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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.

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    $\begingroup$ 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 $x$. $\endgroup$
    – Anonymous
    Jul 22 '21 at 0:08
  • $\begingroup$ 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. $\endgroup$ Jul 22 '21 at 14:16
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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 $x$.

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Comments from Alan Dow:

(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.

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    $\begingroup$ 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, $\mathfrak{p}=\mathfrak{c}$). $\endgroup$ Jul 24 '21 at 15:40

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