Totally Bounded Metric Space is Separable

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Theorem

A totally bounded metric space is separable.


Proof

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


By the definition of total boundedness, we can use the axiom of countable choice to construct a sequence $\left\langle{F_n}\right\rangle_{n \ge 1}$ such that:

For all natural numbers $n \ge 1$, $F_n$ is a finite $\left({1/n}\right)$-net for $M$.

Let $\displaystyle S = \bigcup_{n \mathop \ge 1} F_n$.

From Countable Union of Countable Sets is Countable, it follows that $S$ is countable.

It suffices to prove that $S$ is everywhere dense in $M$.


Let $S^-$ denote the closure of $S$.

Let $x \in X$.

Let $U \subseteq X$ be open in $M$ such that $x \in U$.

By definition, there exists a strictly positive real number $\epsilon$ such that $B_{\epsilon} \left({x}\right) \subseteq U$.

That is, the open $\epsilon$-ball of $x$ in $M$ is contained in $U$.

By the Archimedean Principle, there exists a natural number $n > \dfrac 1 \epsilon$.

That is, $\dfrac 1 n < \epsilon$, and so $B_{1 / n} \left({x}\right) \subseteq B_{\epsilon} \left({x}\right)$.

Since $\subseteq$ is a transitive relation, we have $B_{1/n} \left({x}\right) \subseteq U$.

By the definition of a net, there exists a $y \in F_n$ such that $x \in B_{1/n} \left({y}\right)$.

By axiom $\left({M3}\right)$ for a metric, it follows from the definition of an open ball that $y \in B_{1/n} \left({x}\right)$.

Since $y \in S \cap U$, it follows that $x$ is an adherent point of $S$.

By definition of adherent point, we have $x \in S^-$.

That is, $X \subseteq S^-$, and so $S$ is everywhere dense in $M$.

$\blacksquare$


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.