Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
1  
Voting to close because the question is (in the process of being) answered on MO. @John: Why did you repost? –  Rasmus Oct 28 '11 at 7:35
    
@Rasmus: However, as yet it hasn’t actually been answered on MO, and (barring error) I’ve answered it here. –  Brian M. Scott Oct 28 '11 at 8:12
1  
@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. –  Rasmus Oct 28 '11 at 8:48
1  
@Rasmus: I much prefer to work here; for various reasons I really don’t want to get involved with MO just now. –  Brian M. Scott Oct 28 '11 at 9:13

1 Answer 1

up vote 2 down vote accepted

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 $\mathscr{U}$.

share|improve this answer
    
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 think –  Henno Brandsma Oct 28 '11 at 8:56
    
Isn't the topology as you define not exactly the usual topology on the Cech-Stone? –  Henno Brandsma Oct 28 '11 at 8:57
    
@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. –  Brian M. Scott Oct 28 '11 at 9:10
    
Ah yes, it's like a bigger version of Mrowka's Psi. Nice –  Henno Brandsma Oct 28 '11 at 20:10

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.