Real Quiz 1 Solution

Answer: TRUE.

Step 1 — Understand the statement. We are given a set

\[ U\subseteq\mathbb{R} \]

which is both nonempty and open.

Nonempty means that \(U\) contains at least one point. Therefore choose

\[ x\in U. \]

Step 2 — Use the definition of open set.

Because \(U\) is open, there exists some \(\varepsilon>0\) such that

\[ (x-\varepsilon,x+\varepsilon)\subseteq U. \]

Thus \(U\) contains an entire open interval.

Step 3 — Use density of the rational numbers.

The rational numbers \(\mathbb Q\) are dense in \(\mathbb R\). This means that every nonempty open interval contains a rational number.

Therefore there exists some

\[ q\in\mathbb Q \]

such that

\[ x-\varepsilon<q<x+\varepsilon. \]

Hence

\[ q\in(x-\varepsilon,x+\varepsilon) \subseteq U. \]

Therefore \(U\) contains a rational number.

Hence the statement is TRUE.

Answer: FALSE.

The important topology rules are:

\[ \boxed{ \begin{aligned} &\text{Arbitrary union of open sets is open},\\ &\text{Finite intersection of open sets is open},\\ &\text{Arbitrary intersection of closed sets is closed},\\ &\text{Finite union of closed sets is closed}. \end{aligned}} \]

Notice carefully: an arbitrary union of closed sets need not be closed.

To disprove the statement, it is enough to give one counterexample.

Define

\[ F_n= \left[ \frac1n,1 \right]. \]

Every \(F_n\) is a closed interval, hence every \(F_n\) is closed in \(\mathbb R\).

Now take their union:

\[ \bigcup_{n=1}^{\infty}F_n. \]

The left endpoints are

\[ 1,\frac12,\frac13,\frac14,\ldots \]

and these numbers approach \(0\).

Therefore

\[ \bigcup_{n=1}^{\infty} \left[\frac1n,1\right] =(0,1]. \]

The point \(0\) is a limit point of \((0,1]\), but

\[ 0\notin(0,1]. \]

Therefore \((0,1]\) is not closed.

Hence an arbitrary union of closed sets need not be closed.

Therefore the statement is FALSE.

Answer: FALSE.

The key concept here is subspace topology.

If \(A\subseteq\mathbb Q\), saying that \(A\) is open in \(\mathbb Q\) means that there exists an open set \(V\subseteq\mathbb R\) such that

\[ A=V\cap\mathbb Q. \]

This does not imply that \(A\) itself is open in \(\mathbb R\).

Take the simplest possible example:

\[ A=\mathbb Q. \]

In the subspace \(\mathbb Q\), the whole space \(\mathbb Q\) is open.

Indeed,

\[ \mathbb Q = \mathbb R\cap\mathbb Q, \]

and \(\mathbb R\) is open in itself.

However, \(\mathbb Q\) is not open in \(\mathbb R\).

To see this, choose any rational number \(q\in\mathbb Q\). Every interval

\[ (q-\varepsilon,q+\varepsilon) \]

contains irrational numbers as well.

Therefore no open interval around \(q\) can lie completely inside \(\mathbb Q\).

Hence \(\mathbb Q\) is not open in \(\mathbb R\).

Therefore the statement is FALSE.

Answer: FALSE.

We must distinguish between bounded and Cauchy.

A bounded sequence is only required to remain inside some fixed interval. It may still oscillate forever.

Choose

\[ x_n=1 \]

for every \(n\).

The sequence

\[ 1,1,1,1,\ldots \]

is clearly Cauchy.

Now let

\[ y_n=(-1)^n. \]

Then

\[ y_n=-1,1,-1,1,\ldots \]

depending on the starting convention.

This sequence is bounded because

\[ |y_n|\le1. \]

However, it is not Cauchy.

Now

\[ x_ny_n = 1\cdot(-1)^n = (-1)^n. \]

Consider consecutive terms:

\[ |x_{n+1}y_{n+1}-x_ny_n| = 2. \]

Their distance does not approach \(0\).

Hence the product sequence is not Cauchy.

Therefore the statement is FALSE.

A correct related theorem is: if both \((x_n)\) and \((y_n)\) are convergent, then \((x_ny_n)\) is convergent and hence Cauchy.

Answer: TRUE.

Two important theorems are being used.

