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# 4.3: Closed Sets

• • Dan Sloughter
• Professor (Mathematics) at Furman University
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## Definition: limit points

We call a point $$x \in \mathbb{R}$$ a limit point of a set $$A \subset \mathbb{R}$$ if for every $$\epsilon>0$$ there exists $$a \in A, a \neq x,$$ such that $$a \in(x-\epsilon, x+\epsilon)$$.

## Definition: isolated point

Suppose $$A \subset \mathbb{R} .$$ We call a point $$a \in A$$ an isolated point of $$A$$ if there exists an $$\epsilon>0$$ such that $A \cap(a-\epsilon, a+\epsilon)=\{a\}.$

## Exercise $$\PageIndex{1}$$

Identify the limit points and isolated points of the following sets:

a. $$[-1,1]$$,

b. $$(-1,1)$$,

c. $$\left\{\frac{1}{n}: n \in \mathbb{Z}^{+}\right\}$$,

d. $$\mathbb{Z}$$,

e. $$\mathbb{Q}$$.

## Exercise $$\PageIndex{2}$$

Suppose $$x$$ is a limit point of the set $$A .$$ Show that for every $$\epsilon>0,$$ the set $$(x-\epsilon, x+\epsilon) \cap A$$ is infinite.

We let $$A^{\prime}$$ denote the set of limit points of a set $$A .$$

## Definition: closures

Given a set $$A \subset \mathbb{R},$$ we call the set $$\bar{A}=A \cup A^{\prime}$$ the closure of $$A$$.

## Definition: Closed sets

We call a set $$C \subset \mathbb{R}$$ closed if $$C=\bar{C}$$.

## Proposition $$\PageIndex{1}$$

If $$A \subset \mathbb{R},$$ then $$\bar{A}$$ is closed.

Proof

Suppose $$x$$ is a limit point of $$\bar{A} .$$ We we will show that $$x$$ is a limit point of $$A,$$ and hence $$x \in \bar{A} .$$ Now for any $$\epsilon>0,$$ there exists $$a \in \bar{A}, a \neq x,$$ such that

$a \in\left(x-\frac{\epsilon}{2}, x+\frac{\epsilon}{2}\right).$

If $$a \in A,$$ let $$b=a .$$ If $$a \notin A,$$ then $$a$$ is a limit point of $$A,$$ so there exists $$b \in A,$$ $$b \neq a$$ and $$b \neq x,$$ such that

$b \in\left(a-\frac{\epsilon}{2}, a+\frac{\epsilon}{2}\right).$

In either case

$|x-b| \leq|x-a|+|a-b|<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon .$

Hence $$x \in A^{\prime},$$ and so $$\bar{A}$$ is closed. $$\quad$$ Q.E.D.

## Proposition $$\PageIndex{2}$$

A set $$C \subset \mathbb{R}$$ is closed if and only if for every convergent sequence $$\left\{a_{k}\right\}_{k \in K}$$ with $$a_{k} \in C$$ for all $$k \in K$$,

$\lim _{k \rightarrow \infty} a_{k} \in C.$

Proof

Suppose $$C$$ is closed and $$\left\{a_{k}\right\}_{k \in K}$$ is a convergent sequence with $$a_{k} \in C$$ for all $$k \in K .$$ Let $$x=\lim _{k \rightarrow \infty} a_{k} .$$ If $$x=a_{k}$$ for some integer $$k,$$ then $$x \in C .$$ Otherwise, for every $$\epsilon>0,$$ there exists an integer $$N$$ such that $$\left|a_{N}-x\right|<\epsilon$$. Hence $$a_{N} \neq x$$ and

$a_{N} \in(x-\epsilon, x+\epsilon).$

Thus $$x$$ is a limit point of $$C,$$ and so $$x \in C$$ since $$C$$ is closed.

