Fully normal implies collectionwise normal?

Question

T214 of pi-Base tracks that fully $T_4$ spaces are collectionwise normal, citing an assertion from a diagram in Steen/Seebach's Counterexamples.

Is fully normal sufficient? In any case, how can this be proven?

Answer 1

Yes. To show a space $X$ is collectionwise normal, let $\mathcal H$ be a collection of closed sets such that for each $x\in X$, there exists $U_x$ open that intersects at most one set in $\mathcal H$.

Then $\{U_x:x\in X\}$ is an open cover. When applying fully normal, we obtain an open refinement $\mathcal V$ covering $X$ such that for each $x\in X$ there exists $y\in X$ such that $V\subseteq U_y$ for all $V\in\mathcal V$ with $x\in V$. (That is, we obtain a "barycentric" refinement; note some authors use "star" refinements for full normality, but this is equivalent.)

For $H\in\mathcal H$, let $W_H=\bigcup\{V\in\mathcal V:V\cap H\not=\emptyset\}$. Given $x\in H$, $x\in V$ for some $V\in\mathcal V$, and thus $x\in V\cap H$ and $x\in W_H$.

Suppose $x\in W_H\cap W_L$. Since $x\in W_H$, pick $V_H\in \mathcal V$ where $V_H\cap H\not=\emptyset$ and $x\in V_H$. Since $x\in W_L$, pick $V_L\in \mathcal V$ where $V_L\cap L\not=\emptyset$ and $x\in V_L$. Now from full normality consider the $y\in X$ such that both $V_H\subseteq U_y$ and $V_L\subseteq U_y$. This $U_y$ intersects at most one set in $\mathcal H$. Since $U_y$ intersects both $H$ and $L$, we have that $H=L$, and $\{W_H:H\in\mathcal H\}$ is pairwise disjoint. This proves $X$ is collectionwise normal.