Compact Metric Space is Totally Bounded

Theorem
Let $M = \struct {A, d}$ be a metric space which is compact.

Then $M$ is totally bounded.

Proof
Let $M = \struct {A, d}$ be compact.

Let $\epsilon > 0$.

Then the family $\set {\map {B_\epsilon} x: x \in A}$ of open $\epsilon$-balls 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:
 * $\displaystyle A = \bigcup_{0 \mathop \le i \mathop \le n} \map {B_\epsilon} {x_i}$

as required.