# How to prove that a compact set in a Hausdorff topological space is closed?

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.

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.
• But "$\bigcap_{y\in F} U_y$" would not be out instead of "\bigcap_{x\in F} U_x$"? – MathOverview Nov 18 '11 at 13:12 • @user19266: They say exactly the same thing:$x$(in mine) and$y$(in yours) are dummy variables. You could just as well say$\bigcap_{\xi\in F}U_\xi$if you really wanted to. – Brian M. Scott Nov 18 '11 at 13:30 • @JesterTran: It’s defined in the first sentence of the answer. For a given$y\in\Bbb X\setminus KU_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$. – Brian M. Scott Aug 7 '16 at 7:21 • According to this answer,$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$not an empty set? – nasdaq Jun 25 '18 at 2:22 A "sequential" proof: Let$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\$.