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How to prove that a compact set $K$ in a Hausdorff topological space $\mathbb{X}$ is closed? I seek a proof that is as self contained as possible.

Thank you.

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2 Answers 2

Fix $x\in\mathbb{X}\setminus K$. Since $\mathbb{X}$ is Hausdorff, for each $y\in K$ there are disjoint open sets $U_y$ and $V_y$ such that $x\in U_y$ and $y\in V_y$. $\{V_y:y\in K\}$ is an open cover of $K$, so it has a finite subcover, say $\{V_y:y\in F\}$, where $F$ is some finite subset of $K$. Let $$U=\bigcap_{x\in F}U_x\;;$$ clearly $U$ is an open nbhd of $x$ disjoint from $K$. Since $x$ was an arbitrary point of $\mathbb{X}\setminus K$, $K$ must be closed.

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I thought it would be very laborious. Thanks, Brian. –  Elias Nov 18 '11 at 13:05
    
But "$ \bigcap_{y\in F} U_y$" would not be out instead of "\bigcap_{x\in F} U_x$"? –  Elias Nov 18 '11 at 13:12
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@user19266: They say exactly the same thing: $x$ (in mine) and $y$ (in yours) are dummy variables. You could just as well say $\bigcap_{\xi\in F}U_\xi$ if you really wanted to. –  Brian M. Scott Nov 18 '11 at 13:30

A "sequential" proof: Let $x_\alpha \in K$ be a net with limit $x \in \mathbb{X}$. By compactness of $K$, there exists a subnet $x_{\alpha_{\beta}}$ which converges in $K$. Let $y \in K$ denote its limit. Since it's a subnet of $x_\alpha$, it follows that also $x_\alpha \to y$. Since $\mathbb{X}$ is Hausdorff, nets have unique limits, so $y=x$ and in particular $x \in K$.

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This proof is interesting Mark. I like it. –  Elias Nov 18 '11 at 13:23

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