ANA1_L17

Open cover compactness

Open covers and subcovers

Definition 1(Definition).

Let $E\subset X$. An open cover of $E$ is a collection $\{V_{\alpha }{\}}_{\alpha \in I}$ of open sets in $X$ such that

$$\bigcup _{\alpha \in I}{V}_{\alpha }\supset E.$$

A subcover of $\{V_{\alpha }{\}}_{\alpha \in I}$ is a subset $\{V_{\beta }{\}}_{\beta \in J\subset I}$ with

$$\bigcup _{\beta \in J}{V}_{\beta }\supset E.$$

Absolute and relative definitions of open cover compactness

Definition 2(Absolute definition for a metric space$E$).

$E$ is open cover compact if every open cover of $E$ has a finite subcover.

Definition 3(Relative defintion for$E\subset X$).

$E$ is open cover compact in $X$ is every open cover of $E$ in $X$ has a finite subcover.

Emphasis on “every” in both definitions.

Equivalence

$E$ is open cover compact as per definition 1 as a metric space in its own right $⟺$ $E$ is an open cover compact subset of $X$ as per 2.

Proof of $⟸$
$E$ is an open cover compact subset of $X$. Let $W\equiv \{W_{\alpha }{\}}_{\alpha \in I}$ be an open cover of $E$ by sets open in $E$. We need to show that a finite subset of $W$ covers $E$. We know from our study of subspace topology that for each $W_{\alpha }$ there must exist $V_{\alpha }\subset X$ open in $X$ such that $W_{\alpha }=E\cap V_{\alpha }$. Since $W_{\alpha }\subset V_{\alpha }$ for all $\alpha$, $V\equiv \{V_{\alpha }{\}}_{\alpha \in I}$ covers $E$ in $X$. From the hypothesis, there exists a finite subcover $\{V_{\beta }{\}}_{\beta \in J\subset I}$ of $V$ covering $E$. Thus,

$$\begin{array}{rl}\bigcup _{\beta \in J}{V}_{\beta }& \supset E\\ \left(\bigcup _{\beta \in J}{V}_{\beta }\right)\cap E& \supset E\\ \bigcup _{\beta \in J}(V_{\beta }\cap E)& \supset E\\ \bigcup _{\beta \in J}{W}_{\beta }& \supset E.\end{array}$$

Proof of $⟹$
Start with an open cover $\{V_{\alpha }{\}}_{\alpha \in I}$ of $E$ with sets which are open in $X$. We want to show that a finite subset of this open cover covers $E$.

$$\begin{array}{rl}\bigcup _{\alpha \in I}{V}_{\alpha }& \supset E\\ \left(\bigcup _{\alpha \in I}{V}_{\alpha }\right)\cap E& \supset E\\ \bigcup _{\alpha \in I}(V_{\alpha }\cap E)& \supset E\\ \bigcup _{\alpha \in I}{W}_{\alpha }& \supset E\end{array}$$

Where $W_{\alpha }=V_{\alpha }\cap E$. Note that each $W_{\alpha }$ is open in $E$. Thus, $\{W_{\alpha }{\}}_{\alpha \in I}$ is an open cover of $E$, and from the hypothesis it must have a finite subcover $\{W_{\beta }{\}}_{\beta \in J\subset I}$. It follows that $\{V_{\beta }{\}}_{\beta \in J\subset I}$ also covers $V$, since $W_{\alpha }\subset V_{\alpha }$ for every $\alpha$. ❏

So, as we have seen with the previous forms of compactness, open cover compactness in intrinsic.

An analogy

Consider a policeman in each open set of an open cover of a set $X$ whose can see only inside their open set. If $X$ is compact, regardless of how myopic the policemen are (regardless of how small the open sets are/ how many open sets are present in the cover), a finite number of them will be able to watch the entire set $X$.

Equivalence with limit point compactness

Theorem 4(Claim).

$X$ is open cover compact $⟹$ $X$ is limit point compact.

Proof
We will prove the contrapositive. Consider an infinite subset $S\subset X$. Assume $S$ does not have a limit point. Thus, for every point $p\in X$ there exists $r_p>0$ such that $|B_{r_p}(p,S)|\le 1$. Consider the set $E=\{B_{r_p}(p,X)|p\in X\}$. Clearly, $E$ is an open cover of $X$. However, the union of a finite subset of $E$ will contain only finitely many elements of $S$. So, $X$ cannot be open cover compact. ❏

This will allow us to import all the results we proved for limit point compact sets to open cover compact sets.

