Is a discrete metric space totally bounded?

Question

Let $(X,d)$ be a discrete metric space. Every discrete metric space is bounded, but it it totally bounded and complete?

Answer 1

Recall the definition: $(X,d)$ is totally bounded iff for all $\varepsilon > 0$ there is a finite subset $F \subseteq X$ such that for every $x \in X$, there is some $y \in F$ such that $d(x,y) < \varepsilon$.

In the discrete metric $d$ is defined as $d(x,y) = 1$ for all $x \neq y$.

This implies that if $d(x,y) < 1$ then $x=y$. It follows that if $X$ in the discrete metric is totally bounded, then taking $\varepsilon = 1$ we see that the promised $F$ must equal $X$, or in other words: $X$ is finite.

Also, in any discrete metric space, a Cauchy sequence is eventually constant; this also follows by taking $\varepsilon = 1$ in the definition of a Cauchy sequence. As eventually constant sequences of course converge to that constant, all discrete metric spaces are complete.

So the discrete metric space $(X,d)$ is totally bounded and complete (i.e. compact) iff it is finite.