# Prove Heine-Borel Theorem

Prove Heine-Borel Theorem: "A subset $S$ of $\mathbb{R}$ is compact if and only if every open cover for $S$ has a finite subcover."

Suggestions: Let $S \subset \mathbb{R}$. If every open cover for $S$ has a finite subcover, then $S$ must be compact. Why? Now, assume that $S$ is compact and let $\mathcal O$ be any open cover for $S$. Let $\epsilon > 0$ so that for all x in $S$, there is come $E \in \mathcal O$ such that $D(x,\epsilon) \subset E$. Assume that S is totally bounded.

• I don't understand the question. Is the quoted sentence your definition of "compact"? It's not clear from what follows that you understand the statement of the Heine-Borel theorem, so it might help if you included the exact statement of the theorem in your question. – Trevor Wilson Oct 27 '13 at 17:44

I'm not sure if I understand your question. There seems to be a typo. As you know, it is not the case that every open cover of $\mathbb{R}$ has a finite subcover. If $S$ is unbounded, it is easy to find an open cover (that covers $\mathbb{R}$, not just $S$) with no finite subcollcection covering $S$. If $S$ is not closed, $S$ has a limit point $x$ that doesn't belong to $S$, and it is easy to construct an open cover of $\mathbb{R} \setminus \{x\}$ (which contains $S$) with no finite ssubcollection covering $S$.
The existence of $\epsilon$ that you are using in your argument is far from obvious (since $\epsilon$ does not depend on $x$), though such an $\epsilon$ does exist. Are you allowed to assume this $\epsilon$ exists, or do you need to prove it? You only need it if you are going to use the idea of sequential compactness, which is not necessary for what you seem to be trying to prove.
Proving the other direction, that any closed and bounded subset of $\mathbb{R}$ is compact, is more difficult, but not terribly difficult. You can probably Google it and find a complete, excellent proof faster than you will get one here. You will also find proofs of the easier direction, as I describe above. It would also not surprise me if this question and been asked and answered in this forum before.
• My approach is a mess! Basically I just have to prove the theorem. As a suggestion the exercise says to start by explaining why if every open cover of $\mathbb{R}$ has a finite sub cover, then $S$ must be compact. Then assume $S$ is compact and $\mathcal O$ is an open cover for $S$. Then let "$\epsilon>0$ so that for all $x \in S$, there is come $E \in \mathcal O$ such that $D(x,\epsilon) \subset E$." – user99638 Oct 27 '13 at 17:39
• Thx! Your answer gave me an idea to start writing. I really don't understand what I need $\epsilon$ for, but I will see other proofs to get an idea on how to use it. – user99638 Oct 27 '13 at 18:24
• @OrchidFibio : you only need the $\epsilon$ mentioned in your question if you are going to use the idea of sequential compactness. There is a decent proof that any closed bounded interval is compact (should be easily Googlable, or you can find it on this forum if you ask for it clearly). Then you use the face (easily proven) that a closed subset of a compact set is compact. – Stefan Smith Oct 27 '13 at 19:18