Compact Sets

Open Cover

The open cover of a set $E$ in a Metric Space $X$ is the collection $\{G_{\alpha }\}$ of open subsets of $X$ such that $E\subset {\bigcup }_{\alpha }{G}_{\alpha }$

• think of it as “covering” a set with open subsets

Example:

Let $A=(1/2,1]$, then

$$\left\{\left(-45,\frac{3}{4}\right),\left(\frac{1}{2},2\right)\right\}$$

is a collection of open sets and each element is open, and $A=(1/2,1]\subset (-45,3/4)\cup (1/2,2)$.

Compact

A subset $K$ of a Metric Space $X$ is said to be compact if every open cover of $K$ contains a finite subcover.

More explicitly, if $\{G_{\alpha }\}$ is an open cover of $K$, then there are finitely many indices ${\alpha }_1,\cdots {\alpha }_n$ such that

$$K\subset G_{{\alpha }_1}\cup \cdots \cup G_{{\alpha }_n}$$

Corollary

Every finite set is compact since they all have open covers that have finitely many indices.

Lemma (Compact Implies Bounded)

Suppose $K\subset X$ is compact, nonempty. Then $K$ is bounded.

Let $x\in X$. Now, for every $p\in K,r_p:=d(x,p)$ let $r_p:=d(x,p)+1>0$. Note that $p\in N_{r_p}(x)=\{y\in X\mid d(y,x)<r_p\}$ by construction. Now, $\{N_{r_p}(x)\mid p\in K\}$ is a cover of $K$.

Since $K$ is compact, there exists a finite subcover such that there is a finite $p_1,\cdots p_n\in K$ such that

$$K\subset \bigcup _{i=1}^n{N}_{r_{p_i}}(x)$$

Let $R=\max \{r_{p_i}\mid i=1,\cdots ,n\}>0$. Then $\forall i=1,\cdots ,n$.

$$N_{r_{p_i}}(x)\subset N_R(x)$$

and so $K\subset N_R(x)$ and is bounded. Remember that $x$ is fixed.

Lemma (Compact Implies Closed)

Suppose $K\subset X$ is compact, nonempty. Then $K$ is closed.

Let $p\in K^c$. Then if $p$ is interior, then $K^c$ is open and $K$ is closed. We first find neighborhoods which miss points in $K$.

Let $q\in K$, then $d(p,q)>0$. Let $r_q=\frac{1}{2}d(p,q)$. Then $N_{r_q}(p)\cap N_{r_q}(q)=\varnothing$. Indeed, if $x\in N_{r_q}(p)\cap N_{r_q}(q)$, then

$$2r_q=d(p,q)\le d(p,x)+d(x,q)<2r_q$$

a contradiction.

Now $\{N_{r_q}(q)\mid q\in K\}$ is an open cover of $K$. Since $K$ is compact, $\exists$ a finite subcover $\exists q_1\cdots q_n\in K$ such that

$$K\subset \bigcup _{i=1}^n{N}_{r_{q_i}}(q_i)$$

Let $r=\min \{r_{q_i}\mid i=1,\cdots ,n\}>0$. Then $N_r(p)\subset N_{r_{q_i}}(p),\forall i$

• and $N_{r_{q_i}}(p)\cap N_{r_{q_i}}(q_i)=\varnothing$
• and $N_r(p)\cap N_{r_{q_i}}(q_i)=\varnothing ,\forall i$ Now since $K\subset {\bigcup }_{i=1}^n{N}_{r_{q_i}}(q_i)$ an $N_r(p)\cap K=\varnothing$, $p$ has a neighborhood that misses $K$, and is an interior point. Therefore $K^c$ is open and by lemma, $K$ is closed.

Example

Let

$$A=\left\{\frac{1}{n}|n\in \mathbb{N}\right\}$$

This is not compact. We need to find one open cover with no finite subcover. We will find $I_1,I_2,\cdots$ so that

1. $1/n\in I_n$ and $I_n$ is open
2. $I_n\cap I_m=\varnothing ,\forall n\ne m$
3. $A\subset {\bigcup }_{m\in \mathbb{N}}{I}_m$

Define $I_n:=(1/n-r_n,1/n+r_n)=N_{r_n}(1/n)$.