Now suppose that for every convergent sequence $$\left\{a_{k}\right\}_{k \in K}$$ with $$a_{k} \in C$$ for all $$k \in K, \lim _{k \rightarrow \infty} a_{k} \in C .$$ Let $$x$$ be a limit point of $$C .$$ For $$k=1,2,3, \ldots,$$ choose $$a_{k} \in C$$ such that $$a_{k} \in\left(x-\frac{1}{k}, x+\frac{1}{k}\right) .$$ Then clearly

$\boldsymbol{x}=\lim _{k \rightarrow \infty} a_{k},$

so $$x \in C .$$ Thus $$C$$ is closed. $$\quad$$ Q.E.D.

## Exercise $$\PageIndex{3}$$

Show that every closed interval $$I$$ is a closed set.

## Proposition $$\PageIndex{3}$$

Suppose $$A$$ is a set and, for each $$\alpha \in A, C_{\alpha}$$ is a closed set. Then

$\bigcap_{\alpha \in A} C_{\alpha}$

is a closed set.

Proof

Suppose $$x$$ is a limit point of $$\bigcap_{\alpha \in A} C_{\alpha} .$$ Then for any $$\epsilon>0,$$ there exists $$y \in \bigcap_{\alpha \in A} C_{\alpha}$$ such that $$y \neq x$$ and $$y \in(x-\epsilon, x+\epsilon) .$$ But then for any $$\alpha \in A,$$ $$y \in C_{\alpha},$$ so $$x$$ is a limit point of $$C_{\alpha}$$. Since $$C_{\alpha}$$ is closed, it follows that $$x \in C_{\alpha}$$ for every $$\alpha \in A .$$ Thus $$x \in \bigcap_{\alpha \in A} C_{\alpha}$$ and $$\bigcap_{\alpha \in A} C_{\alpha}$$ is closed. $$\quad$$ Q.E.D.

## Proposition $$\PageIndex{4}$$

Suppose $$C_{1}, C_{2}, \ldots, C_{n}$$ is a finite collection of closed sets. Then

$\bigcup_{i=1}^{n} C_{i}$

is closed.

Proof

Suppose $$\left\{a_{k}\right\}_{k \in K}$$ is a convergent sequence with $$a_{k} \in \bigcup_{i=1}^{n} C_{i}$$ for every $$k \in K .$$ Let $$L=\lim _{k \rightarrow \infty} a_{k} .$$ Since $$K$$ is an infinite set, there must an integer $$m$$ and a subsequence $$\left\{a_{n_{j}}\right\}_{j=1}^{\infty}$$ such that $$a_{n_{j}} \in C_{m}$$ for $$j=1,2, \ldots$$. Since every subsequence of $$\left\{a_{k}\right\}_{k \in K}$$ converges to $$L,\left\{a_{n_{j}}\right\}_{j=1}^{\infty}$$ must converge to $$L .$$ Since $$C_{m}$$ is closed,

$L=\lim _{j \rightarrow \infty} a_{n_{j}} \in C_{m} \subset \bigcup_{i=1}^{n} C_{i}.$

Thus $$\bigcup_{i=1}^{n} C_{i}$$ is closed. $$\quad$$ Q.E.D.

Note that both $$\mathbb{R}$$ and $$\emptyset$$ satisfy the definition of a closed set.

## Proposition $$\PageIndex{5}$$

A set $$C \subset \mathbb{R}$$ is closed if and only if $$\mathbb{R} \backslash C$$ is open.

Proof

Assume $$C$$ is closed and let $$U=\mathbb{R} \backslash C .$$ If $$C=\mathbb{R},$$ then $$U=\emptyset,$$ which is open; if $$C=\emptyset,$$ then $$U=\mathbb{R},$$ which is open. So we may assume both $$C$$ and $$U$$ are nonempty. Let $$x \in U .$$ Then $$x$$ is not a limit point of $$C,$$ so there exists an $$\epsilon>0$$ such that

$(x-\epsilon, x+\epsilon) \cap C=\emptyset.$

Thus

$(x-\epsilon, x+\epsilon) \subset U,$

so $$U$$ is open.

