Equivalence of Definitions of Totally Bounded Metric Space

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

So, let $$x \in S$$.

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

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

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

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

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

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