Paracompact Hausdorff implies normal

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

Related facts

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_ys, which are disjoint from the corresponding U_ys. 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_as 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_bs, b \in B. Then, C and D are the required disjoint open subsets: This follows from the previous step.