<--(|)----(|)----(|)-- ... --(|)----(|)----(|)--->
  1/n+1   1/n    1/n-1       1/3    1/2     1 

Then we can let $r_n=\frac{1}{2}(1/n-1/(n-1))$ such that $I_n\cap I_m=\varnothing$ and $A\subset {\bigcup }_{n\in \mathbb{N}}{I}_n$.

---

Theorem (Relative Compactness)

Suppose $K\subset Y\subset X$. Then $K$ is compact relative to $Y$ if and only if $K$ is compact in $X$.

Proof:

$(⟹):$ Suppose $K$ is compact relative to $X$, and let $\{O_{\alpha }\}$ be a collection of sets, open relative to $X$, and an open cover of $K$.

We need to find $O_{{\alpha }_1},\cdots O_{{\alpha }_n}$ such that

$$K\subset \bigcup _{\alpha \in I}{O}_{\alpha }$$

Let $V_{\alpha }=O_{\alpha }\cap Y$ for $\alpha \in I$. Then $V_{\alpha }$ is open relative $Y$ by the lemma. Note that $K\subset Y$ so

$$\begin{array}{rl}K& \subset \left(\bigcup _{\alpha \in I}{O}_{\alpha }\right)\cap Y\\ & =\bigcup _{\alpha \in I}(O_{\alpha }\cap Y)\\ & =\bigcup _{\alpha \in I}{V}_{\alpha }\end{array}$$

Since $K$ is compact relative $Y$ an $\{V_{\alpha }\mid \alpha \in I\}$ is an open relative $Y$ cover of $K$. Then $\exists {\alpha }_1,\cdots ,{\alpha }_n$ such that

$$K\subset \bigcup _{i=1}^n{V}_{{\alpha }_i}⟹K\subset \bigcup _{i=1}^n{O}_{{\alpha }_i}$$

$(⟸):$ Suppose $K$ is compact in $X$.

Suppose $\{V_{\alpha }\mid \alpha \in I\}$ is an open relative $Y$ cover of $K$. We need to extract a finite subcover. Note that by lemma we have that

$$V_{\alpha }=O_{\alpha }\cap Y$$

for some $O_{\alpha }$. Then

$$K\subset \bigcup _{\alpha \in I}{V}_{\alpha }=\left(\bigcup _{\alpha \in I}{O}_{\alpha }\cap Y\right)\subset \bigcup _{\alpha \in I}{O}_{\alpha }$$

so $K$ has an open cover $\{O_{\alpha }\mid \alpha \in I\}$ since $K$ is compact then $\exists$ finite subcover, say ${\alpha }_1\cdots {\alpha }_n$ such that

$$K\subset \bigcup _{i=1}^n{O}_{\alpha }$$

Recall $K\subset Y$ so take the intersection of the above with $Y$. So,

$$\begin{array}{rl}K& =K\cap Y\\ & \subset \left(\bigcup _{i=1}^n{O}_{{\alpha }_i}\right)\cap Y\\ & =\bigcup _{i=1}^n(O_{{\alpha }_i}\cap Y)\\ & =\bigcup _{i=1}^n{V}_{{\alpha }_i}\end{array}$$

As we found a finite subcover, we are done.

Lemma (Closed x Compact)

• Let $K\subset X$ be compact and $F$ be closed. Then $F\cap X$ is compact.
• In particular, if $K$ is compact, then for $A\subset X$, $A\cap K\subset K$ is compact.

We prove the second statement. Let $A\subset X$ be closed, and by the lemma, $K$ is closed and $A\cap K$ is closed. hence $A\cap K\subset K$ and some $K$ compact. We have $A\cap K$ is compact.

Now the first claim. WTS $F$ is compact. Let $\{O_{\alpha }\mid \alpha \in I\}$ be an open cover of $F$. $F$ is closed, so $F^c$ is open. Consider

$$\{O_{\alpha }\mid \alpha \in I\}\cup \{F^c\}$$

Note $K\subset F^c\cup {\bigcup }_{\alpha \in I}$, which covers $K$ and some extra elements. Since $K$ is compact we can extract finite indices ${\alpha }_1,\cdots ,{\alpha }_n\in I$ such that

$$K\subset O_{{\alpha }_1}\cup \cdots \cup O_{{\alpha }_n}\cup F^c$$

Since $F\subset K\subset F^c\cup {\bigcup }_{i=1}^n{O}_{{\alpha }_i}$. Then as $F\cap F^c=\varnothing$, then $F={\bigcup }_{i=1}^n{O}_{{\alpha }_i}$ and $F$ is compact.

