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Let X be a Hausdorff space and Difine the Property A as following: 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 opensets $U_1$ and $U_2$ $\in \mathscr{U}$, st $x \in U_1$ 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?

edited Oct 28 2011 at 5:01

asked Oct 27 2011 at 2:17

Paul
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I'm not sure if I follow what you're asking; it looks to me like some quantifiers have been left out. Is this condition supposed to hold for every such family $\mathcal{U}'$ in a given base $\mathcal{U}$? – Todd Eisworth Oct 27 2011 at 3:22

Yes, $\mathscr{U'}$ is asked for every. – Paul Oct 27 2011 at 5:34

"$\cal{U}'$ is asked for every" by whom? – Ramiro de la Vega Oct 27 2011 at 12:05

I'm sorry. I have edited the question. Now It may be more clearly. – Paul Oct 28 2011 at 0:28

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This question was cross-posted to math.SE: math.stackexchange.com/questions/76581 – Rasmus Oct 28 2011 at 8:51

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Nathan · Answer 1 · 2011-10-27 04:29:04Z UTC

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Hodel defined the Hausdorff pseudo-character $H\psi(X)$ as the smallest infinite cardinal so that each point lies in at most $H\psi(X)$ elements of an open family $\mathcal{U'}$ that witnesses the Hausdorff property of $X$. Any $X$ such that $\omega = \psi(X) < H\psi(X)$ would seem to answer your question. I don't know of any specific example, but I guess there must be one. Why define $H\psi$ if it's really the same as $\psi$?

answered Oct 27 2011 at 4:29

Nathan
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	I guess John´s question is a little easier since $\cal{U}'$ is restricted to be a subset of some fixed base, although it is hard to tell since all quantifiers are messed up. For instance take $X=\omega_1$ with the order topology; then $H\psi(X)=\psi(X)=\omega$ BUT if $\cal{U}$ is the base consisting of bounded intervals then you have "property A". – Ramiro de la Vega Oct 27 2011 at 12:40
	Same sort of thing happens if you let $X$ be any infinite discrete space and let $\mathcal{U}=\mathcal{U}'=\mathcal{P}(X)$. – Todd Eisworth Oct 27 2011 at 18:58
	@Ramiro de la Vega and Todd Eisworth, I'm sorry that I haven't said clearly. $\cal{U}'$ is any subcollection of $\cal{U}$ that witnesses the Hausdorff property of X and satisfy the Property A. It is "any" ! – Paul Oct 28 2011 at 0:24

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