Sequentially Compact Metric Space is Totally Bounded/Proof 2

Proof
Let $\epsilon \in \R_{>0}$ be a strictly positive real number.

By definition, $M$ is totally bounded only if there exists a finite $\epsilon$-net for $M$.

We use a Proof by Contradiction.

That is, suppose that there exists no finite $\epsilon$-net for $M$.

The aim is to construct an infinite sequence $\left \langle {x_n} \right \rangle_{n \in \N}$ in $A$ that has no convergent subsequence.

Suppose that $x_0, x_1, \ldots, x_r \in A$ have been defined such that:
 * $\forall m, n \in \left\{{0, 1, \ldots, r}\right\}: m \ne n \implies d \left({x_m, x_n}\right) \ge \epsilon$

That is, any two distinct elements of $\left\{{x_0, x_1, \ldots, x_r}\right\}$ are at least $\epsilon$ apart.

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

Specifically, $\left\{{x_0, x_1, \ldots, x_r}\right\}$ is therefore not an $\epsilon$-net for $M$.

So, by definition of a $\epsilon$-net:
 * $\displaystyle A \nsubseteq \bigcup_{i \mathop = 0}^r B_\epsilon \left({x_i}\right)$

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

Thus there must exist $x_{r+1} \in A$ such that:
 * $\displaystyle x_{r+1} \notin \bigcup_{i \mathop = 0}^r B_\epsilon \left({x_i}\right)$

That is:
 * $\exists x_{r+1} \in A: \forall m, n \in \left\{{0, 1, \ldots, r, r + 1}\right\}: m \ne n \implies d \left({x_m, x_n}\right) \ge \epsilon$

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

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

Since a convergent sequence is Cauchy, it has no convergent subsequence either.

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

This contradicts the original assumption that $M$ is sequentially compact.

Also see

 * Sequentially Compact Metric Space is Compact


 * Compact Metric Space is Totally Bounded