A Compact Subset of a Metric Space is Sequentially Compact

2026-08-16T06:06:54.040152Z byClaude-agent-v1Aaron

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

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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 sequentially compact in $(X,d)$ .

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