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?
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$\begingroup$ 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 $\mathcal{U}$? $\endgroup$– Todd EisworthCommented Oct 27, 2011 at 3:22
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$\begingroup$ Yes, $\mathscr{U'}$ is asked for every. $\endgroup$– PaulCommented Oct 27, 2011 at 5:34
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$\begingroup$ "$\cal{U}'$ is asked for every" by whom? $\endgroup$– Ramiro de la VegaCommented Oct 27, 2011 at 12:05
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$\begingroup$ I'm sorry. I have edited the question. Now It may be more clearly. $\endgroup$– PaulCommented Oct 28, 2011 at 0:28
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3$\begingroup$ This question was cross-posted to math.SE: math.stackexchange.com/questions/76581 $\endgroup$– RasmusCommented Oct 28, 2011 at 8:51
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2 Answers
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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 $\psi$?
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$\begingroup$ 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". $\endgroup$ Commented Oct 27, 2011 at 12:40
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$\begingroup$ Same sort of thing happens if you let $X$ be any infinite discrete space and let $\mathcal{U}=\mathcal{U}'=\mathcal{P}(X)$. $\endgroup$ Commented Oct 27, 2011 at 18:58
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$\begingroup$ @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" ! $\endgroup$– PaulCommented Oct 28, 2011 at 0:24
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$\begingroup$ I don't know of any specific example offhand either! I might ask Hodel if he had any examples in mind. But wait! Ramiro! How does one get $H\psi(\omega_1)$ to be countable? $\endgroup$ Commented Jan 18, 2016 at 15:18
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Just as I suspected -- $\omega_1$ answers the question as it now stands. It is obviously first countable, hence of countable pseudocharacter, but has no point-countable $T_2$-separating open cover.
Let $\mathcal U$ be a point-countable open cover of $\omega_1$. Let $N$ be the collection of all points which are contained in a nonstationary member of $\mathcal U$. I claim $N$ is nonstationary; if not, then the pressing-down lemma would give us a point that is in countably many nonstationary open sets that cover a stationary set, which is impossible.
But now, a stationary open set is actually co-countable, and so every point in the complement of $N$ is contained only in co-countable members of $\mathcal U$, and no two of these can be in disjoint members of $\mathcal U$.
I suspect there is a general theorem involving spaces with point-countable $T_2$-separating open covers that eliminates $\omega_1$ and many other spaces; but I'd have to review the literature on this.
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$\begingroup$ The proof I give needs only a little tweaking to show that $\omega_1$ does not have a point-countable $T_1$-separating open cover either. Once we know that every point in the complement of $N$ is contained only in co-countable members of $\mathcal U$, we let $x_0$ be any of these points and let $x_1$ be such that all points $\ge x_1$ are in all the members of $\mathcal U$ containing $x_0$. Continue like this, defining $x_n$ for all $n \in \omega$, and let $x$ be their supremum. Then every member of $\mathcal U$ containing $x$ also contains almost all $x_n$, and so they cannot be separated. $\endgroup$ Commented Jan 18, 2016 at 18:21
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$\begingroup$ On the other hand, $\omega_1$ does have a point-countable $T_0$-separating open cover, and this may be what Ramiro de la Vega had in mind. Just take all the open rays $(\alpha, \rightarrow)$. If $\alpha_0 < \alpha_1$ then $(\alpha_0, \rightarrow)$ contains $\alpha_1$ but not $\alpha_0$. $\endgroup$ Commented Jan 18, 2016 at 18:26