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In this page of Dan Ma's topology blog about collectionwise normal spaces he proves this result:

Proposition: Any (Hausdorff) normal and countably compact space is collectionwise normal.

The blog assumes that spaces are Hausdorff (or T1 here), but I am interested to know what happens without the T1 assumption.

More specifically, the proposition follows from the following:

Lemma: If $X$ is a T1 space, the following are equivalent:

  • (A) $X$ has countable extent;
  • (B) All discrete families of nonempty closed subsets of X are at most countable.

Here the extent of a space $X$ is the supremum of the cardinalities of closed discrete subsets of $X$. A discrete family of subsets of $X$ is a family such that each point of $X$ has a nbhd meeting at most one set in the family.

The proof of the lemma is not difficult (see below for completeness). In fact, (B) implies (A) always, even without the T1 assumption.

Question: Does (A) imply (B) without the T1 assumption?

I can't prove it, but I don't see a counterexample either. If it were true, we could generally conclude that every normal limit point compact space is collectionwise normal (since limit compact spaces have countable extent). For T1 spaces it is not a generalization since limit compact is equivalent to countably compact in that case.

Proof of (A) implies (B) assuming T1: Let $\mathscr{F}$ be a discrete family of nonempty closed subsets. For each $F\in\mathscr{F}$ pick some $x_F\in F$. Then $\mathscr{G}=\{\{x_F\}:F\in\mathscr{F}\}$ is a discrete family of closed subsets (singletons). Hence $A=\cup\mathscr{G}=\{x_F:F\in\mathscr{F}\}$ is closed and discrete. Hence $A$ is at most countable and the same is true of $\mathscr{F}$.

Proof of (B) implies (A) without extra assumption: Let $A$ be a closed and discrete subset of $X$. For any $x\in A$, the singleton $\{x\}$ is closed in $A$ because $A$ is discrete, and $A$ is closed in $X$, therefore $\{x\}$ is closed in $X$. So the family $\mathscr{F}=\{\{x\}:x\in A\}$ is a discrete family of nonempty closed subsets of $X$. Hence $\mathscr{F}$ is at most countable and so is $A$.

PatrickR
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  • The edit clarifies. Thx. BTW all standard texts and papers include $T_1$ in the definition of collectionwise normal, so the question is almost never relevant. – Henno Brandsma Aug 22 '20 at 07:08

2 Answers2

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Brian has already shown that the lemma (A) implies (B) is false if one does not assume $T_1$.

Here is a proof that the result in the original title is true, bypassing the lemma entirely.

Proposition: (without assuming $T_1$) Any normal countably compact space $X$ is collectionwise normal.

Proof: Take a discrete family of closed subsets of $X$. Since $X$ is countably compact, the family must be finite. This is Lemma 2 in this answer. Now we have a finite number of pairwise disjoint nonempty closed subsets, and we can enclose each set in the family in a open set with the open sets pairwise disjoint, by normality of $X$.

