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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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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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    $\begingroup$ I thought it would be very laborious. Thanks, Brian. $\endgroup$ – MathOverview Nov 18 '11 at 13:05
  • $\begingroup$ But "$ \bigcap_{y\in F} U_y$" would not be out instead of "\bigcap_{x\in F} U_x$"? $\endgroup$ – MathOverview Nov 18 '11 at 13:12
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    $\begingroup$ @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. $\endgroup$ – Brian M. Scott Nov 18 '11 at 13:30
  • $\begingroup$ @JesterTran: It’s defined in the first sentence of the answer. For a given $y\in\Bbb X\setminus K$ $U_y$ is just a name for a specific open set satisfying certain conditions: $x\in U_y$, and there is an open nbhd $y$, which I’m calling $V_y$, that is disjoint from $U_y$. $\endgroup$ – Brian M. Scott Aug 7 '16 at 7:21
  • $\begingroup$ According to this answer, $F$ is a finite subset of K. $x \in X - K$. And $U_x$ is a neighborhood of $x$. Wouldn't that mean that there is no $x$ in $F$? How is $\bigcap_{x\in F} U_x$ not an empty set? $\endgroup$ – nasdaq Jun 25 '18 at 2:22
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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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  • $\begingroup$ This proof is interesting Mark. I like it. $\endgroup$ – MathOverview Nov 18 '11 at 13:23

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