# 4.4: Compact Sets

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##### Definition

Suppose $$T \subset \mathbb{R} .$$ If $$A$$ is a set, $$U_{\alpha}$$ is an open set for every $$\alpha \in A,$$ and

$T \subset \bigcup_{\alpha \in A} U_{\alpha},$

then we call $$\left\{U_{\alpha}: \alpha \in A\right\}$$ an open cover of $$T$$.

##### Example $$\PageIndex{1}$$

For $$n=3,4,5, \dots,$$ let

$U_{n}=\left(\frac{1}{n}, \frac{n-1}{n}\right).$

Then $$\left\{U_{n}: n=3,4,5, \ldots\right\}$$ is an open cover of the open interval $$(0,1)$$.

##### Definition

Suppose $$\left\{U_{\alpha}: \alpha \in A\right\}$$ is an open cover of $$T \subset \mathbb{R} .$$ If $$B \subset A$$ and

$T \subset \bigcup_{\beta \in B} U_{\beta},$

then we call $$\left\{U_{\beta}: \beta \in B\right\}$$ a subcover of $$\left\{U_{\alpha}: \alpha \in A\right\} .$$ If $$B$$ is finite, we call $$\left\{U_{\beta}: \beta \in B\right\}$$ a finite subcover of $$\left\{U_{\alpha}: \alpha \in A\right\}$$.

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

Show that the open cover of $$(0,1)$$ given in the previous example does not have a finite subcover.

##### Definition

We say a set $$K \subset \mathbb{R}$$ is compact if every open cover of $$K$$ has a finite sub cover.

##### Example $$\PageIndex{2}$$

As a consequence of the previous exercise, the open interval $$(0,1)$$ is not compact.

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

Show that every finite subset of $$\mathbb{R}$$ is compact.

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

Suppose $$n \in \mathbb{Z}^{+}$$ and $$K_{1}, K_{2}, \ldots, K_{n}$$ are compact sets. Show that $$\bigcup_{i=1}^{n} K_{i}$$ is compact.

##### proposition $$\PageIndex{1}$$

If $$I$$ is a closed, bounded interval, then $$I$$ is compact.

Proof

Let $$a \leq b$$ be finite real numbers and $$I=[a, b] .$$ Suppose $$\left\{U_{\alpha}: \alpha \in A\right\}$$ is an open cover of $$I .$$ Let $$\mathcal{O}$$ be the set of sets $$\left\{U_{\beta}: \beta \in B\right\}$$ with the properties that $$B$$ is a finite subset of $$A$$ and $$a \in \bigcup_{\beta \in B} U_{\beta} .$$ Let

$(s-\epsilon, s+\epsilon) \subset U_{\alpha}.$

Moreover, there exists a $$\left\{U_{\beta}: \beta \in B\right\} \in \mathcal{O}$$ for which

$\left[a, s-\frac{\epsilon}{2}\right] \subset \bigcup_{\beta \in B} U_{\beta}.$

But then

$\left\{U_{\beta}: \beta \in B\right\} \cup\left\{U_{\alpha}\right\} \in \mathcal{O}$

and

$\left[a, s+\frac{\epsilon}{2}\right] \subset\left(\bigcup_{\beta \in B} U_{\beta}\right) \cup U_{\alpha},$

contradicting the definition of $$s .$$ Hence we must have $$s=b .$$ Now choose $$U_{\alpha}$$ such that $$b \in U_{\alpha} .$$ Then, for some $$\epsilon>0$$,

$(b-\epsilon, b+\epsilon) \subset U_{\alpha}.$

Moreover, there exists $$\left\{U_{\beta}: \beta \in B\right\} \in \mathcal{O}$$ such that

$\left[a, b-\frac{\epsilon}{2}\right] \subset \bigcup_{\beta \in B} U_{\beta}.$

Then

$\left\{U_{\beta}: \beta \in B\right\} \cup\left\{U_{\alpha}\right\} \in \mathcal{O}$

is a finite subcover of $$I .$$ Thus $$I$$ is compact. $$\quad$$ Q.E.D.

##### proposition $$\PageIndex{2}$$

If $$K$$ is a closed, bounded subset of $$\mathbb{R},$$ then $$K$$ is compact.

Proof

Since $$K$$ is bounded, there exist finite real numbers $$a$$ and $$b$$ such that $$K \subset[a, b] .$$ Let $$\left\{U_{\alpha}: \alpha \in A\right\}$$ be an open cover of $$K .$$ Let $$V=\mathbb{R} \backslash K .$$ Then

$\left\{U_{\alpha}: \alpha \in A\right\} \cup\{V\}$

in the latter case, we have

$K \subset[a, b] \backslash V \subset \bigcup_{\beta \in B} U_{\beta}.$

In either case, we have found a finite subcover of $$\left\{U_{\alpha}: \alpha \in A\right\}$$. $$\quad$$ Q.E.D.

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

Show that if $$K$$ is compact and $$C \subset K$$ is closed, then $$C$$ is compact.

##### proposition $$\PageIndex{3}$$

If $$K \subset \mathbb{R}$$ is compact, then $$K$$ is closed.

