Totally Bounded Metric Space is Second-Countable/Proof 1

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Theorem

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

Then $M$ is second-countable.


Proof

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

Let $\epsilon = 1, \dfrac 1 2, \dfrac 1 3, \ldots$

As $M$ is totally bounded, for each $\epsilon$ there exists a finite $\epsilon$-net $\mathcal C$ for $M$.

From Net forms Basis for Metric Space, $\mathcal C$ is a countable basis for $M$.

That is, $M$ is second-countable.

$\blacksquare$


Axiom of Countable Choice

This theorem depends on the Axiom of Countable Choice.

Although not as strong as the Axiom of Choice, the Axiom of Countable Choice is similarly independent of the Zermelo-Fraenkel axioms.

As such, mathematicians are generally convinced of its truth and believe that it should be generally accepted.



Sources