Monotonically normal implies collectionwise normal

Statement
Any monotonically normal space is a collectionwise normal space.

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 $$A,A',B,B'$$, if $$A \subseteq A'$$ and $$B' \subseteq B$$, with $$A,B$$ disjoint and $$A',B'$$ disjoint, we have $$G(A,B) \subseteq G(A',B')$$.

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

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:

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

We note the following:


 * 1) The $$U_\alpha$$s are all open: This is because $$U_\alpha$$ is the set difference between an open subset and a closed subset, hence it is the intersection of two open subsets, hence it is open.
 * 2) Each $$U_\alpha$$ contains the corresponding set $$A_\alpha$$: This is because the set $$G(P_\alpha \cup A_\alpha, Q_\alpha)$$ contains $$A_\alpha$$ by definition, whereas the closure of $$G(P_\alpha,A_\alpha \cup Q_\alpha)$$ is disjoint from $$A_\alpha$$ by definition.
 * 3) $$U_\alpha$$ does not intersect $$U_\beta$$ for $$\alpha < \beta$$: $$U_\beta$$ is disjoint from $$G(P_\beta,A_\beta \cup Q_\beta)$$, which in turn contains $$G(P_\alpha \cup A_\alpha, Q_\alpha)$$ by the monotonicity of $$G$$. Thus, $$U_\beta$$ and $$U_\alpha$$ are disjoint.