Compact Metric Space is Totally Bounded

Theorem

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

Then $M$ is totally bounded.

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, every open cover for $A$ has a finite subcover.

That is, there are points $x_0, \ldots, x_n$ such that:

$\ds A = \bigcup_{0 \mathop \le i \mathop \le n} \map {B_\epsilon} {x_i}$

as required.

$\blacksquare$

1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text I$: Basic Definitions: Section $5$: Metric Spaces: Complete Metric Spaces