Totally Bounded Metric Space is Bounded

Theorem
Let $M = \left({A, d}\right)$ be a totally bounded metric space.

Then $M$ is bounded.

Proof
Let $M = \left({A, d}\right)$ be totally bounded.

Then there exist $n \in \N$ and points $x_0, \dots, x_n \in A$ such that:
 * $\displaystyle \inf_{0 \mathop \le i \mathop \le n} d \left({x_i, x}\right) \le 1$

for all $x \in A$.

Let us set:
 * $a := x_0$
 * $\displaystyle D := \max_{0 \mathop \le i \mathop \le n} d \left({x_0, x_i}\right)$
 * $K := D + 1$

Now let $x \in A$ be arbitrary.

Then by assumption there exists $i$ such that $d \left({x_i, x}\right) \le 1$.

Hence:
 * $ d \left({a, x}\right) \le d \left({a, x_i}\right) + d \left({x_i, x}\right) \le 1 + D = K$

So $M$ is bounded, as claimed.