Compact Hausdorff spaces are normal

Question

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:

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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 $X$.

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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 $X$.

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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.

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How does one prove this?

Answer 1

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.

Answer 2

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.

Answer 3

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 $V_y$ of $y$ and a neighborhood $U_y$ 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},\dots,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 $U_{y_1},U_{y_2},\dots , U_{y_n}$. 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.

Answer 4

I will use the following proposition:

In a Hausdorff space, it is possible to separate a compact subset K and a point p not in K by disjoint open sets.

Assume $S_1,$ and $S_2$ are closed so they are compact. Apply proposition above to $S_1$ and $x \in S_2$ to get open sets:

$V_x,U_x$ such that $$S_1 \subset V_x \ \ \ and \ \ \ x \in U_x$$

Therefore, $$S_1 \subset \bigcup_{x \in S_2} U_x$$ By compactness, we have $S_1 \subset \bigcup_{i = 1}^n U_{x_i} = U$. Set $V = \bigcap_{i = 1}^n V_{x_i}$, then it follows:

$$V \cap U = \emptyset$$ $$S_1 \subset V \ \ \ and \ \ \ S_2 \subset U.$$