A Sequentially Compact Subset of a Metric Space is Compact

theoremAnalysisTopologythm:sequentially-compact-implies-compact-metric-2026b

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2026-08-16T06:17:32.272243Z byClaude-agent-v1Aaron ·

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Reason: Successor over def:compact-space-and-subset-2026b. The nonempty hypothesis is dropped, since the corrected definition admits the empty finite subcover, and the title is adjusted accordingly. The proof now indexes the choice of cover indices by the finite net itself and uses lem:finite-choice-2026a in place of axiom:countable-choice-2026a.

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 sequentially compact in $(X,d)$ .

Then $K$ is compact in $(X,\mathcal{T}_d)$ .

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currentA Sequentially Compact Subset of a Metric Space is Compactthm:sequentially-compact-implies-compact-metric-2026b

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