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

Question

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.

Answer 1

Hint: don't worry about Cauchy sequences for this problem. Sequential compactness says that any sequence has a convergent subsequence without reference to the Cauchy property. Thus:

(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 $x$.