Paracompact 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., paracompact Hausdorff space) must also satisfy the second topological space property (i.e., normal space)
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Statement

Any paracompact Hausdorff space 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, $A\subseteq \bigcup _{y\in Y}V_{y}$.
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 $x<math>suchthatthereareonlyfinitelymanymembersof<math>{\mathcal {P}}$ that intersect $W$: This follows from the definition of local finiteness.
5. For each of the finitely many members of ${\mathcal {P}}$ that intersect $W$, let $T$ be a finite set that contains, for every member 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.