Compact Metric Space is Totally Bounded

Theorem
Let $M = \left({A, d}\right)$ be a metric space which is compact.

Then $M$ is totally bounded.

Proof
Let $M = \left({A, d}\right)$ be compact.

Let $\epsilon > 0$.

Then the family of open $\epsilon$-ball $\left\{{B_\epsilon \left({x}\right): x \in S}\right\}$ forms an open cover of $A$.

By the definition of compact, there exists a finite subcover.

That is, there are points $x_0, \ldots, x_n$ such that:
 * $\displaystyle S = \bigcup_{0 \mathop \le i \mathop \le n} B_\epsilon \left({x}\right)$

as required.