Timeline for Is there a countable pseudocharacter Hausdorff space，such that...?

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Jan 18, 2016 at 18:26	comment	added	Peter Nyikos		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 $\alpha_0$.
Jan 18, 2016 at 18:21	comment	added	Peter Nyikos		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.
Jan 18, 2016 at 17:52	history	answered	Peter Nyikos	CC BY-SA 3.0