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.

$\blacksquare$

Also see

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

Sources

• 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text I$: Basic Definitions: Section $5$: Metric Spaces: Complete Metric Spaces