Sequential compactness implies compactness

Question

I'm currently trying to prove that if $X$ is a metric space and $X$ is sequentially compact, then $X$ itself is compact. My current strategy is to proceed by contraposition. We assume $X$ is not compact. Then it has an open cover $\mathcal{O}$ for which no finite subcover contains all of $X$. Next, I think it would be easier to proceed by contradiction: we assume that a sequence $\{x_n\}$ has a subsequence $\{x_{n_k}\}$ that converges to a point $L$. We know that since $\mathcal{O}$ covers $X$, there is some open set $O\in \mathcal{O}$ such that $L\in O$. Then there is some $N\in\mathbb{N}$ such that for all $k\geq N$, $x_{n_k}\in O$. I'm not sure how to proceed from here. I'm expecting to be able to derive that the open cover $\mathcal{O}$ has a finite subcover, contradicting our assumption that $X$ is not compact.

Any advice will be helpful. Thanks!

Answer 1

A few notes to make: Firstly if you are proceeding by contradiction you need to assume that every sequence $\{x_n\}$ has a convergent subsequence, not just one (the negation of not sequentially compact is compact).

Secondly, you haven't used the fact that $X$ is a metric space, which is crucial here, so to proceed you'd definitely need to use some property only possessed by metrisable spaces.

It is not a trivial fact that sequentially compact metric spaces are compact, and I think you will struggle to proceed without some lemmas. I recommend looking at Lebesgue's Number Lemma or the fact that $X$ has the Lindelof property, as well as the fact that sequentially compact spaces are totally bounded and that sequentially compact metric spaces satisfy the finite intersection property for closed sets.