Compact Metric Space is Totally Bounded
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Theorem
Let $M = \struct {A, d}$ be a metric space which is compact.
Then $M$ is totally bounded.
Proof
Let $M = \struct {A, d}$ be compact.
Let $\epsilon > 0$.
Then the family $\set {\map {B_\epsilon} x: x \in A}$ of open $\epsilon$-balls forms an open cover of $A$.
By the definition of compact, there exists a finite subcover.
That is, there are points $x_0, \ldots, x_n$ such that:
- $\ds A = \bigcup_{0 \mathop \le i \mathop \le n} \map {B_\epsilon} {x_i}$
as required.
$\blacksquare$
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