Totally Bounded Metric Space is Bounded

Theorem
Let $M = \struct {A, d}$ be a totally bounded metric space.

Then $M$ is bounded.

Proof
Let $M = \struct {A, d}$ be totally bounded.

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

for all $x \in A$.

Let us set:
 * $a := x_0$
 * $\ds D := \max_{0 \mathop \le i \mathop \le n} \map d {x_0, x_i}$
 * $K := D + 1$

Now let $x \in A$ be arbitrary.

Then by assumption there exists $i$ such that $\map d {x_i, x} \le 1$.

Hence:
 * $\map d {a, x} \le \map d {a, x_i} + \map d {x_i, x} \le 1 + D = K$

So $M$ is bounded, as claimed.

Also see

 * Bounded Metric Space is not necessarily Totally Bounded, showing that the converse does not necessarily hold.