# Prove “K is compact i.e. K is closed and bounded ” implies “Any open cover for K has a finite subcover”

Definition: Let $$A\subseteq\mathbb{R}$$. An open cover for A is a $$(\text{possibly infinite})$$ colection of open sets $$\left\lbrace O_{\lambda}:\lambda\in Λ \right\rbrace$$ whose union contains the set A; that is, $$A\subseteq \bigcup_{\lambda\in Λ}O_{\lambda}$$. Given an open cover for A, a finite subcover is a finite sub collection of open sets from the original open cover whose union still manages to completely contain A.

Exercise asks me to prove it by contradiction but I couldn't follow the specified steps. I still got to this : Assume $$K\subset \mathbb{R}$$ is closed and bounded and $$\exists$$ an open cover {$${O_j: j\in I}$$}such that it's impossible to extract a finite subcover for $$K$$ from my open cover . No finite subcover implies that for every candidate for a finite subcover there is some element of $$K$$ that isn't contained. Form a sequence with some of these elements, $$x_n$$ . Because my set is compact, this sequence has a subsequence that converges to a limit $$x$$ in my set. Because $$x$$ is the limit of my sequence, it's excluded from every finite subcover but because it's in $$K$$, there is an open set $$O_M$$ from my open cover for $$K$$ that contains my $$x$$. You get far enough into the sequence such that the intersection of $$O_M$$ and the finite subcover candidate associated to $$x_n$$ isn't empty, you unite both sets and then you have a finite subcover (?). And there is your contradiction. Could I have a correction please?

The book is Understanding analysis, Stephen Abbott PS: I'm sorry I can't format my text the right way for the moment, I'll do it as soon as I'm able to.

• Hello! An advice: you should try to format it properly (this link could be useful: meta.math.stackexchange.com/questions/5020/…). Moreover, it is strongly encouraged to actually write down (still using the link above) the text of the book you refer to, instead of using a – not so readable – picture. :) – Kolmin Sep 30 '16 at 18:35
• Hi ! I'm gonna check your link, I thought I couldn't write it with the symbols because I'm on my phone :D On the picture only the definition part matters, I'm gonna specify that. Thank you for checking my question :) – Mehdi Slimani Sep 30 '16 at 18:40
• This is the Heine–Borel theorem. The linked article has a proof. – arkeet Sep 30 '16 at 18:46
• Let's say that $\mathcal U$ is your open cover and let's say that it is uncountable. How many different ways are there to pick just a single element out of $\mathcal U$? Would this form a sequence? – Ennar Sep 30 '16 at 18:49
• You didn't mention it, but I suppose you want $K$ to be a subset of some $\Bb R^n$ – Hagen von Eitzen Sep 30 '16 at 19:10

One problem with this approach is that the union of $\{O_M\}$ with the finite subset $S$ that is associated with $x_n$ now covers the point $x_n$ but that does not imply that it covers all of $K$ as there may be much more of $K$ that is not covered by $S$ nor by $O_M.$ Another problem is the unwarranted claim that x is excluded from any finite sub-cover. This does not follow from the fact that x is the limit of "my sequence."
The def'n of compact is that $K$ is compact iff every open cover of K has a finite sub-cover. It is a theorem that a subset of the reals is compact iff it is closed and bounded.