Proof

Suppose $$x$$ is a limit point of $$K$$ and $$x \notin K .$$ For $$n=1,2,3, \ldots,$$ let

$U_{n}=\left(-\infty, x-\frac{1}{n}\right) \cup\left(x+\frac{1}{n},+\infty\right).$

Then

$\bigcup_{n=1}^{\infty} U_{n}=(-\infty, x) \cup(x,+\infty) \supset K .$

However, for any $$N \in \mathbb{Z}^{+},$$ there exists $$a \in K$$ with

$a \in\left(x-\frac{1}{N}, x+\frac{1}{N}\right),$

and hence

$a \notin \bigcup_{n=1}^{N} U_{n}=\left(-\infty, x-\frac{1}{N}\right) \cup\left(x+\frac{1}{N},+\infty\right).$

Thus the open cover $$\left\{U_{n}: n \in \mathbb{Z}^{+}\right\}$$ does not have a finite subcover, contradicting the assumption that $$K$$ is compact. $$\quad$$ Q.E.D.

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

Suppose that for each $$\alpha$$ in some set $$A, K_{\alpha}$$ is compact. Show that $$\bigcap_{\alpha \in A} K_{\alpha}$$ is compact.

##### proposition $$\PageIndex{4}$$

If $$K \subset \mathbb{R}$$ is compact, then $$K$$ is bounded.

Proof

Suppose $$K$$ is not bounded. For $$n=1,2,3, \ldots,$$ let $$U_{n}=(-n, n) .$$ Then

$\bigcup_{n=1}^{\infty} U_{n}=(-\infty, \infty) \supset K .$

But, for any integer $$N,$$ there exists $$a \in K$$ such that $$|a|>N,$$ from which it follows that

$a \notin \bigcup_{n=1}^{N} U_{n}=(-N, N).$

Thus the open cover $$\left\{U_{n}: n \in \mathbb{Z}^{+}\right\}$$ does not have a finite subcover, contradicting the assumption that $$K$$ is compact. $$\quad$$ Q.E.D.

Taken together, the previous three propositions yield the following fundamental result:

##### Theorem $$\PageIndex{5}$$

A set $$K \subset \mathbb{R}$$ is compact if and only if $$K$$ is closed and bounded.

##### proposition $$\PageIndex{6}$$

If $$K \subset \mathbb{R}$$ is compact and $$\left\{x_{n}\right\}_{n \in I}$$ is a sequence with $$x_{n} \in K$$ for every $$n \in I,$$ then $$\left\{x_{n}\right\}_{n \in I}$$ has a convergent subsequence $$\left\{x_{n_{k}}\right\}_{k=1}^{\infty}$$ with

$\lim _{k \rightarrow \infty} x_{n_{k}} \in K .$

Proof

Since $$K$$ is bounded, $$\left\{x_{n}\right\}_{n \in I}$$ has a convergent subsequence $$\left\{x_{n_{k}}\right\}_{k=1}^{\infty}$$. Since $$K$$ is closed, we must have $$\lim _{k \rightarrow \infty} x_{n_{k}} \in K$$.

##### proposition $$\PageIndex{7}$$

Suppose $$K \subset \mathbb{R}$$ is such that whenever $$\left\{x_{n}\right\}_{n \in I}$$ is a sequence with $$x_{n} \in K$$ for every $$n \in I,$$ then $$\left\{x_{n}\right\}_{n \in I}$$ has a subsequence $$\left\{x_{n_{k}}\right\}_{k=1}^{\infty}$$ with $$\lim _{k \rightarrow \infty} x_{n_{k}} \in K .$$ Then $$K$$ is compact.

Proof

Suppose $$K$$ is unbounded. Then we may construct a sequence $$\left\{x_{n}\right\}_{n=1}^{\infty}$$ such that $$x_{n} \in K$$ and $$\left|x_{n}\right|>n$$ for $$n=1,2,3, \ldots$$ Hence the only possible subsequential limits of $$\left\{x_{n}\right\}_{n=1}^{\infty}$$ would be $$-\infty$$ and $$+\infty,$$ contradicting our assumptions. Thus $$K$$ must be bounded.

Now suppose $$\left\{x_{n}\right\}_{n \in I}$$ is a convergent sequence with $$x_{n} \in K$$ for all $$n \in I .$$ If $$L=\lim _{n \rightarrow \infty} x_{n},$$ then $$L$$ is the only subsequential limit of $$\left\{x_{n}\right\}_{n \in I} .$$ Hence, by the assumptions of the proposition, $$L \in K .$$ Hence $$K$$ is closed.

Since $$K$$ is both closed and bounded, it is compact. $$\quad$$ Q.E.D.

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

Show that a set $$K \subset \mathbb{R}$$ is compact if and only if every infinite subset of $$K$$ has a limit point in $$K$$.

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

Show that if $$K$$ is compact, then $$\sup K \in K$$ and $$\inf K \in K$$.

##### Theorem $$\PageIndex{8}$$

Given a set $$K \subset \mathbb{R},$$ the following are equivalent:

1. Every open cover of $$K$$ has a finite subcover.

2. Every sequence in $$K$$ has a subsequential limit in $$K$$.

3. Every infinite subset of $$K$$ has a limit point in $$K$$.

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

Suppose $$K_{1}, K_{2}, K_{3}, \ldots$$ are nonempty compact sets with

$K_{n+1} \subset K_{n}$

for $$n=1,2,3, \ldots$$ Show that

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

is nonempty.

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

We say a collection of sets $$\left\{D_{\alpha}: \alpha \in A\right\}$$ has the finite intersection property if for every finite set $$B \subset A$$,

$\bigcap_{\alpha \in B} D_{\alpha} \neq \emptyset .$

Show that a set $$K \subset \mathbb{R}$$ is compact if and only for any collection

$\left\{E_{\alpha}: \alpha \in A, E_{\alpha}=C_{\alpha} \cap K \text { where } C_{\alpha} \subset \mathbb{R} \text { is closed }\right\}$

which has the finite intersection property we have

$\bigcap_{\alpha \in A} E_{\alpha} \neq \emptyset .$

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