# Equivalence of Definitions of Totally Bounded Metric Space

## Theorem

The following definitions of the concept of Totally Bounded Metric Space are equivalent:

### Definition 1

A metric space $M = \struct {A, d}$ is totally bounded if and only if:

for every $\epsilon \in \R_{>0}$ there exists a finite $\epsilon$-net for $M$.

That is, $M$ is totally bounded if and only if:

for every $\epsilon \in \R_{>0}$ there exists a finite set of points $x_1, \ldots, x_n \in A$ such that:
$\displaystyle A = \bigcup_{i \mathop = 1}^n \map {B_\epsilon} {x_i}$
where $\map {B_\epsilon} {x_i}$ denotes the open $\epsilon$-ball of $x_i$.

That is: $M$ is totally bounded if and only if, given any $\epsilon \in \R_{>0}$, one can find a finite number of open $\epsilon$-balls which cover $A$.

### Definition 2

A metric space $M = \struct {A, d}$ is totally bounded if and only if:

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

## Proof

### Definition 1 implies Definition 2

Let $M$ be totally bounded by definition 1.

That is:

$\forall \epsilon \in \R_{>0}: \left({A, d}\right)$ has a finite $\epsilon$-net.

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

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

By definition:

$\displaystyle A \subseteq \bigcup_{x_i \mathop \in A'} B_\epsilon \left({x_i}\right)$

Now let $x \in A$, and so:

$\displaystyle x \in \bigcup_{x_i \mathop \in A'} B_\epsilon \left({x_i}\right)$

Thus:

$\exists i: 0 \le i \le n: x \in B_\epsilon \left({x_i}\right)$

and so:

$d \left({x_i, x}\right) < \epsilon$

But:

$\displaystyle \inf_{0 \mathop \le i \mathop \le n} d \left({x_i, x}\right) \le d \left({x_i, x}\right)$

Thus it follows that there exist finitely many points $x_0, \dots, x_n \in x$ such that:

$\displaystyle \inf_{0 \mathop \le i \mathop \le n} d \left({x_i, x}\right) \le \epsilon$

for all $x \in A$.

Thus $M$ is totally bounded by definition 2.

$\Box$

### 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 $\displaystyle \inf_{0 \mathop \le i \mathop \le n} d \left({x_i, x}\right) \le \epsilon$ for all $x \in A$.

So, let $x \in A$.

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

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

Hence:

$x \in B_{\epsilon'} \left({x_i}\right)$

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

So $\displaystyle x \in \bigcup_{x_i \mathop \in A'} B_{\epsilon'} \left({x_i}\right)$ and hence $\displaystyle A \subseteq \bigcup_{x_i \mathop \in A'} B_{\epsilon'} \left({x_i}\right)$.

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

Thus $M$ is totally bounded by definition 1.

$\blacksquare$