Equivalence of Definitions of Totally Bounded Metric Space

Definition 1 implies Definition 2
Let $M$ be totally bounded by definition 1.

That is:
 * $\forall \epsilon \in \R_{>0}: \struct {A, d}$ has a finite $\epsilon$-net.

So, let $\epsilon \in \R_{>0}$.

Let $A' = \set {x_0, x_1, \ldots, x_n}$ be such a finite $\epsilon$-net of $A$.

By definition:
 * $\ds A \subseteq \bigcup_{x_i \mathop \in A'} \map {B_\epsilon} {x_i}$

Now let $x \in A$, and so:
 * $\ds x \in \bigcup_{x_i \mathop \in A'} \map {B_\epsilon} {x_i}$

Thus:
 * $\exists i: 0 \le i \le n: x \in \map {B_\epsilon} {x_i}$

and so:
 * $\map d {x_i, x} < \epsilon$

But:
 * $\ds \inf_{0 \mathop \le i \mathop \le n} \map d {x_i, x} \le \map d {x_i, x}$

Thus it follows that there exist finitely many points $x_0, \dots, x_n \in x$ such that:
 * $\ds \inf_{0 \mathop \le i \mathop \le n} \map d {x_i, x} \le \epsilon$

for all $x \in A$.

Thus $M$ is totally bounded by definition 2.

Definition 2 implies Definition 1
Let $M$ be totally bounded by definition 2.

That is:
 * for every $\epsilon > 0$ there exist finitely many points $x_0, \dots, x_n \in A$ such that $\ds \inf_{0 \mathop \le i \mathop \le n} \map d {x_i, x} \le \epsilon$ for all $x \in A$.

So, let $x \in A$.

Let $\epsilon' = \dfrac \epsilon 2$.

Then by definition $\exists n \in \N: A' = \set {x_0, x_1, \ldots, x_n}$ such that $\forall x \in A: \exists x_i \in A': \map d {x_i, x} \le \epsilon'$.

Hence:
 * $x \in \map {B_{\epsilon'} } {x_i}$

where $\map {B_{\epsilon'} } {x_i}$ is the open $\epsilon'$-ball of $x_i$.

So:
 * $\ds x \in \bigcup_{x_i \mathop \in A'} \map {B_{\epsilon'} } {x_i}$

and hence:
 * $\ds A \subseteq \bigcup_{x_i \mathop \in A'} \map {B_{\epsilon'} } {x_i}$

Thus by definition, $A'$ is a finite $\epsilon'$-net of $A$.

Thus $M$ is totally bounded by definition 1.