusion of Question 7(e), any arbitrary open set \( U \subseteq \mathbb{R} \) can be expressed as a countable union of disjoint open intervals. Let \( U = \bigcup_{\alpha} I_\alpha \), where each \( I_\alpha = (a_\alpha, b_\alpha) \).
On each individual interval \( I_\alpha \), we define the continuous, unbounded function we constructed in Part 1:
\( f_\alpha(x) = \frac{1}{x - a_\alpha} \) for \( x \in I_\alpha \).
Now we want to glue these functions together to form a single function \( f \) defined on all of \( U \), such that \( f(x) = f_\alpha(x) \) whenever \( x \in I_\alpha \).
According to Question 3 (The Pasting Lemma), this combined function \( f \) is continuous on \( U \) as long as the functions agree on the overlapping regions. Because our intervals \( I_\alpha \) are strictly disjoint, there is absolutely no overlap (\( I_\alpha \cap I_\beta = \emptyset \)). Therefore, the condition from Q3 is vacuously satisfied.
Consequently, by Q3, the combined function \( f \) is perfectly continuous across the entire open set \( U \). Furthermore, because the function is unbounded on every sub-interval \( I_\alpha \), the overall function \( f \) is unbounded on the set \( U \).

Hinglish Summary: Interval \( (a, b) \) ke andar hum \( f(x) = \frac{1}{x - a} \) banate hain jo \( a \) ke paas aate aate anant (\( \infty \)) tak bhagta hai. Q7 ne bataya tha ki koi bhi bada open set \( U \) chote-chote disjoint intervals se banta hai. Hum har interval mein yehi infinity ki taraf bhagne wala function laga dete hain. Q3 (Pasting Lemma) kehta hai ki agar functions overlap (takrate) nahi hain, toh unko jodkar banne wala naya function continuous hi rehta hai. Kyunki intervals bilkul door-door (disjoint) hain, function continuous bhi rehta hai aur unbounded bhi.