Totally Bounded Metric Space is Separable

Theorem
A totally bounded metric space is separable.

Proof
Let $M = \struct {A, d}$ be a totally bounded metric space.

By the definition of total boundedness, we can use the axiom of countable choice to construct a sequence $\sequence {F_n}_{n \mathop \ge 1}$ such that:
 * For all natural numbers $n \ge 1$, $F_n$ is a finite $\paren {1/n}$-net for $M$.

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

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 $\map {B_\epsilon} x \subseteq U$.

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 $\map {B_{1 / n} } x \subseteq \map {B_\epsilon} x$.

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

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

By, it follows from the definition of an open ball that $y \in \map {B_{1/n} } x$.

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 $M$.