Theorem 1: The continuous image of a connected set is connected.

\[ A\text{ connected} \quad\text{and}\quad f\text{ continuous} \Longrightarrow f(A)\text{ connected}. \]

Theorem 2: The connected subsets of \(\mathbb R\) are exactly the intervals.

Therefore, because \(A\) is connected and \(f\) is continuous,

\[ f(A) \]

must be connected.

Since \(f(A)\subseteq\mathbb R\), every connected subset of \(\mathbb R\) is an interval.

Hence \(f(A)\) is an interval.

Therefore the statement is TRUE.

First split the set into two parts:

\[ A=A_1\cup A_2, \]

where

\[ A_1= \left\{ \frac{n}{n+1}:n\in\mathbb N \right\} \]

and

\[ A_2=(2,3). \]

This makes the problem much easier.

Step 1 — Understand the sequence part

Rewrite

\[ \frac{n}{n+1} = 1-\frac{1}{n+1}. \]

Hence

\[ \frac{n}{n+1}<1 \]

for every \(n\), and

\[ \frac{n}{n+1}\to1. \]

The sequence begins

\[ \frac12,\frac23,\frac34,\frac45,\ldots \]

and increases toward \(1\).

Step 2 — Find the infimum

The smallest sequence term occurs at \(n=1\):

\[ \frac{1}{1+1} = \frac12. \]

Every point of \((2,3)\) is much larger than \(1/2\).

Therefore

\[ \boxed{\inf A=\frac12}. \]

Since

\[ \frac12\in A, \]

\(1/2\) is also the minimum of \(A\).

Step 3 — Find the supremum

The largest values of \(A\) come from the interval

\[ (2,3). \]

Every element of this interval is less than \(3\), so \(3\) is an upper bound.

However, for every \(\varepsilon>0\), there are elements of \(A\) between

\[ 3-\varepsilon \quad\text{and}\quad 3. \]

Thus no number smaller than \(3\) can be an upper bound.

Hence

\[ \boxed{\sup A=3}. \]

But

\[ 3\notin A, \]

so \(A\) has no maximum.

Step 4 — Find the closure

Closure means:

\[ \boxed{\text{set + all missing limit points}} \]

The sequence

\[ \frac{n}{n+1} \]

approaches \(1\), but \(1\) is not one of the sequence terms. Therefore \(1\) must be added.

Also,

\[ \overline{(2,3)}=[2,3]. \]

Therefore

\[ \boxed{ \overline A = \left\{ \frac{n}{n+1}:n\in\mathbb N \right\} \cup\{1\} \cup[2,3] }. \]

Step 5 — Is \(A\) compact?

In \(\mathbb R\), by the Heine–Borel theorem,

\[ A\text{ compact} \iff A\text{ closed and bounded}. \]

\(A\) is bounded because

\[ \frac12\le x<3 \]

for every \(x\in A\).

But \(A\) is not closed because it does not contain the limit points

\[ 1,\;2,\;3. \]

Therefore

\[ \boxed{A\text{ is not compact}.} \]

(i) Nested closed sets

The statement is false.

Take

\[ F_n=[n,\infty). \]

For example,

\[ F_1=[1,\infty), \] \[ F_2=[2,\infty), \] \[ F_3=[3,\infty), \]

and so on.

These satisfy

\[ F_1\supseteq F_2\supseteq F_3\supseteq\cdots. \]

Each \(F_n\) is nonempty and closed.

We claim that

\[ \bigcap_{n=1}^{\infty}F_n=\varnothing. \]

Suppose \(x\in\mathbb R\). Choose an integer \(n>x\).

Since

\[ x<n, \]

we have

\[ x\notin[n,\infty)=F_n. \]

Thus no real number can belong to every \(F_n\).

Therefore the intersection is empty.

How to repair the statement: Require the sets to be compact.

A standard nested compact set theorem says that a decreasing sequence of nonempty compact sets has nonempty intersection.

In \(\mathbb R\), we may equivalently require the sets to be closed and bounded.

(ii) Continuous image of an open set

This statement is also false.

Let

\[ f(x)=x^2 \]

and let

\[ U=(-1,1). \]

\(U\) is open.

Its image under \(f\) is

\[ f(U)=[0,1). \]

Why is \(0\) included? Because

\[ f(0)=0. \]

Why is \(1\) not included? Because \(x=\pm1\) are not in \(U\).

The set

\[ [0,1) \]

is not open in \(\mathbb R\).

Hence continuous functions do not necessarily send open sets to open sets.

