Even covers and collectionwise normal spaces

Question

We call $X$ strongly collectionwise normal if the set $\mathcal{U}_\Delta$ of all neighbourhoods of the diagonal $\Delta_X$ of $X\times X$ is a uniformity. This is equivalent to the property that for all $U\in\mathcal{U}_\Delta$ there exists $V\in\mathcal{U}_\Delta$ with $V\circ V\subseteq U$.

In the article Even covers and collectionwise normal spaces by Shapiro and Smith the following theorem is proven:

Theorem 4.8. If $X$ is a completely regular space then the following are equivalent:

1. $X$ is strongly collectionwise normal

2. Every even open cover of $X$ is normal

Here a cover $\mathcal{V}$ of $X$ is even if there is $W\in\mathcal{U}_\Delta$ such that $\{W(x) : x\in X\}$ refines $\mathcal{V}$, where $W(x) = \{y : (x, y)\in W\}$.

An open cover $\mathcal{V}$ is normal if there exists a sequence $(\mathcal{V}_n)$ of open covers such that $\mathcal{V}_{n+1}$ is a star-refinement of $\mathcal{V}_n$ and $\mathcal{V}_1 = \mathcal{V}$, and what authors of the article show to be equivalent to the property that $\mathcal{V}$ is $\aleph_0$-even, that is there exists a sequence $W_n\in\mathcal{U}_\Delta$ such that $W_{n+1}\circ W_{n+1}\subseteq W_n$ and $\{W_1(x) : x\in X\}$ is a refinement of $\mathcal{V}$.

The issue I have is with the proof that $(2)$ implies $(1)$. It is claimed that if $W\in\mathcal{U}_\Delta$ then $\mathcal{W} = \{W(x) : x\in X\}$ is an even cover, and that implies that there is $U\in\mathcal{U}_\Delta$ such that $U\circ U\subseteq W$. However, I don't see how it would imply this. Clearly there is a sequence $W_n\in\mathcal{U}_\Delta$ such that $W_{n+1}\circ W_{n+1}\subseteq W_n$ and $\{W_1(x) : x\in X\}$ refines $\mathcal{V}$. However, I don't think we can guarantee that there is $n$ with $W_n(x)\subseteq W(x)$ for all $x\in X$, which would imply that $W_n\subseteq W$, and hence allow us to take $U = W_{n+1}$.

So, can this proof somehow be fixed? And, what I am interested in the most, is if we can nonetheless show that a fully normal space is strongly collectionwise normal, even if this particular theorem is wrong.

Answer 1

As implied by Tyrone, the claim that a fully normal space is strongly collectionwise normal is true. See, for example, the result in H. J. Cohen's paper Sur un problème de M. Dieudonné (translated into English and paraphrased).

Proposition [1, Thm. 2]. A topological space $X$ is strongly collectionwise normal if and only if for every open covering $\mathscr{U}$, there exists a refinement $\mathscr{V} \leq \mathscr{U}$ with the following property: if $x, y \in X$ are contained in a same set $V \in \mathscr{V}$, or the union of two such sets $V_1, V_2 \in \mathscr{V}$ with a common point $z \in V_1 \cap V_2$, then they are both contained in some set $U \in \mathscr{U}$.

A space with this property is said to be almost 2-fully normal, as defined by M. Mansfield. It is immediate from the definitions that a space which is fully normal is also almost 2-fully normal.

However, the statement and proof of Theorem 4.8 are incorrect; there is no way to deduce that the self-composite of any $W_n$ must lie in $W$. In a paper by H. Künzi and P. Fletcher it is noted that the assertion of Theorem 4.8. must be false due to an earlier counterexample¹ of a collectionwise normal, evenly paracompact space which is not almost 2-fully normal. This was initially given by R. H. Bing as a counterexample to a different problem which can also be found in the paper of H. J. Cohen.

References

1. Cohen, Herman J., Sur une problème de M. Dieudonné, C. R. Acad. Sci., Paris 234, 290-292 (1952). ZBL0046.16403.

2. Künzi, Hans-Peter; Fletcher, Peter, Even covering properties and somewhat normal spaces, Can. Math. Bull. 29, 154-159 (1986). ZBL0593.54020.

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¹Note that condition $(2)$ of Theorem 4.8 is equivalent to $X$ being somewhat normal (as defined in [2]), and that a normal space is somewhat normal if and only if it is evenly paracompact [2, Proposition 2.2].