Now suppose $$U=\mathbb{R} \backslash C$$ is open. If $$U=\mathbb{R},$$ then $$C=\emptyset,$$ which is closed; if $$U=\emptyset,$$ then $$C=\mathbb{R},$$ which is closed. So we may assume both $$U$$ and $$C$$ are nonempty. Let $$x$$ be a limit point of $$C .$$ Then, for every $$\epsilon>0$$,

$(x-\epsilon, x+\epsilon) \cap C \neq \emptyset .$

Hence there does not exist $$\epsilon>0$$ such that

$(x-\epsilon, x+\epsilon) \subset U.$

Thus $$x \notin U,$$ so $$x \in C$$ and $$C$$ is closed. $$\quad$$ Q.E.D.

## Exercise $$\PageIndex{4}$$

For $$n=1,2,3, \ldots,$$ let $$I_{n}=\left(-\frac{1}{n}, \frac{n+1}{n}\right) .$$ Is

$\bigcap_{n=1}^{\infty} I_{n}$

open or closed?

## Exercise $$\PageIndex{5}$$

For $$n=3,4,5, \ldots,$$ let $$I_{n}=\left[\frac{1}{n}, \frac{n-1}{n}\right] .$$ Is

$\bigcup_{n=3}^{\infty} I_{n}$

open or closed?

## Exercise $$\PageIndex{6}$$

Suppose, for $$n=1,2,3, \ldots,$$ the intervals $$I_{n}=\left[a_{n}, b_{n}\right]$$ are such that $$I_{n+1} \subset I_{n} .$$ If $$a=\sup \left\{a_{n}: n \in \mathbb{Z}^{+}\right\}$$ and $$b=\inf \left\{b_{n}: n \in \mathbb{Z}^{+}\right\},$$ show that

$\bigcap_{n=1}^{\infty} I_{n}=[a, b].$

## Exercise $$\PageIndex{7}$$

Find a sequence $$I_{n}, n=1,2,3, \ldots,$$ of closed intervals such that $$I_{n+1} \subset I_{n}$$ for $$n=1,2,3, \ldots$$ and

$\bigcap_{n=1}^{\infty} I_{n}=\emptyset.$

## Exercise $$\PageIndex{8}$$

Find a sequence $$I_{n}, n=1,2,3, \ldots,$$ of bounded, open intervals such that $$I_{n+1} \subset I_{n}$$ for $$n=1,2,3, \ldots$$ and

$\bigcap_{n=1}^{\infty} I_{n}=\emptyset .$

## Exercise $$\PageIndex{9}$$

Suppose $$A_{i} \subset \mathbb{R}, i=1,2, \ldots, n,$$ and let $$B=\bigcup_{i=1}^{n} A_{i} .$$ Show that

$\overline{B}=\bigcup_{i=1}^{n} \overline{A_{i}}.$

## Exercise $$\PageIndex{10}$$

Suppose $$A_{i} \subset \mathbb{R}, i \in \mathbb{Z}^{+},$$ and let

$B=\bigcup_{i=1}^{\infty} A_{i}.$

Show that

$\bigcup_{i=1}^{\infty} \overline{A_{i}} \subset \overline{B} .$

Find an example for which

$\overline{B} \neq \bigcup_{i=1}^{\infty} \overline{A_{i}}.$

## Exercise $$\PageIndex{11}$$

Suppose $$U \subset \mathbb{R}$$ is a nonempty open set. For each $$x \in U,$$ let

$J_{x}=\bigcup(x-\epsilon, x+\delta),$

where the union is taken over all $$\epsilon>0$$ and $$\delta>0$$ such that $$(x-\epsilon, x+\delta) \subset U$$.

a. Show that for every $$x, y \in U,$$ either $$J_{x} \cap J_{y}=\emptyset$$ or $$J_{x}=J_{y}$$.

b. Show that

$U=\bigcup_{x \in B} J_{x},$

where $$B \subset U$$ is either finite or countable.