# Compact subset in a Hausdorff space, is closed.

Let $$(X, \tau)$$ be a topological Hausdorff space and $$K\subseteq X$$ compact.

If $$(X,\tau)$$ is a Hausdorff space, then is $$K$$ closed.

For my proof I want to show, that $$K=K\cup\partial K=\overline{K}$$. $$K\subseteq\overline{K}$$ is trivial, so I need to show $$K\supseteq K\cup\partial K$$:

Let $$\overline{x}\in K\cup\partial K$$. If $$\overline{x}\in K$$, there is nothing to show. Hence, suppose $$\overline{x}\in\partial K$$. Then we have for every neighborhood $$U$$ of $$\overline{x}$$ that $$U\cap K\neq\emptyset$$ and $$U\cap (X\setminus K)\neq\emptyset.$$

Since $$\overline{x}\notin K$$, $$\overline{x}\in U\cap (X\setminus K)$$

For $$x\in K$$ let $$U_x$$ be a neighborhood of $$x$$. Then $$K\subseteq \bigcup_{x\in K} U_x.$$ Since $$K$$ is compact there exist $$x_0,\dotso, x_n$$ such that $$K\subseteq\bigcup_{i=0}^n U_{x_i}$$ open.

Thus $$U\cap\bigcup_{i=0}^n U_{x_i}\neq\emptyset$$ and hence there exists $$x_l\neq\overline{x}$$ such that $$U_{x_l}\cap U\neq\emptyset$$ for every neighborhood $$U$$ of $$\overline{x}$$, which contradicts to $$X$$ beeing a Hausdorff space. Therefore, $$\overline{x}\in K$$ and $$K$$ is indeed closed.

• How do you go from “every neighborhood $U$ of $\overline x$ intersects $\bigcup_{i=0}^nU_{x_i}$” to “it exists $x_l\neq\overline x$ such that $U_{x_l}\cap U\neq\emptyset$ for every neighborhood $U$ of $\overline x$”? Jul 21 '18 at 16:28
• @JoséCarlosSantos Yes, I think I misphrased it at that point. My thought is/was that $U$ can be choosen arbitrarly. And for every neighborhood $U$ you get to this point, where you can claim, that it has to intersect with some neighborhood in the set $\bigcup_{i=0}^n U_{x_i}$ which contradicts the property of the Hausdorff space. Jul 21 '18 at 16:35
• @Sou I assumed, that $\overline{x}\notin K$ since I make a distinction of cases. Jul 21 '18 at 16:37
• Your argument "$\exists x_l$ such that $U_{x_l} \cap U \neq \emptyset$ for every nbhd $U$ of $\bar{x}$" is not enough to imply $X$ is not hausdorff. You have to find a pair of points, in your case you want $x_l,\bar{x}$, such that for every neighbourhoods $U$ of $x_l$ $\textbf{ and }$ $V$ of $\bar{x}$, you have $U \cap V \neq \emptyset$. So you have to show not only every $U$ of $\bar{x}$, but also for every nbhd $U_{x_l}$. Jul 21 '18 at 16:57
To show $K^c$ is open: Let $x\in K^c$. For each $y\in K$, let $U_y$ and $V_y$ be such that $x\in U_y, y\in V_y$ and $U_y\cap V_y=\emptyset$, by Hausdorffness. Now $\{V_y\}_{y\in K}$ is an open cover of $K$. Take a finite subcover, $V_{y_1},\dots, V_{y_n}$, by compactness. Then consider $U=U_{y_1}\cap U_{y_2} \dots \cap U_{y_n}$. It is easy to see that $U$ is a nbhd of $x$ with $U\subset K^c$. $\square$