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.

I appreciate your thoughts on my proof. Thanks in advance.

  • $\begingroup$ 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 $\overline x$”? $\endgroup$ Jul 21 '18 at 16:28
  • $\begingroup$ @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. $\endgroup$
    – Cornman
    Jul 21 '18 at 16:35
  • $\begingroup$ @Sou I assumed, that $\overline{x}\notin K$ since I make a distinction of cases. $\endgroup$
    – Cornman
    Jul 21 '18 at 16:37
  • $\begingroup$ 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 $U_{x_l}$. $\endgroup$ Jul 21 '18 at 16:57
  • $\begingroup$ @Sou Do you see a way to fix my proof, or do you think that my attempt is not usefull? $\endgroup$
    – Cornman
    Jul 21 '18 at 17:01

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 $


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