# If (X,d) is a metric space and $K \subseteq X$ is compact. Show that $K$ is complete.

I've seen other proofs of this on here but all of them seem to rely on showing compact iff sequentially compact in a metric space but instead I want to use only the definition of compactness, that every open cover of a compact space admits a finite subcover to prove this.

Proof by contradiction.

Suppose K is compact and not complete. Then there is a Cauchy sequence in K that is not convergent in K. Let $$(x_n)$$ be this sequence. I want to somehow show that this will give an open cover with no finite subcover but I can't see how I do that. Or I want to show that somehow the cluster point for the cauchy sequence must be in K. But I can't see how I can show that.

• What is "this" in "but this is supposed to use only the fact that every open cover of a compact space admits a finite subcover."? – Michael Greinecker Oct 24 '18 at 23:05
• I meant the question. Basically to prove this using just the definition of compactness instead of using that compactness implies sequentially compact and then using the definition of sequentially compact to give that the cauchy sequence has a convergent subsequence. – AColoredReptile Oct 24 '18 at 23:08
• Are you willing to use the fact that every metric space has a completion? – Michael Greinecker Oct 24 '18 at 23:09

## 1 Answer

It is not hard to prove that if a Cauchy sequence has a convergent subsequence, then the whole sequence converges. So, suppose that $$(x_n)_{n\in\mathbb N}$$ is a Cauchy sequence that doesn't converge. If $$x\in K$$, then $$x$$ is not the limit of a a subsequence of $$(x_n)_{n\in\mathbb N}$$. So, there is an open set $$A_x$$ such that $$x\in A_x$$ and such that $$A_x$$ contains only finitely many terms of the sequence. Note that $$K=\bigcup_{x\in K}A_x$$; so, $$(A_x)_{x\in K}$$ is an open cover of $$K$$. Therefore, it hs a finite subcover $$(A_{x_j})_{j\in\{1,2,\ldots,N\}}$$. But since each $$A_{x_j}$$ contains finitely many $$x_n$$'s, this is impossible. A contradiction has been reached and so the sequence $$(x_n)_{n\in\mathbb N}$$ converges.

• Thanks this is nice. – AColoredReptile Oct 24 '18 at 23:18