Corollary (Chain of Inclusion)

Let

$$K_1\supset K_2\supset K_3\supset \cdots$$

be a collection of nonempty compact sets. Then ${\bigcap }_{n\in \mathbb{N}}{K}_n$ is nonempty.

Suppose not. Then ${\bigcap }_{n\in \mathbb{N}}{K}_n=\varnothing$. Then lemma says $\exists n_1<n_2<\cdots <n_{\ell }$ such that

$$\bigcap _{i=1}^{\ell }{K}_{n_i}=\varnothing$$

Note that $K_{n_1}\supset K_{n_2}\supset \cdots \supset K_{n_{\ell }}$. But then

$$K_{n_{\ell }}=\bigcap _{i=1}^{\ell }{K}_{n_i}=\varnothing$$

which is a contradiction.

Lemma (Intersection of Compact)

Suppose $\{K_{\alpha }{\}}_{\alpha \in I}$ is a collection of compact subsets of $X$. Assume that

$$\bigcap _{\alpha \in I}{K}_{\alpha }=\varnothing$$

Then $\exists$ finite subcollection whose intersection is $\varnothing$.

Proof: W will use compactness of one set and closedness of the rest. Fix some $K_{{\alpha }_0}$ in the collection. Because

$$\begin{array}{rl}\bigcap _{\alpha }{K}_{\alpha }=\varnothing & ⟹K_{{\alpha }_0}\cap \left(\bigcap _{\alpha \in I,\alpha \ne {\alpha }_0}\right)=\varnothing \\ & ⟺K_{{\alpha }_0}\subset {\left(\bigcap _{\alpha \in I,\alpha \ne {\alpha }_0}{K}_{\alpha }\right)}^c=\bigcup _{\alpha \in I,\alpha \ne {\alpha }_0}{K}_{\alpha }^c\end{array}$$

Now this is an open cover of $K_{{\alpha }_0}$ since $K_{\alpha }$ closed for $\alpha \in I$. Hence, $\exists {\alpha }_1\cdots {\alpha }_n\in I,{\alpha }_i\ne {\alpha }_0$ such that

$$K_{{\alpha }_0}\subset \bigcup _{i=1}^n{K}_{{\alpha }_i}^c={\left(\bigcap _{i=1}^n{K}_{{\alpha }_i}\right)}^c$$

which implies

$$K_{{\alpha }_0}\cap \left(\bigcap _{i=1}^n{K}_{{\alpha }_i}\right)=\varnothing$$

Theorem (Criterion for Compactness)

Let $(X,d)$ be a metric space. Then $K\subset X$ is compact $⟺$ every infinite subset $A$ of $K$ has a limit point in $K$.

Proof: We only show the forward direction via contradiction. Assume $\exists A\subset K$ where $A$ is infinite such that no limit points are in $K$. Then $\forall p\in K,\exists r_p>0$ such that

$$N_{r_p}(p)\cap K=\{\begin{array}{ll}\varnothing & p\in K-A\\ \{p\}& p\in A\end{array}$$

(Every neighborhood on $p$ cannot intersect with $A$ at all.)

Now, $\{N_{r_p}\mid p\in K\}$ is an open cover of $K$. Since $A$ is infinite, then we need infinitely many neighborhoods $N_{r_p}(p)$ to cover $A$. Namely, one for every $p\in A$. So, no finite subcover, and a contradiction.

Backwards: From Homework 4.1

Lemma ($\mathbb{R}$ is not Compact)

$\mathbb{R}$ is not compact. This is done by constructing a single open cover with no finite subcover.

$$\forall n\in \mathbb{Z},I_n=(n,n+z)\ni n+1$$

Note $\mathbb{R}\subset {\bigcup }_{n\in \mathbb{Z}}{I}_n$ but no finite sub cover exists. This is by consequence of Archimedean Property.

Theorem 2.34

Compact subsets of metric spaces are closed.

Let $K$ be a compact subset of metric space $X$. Proof follows from showing that $K^c$ is open relative to $X$.

Theorem 2.35

Closed subsets of compact sets are compact.