Compact sets in a metric space are bounded

Let $(𝑋, 𝑑)$ be a metric space and $𝐾 \subseteq 𝑋$ be compact. Then $𝐾$ is bounded. #m/thm/anal

Proof

The set ${B 1 (𝑥)} 𝑥 \in 𝑋$ forms an open cover of $𝐾$ , so it has a finite subcover ${B 1 (𝑥 𝑖)} 𝑛 𝑖 = 1$ . Let

𝑀 = m a x {𝑑 (𝑥 𝑖, 𝑥 𝑗) : 𝑖, 𝑗 = 1, \dots, 𝑛}

Then for any $𝑝, 𝑞 \in 𝑋$ , $𝑝 \in B 1 (𝑥 𝑖)$ and $𝑞 \in B 1 (𝑥 𝑗)$ for some $𝑖, 𝑗$ , hence

𝑑 (𝑝, 𝑞) \leq 𝑑 (𝑝, 𝑥 𝑖) + 𝑑 (𝑥 𝑖, 𝑥 𝑗) + 𝑑 (𝑥 𝑗, 𝑞) \leq 1 + 𝑀 + 1 \leq 𝑀 + 2

therefore $𝐾$ is bounded.

#state/tidy | #lang/en | #SemBr