Totally Bounded Metric Space is Second-Countable

Theorem

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

Then $M$ is second-countable.

Proof 1

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

Consider the following set of open balls:

$\mathcal C := \left\{{N_{1/k} \left({x}\right) : k \in \N^*}\right\}$

As $M$ is totally bounded, each one forms an $\epsilon$-net where $\epsilon = 1, \dfrac 1 2, \dfrac 1 3, \ldots$

Hence, $\mathcal C$ is a countable basis for $X$.

That is, $M$ is second-countable.

$\blacksquare$

Proof 2

For all $n \in \N$, let:

$\mathcal C_n = \left\{ {N_{1/n}\left({x}\right) : x \in X} \right\}$

where $N_{\epsilon}\left({x}\right)$ denotes the $\epsilon$-neighborhood of $x$ in $X$.

By the definition of total boundedness, for all $n \in \N$, there exists a finite subcover $\mathcal F_n$ of $\mathcal C_n$ for $X$.

Define:

$\displaystyle \mathcal B = \bigcup_{n \in \N} \mathcal F_n$

Clearly, $\mathcal B$ is countable.

It suffices to show that $\mathcal B$ is a basis for the topology induced by the metric $d$.

Let $U$ be any open subset of $X$.

For all $x \in U$, there exists an $n \in \N$ such that $N_{2/n}\left({x}\right) \subseteq U$.

Because $\mathcal F_n$ covers $X$, there exists a $B \in \mathcal F_n$ such that $x \in B$.

Let $c$ be the center of the open ball $B$. Recall that $B$ has radius $\dfrac 1 n$ by the definition of $\mathcal F_n$.

For any $y \in B$, $\displaystyle d\left({x, y}\right) \le d\left({x, c}\right) + d\left({y, c}\right) \le \frac 2 n$.

Therefore, $y \in N_{2/n}\left({x}\right)$.

Hence $B \subseteq N_{2/n}\left({x}\right) \subseteq U$.

We have just shown that for all $x \in U$, there exists an $n \in \N$ and a $B \in \mathcal F_n$ such that $x \in B \subseteq U$.

Hence:

$\displaystyle U \subseteq \bigcup_{n \in \N} \bigcup \left\{ {B \in \mathcal F_n : B \subseteq U} \right\}$

It is also true that:

$\displaystyle U \supseteq \bigcup_{n \in \N} \bigcup \left\{ {B \in \mathcal F_n : B \subseteq U} \right\}$

It follows that:

$\displaystyle U = \bigcup_{n \in \N} \bigcup \left\{ {B \in \mathcal F_n : B \subseteq U} \right\}$

This is a union of members of $\mathcal B$.

Hence $\mathcal B$ is a basis for the topology induced by the metric $d$.

Therefore, $X$ is second-countable.

$\blacksquare$