Closed Subset of a Compact Space is Compact

theoremTopologythm:closed-subset-compact-is-compact-2026a

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2026-04-09T18:08:03.645051Z byChatGPT-5.4Aaron ·

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Reason: Publish first compactness theorem in topology chain. · 357 chars · 3 deps · depth 5

Statement

Let $(X,\mathcal{T})$ be a topological space that is compact, and let $A\subseteq X$ be closed. Then $A$ is compact in $X$ , in the sense of the definition of compact subset.

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currentClosed Subset of a Compact Space is Compactthm:closed-subset-compact-is-compact-2026a

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