# Why "any covering of a compact set by a collection of open sets contains a finite subcovering"?

I'm reading Stein's textbook, and this is his version of the problem:

A set $$E$$ is bounded if it is contained in some ball of finite radius. A bounded set is compact if it is also closed. Compact sets enjoy the Heine-Borel covering property:

• Assume $$E$$ is compact, $$E\subset\bigcup_\alpha\mathcal{O}_\alpha$$, and each $$\mathcal{O}_\alpha$$ is open. Then there are finitely many of the open sets, $$\mathcal{O}_{\alpha_1},\mathcal{O}_{\alpha_2},\ldots,\mathcal{O}_{\alpha_N}$$ such that $$E\subset\bigcup_{j=1}^N\mathcal{O}_{\alpha_j}$$.

In words, any covering of a compact set by a collection of open sets contains a finite subcovering.

I can't see how to prove it. Let's consider $$\mathbb{R}$$ and let $$E$$ be the closed interval $$[0,1]$$.

• Are you considering subsets $E$ of $\mathbb{R}^n$ or just in $\mathbb{R}$? What work have you done to try to solve the problem so far? Sep 24, 2018 at 16:39
• Two things are important to note: 1) your text is unusual in that the definition of compact is usually defined to mean every open cover has a finite subcover and you then prove that for $\mathbb R^n$ this is true if and only if it is closed and bounded. 2) there are spaces other then $\mathbb R^n$ where being closed and bounded does not mean every open cover has a finite subcover but (correct me if I'm wrong) if it is true every open cover has finite means closed and bounded. Sep 24, 2018 at 17:01
• Google Heine-Borel Theorem. Bear in mind nearly everything you find will use "compact" to mean that every open cover has a finite subcover. Sep 24, 2018 at 17:04
• which book by Stein? Surely he offers a proof of this! Sep 24, 2018 at 17:15
• The only way to answer his is to know the preceding material in that book. Stein would not simply include this problem without considerable work leading up to it. No student should be expected to prove this from scratch without that preceding work. Or maybe Stein is stating this here, and will do the work of proving it later in the book. Sep 25, 2018 at 13:23

With the help of a classmate, I came up with a proof. Suppose U is an infinite collection of open intervals that covers [0,1]. Consider the following statement for the open interval I:

S(I): There exist a finite collection of open intervals in U that covers I.

We want to prove S([0,1]) is true.

• Basically right. Details: Let $I_1=[0,1].$ Assume $\neg S(I_1).$ If $S_n=[a,b]$ and $\neg S(I_n),$ let $I_{n+1}=[a, (a+b)/2]$ if $\neg S([a,(a+b)/2]),$ otherwise let $I_n=[(a+b)/2,b].$ By induction on $n\in \Bbb N$ we have, for all $n\in \Bbb N,$ that $\neg S(I_n)$ and $|\min I_n-\min I_{n+1}|\leq 2^{n-1}.$ Now $(\min I_n)_{n\in \Bbb n}$ is a Cauchy sequence, so it converges to some $x\in [0,1].$ If $U$ is $any$ open family with $\cup U\supset [0,1]$ there exists $e>0$ and $u\in U$ such that $[0,1]\cap (-e+x,e+x)\subset u$..... (continued).... Sep 25, 2018 at 2:22
• (Typo in prior comment: That should say $|\min I_n-\min I_{n+1}|\leq 2^{-n}$ ).... (continued)... Now $\max I_n-\min I_n=2^{-(n-1)}$ (also, obviously, by induction on $n$). So if $n$ is large enough that $|x-\min I_n|<e/2$ and $2^{-(n-1)}<e/2,$ then $I_n\subset (-e+x,e+x)\subset u \in U,$ so $I_n$ is covered by the one member $u$ of $U,$ contrary to $\neg S(I_n)$... I have assumed throughout that $\neg S(A)$ means that $A$ cannot be covered by a finite subset of $U.$ (Note that the $U$ of my comments is not the same as the $U$ of your answer, but is more general) Sep 25, 2018 at 2:41
• Another method: Let $C$ be an open cover of $[0,1].$ For $x\in [0,1]$ let $x\in S$ iff $[0,x]$ can be covered by finitely many members of $C.$ Prove that (I) $S$ is not empty, (II) $\sup S \in S,$ and (III) if $1\ne x\in S$ then $x\ne \sup S.$ Sep 27, 2018 at 11:18