# Compact Hausdorff spaces are normal

I want to show that compact Hausdroff spaces are normal.

To be honest, I have just learned the definition of normal, and it is a past exam question, so I want to learn how to prove this:

I believe from reading the definition, being a normal space means for every two disjoint closed sets of $X$ we have two disjoint open sets of $X$.

So as a Hausdorff space, we know that $\forall x_1,x_2\in X,\exists B_1,B_2\in {\Large{\tau}}_X|x_1\in B_1, x_2\in B_2$ and $B_1\cap B_2=\emptyset$

Now compactness on this space, means we also have for all open covers of $X$ we have a finite subcover of $X$.

Now if we take all of these disjoint neighborhoods given by the Hausdorff condition, we have a cover of all elements, I am not sure how to think of this in terms of openness, closedness.

How does one prove this?

We'll do this in two parts.

$$\textbf{Lemma.}$$ Let $$T$$ be Hausdorff. Suppose $$x\in T$$ and $$Y \subset T$$ be compact and let $$x \notin Y$$. Then there are open sets $$U$$, $$V$$ separating x and Y, i.e., $$x\in U$$ and $$Y\subset V$$ such that $$U\cap V=\varnothing$$.

Proof of the lemma: Since the space is Hausdorff, for every $$y\in Y$$ there is a neighborhood $$U_y$$ of $$y$$ and a neighborhood $$V_y$$ of $$x$$ such that $$U_y\cap V_y=\varnothing$$. We have that $$\bigcup_{y\in Y} V_y$$ is an open cover of Y and since compact, there exists a finite subcover $$V_{y_1},V_{y_2},...V_{y_n}$$ for Y. Now let \begin{align*} U= \bigcap_{j=1}^n U_{y_j}, \qquad V= \bigcup_{j=1}^n V_{y_j}. \end{align*} Then $$x\in U$$ and $$Y\subset V$$ where $$U\cap V =\varnothing$$. To show that they really are disjoint assume there is $$z\in U\cap V$$. Then there exists $$z\in V_{y_j}$$ for some $$j$$ and $$z\in U_{y_j}$$ for all $$j$$. But $$U_{y_j}\cap V_{y_j} =\varnothing$$.

$$\textbf{Main proof:}$$ Let $$T$$ be a compact Hausdorff topological space and let $$X$$ and $$Y$$ be disjoint closed sets in $$T$$. By the lemma, for any $$y\in Y$$ there exists a neighborhood $$U_y\ni y$$ and an open set $$O_y$$ containing $$X$$ such that $$U_y\cap O_y =\varnothing$$.

Since $$Y$$ is a closed subset of a compact set it is itself compact, and therefore the cover $$\left\lbrace U_y\right\rbrace _{y\in Y}$$ of $$Y$$ has a finite subcover $$U_{y_1},U_{y_2},\cdots , U_{y_n}$$. The open sets

\begin{align*} O^1=\bigcap_{j=1}^n O_{y_j} \supset X, \qquad O^2=\bigcup_{j=1}^n U_{y_j} \supset Y, \end{align*} are then disjoint open sets containing $$X$$ and $$Y$$, proving that every compact Hausdorff space is normal.

• Could it be that you need to write $O^1=\bigcap_{j=1}^n O_{y_j} \supset X$ instead of what you wrote, becaus otherwise i don't see how $O^1 \cap O^2$ can be disjoint – DeltaChief Jun 19 '16 at 10:05
• You are right @DeltaChief. I have fixed it! – D. ex-Machina Jul 6 '16 at 9:26
• you probably meant $Y \subset V$, not $Y \in V$ – user285001 Jul 15 '18 at 0:56

The standard approach is to first show that such a topological space is regular. The argument to show that the space is normal is an easy adjustment of this argument, so I will present this proof instead.