PatrickR
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    It does appear to be correct. – Brian M. Scott Aug 23 '20 at 02:17
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    Thank you for checking. – PatrickR Aug 23 '20 at 02:32
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    I think (but check me on this) that $G \times \omega^\downarrow$, where $G$ is Bing's $G$ and $\omega^\downarrow$ is $\omega=\Bbb N$ in the down-topology with non-trivial open sets $U_n={m: m < n}, n \in \omega$ is limit point compact, normal and not collectionwise normal, showing that we cannot weaken countably compact, even for a $T_0$ space. The countably compact part is fine. – Henno Brandsma Aug 23 '20 at 06:19
  • @HennoBrandsma I can show that $X=G\times\omega^\downarrow$ is limit point compact and not collectionwise normal. For showing $X$ is normal, if one could show that the projection $\pi$ of $X$ onto $G$ is closed, it should be enough, because disjoint closed sets in $X$ must have disjoint projections due to the topology of $\omega^\downarrow$, and then the rest is easy. But not sure how to make it work at the non-isolated points of $G$. – PatrickR Aug 23 '20 at 07:45
  • @PatrickR If $A$ and $B$ are disjoint and closed in the product, consider stalks $A_x=({x} \times \omega^\downarrow)\cap A$ (and similarly for $B$). These are essentially closed and disjoint in $\omega^\downarrow$ where there are no disjoint closed sets except when one is empty. So for each $x \in G$ either $A$ or $B$ has points in the $x$-"stalk", but not both. Separate the $\pi_G[A]$ and $\pi_G[B]$ ($G$ is completely normal, so I don't worry about closedness, they're for sure separated) in $G$ and multiply by the whole space to separate $A$ and $B$. – Henno Brandsma Aug 23 '20 at 07:59
  • @HennoBrandsma Why are $\pi_G[A]$ and $\pi_G[B]$ separated? In particular, taking the case where $B$ is a single stalk over $x\in G$, how do we know $x$ is not in the closure of $\pi_G[A]$? It's easy to see if $x$ is an isolated point of $G$, but the case of $x$ non-isolated ($x\in F_P$ in Dan Ma's blog notation) is not so clear. (This particular case of $B$ in a singe stalk is enough to show $\pi$ is a closed map, which would allow to conclude the argument.) – PatrickR Aug 23 '20 at 08:15
  • @HennoBrandsma I now think $X$ is not normal, but please double check. (following Dan Ma's blog notation). Take $x=f_p\in F_P$ and $B={x} \times \omega^\downarrow$ (the full stalk). In $G$, take a shrinking sequence of nbhds of $x$: $U_1\supseteq U_2\supseteq\cdots$ (starting sufficiently close to $x$ to avoid any other point of $F_P$). For each $n$ pick an elements $g_n\in U_n\setminus U_{n+1}$ and let $A=\bigcup_n ({g_n}\times[n,\infty))$. I think that $A$ and $B$ are closed and disjoint, but cannot be separated by disjoint open sets. – PatrickR Aug 23 '20 at 09:19
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    I indeed think $\pi_G[A]$ and $\pi_G[B]$ might not be separated always. But $G \times I_2$ (indiscrete two-point space) is limit point compact, normal and not collectionwise normal. But a $T_0$ example would have been nicer, granted. But like I said, normally $T_1$ is assumed throughout; there is not much interest in these properties (countable compactness, collectionwise normality etc) without at least $T_1$ or Hausdorff anyway. – Henno Brandsma Aug 23 '20 at 15:10
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For $n\in\Bbb N$ let $U_n=\{k\in\Bbb N:k<n\}$, and let $Y$ be $\Bbb N$ with the topology

$$\{U_n:n\in\Bbb N\}\cup\{\Bbb N\}\,,$$

and let $D$ be an uncountable space with the discrete topology. Let $X=D\times Y$; clearly

$$\mathscr{F}=\big\{\{x\}\times Y:x\in D\big\}$$

is an uncountable discrete family of closed sets in $X$. Let $A\subseteq X$. If $|A\cap(\{x\}\times Y)|>1$ for some $x\in D$, then $A$ is not discrete, and if $|A\cap(\{x\}\times Y)|=1$ for some $x\in D$, then $A$ is not closed in $X$, so $X$ has no non-empty closed discrete subsets.

Brian M. Scott
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  • Nice! I was also looking for a counterexample.. – Henno Brandsma Aug 22 '20 at 07:27
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    Note to the OP: this example is limit point compact, normal, $T_0$ and also collectionwise normal, so it does not form a counterexample to your proposed "limit point compact + normal implies collectionwise normal". It's a "local counterexample" (à la Lakatos), it disproves one of the lemmata in the old proof. – Henno Brandsma Aug 22 '20 at 08:09
  • Nice example. Expanding on $Y$ above, any space where no singleton ${x}$ is closed is automatically limit point compact. And the same for the product of any space with such a $Y$. So starting with $X$ = Bing's example G (https://dantopology.wordpress.com/2012/11/18/bings-example-g/) and taking the product with $Y$ = the two point indiscrete space, I think it would be an example of normal limit point compact and not collectionwise normal space. Would you agree? – PatrickR Aug 23 '20 at 00:30
  • @HennoBrandsma adding you, in case you don't see this. Bing's G product with $Y$ = two point indiscrete seems to be a countereample, but it's not $T_0$. Not sure if the product with Brian's $Y$ would also work, which would give a $T_0$ example, but not worth worrying about. – PatrickR Aug 23 '20 at 00:33
  • In any case, I don't think Bing's G is countably compact (and same when taking the product with $Y$), so still not invalidating the question in the title. – PatrickR Aug 23 '20 at 00:36
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    @PatrickR: Bing’s Example G is not countably compact: the non-isolated points are an uncountable closed discrete set. – Brian M. Scott Aug 23 '20 at 02:12