The converse is also true, but harder to prove.

Examples

It follows form the previous claim that every set we showed to be not limit point compact in the previous lecture is not open cover compact either.

For $(0,1)$, consider the open cover $\left\{\left(\frac{1}{n},1\right)|n\in \mathbb{N}\right\}$.
For $\mathbb{R}$, consider the open cover $\{(-n,n)|n\in \mathbb{N}\}$.

Closed intervals in R are open cover compact

Note this is not implied by any of our previous work - it does need to be proved.

Proof
Let $E=\{V_{\alpha }{\}}_{\alpha \in I}$ be an open cover of $[a,b]$. Suppose there does not exist a finite subcover of $E$ covering $[a,b]$. Starting with $[a,b]$, successively bisect the interval, at each step choosing the half which is not covered by a finite subcover of $E$ (at least one of the two is guaranteed to have this property). This should give us nested intervals of the form $[a_0,b_0],[a_1,b_1],\dots$ where the length of each interval is $(a_0-b_0)/2^n$ and each $[a_i,b_i]$ is not covered by a finite subcover of $E$. We know from the nested interval property that ${\bigcap }_{i\ge 0}[a_i,b_i]=\{p\}$. Let $p\in V_{\beta }$ for some $\beta$. Since $V_{\beta }$ is open, there must exist an $ϵ>0$ such that $(p-ϵ,p+ϵ)\subset V_{\beta }$. Since the lengths of $[a_i,b_i]$ converge to $0$ and $p\in [a_i,b_i]$, we can find an $[a_k,b_k]$ such that $[a_k,b_k]\subset (p-ϵ,p+ϵ)$. But this implies the finite subcover $\{V_{\beta }\}$ of $E$ covers $[a_k,b_k]$. $\Rightarrow \Leftarrow$ ❏

Important

Do this dance in ${\mathbb{R}}^k$, and you get that all closed boxes in ${\mathbb{R}}^k$ are open cover compact.

Note that closed intervals in $\mathbb{Q}$ are not compact. For example, $\left\{\left(-1,\frac{1}{\sqrt{2}}-\frac{1}{n}\right)\cup \left(\frac{1}{\sqrt{2}}+\frac{1}{n},2\right):n\in \mathbb{N}\right\}$ is an open cover of $\mathbb{Q}\cap [0,1]$, but does not have a finite subcover.

Exercises in working with open cover compactness

Exercise 1

In the previous lecture, we proved that every (limit point/sequentially) compact set $C$ in a metric space $X$ is closed in $X$ and bounded. We do not need to prove this separately for open cover compact sets, because of the equivalence with limit point compact sets we proved above. However, it is a good exercise.

Theorem 5(Theorem).

Every open cover compact set $C$ is bounded.

Proof
Consider a point $p\in C$. Consider the set $E=\{B_n(p,C)|n\in \mathbb{N}\}$. Clearly, it is an open cover of $C$. Since $C$ is compact, $E$ must have a finite subcover. Thus, there exists $n\in \mathbb{N}$ such that $C=B_n(p,C)$. ❏

Exercise 2

Theorem 6(Theorem).

The arbitrary intersection of compact sets is compact.

Proof
Let $\{C_{\alpha }|\alpha \in I\}$ be a collection of compact sets in $X$. Let $C\equiv {\bigcap }_{\alpha \in I}{C}_{\alpha }$. Note that every $C_{\alpha }$ is closed in $X$ and bounded, and that an arbitrary intersection of closed sets is closed. Thus, $C$ is closed in $X$. Now, consider an open cover $\{V_{\alpha }|\alpha \in J\}$ of $C$. Pick any $C_{\alpha }$. Since $X\setminus C$ is open in $X$, and $C_{\alpha }\setminus C\subset X\setminus C$, every element $p\in C_{\alpha }\setminus C$ has a neighborhood $W_p\subset X\setminus C$. It follows that $\Omega =\{V_{\alpha }|\alpha \in J\}\cup \{W_p|p\in C_{\alpha }\setminus C\}$ is an open cover of $C_{\alpha }$. Since $C_{\alpha }$ is compact, $\Omega$ must have a finite subcover ${\Omega }^{\prime }\subset \Omega$. Since the $W_p$‘s do not contain any points from $C$, it must be that $\Lambda ={\Omega }^{\prime }\cap \{V_{\alpha }|\alpha \in J\}$ must cover $C$. Observe that $|\Lambda |$ is finite and $\Lambda \subset \{V_{\alpha }|\alpha \in J\}$, and we are done. ❏

Alternatively, notice that $X\setminus C$ is open in $X$, and use it instead of $\{W_p|p\in C_{\alpha }\setminus C\}$.

