Equivalence of Definitions of Total Boundedness

Theorem

Let $\left({S, d}\right)$ be a metric space.

The two definitions of total boundedness of $\left({S, d}\right)$ are equivalent:

• $\left({S, d}\right)$ is totally bounded iff for every $\epsilon > 0$ it has a finite $\epsilon$-net.

• $\left({S, d}\right)$ is totally bounded iff for every $\epsilon > 0$ there exist finitely many points $x_0, \dots, x_n \in x$ such that:

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

for all $x \in S$.

Proof

• Suppose that for every $\epsilon > 0$ there exist finitely many points $x_0, \dots, x_n \in S$ such that $\displaystyle \inf_{0 \le i \le n} d \left({x_i, x}\right) \le \epsilon$ for all $x \in S$.

So, let $x \in S$.

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

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

Hence $x \in N_{\epsilon'} \left({x_i}\right)$, where $N_{\epsilon'} \left({x_i}\right)$ is the $\epsilon'$-neighborhood of $x_i$.

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

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

• Now, suppose that for every $\epsilon > 0$, $\left({S, d}\right)$ has a finite $\epsilon$-net.

So, let $\epsilon > 0$.

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

By definition, $\displaystyle S \subseteq \bigcup_{x_i \in S'} N_{\epsilon} \left({x_i}\right)$.

Now let $x \in S$, and so $\displaystyle x \in \bigcup_{x_i \in S'} N_{\epsilon} \left({x_i}\right)$.

Thus $\exists i: 0 \le i \le n: x \in N_{\epsilon} \left({x_i}\right)$, and so $d \left({x_i, x}\right) < \epsilon$.

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

Thus it follows that there exist finitely many points $x_0, \dots, x_n \in x$ such that $\displaystyle \inf_{0 \le i \le n} d \left({x_i, x}\right) \le \epsilon$ for all $x \in S$.

$\blacksquare$