I am looking for a $T_4$ space (normal + $T_1$) that is not $T_5$ (completely normal +$T_1$), but such that every singleton is the countable intersection of open sets. Or, alternatively, a proof that none exists.
Some background:
A $G_{\delta}$ normal space, i.e. a normal space where every closed set is the countable intersection of open sets, is perfectly normal and therefore is also completely normal.
What about a normal space where every singleton is the countable intersection of open sets? i.e. a normal space with a countable pseudo-character (if I understand correctly the term "pseudo-character"). Does such space have to be completely normal?
Best Answer
To re-use an example I used recently for another problem, use $X:=A \times A$ (in the product topology) where $A$ is Aleksandrov’s double arrow space (see here e.g. or here) which is compact, $T_5$, hereditarily Lindelöf, hereditarily separable but not metrisable and contains the Sorgenfrey line $\Bbb S$ as a subspace.
$X$ is compact Hausdorff, as the square of a compact Hausdorff space, so certainly $T_4$. It’s not $T_5$ because the subspace $\Bbb S \times \Bbb S$ is not normal. $A$ is first countable and hence so is $X$, so all points are $G_\delta$ sets, but e.g. $\Delta = \{(x,x): x \in A\}$ is an example of a compact and closed subset that is not a $G_\delta$ set, e.g. So $A$ is an example of a $T_6$ space whose product is not $T_6$ as well.