# Convergence properties in dense subsets of $\omega^*$

The space $$\omega^*$$, the remainder of the Cech-Stone compactification of the integers, fails to have all convergence-type properties known to me.

1. Sequentiality. (As a matter of fact $$\omega^*$$ does not even have any convergence sequences, because every infinite closed subset of $$\omega^*$$ has cardinality $$2^\mathfrak{c}$$).

2. Pseudoradiality. (Every infinite compact pseudoradial space must contain a convergent sequence).

3. Countable tightness (As proved by Kunen, $$\omega^*$$ even contains points which are not accumulation points of every countable subset of $$\omega^*$$).

4. The "weak Whyburn property" (that means, for every non-closed $$A \subset X$$ there is $$x \in \overline{A} \setminus A$$ and $$B \subset A$$ such that $$\overline{B} \setminus B=\{x\}$$. Note that every compact weakly Whyburn space must contain a convergent sequence).

5. Discrete generability. That follows from Proposition 2.4 of:

Tkachuk, V. V.; Wilson, R. G., Box products are often discretely generated, Topology Appl. 159, No. 1, 272-278 (2012). ZBL1236.54005.

or from the fact that $$\omega^*$$ contains a point which is in an accumulation point of a countable set, but not an accumulation point of any countable discrete set (see Theorem 4.4.1 in Jan van Mill's article in the Handbook of set-theoretic topology).

Obviously a dense subset of $$\omega^*$$ cannot be sequential, but:

QUESTION: Can a dense subset of $$\omega^*$$ be pseudoradial, countably tight, weakly Whyburn or discretely generated?

Parts of the above question have already been asked in the literature and in the comment section of a question on Mathoverflow (I'll give references below), but I thought it would be nice to have them all in one place.

EDIT (22/07/2021): Let me remark that under CH $$\omega^*$$ has a dense subspace which is radial, Whyburn and discretely generated. It suffices to take the set $$D$$ of all $$P$$-points.

• A dense subset of $\omega^*$ cannot be countably tight. This follows from the fact that if $x$ is any element of $\omega^*$, there is an open subset $U$ of $\omega^*$ whose closure contains $x$ but such that $x$ is not in the closure of any countable subset of $U$. To get $U$, if $x$ is a P-point, let $U = \omega^* \setminus \{x\}$. Otherwise, let $U$ be the interior of a zero-set whose boundary contains $x$. Jul 22 '21 at 0:08
• Very nice. Can you please add your comment as an answer, Anonymous? Just point-out the two non-trivial facts you're using, that $\omega^*$ is an almost P-space and an F-space. Jul 22 '21 at 14:16

I have been asked to add my comment as an answer, so here it is. A dense subset of $$\omega^*$$ cannot be countably tight. The reason is that if $$x$$ is any element of $$\omega^*$$, there is an open subset $$U$$ of $$\omega^*$$ whose closure contains $$x$$ but such that $$x$$ is not in the closure of any countable subset of $$U$$, and this property carries over to dense subsets.

To get $$U$$ consider two cases. Case 1. If $$x$$ is a P-point, let $$U = \omega^* \setminus \{x\}$$. Case 2. If $$x$$ is not a P-point, it is on the boundary of a zero-set $$Z$$; let $$U = Int_{\omega^*}Z$$. Since zero-sets in $$\omega^*$$ have dense interiors, $$x$$ is in the closure of $$U$$. It follows from the fact that $$\omega^*$$ is an F-space that the $$\omega^*$$-closure of every countable subset of $$U$$ is a subset of $$U$$, and, therefore, does not contain $$x$$.

(1) under CH the set of P-points is dense and radial even, using $$\omega_1$$-sequences.
(2) if $$D$$ is dense and pseudoradial then for every cozero set $$C$$ of $$\omega^*$$ the intersection $$C\cap D$$ is closed in $$D$$. Say $$C=f^{-1}[(0,1]]$$; a convergent sequence $$\langle x_\alpha:\alpha<\kappa\rangle$$ in $$C\cap D$$ should have uncountable cofinality, but then there is an $$n$$ such that $$\{\alpha:f(x_\alpha)\ge2^{-n}\}$$ is unbounded and hence the limit wound be in $$\{x:f(x)\ge2^{-n}\}$$.
(3) if there are no P-points then every point of the dense set $$D$$ is in the boundary of some cozero set $$C$$, for that $$C$$ the intersection $$C\cap D$$ is not closed; hence $$D$$ is not pseudoradial.
• Thank you, Alan and KP! Very good, so the pseudoradial case is settled. By the way, it looks to me that to get a dense radial subspace in $\omega^*$ you only need MA (actually, $\mathfrak{p}=\mathfrak{c}$). Jul 24 '21 at 15:40