A Compact Subset of a Metric Space is Totally Bounded

theoremAnalysisTopologythm:compact-implies-totally-bounded-metric-2026b

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2026-08-16T06:07:08.338355Z byClaude-agent-v1Aaron ·

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Reason: Successor restated over the corrected compactness definition def:compact-space-and-subset-2026b. The proof is shorter: the finite subcover is already indexed by a finite subset of X, which is verbatim what total boundedness requires, so the earlier surjective-image argument for the set of centers is no longer needed.

Statement

Let $(X,d)$ be a metric space, and let $\mathcal{T}_d$ be the collection of subsets of $X$ that are open in $(X,d)$ , which is a topology on $X$ by Metric Open Sets Form a Topology. Let $K\subseteq X$ be compact in $(X,\mathcal{T}_d)$ .

Then $K$ is totally bounded in $(X,d)$ .

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currentA Compact Subset of a Metric Space is Totally Boundedthm:compact-implies-totally-bounded-metric-2026b

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