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Let X be a Hausdorff space and define the Property A as follows:

If $\mathscr{U}$ is a collection of open sets of $X$ that witnesses Hausdorff property of $X$ (= $\forall x,y \in X$, there exist two disjoint open sets $U_1$ and $U_2$ $\in \mathscr{U}$ s.t. $x \in U_1$ and $y \in U_2$), then there is a point $x\in X$ such that $|\{U \in \mathscr{U}: x \in U\}|>\omega$.

Is there a countable pseudocharacter Hausdorff space $X$ with the property A? The same question is also asked here.

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    Voting to close because the question is (in the process of being) answered on MO. @John: Why did you repost?2011-10-28
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    @Rasmus: However, as yet it *hasn’t* actually been answered on MO, and (barring error) I’ve answered it here.2011-10-28
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    @Brian M. Scott: Wouldn't it have been better to close this thread and add answers to the MO thread, where already several comments and an answer have been given? Now we have two simultanious threads on the same topic.2011-10-28
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    @Rasmus: I much prefer to work here; for various reasons I really don’t want to get involved with MO just now.2011-10-28

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Let $X = \beta\omega$ with the following topology: points of $\omega$ are isolated, and each $p\in\beta\omega\setminus\omega$ has $\{\{p\}\cup U:U\in p\}$ as a local base. (I think of the points of $\beta\omega\setminus\omega$ as free ultrafilters on $\omega$.)

This topology refines the Čech-Stone topology on $\beta\omega$, so it’s certainly Hausdorff, and since it has a base of countable sets, it must have countable pseudocharacter as well. But $\beta\omega\setminus\omega$ is uncountable (in fact of cardinality $2^{2^\omega}$), so any family $\mathscr{U}$ that witnesses the Hausdorffness of $X$ must be uncountable.

Added: That actually doesn’t follow without further argument. However, $|\beta\omega|>2^\omega$, so if $\mathscr{U}$ is a countable family of subsets of $\beta\omega$, there are distinct $p,q\in\beta\omega$ such that $$\{U\in\mathscr{U}:p\in U\}=\{U\in\mathscr{U}:q\in U\},$$ and $\mathscr{U}$ cannot then witness the Hausdorffness of $X$.

Each member of $\mathscr{U}$ has non-empty intersection with $\omega$, so some $n\in\omega$ must belong to uncountably many members of $\mathscr{U}$.

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    The example probably works, but one step in the reasoning is incorrect: in the reals (uncountable) the rational intervals (a countable family) witness the Hausdorffness. But the remainder of C-S is probably so big that you're right in the end, I think2011-10-28
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    Isn't the topology as you define not exactly the usual topology on the Cech-Stone?2011-10-28
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    @Henno: Yes, I caught and fixed the oversight just before I refreshed and saw your note. No, it’s not the usual topology on $\beta\omega$, because I’ve made $\beta\omega\setminus\omega$ a closed, discrete set.2011-10-28
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    Ah yes, it's like a bigger version of Mrowka's Psi. Nice2011-10-28