Compact Hausdorff implies normal

This article gives the statement and possibly, proof, of an implication relation between two topological space properties. That is, it states that every topological space satisfying the first topological space property (i.e., compact Hausdorff space) must also satisfy the second topological space property (i.e., normal space)
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Statement

Any compact Hausdorff space (a topological space that is both a compact and Hausdorff) is normal.

Related facts

Intermediate properties

• Paracompact Hausdorff space: Further information: paracompact Hausdorff implies normal

Other related facts

• any closed subset of a compact space is compact
• any compact subset of a Hausdorff space is closed: The proof of this uses a very similar argument.
• Any locally compact Hausdorff space is completely regular

Facts used

1. Compactness is weakly hereditary: Any closed subset of a compact space is compact.
2. A union of arbitrarily many open subsets is open.
3. An intersection of finitely many open subsets is open.

Proof

Suppose $X$ is a compact Hausdorff space. We need to show that $X$ is normal. We will proceed in two steps: we will first show that $X$ is a regular space, and then show that $X$ is normal.

Proof of regularity

Given: $x \in X$ is a point and $A$ is a closed subset of $X$ not containing $x$.

To find: Disjoint open subsets containing $A$ and $x$ respectively.

Solution:

Step no.	Assertion/construction	Given data used	Facts used	Previous steps used	Explanation
1	For every $y \in A$, construct disjoint open subsets $V_y \ni x$ and $U_y \ni y$. In particular, the union of the $U_y$s contains $A$	$X$ is Hausdorff	--	--	--
2	$A$ is compact	$X$ is compact, $A$ is closed in $X$	Fact (1)
3	The $U_y \cap A$ have a finite subcover (as an open cover of $A$). Thus, there is a finite set $y_1,y_2,\ldots, y_r$ of points such that the set $U = \bigcup_{i=1}^n U_{y_i}$ contains $A$.	--	--	Step (2)
4	$U = \bigcup_{i=1}^r U_{y_i}$ is an open set	--	Fact (2)	Step (3)
5	$V = \bigcap_{i=1}^r V_{y_i}$ is an open set containing $x$	--	Fact (3)	Steps (1),(3)	[SHOW MORE] The crucial point is that there are finitely many open sets ($n$ of them) being intersected, and each contains $x$.
6	$U$ and $V$ are disjoint	--	--	Step (1)	[SHOW MORE] Each $U_{y_i}$ is disjoint from the corresponding $V_{y_i}$. For $z \in U$, we must have $z \in U_{y_i}$ for some $i$. But then, $z \notin V_{y_i}$, hence $z \notin V$. Thus, $U \cap V$ is empty.
7	The sets $U$ and $V$ are disjoint open subsets containing $A$ and $x$ respectively.	--	--	Steps (3)--(6)

Proof of normality

Given: Disjoint closed subsets $A$ and $B$ of $X$.

To find: Disjoint open subsets $U$ and $V$ of $X$ such that $A \subseteq U$ and $B \subseteq V$.

Solution:

Step no.	Assertion/construction	Given data used	Facts used	Previous steps used	Explanation
1	For every point $x \in B$, we can define disjoint open subsets $U_x \supset A$ and $V_x \ni x$. In particular, $V_x$ form an open cover of $B$.	--	--	$X$ is regular, by the previous half of the proof.
2	$B$ is compact	$X$ is compact, $B$ is closed in $X$	Fact (1)
3	The $V_x \cap B$ have a finite subcover, say corresponding to points $x_1, x_2, \dots, x_n$. Thus, the union $V = \bigcup_{i=1}^n V_{x_i}$ contains $B$	--	--	Step (2)
4	$V = \bigcup_{i=1}^n V_{x_i}$ is an open subset of $X$	--	Fact (2)	Step (3)
5	$U = \bigcap_{i=1}^n U_{x_i}$ is an open subset of $X$	--	Fact (3)	Step (3)
6	$U$ and $V$ are disjoint	--	--	Step (1)	[SHOW MORE] Each $U_{x_i}$ is disjoint from the corresponding $V_{x_i}$. For $z \in V$, we must have $z \in V_{x_i}$ for some $i$. But then, $z \notin U_{x_i}$, hence $z \notin U$. Thus, $U \cap V$ is empty.
7	$U$ and $V$ are disjoint open subsets of $X$ containing $A$ and $B$ respectively.	--	--	Steps (3)--(6)

References

Textbook references

• Topology (2nd edition) by James R. Munkres, Page 202, Theorem 32.3, ^{More info}
• Lecture Notes on Elementary Topology and Geometry (Undergraduate Texts in Mathematics) by I. M. Singer and J. A. Thorpe, Page 29, ^{More info}