Compact subsets of a Hausdorff space are closed

Let $𝑋$ be a Hausdorff space. If $𝐾 \subseteq 𝑋$ is compact, then $𝐾$ is closed topology

Proof

Let $𝑏 \in 𝑋 ∖ 𝐾$ . For each $𝑥 \in 𝐾$ assign an open neighbourhood $𝑈 𝑥$ of $𝑥$ and likewise an open neighbourhood $𝑉 𝑥$ of $𝑏$ such that $𝑈 𝑥 \cap 𝑉 𝑥 = \emptyset$ for all $𝑥 \in 𝐾$ (the Hausdorff property). Then $(𝑈 𝑥) 𝑥 \in 𝐾$ is an open cover of $𝐾$ and thus has a finite subcover $(𝑈 𝑥 𝑖) 𝑛 𝑖 = 1$ . Then $𝑉 = ⋂ 𝑛 𝑖 = 1 𝑉 𝑖$ is an open subset of $𝑋 ∖ 𝐾$ . Since every $𝑏 \in 𝑋 ∖ 𝐾$ has an open neighbourhood $𝑉 \subseteq 𝑋 ∖ 𝐾$ , $𝑋 ∖ 𝐾$ is open, whence $𝐾$ is closed.

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Compact subsets of a Hausdorff space are closed

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