Totally Bounded Metric Space is Bounded

Lemma

A totally bounded metric space $\left({S, d}\right)$ is bounded.

Proof

Suppose that $\left({S, d}\right)$ is totally bounded.

Then there exist $n \in \N$ and points $x_0, \dots, x_n \in S$ such that:

$\displaystyle \inf_{0 \le i \le n} d \left({x_i, x}\right) \le 1$

for all $x \in S$.

Let us set:

$a := x_0$

$\displaystyle D := \max_{0 \leq i \leq n} d \left({x_0, x_i}\right)$

$K := D + 1$.

Now let $x \in S$ 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 $\left({S, d}\right)$ is bounded, as claimed.

$\blacksquare$

Sources

• Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (1970)... (previous)... (next): $\text{I}: \ \S 5$: Complete Metric Spaces