Sequentially Compact Metric Space is Totally Bounded/Proof 2

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

Then $M$ is totally bounded.

Proof
Let $M = \left({A, d}\right)$ be a sequentially compact metric space.

By definition, a metric space is totally bounded iff, for every $\epsilon \in \R_{>0}$, there exists a finite $\epsilon$-net of $M$.

Aiming for a contradiction, let there exist no finite $\epsilon$-net for $M$.

The aim is to construct a sequence $\left \langle {x_n} \right \rangle$ in $M$ with no convergent subsequence.

Suppose that $x_1, x_2, \ldots, x_r \in A$ have been chosen such that:
 * $\forall m, n \in \left\{{1, 2, \ldots, r}\right\}: m \ne n \implies d \left({x_m, x_n}\right) \ge \epsilon$

That is, every pair of distinct elements of $\left\{{x_1, x_2, \ldots, x_r}\right\}$ is at least $\epsilon$ apart.

By hypothesis, there exists no finite $\epsilon$-net for $M$.

Specifically, $\left\{{x_1, x_2, \ldots, x_r}\right\}$ is not therefore an $\epsilon$-net.

So by definition of finite $\epsilon$-net:
 * $\displaystyle A \nsubseteq \bigcup_{i \mathop = 1}^r B_\epsilon \left({x_i}\right)$

where $B_\epsilon \left({x_i}\right)$ is the open $\epsilon$-ball of $x_i$.

Thus there must exist $x_{r+1} \in A$ such that:
 * $\displaystyle x_{r+1} \notin \bigcup_{i \mathop = 1}^r B_\epsilon \left({x_i}\right)$

That is:
 * $\exists x_{r+1} \in A: \forall m, n \in \left\{{1, 2, \ldots, r, r+1}\right\}: m \ne n \implies d \left({x_m, x_n}\right) \ge \epsilon$

Thus, by induction, the sequence $\left \langle {x_n} \right \rangle \subseteq A$ has been created such that:
 * $\forall m, n \in \N: m \ne n: d \left({x_m, x_n}\right) \ge \epsilon$

Thus $\left \langle {x_n} \right \rangle$ has no Cauchy subsequence.

So by Convergent Sequence is Cauchy Sequence, it has no convergent subsequence either.

Thus, by definition, $M$ is not sequentially compact.

The result follows from Proof by Contradiction.

Also see

 * Sequentially Compact Metric Space is Compact


 * Compact Metric Space is Totally Bounded