Compact Metric Space is Totally Bounded

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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 $\left\{{B_\epsilon \left({x}\right): x \in A}\right\}$ 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:

$\displaystyle A = \bigcup_{0 \mathop \le i \mathop \le n} B_\epsilon \left({x_i}\right)$

as required.

$\blacksquare$


Sources