# real analysis – Subset of discrete metric space is sequentially compact iff it’s finite

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 $$x$$.