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Equivalence of the three versions of compactness

While we have proved a lot of the preceding results in general metric spaces, we will mostly be restricting ourselves to ${\mathbb{R}}^k$ in this course.

Let $C\subset {\mathbb{R}}^k$. Consider the following statements.

1. $C$ is open cover compact.
2. $C$ is limit point/sequentially compact.
3. $C$ is closed in ${\mathbb{R}}^k$ and bounded.

Here’s everything we’ve proved so far:

• 1 implies 2
• 2 implies 3
• 3 implies 2

If we show $2⟹1$ or $3⟹1$, we will have shown that the three versions of compactness are equivalent. We will do the latter, since it is easier.

Our proof of $3⟹1$ will look very similar to our proof of $3⟹2$. We have already shown that closed boxes in ${\mathbb{R}}^k$ are open cover compact. If we prove that that closed subsets of open cover compact sets are open cover compact (we proved an analogue of this in the previous lecture for limit point compactness), $3⟹1$ should follow.

Closed subsets of compact sets are compact

Theorem 7(Lemma).

If $X$ is open cover compact, $C$ is closed in $X$ $⟹$ $C$ is open cover compact.

Proof
Take an open cover $\{V_{\alpha }{\}}_{\alpha \in I}$ of $C$, $V_{\alpha }$ open in $X$. Since $C$ is closed in $X$, $X\setminus C$ is open in $X$. Thus, $E=\{V_{\alpha }{\}}_{\alpha \in I}\cup \{X\setminus C\}$ is an open cover of $X$. Since $X$ is open cover compact, a finite subcover $E^{\prime }$ of $E$ covering $X$ exists. In particular, $E^{\prime }$ covers $C$. If $E^{\prime }$ covers $C$, $E^{\prime }\setminus (X\setminus C)$ also covers $C$ (basically, if $X\setminus C$ is in $E^{\prime }$, we can throw it away while still having $E^{\prime }$ cover $C$). Thus, $E^{\prime }$ is a finite subcover of $\{V_{\alpha }{\}}_{\alpha \in I}$. ❏

The equivalence

Here’s the proof of $3⟹1$. Notice its similarity with the proof of $3⟹2$ ($⟸$ of Heine Borel).

Proof
Since $C$ is bounded, $C$ lies in some closed box. Because $C$ is closed in ${\mathbb{R}}^k$, $C$ is closed in that box ($C=(\text{Box})\cap C$). We know that closed boxes in ${\mathbb{R}}^k$ are open cover compact, so our box is too. It follows from the previous lemma that $C$ is compact. ❏

With this, we have shown that open cover, sequential, and limit point compactness are equivalent for ${\mathbb{R}}^k$.

Warning

The pathway we used to prove the equivalence of the three forms of compactness fails in general metric spaces. As noted in the previous lecture, 3 does not imply 1 in this case. Thus, we will be left with no choice but to prove 2 implies 1.

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Epilogue

Tasks:

• In the above section, try to prove $1⟹3$ (for general metric spaces) directly without using $2$ as an intermediary. To be precise, prove this statement: If $C$ is an open cover compact subset of $X$, then $C$ is closed in $X$ (We have already proven the bounded part).

Briefly discussed uniform continuity.

Another exercise in working with open cover compactness

Prove this theorem (Image of a compact set under a continuous function is compact), for open cover compact sets sets without using the equivalence.

Proof
Consider an open cover $\{V_{\alpha }{\}}_{\alpha \in I}$ of $f(X)$.

$$\begin{array}{rl}f(X)& \subset \bigcup _{\alpha \in I}{V}_{\alpha }\\ X& \subset f^{-1}\left(\bigcup _{\alpha \in I}{V}_{\alpha }\right)=\bigcup _{\alpha \in I}{f}^{-1}(V_{\alpha })\end{array}$$

Now, since $f$ is continuous, $f^{-1}(V_a)$ is open in $X$ for all $\alpha$. Thus, $\{f^{-1}(V_{\alpha }){\}}_{\alpha \in I}$ is an open cover of $X$. Since $X$ is compact, there exists a finite subcover.

$$\begin{array}{rl}X& \subset \bigcup _{\beta \in J\subset I}{f}^{-1}(V_{\beta })\\ f(X)& \subset \bigcup _{\beta \in J\subset I}{V}_{\beta }\end{array}$$

❏