# Subset of discrete metric space is sequentially compact iff it's finite

I have the following problem:

Let $$(X,d)$$ a metric space with discrete metric and $$K \subset X$$. Show that $$K$$ is sequentially compact iff $$K$$ is a finite set.

First, I searched for what a sequentially compact set is, which is when every sequence in a set has a convergent subsequence. I know that the discrete metric space is complete, thus it's Cauchy sequences converge and any subsequence from those Cauchy sequences converge. But I don't know if I can use this to prove that it is sequentially compact and then how can I link it with the idea of the set being finite. Any help is much appreciated, thanks.

• Do you know what convergent sequences in a discrete space look like? Commented Sep 6, 2021 at 21:05
• It means that it eventually becomes a constant. Commented Sep 6, 2021 at 21:11

(1) If $$(X, d)$$ is a discrete metric space and $$X$$ is infinite, you can choose an infinite sequence $$x_1, x_2, \ldots \in X$$ such that, for any $$i, j$$, $$x_i \neq x_j$$ unless $$i = j$$. Now you can show that $$x_1, x_2, \ldots$$ has no convergent subsequence, because the only way a sequence in a discrete space can converge is if it is eventually constant.
(2) if $$(X, d)$$ is a discrete metric space and $$X$$ is finite, then any sequence $$x_1, x_2 \ldots \in X$$, must visit some $$x$$ in $$X$$ infinitely many times. so $$x_1, x_2 \ldots$$ has a constant subsequence whose elements are equal to $$x$$ and therefore converges to $$x$$.