Correct theorem:

\[ \boxed{ f\text{ continuous} \iff f^{-1}(V)\text{ is open whenever }V\text{ is open} } \]

Continuity controls preimages of open sets, not images of open sets.

Part (i)

We want to prove

\[ f(\overline A) \subseteq \overline{f(A)}. \]

To prove one set is contained in another, take an arbitrary element of the first set and prove that it belongs to the second set.

Let

\[ y\in f(\overline A). \]

By the definition of image, there exists some

\[ x\in\overline A \]

such that

\[ y=f(x). \]

Now use the sequential characterization of closure:

\[ x\in\overline A \]

means that there exists a sequence

\[ (x_n)\subseteq A \]

satisfying

\[ x_n\to x. \]

Since \(f\) is continuous,

\[ x_n\to x \Longrightarrow f(x_n)\to f(x). \]

But every \(x_n\in A\). Therefore

\[ f(x_n)\in f(A). \]

Thus we have a sequence of points from \(f(A)\) converging to

\[ f(x)=y. \]

Hence, by the definition of closure,

\[ y\in\overline{f(A)}. \]

Since \(y\) was arbitrary,

\[ \boxed{ f(\overline A) \subseteq \overline{f(A)} }. \]

Part (ii)

We need an example where the inclusion in part (i) is strict.

Choose

\[ f(x)=\frac{1}{1+x^2}. \]

This function is continuous on all of \(\mathbb R\).

Since \(\mathbb R\) is closed,

\[ \overline{\mathbb R} = \mathbb R. \]

Therefore

\[ f(\overline{\mathbb R}) = f(\mathbb R). \]

Now

\[ 0< \frac{1}{1+x^2} \le1. \]

The maximum value \(1\) occurs at

\[ x=0. \]

Therefore \(1\) belongs to the image.

However,

\[ \frac{1}{1+x^2}>0 \]

for every real \(x\), so the function never equals \(0\).

Also,

\[ \lim_{|x|\to\infty} \frac{1}{1+x^2} = 0. \]

Therefore values of the function can approach \(0\) arbitrarily closely.

Hence

\[ f(\mathbb R)=(0,1]. \]

Consequently,

\[ \overline{f(\mathbb R)} = [0,1]. \]

Therefore

\[ f(\overline{\mathbb R}) = (0,1] \]

while

\[ \overline{f(\mathbb R)} = [0,1]. \]

Hence

\[ \boxed{ f(\overline{\mathbb R}) \ne \overline{f(\mathbb R)} }. \]

Part (i)

The key theorem is the Extreme Value Theorem.

It states:

\[ \boxed{ \text{A continuous real-valued function on a nonempty compact set} \text{ attains its maximum and minimum.} } \]

Since \(K\) is nonempty and compact and \(f\) is continuous, there exists some point

\[ p\in K \]

at which \(f\) attains its minimum.

Therefore

\[ f(p)\le f(x) \]

for every \(x\in K\).

Define

\[ c=f(p). \]

The hypothesis tells us that

\[ f(x)>0 \]

for every \(x\in K\).

In particular,

\[ f(p)>0. \]

Hence

\[ c>0. \]

Since \(f(p)\) is the minimum,

\[ f(x)\ge f(p)=c \]

for every \(x\in K\).

Therefore

\[ \boxed{ \exists c>0 \text{ such that } f(x)\ge c \text{ for every }x\in K }. \]

Part (ii)

Since \(A\) is not closed, it fails to contain at least one of its limit points.

Therefore there exists

\[ a\in\overline A\setminus A. \]

Define

\[ f(x)=|x-a| \]

for \(x\in A\).

This function measures the distance from \(x\) to the missing limit point \(a\).

The absolute-value function is continuous, and therefore \(f\) is continuous on \(A\).

Since

\[ a\notin A, \]

every \(x\in A\) satisfies

\[ x\ne a. \]

Therefore

\[ |x-a|>0. \]

Hence

\[ f(x)>0 \]

for every \(x\in A\).

But since

\[ a\in\overline A, \]

there exists a sequence

\[ x_n\in A \]

such that

\[ x_n\to a. \]

Therefore

\[ f(x_n) = |x_n-a| \to0. \]

Thus values of \(f\) can become arbitrarily close to \(0\).

Since

\[ f(x)>0 \]

for every \(x\in A\), \(0\) is a lower bound.

And since values can approach \(0\) arbitrarily closely, no positive number can be a larger lower bound.

Hence

\[ \boxed{ \inf_{x\in A}f(x)=0 }. \]