# Prove that every compact Hausdorff space is normal

A topological space $X$ is normal if every singleton is closed and for any disjoint closed subsets $A, B \subseteq X$ there are disjoint open subsets $U,V$ with $A \subseteq U$, $B \subseteq V$.

Prove that if $X$ is a compact Hausdorff space then $X$ is normal.

## Answer

We first prove the following Lemma.

Lemma. Let $X$ be a Hausdorff space and $A \subset X$ be a compact subset with $b \notin A$. Then there exist open sets $U$ and $V$ such that $A \subset U$, $b \in V$, and $U \cap V=\emptyset$.

Proof. Since the space is Hausdorff, for every $a \in A$ there exist open sets $U_a$ and $V_a$ such that $a \in U_a$, $b \in V_a$ and $U_a \cap V_a=\emptyset$. Note that $\{U_a\}_{a\in A}$ is an open cover for the compact set $A$. Hence there exists a finite sub-cover $U_{a_1}, U_{a_2},...,U_{a_n}$ for $A$, i.e. \[ A \subset \bigcup\limits_{i=1}^n U_{a_i}. \] Let $V=\bigcap\limits_{i=1}^n V_{a_i}$ and $U=\bigcup\limits_{i=1}^n U_{a_i}$, then $A \subseteq U$, $b \in V$ and $U \cap V=\emptyset$. Indeed if $c \in U\cap V$, then $c \in U_{a_j}$ for some $j \in \{1,2,...,n\}$ and $c \in V_{a_i}$ for all $i \in \{1,2,...,n\}$. Hence $c \in U_{a_j}\cap V_{a_j}$ which is a contradiction.

We also need the following proposition.

Proposition A. Every closed subset of a compact space is compact.

Proof. Suppose $A \subset X$ be closed and let $ \{U_i\}_{i \in I} $ be an open cover for $A$. Since $X\setminus A$ is open, $ X \subset (\bigcup\limits _{i\in I} U_i) \cup (X \setminus A)$ is an open cover for the compact space $X$. Therefore there is a finite sub-cover and $X \subset (\bigcup\limits _{k=1}^n U_{i_k}) \cup (X\setminus A)$ for some $n\in \N$. Then $A\subseteq \bigcup \limits_{k=1}^n U_{i_k}$ and hence $A$ is compact.

Main Proof of the problem. First note that by Proposition A the closed subsets $A,B$ are also compact. By above lemma, for every $b \in B$ there exists open sets $U_b$ and $V_b$ such that $b\in V_b$, $A \subset U_b$, and $V_b \cap U_b=\emptyset $. Note that \[ B \subset \bigcup \limits_{b \in B}V_b \] is an open cover for $B$. Since $B$ is compact, there is a finite subcover \[ B \subset \bigcup \limits _{i=1}^m V_{b_i} \text{for some} m \in \N. \] Now define $U=\bigcap \limits_{i=1}^m U_{b_i}$ and $V=\bigcup \limits_{i=1}^m V_{b_i}$. Then $A \subseteq U$, $B \subseteq V$ and $U \cap V=\emptyset$. Indeed if $c \in U\cap V$, then there exist $j \in \{1,2,...,m\}$ with $c \in U_{b_j}\cap V_{b_j}=\emptyset$, which is a contradiction. Finally we show that every singleton $\{x\}$ is a closed set. Since $X$ is Hausdorff, for every $y\neq x$ there exists an open set $V_y$ such that $y\in V_y$ and $x \not\in V_y$. Hence \[\{x\}^c = \bigcup_{y \neq x} V_y.\] Thus $\{x\}^c$ is open and hence $\{x\}$ is closed.

- answered
- 413 views
- $50.00

### Related Questions

- Prove Holder-continuity for $\mu_\lambda (x) = \sum\limits_{n=1}^\infty \frac{ \cos(2^n x)}{2^{n \lambda} }$
- A lower bound on infinite sum of exponential functions (corrected version)
- Assume there is no $x ∈ R$ such that $f(x) = f'(x) = 0$. Show that $$S =\{x: 0≤x≤1,f(x)=0\}$$ is finite.
- real analysis
- Uniform convergence of functions
- Banach fixed-point theorem and the map $Tf(x)=\int_0^x f(s)ds $ on $C[0,1]$
- Measure Theory and the Hahn Decomposition Theorem
- The space of continuous functions is a normed vector space