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 Contraposition.

To that end, suppose that there exists no finite $\epsilon$-net for $M$.

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

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

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

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

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

So, by definition of a $\epsilon$-net:
 * $\ds A \nsubseteq \bigcup_{i \mathop = 0}^r \map {B_\epsilon} {x_i}$

where $\map {B_\epsilon} {x_i}$ denotes the open $\epsilon$-ball of $x_i$ in $M$.

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

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

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

Thus $\sequence {x_n}$ has no Cauchy subsequence.

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

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

From Rule of Transposition, if $M$ is sequentially compact, then there exists no finite $\epsilon$-net for $M$.

So if $M$ is sequentially compact, it is certainly not totally bounded.

Also see

 * Sequentially Compact Metric Space is Compact


 * Compact Metric Space is Totally Bounded