Timeline for Is there a countable pseudocharacter Hausdorff space，such that...?
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| when toggle format | what | by | license | comment | |
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| Jan 18, 2016 at 15:18 | comment | added | Peter Nyikos | 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? | |
| Oct 28, 2011 at 0:24 | comment | added | Paul | @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" ! | |
| Oct 27, 2011 at 18:58 | comment | added | Todd Eisworth | Same sort of thing happens if you let $X$ be any infinite discrete space and let $\mathcal{U}=\mathcal{U}'=\mathcal{P}(X)$. | |
| Oct 27, 2011 at 12:40 | comment | added | Ramiro de la Vega | 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". | |
| Oct 27, 2011 at 4:29 | history | answered | Nathan | CC BY-SA 3.0 |