Let $X$ be a compact Hausdorff space and let $F\subset X$ be a closed set. Let $y\in X\setminus F$. We need to construct open sets $U$ and $V$ with $F\subseteq U$ and $y\in V$ such that $U\cap V=\emptyset$. Since $X$ is Hausdorff, for each $x\in F$ there exists neighbourhoods $U_x$ of $x$ and $V_x$ of $y$ such that $x\in U_x$ and $y\in V_x$ and $U_x\cap V_x=\emptyset$. Let $\mathcal F$ be the collection $$\mathcal F=\{U_x:x\in F\}.$$ Then $\mathcal F$ is an open cover of $F$ and $F$ is compact (why?) so there exists finitely many points $x_1,\dotsc,x_n\in F$ such that $F\subseteq \bigcup_{j=1}^n U_{x_j}$. Set $U=\bigcup_{j=1}^n U_{x_j}$ and let $V=\bigcap_{j=1}^n V_{x_j}$. Now $U$ and $V$ are open sets and $F\subseteq U$ and $y\in V$. Moreover, since $U_x\cap V_x=\emptyset$ for each $x\in F$, it follows that $U\cap V=\emptyset$. This shows that $X$ is regular.

Outline: Start with proving regularity i.e. for $x\in X$ and a closed subset $A\subset X$ not containing $x$, there are disjoint open subsets $U,V\subset X$, such that $x\in U$ and $A\subset V$. For this, note that $A$ is compact, as a closed subset of a compact space. Since $X$ is Hausdorff, for every $a\in A$ there are disjoint open subset $U_a,V_a$, with $x\in U_a$ and $a\in V_a$. The $V_a$'s cover $A$, and a simple argument shows regularity.

Then, using regularity, a process very similar to the one in the above paragraph shows normality.

$$(X,\tau ),$$ compact space; $$\left( X,\tau \right) ,$$ Hausdorff; $$A\in \mathcal{C}\left( X,\tau \right) ;$$ $$B\in \mathcal{C}\left( X,\tau \right) ,$$ $$A\cap B=\emptyset,$$ $$x\in A$$ and $$y\in B.$$

$$\left.\begin{array}{r}\left( x\in A\right) \left( y\in B\right) \\ \\ A\cap B=\emptyset \end{array}\right\}\Rightarrow \!\!\!\!\!\begin{array}{c}\mbox{} \\ \mbox{} \\ \left.\begin{array}{c} x\neq y \\ \mbox{} \\ {\left( X,\tau \right) ,\text{ }T_{2}}\end{array}\right\}\Rightarrow \!\!\!\!\!\end{array}$$

$$\left.\begin{array}{r}\Rightarrow \left( \exists U_{x}\in \mathcal{U}\left( x\right) \right) \left( \exists V_{y}\in \mathcal{U}\left( y\right) \right) \left( U_{x}\cap V_{y}=\emptyset \right) \\ \\ \left( \mathcal{A}:=\left\{ U_{x}|x\in A\right\} \right) \left( \mathcal{B}:=\left\{ V_{y}|y\in B\right\} \right) \end{array}\right\} \Rightarrow$$

$$\left.\begin{array}{r}\Rightarrow \left( \mathcal{A}\subseteq \tau \right) \left( A\subseteq \cup \mathcal{A}\right)\left( \mathcal{B}\subseteq \tau \right) \left( B\subseteq \cup \mathcal{B}\right) \\ \\ \left( (X,\tau )\text{ is compact space}\right) \left( A,B\in \mathcal{C}\left( X,\tau \right) \right) \Rightarrow \left( A\text{ is }\tau \text{-compact}\right)\left( B\text{ is }\tau \text{-compact}\right)\end{array}\right\} \Rightarrow$$

$$\left.\begin{array}{r}\Rightarrow \left( \exists \mathcal{A}^{\ast }\subseteq \mathcal{A}\right) \left( \left\vert \mathcal{A}^{\ast} \right\vert <\aleph _{0}\right) \left( A\subseteq \cup \mathcal{A}^{\ast}\right) \left( \exists \mathcal{B}^{\ast }\subseteq \mathcal{B}\right) \left( \left\vert \mathcal{B}^{\ast }\right\vert <\aleph _{0}\right) \left( B\subseteq \cup \mathcal{B}^{\ast}\right)\\ \\ \left( U:=\cup \mathcal{A}^{\ast }\right) (V:=\cup \mathcal{B}^{\ast })\end{array}\right\} \Rightarrow$$

$$\left.\begin{array}{c}\Rightarrow \left( U\in \mathcal{U}(A)\right)\left( V\in \mathcal{U}(B)\right)\left( U\cap V=\emptyset \right).\end{array}\right.$$

NOTE : $$\mathcal{U}(A):=\{U|(U\in \tau)(A\subseteq U)\}$$