Heine-Borel theorem

The Heine-Borel characterizes compact subsets of Real coördinate space. Let $𝑆 \subseteq ℝ 𝕟$ . Then $𝑆$ is compact iff it is closed and bounded. topology

Proof

The forward direction follows from Compact subsets of a Hausdorff space are closed and Compact sets in a metric space are bounded.

For the converse, let $𝐾 \subseteq 𝑋$ be closed and bounded. Then it can be enclosed with an $𝑛$ -box $[- 𝑎, 𝑎] 𝑛$ . Since Closed subsets of a compact space are compact, it is enough to prove $𝑇 0 = [- 𝑎, 𝑎] 𝑛$ is compact.

Suppose $𝑇 0$ is not compact. Then there exists an open cover ${𝑈 𝛼} 𝛼 \in 𝐴$ with no finite subcover. $𝑇 0$ can be broken into $2 𝑛$ sub-boxes of half its side length, at least one of which must require an infinite subcover of ${𝑈 𝛼} 𝛼 \in 𝐴$ . Call this $𝑇 1$ . Continuing this argument iteratively, one obtains a sequence of shrinking $𝑛$ -boxes
$𝑇 0 ⊋ 𝑇 1 ⊋ \dots ⊋ 𝑇 𝑘 ⊋ \dots$
each requiring infinite subcovers, where $𝑇 𝑘$ has side length $2 1 - 𝑘 𝑎$ . One may construct a sequence $(𝑥 𝑘) \infty 𝑘 = 1$ such that $𝑥 𝑘 \in 𝑇 𝑘$ , which is clearly a Cauchy sequence and thus converges to some $𝑥$ by completeness of $ℝ 𝑛$ . By sequential closedness $𝑥 \in 𝑇 𝑘$ for all $𝑘 \in ℕ 0$ . Now since ${𝑈 𝛼} 𝛼 \in 𝐴$ is a cover there exists some $𝛽 \in 𝐴$ such that $𝑥 \in 𝑈 𝛽$ , and by openness there exists open ball $B 𝜖 (𝑥) \subseteq 𝑈 𝛽$ . For sufficiently large $𝑘$ , $𝑇 𝑘 \subseteq B 𝜖 (𝑥) \subseteq 𝑈 𝛽$ , whence ${𝑈 𝛽}$ is a finite subcover of $𝑇 𝑘$ , a contradiction. Therefore $𝑇 0$ is compact, so $𝐾$ is compact.

An alternate proof follows from Tikhonov’s theorem.

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Heine-Borel theorem

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