Monotonically normal implies collectionwise 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., monotonically normal space) must also satisfy the second topological space property (i.e., collectionwise normal space)
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Statement

Any monotonically normal space is a collectionwise normal space.

Definitions used

Monotonically normal space

Further information: monotonically normal space

A topological space $X$ is termed a monotonically normal space if there exists an operator $G$ from pairs of disjoint closed subsets to open subsets such that

• For all disjoint closed subsets $A,B$, $G(A,B)$ is an open subset whose closure is disjoint from $B$
• For closed subsets $A,A^{\prime },B,B^{\prime }$, if $A\subseteq A^{\prime }$ and $B^{\prime }\subseteq B$, with $A,B$ disjoint and $A^{\prime },B^{\prime }$ disjoint, we have $G(A,B)\subseteq G(A^{\prime },B^{\prime })$.

Such a $G$ is termed a monotone normality operator.

Collectionwise normal space

Further information: collectionwise normal space

A topological space $X$ is termed a collectionwise normal space if, given any discrete colleciton of closed subsets of $X$ (i.e., a collection of pairwise disjoint closed subsets such that the union of any subcollection is closed), there exist pairwise disjoint open subsets containing them.

Proof

Given: A monotonically normal space $X$ with a monotone normality operator $G$. A discrete collection of closed subsets $A_i,i\in I$.

To prove: There exist pairwise disjoint open subsets $U_i,i\in I$, such that $A_i\subseteq U_i$.

Proof: By the well-ordering principle, we can well-order $I$. Then, for any $\alpha \in I$, let $P_{\alpha }={\cup }_{i<\alpha }{A}_i$, $Q_{\alpha }={\cup }_{i>\alpha }{A}_i$. Define:

$U_{\alpha }:=G(P_{\alpha }\cup A_{\alpha },Q_{\alpha })\setminus \overline{G(P_{\alpha },A_{\alpha }\cup Q_{\alpha })}$

Clearly, the $U_{\alpha }$s are all open, and they are pairwise disjoint, and $U_{\alpha }$ contains $A_{\alpha }$.