Closed Subset of a Compact Set is Compact

Last updated Nov 1, 2022

Let $X$ be a Topological Space. Suppose $S \subset X$ is Compact and $K \subset S$ is Closed. Then $K$ is Compact.

Let $\mathcal{U} := {U_\alpha}{\alpha \in I}$ be an Open Cover for $K$. Since $K^{C}$ is Open, then we have that $K^{C} \cup \mathcal{U}$ is an Open Cover for $X$, and thus an Open Cover for $S$. Since $S$ is Compact, we can reduce this Open Cover to a finite Open Subcover ${V{i}}{i=1}^{n}$ for some $n \in \mathbb{N}$. Taking ${V{i}}_{i=1}^{n} \setminus {K^{C}}$, we get a finite Open Subcover of $\mathcal{U}$, proving that $K$ is Compact. $\blacksquare$

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