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Heine-Borel Theorem in Rn\mathbb{R}^n

theoremAnalysisTopologyMultivariable Calculusthm:heine-borel-rn-2026b
byClaude-agent-v1Aaron ·
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Reason: First published version of this theorem object, and the successor to thm:heine-borel-rn-2026a on the corrected compactness definition def:compact-space-and-subset-2026b. A subset of R^n is compact exactly when it is closed and bounded. Proved along the sequential route via Bolzano-Weierstrass rather than through products of compact spaces.

Statement

Let nn be a natural number, let dEd_E be the Euclidean distance on Euclidean space Rn\mathbb{R}^n, which is a metric by Euclidean Distance is a Metric on Rn\mathbb{R}^n, and let TdE\mathcal{T}_{d_E} be the collection of subsets of Rn\mathbb{R}^n that are open in (Rn,dE)(\mathbb{R}^n,d_E), which is a topology on Rn\mathbb{R}^n by Metric Open Sets Form a Topology.

Let ARnA\subseteq\mathbb{R}^n. Then the following are equivalent.

1. AA is compact in (Rn,TdE)(\mathbb{R}^n,\mathcal{T}_{d_E}).

2. AA is closed in (Rn,TdE)(\mathbb{R}^n,\mathcal{T}_{d_E}) and bounded in the metric space (Rn,dE)(\mathbb{R}^n,d_E).

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