13
$\begingroup$

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?

$\endgroup$
15
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 $\endgroup$ – DeltaChief Jun 19 '16 at 10:05
  • $\begingroup$ You are right @DeltaChief. I have fixed it! $\endgroup$ – D. ex-Machina Jul 6 '16 at 9:26
  • $\begingroup$ you probably meant $Y \subset V$, not $Y \in V$ $\endgroup$ – user285001 Jul 15 '18 at 0:56
7
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.