Closed subsets of a compact space are compact

Let $𝑋$ be a compact space. Then if $𝐴 \subseteq 𝑋$ is closed it is also compact. #m/thm/topology

Proof

Let ${𝐴 𝛼} 𝛼 \in 𝐼 \subseteq 𝐴$ be an open cover of $𝐴$ . Then ${𝑋 ∖ 𝐴} \cup {𝐴 𝛼} 𝛼 \in 𝐼$ is an open cover of $𝑋$ , so by compactness it has a finite subcover ${𝑋 ∖ 𝐴} \cup {𝐴 𝛼 𝑖} 𝑛 𝑖 = 1$ . But it follows that ${𝐴 𝛼 𝑖} 𝑛 𝑖 = 1$ is a finite subcover of ${𝐴 𝛼} 𝛼 \in 𝐼$ . Hence $𝐴$ is compact.

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