Totally Bounded Metric Space is Bounded
Jump to navigation
Jump to search
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