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?
Best Answer
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.