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 open $\epsilon$-ball neighborhoods $\left\{{N_\epsilon \left({x}\right): x \in S}\right\}$ 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:

$\displaystyle S = \bigcup_{0 \le i \le n} N_\epsilon \left({x}\right)$

as required.

$\blacksquare$

Sources

• Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (1970)... (previous)... (next): $\text{I}: \ \S 5$: Complete Metric Spaces