Paracompact Hausdorff implies normal

Statement
Any paracompact Hausdorff space (i.e., a space that is both paracompact and Hausdorff) is a normal space.

Related facts

 * Compact Hausdorff implies normal
 * Paracompactness is weakly hereditary: Any closed subspace of a paracompact space is paracompact.

Proof
Given: A paracompact Hausdorff space $$X$$.

To prove: $$X$$ is a normal space.

Proof: We first prove that $$X$$ is a regular space, and then prove that $$X$$ is normal.

Proof of regularity
To prove: If $$x \in X$$ and $$A$$ is a closed set not containing $$x$$, there exist open subsets $$U,V \subseteq X$$ such that $$x \in U, A \subseteq V$$, and $$U \cap V$$ is empty.

Proof:


 * 1) By Hausdorffness, we can define, for every $$y \in A$$, open subsets $$U_y, V_y$$ such that $$x \in U_y, y \in V_y$$, and $$U_y \cap V_y$$ is empty.
 * 2) The open subsets $$V_y, y \in A$$, cover $$A$$. In other words, $$A \subseteq \bigcup_{y \in A} V_y$$.
 * 3) The sets $$V_y$$ and $$X \setminus A$$ form an open cover of $$X$$. Thus, by paracompactness of $$X$$, there is a locally finite open refinement. Throwing out from this any open subset not intersecting $$A$$, we still get a locally finite collection $$\mathcal{P}$$ of open subsets, each contained in some $$V_y$$, that cover $$A$$.
 * 4) There exists an open set $$W$$ containing $$x$$ such that there are only finitely many members of $$\mathcal{P}$$ that intersect $$W$$: This follows from the definition of local finiteness.
 * 5) Let $$T$$ be a finite subset of $$A$$ that contains, for each of this finite list of members of $$\mathcal{P}$$, a point $$y$$ such that that member is contained in $$V_y$$.
 * 6) Define $$U = W \cap \bigcap_{y \in T} U_y$$ and $$V$$ to be the union of all the members of $$\mathcal{P}$$. Then, $$x \in U, A \subseteq V$$, and $$U$$ and $$V$$ are disjoint: For this, note that all the members of $$\mathcal{P}$$ that intersect $$W$$ are contained in $$V_y$$s, which are disjoint from the corresponding $$U_y$$s. So, $$U$$ is disjoint from $$V$$. Finally, note that $$U$$ is open since it is an intersection of finitely many open subsets, and $$V$$ is open since it is a union of open subsets.

Proof of normality
To prove: If $$A,B \subseteq X$$ are disjoint closed subsets, there exist open sets $$C,D$$ of $$X$$ containing $$A$$ and $$B$$ respectively such that $$C$$ and $$D$$ are disjoint.

Proof:


 * 1) For every $$a \in A$$, there exist open sets $$U_a \ni a, V_a$$ containing $$B$$, such that $$U_a$$ and $$V_a$$ are disjoint. This follows from regularity.
 * 2) The $$U_a$$s form a collection of open subsets of $$X$$ covering $$A$$. Along with $$X \setminus A$$, these form an open cover of $$X$$. This has a locally finite open refinement. Throwing out from this any open subset not intersecting $$A$$, we still get a locally finite collection $$\mathcal{Q}$$ of open subsets, each contained in some $$U_a$$, that cover $$A$$. Let $$C$$ be the union of all members of $$\mathcal{Q}$$.
 * 3) For any $$b \in B$$, there exists an open subset $$D_b$$ around $$b$$ that does not intersect $$C$$: First, there exists an open subset $$W_b$$ around $$b$$ intersecting only finitely many members of $$\mathcal{Q}$$. Let $$T$$ be a finite subset of $$A$$ that contains, for each of this finite list of members of $$\mathcal{Q}$$, a point $$a$$ such that that member is contained in $$U_a$$. Then, $$D_b = W_b \cap \bigcap_T V_a$$ works.
 * 4) Let $$D$$ be the union of all $$D_b$$s, $$b \in B$$. Then, $$C$$ and $$D$$ are the required disjoint open subsets: This follows from the previous step.