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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?

Steven Clontz
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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.

Steven Clontz
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    Every $V\in\mathcal V$ intersects at most one set in $\mathcal H$. But that does not seem to be enough to derive a contradiction in the last paragraph if $H\ne L$. Shouldn't we use that $\mathcal V$ is a star refinement of $\mathcal U$? So $V_L$ cannot meet $V_H$ if $H\ne L$, otherwise it would be contained in $\operatorname{st}(V_H,\mathcal V)$, which is itself contained in some $U_y$ that only meets $H$ and not $L$. – PatrickR Apr 09 '23 at 17:33
  • I think you may be right; I'll make an appropriate edit. – Steven Clontz Apr 09 '23 at 19:46
  • Corrected, thanks! – Steven Clontz Apr 09 '23 at 20:15
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We call a space $X$ strongly collectionwise normal, if for each neighbourhood $U$ of the diagonal, there exists neighbourhood $V$ of the diagonal with $V\circ V\subseteq U$.

Strongly collectionwise normal implies collectionwise normal as shown here (the reader should notice that the assumption of complete regularity is not needed for the proof).

We can show that fully normal implies strongly collectionwise normal. The proof is based on Some generalizations of full normality by M. Mansfield.

If $U$ is a neighbourhood of the diagonal, then for each $x\in X$ take open neighbourhood $W_x$ of $x$ such that $W_x\times W_x\subseteq U$. Consider the cover $\mathcal{W} = \{W_x : x\in X\}$ of $X$. Let $\mathcal{V}$ be an open star-refinement of $\mathcal{W}$, and define $V = \bigcup\{A\times A : A\in\mathcal{V}\}$. If $(x, z)\in V\circ V$ then there exists $y$ with $(x, y)\in V, (y, z)\in V$. So there exist $A_1, A_2\in\mathcal{V}$ such that $x, y\in A_1$ and $y, z\in A_2$. Then $\text{St}(y, \mathcal{V})\subseteq W_t$ for some $t\in X$ since $\mathcal{V}$ being a star-refinement of $\mathcal{W}$ implies its a barycentric refinement of $\mathcal{W}$. So $x, z\in W_t$ and so $(x, z)\in W_t\times W_t\subseteq U$. This shows that $V\circ V\subseteq U$.

Jakobian
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