Let $X$ be a topological space. An open set $U$ in $X$ is called regular open if it equals to the interior of its closure, namely $\mathrm{int}(\mathrm{cl}(U))=U$.
$X$ is called semiregular if its open regular sets form a base for the topology.
Clearly, in a semiregular space, every open set is a union of regular open sets.
However, I don't know wether it is possible to view open sets as an increasing union of regular open sets?
The usual trick wouldn't work, as union of regular open is not regular open.
I can moreover assume: $X$ compact metrizable and the base of the topology is countable. I hope that at least in this case it's possible, but I don't know how to show it.
Thanks.
Best Answer
In a regular space, every open set $U$ is a union of regular open sets whose closures are in $U$. And for two regular open sets $U$, $V$, we have $U ∪ V ⊆ W := \operatorname{int}(\overline{U ∪ V}) ⊆ \overline{U} ∪ \overline{V}$. So you may obtain the increasing union at least in regular hereditarily Lindelöf spaces (so separable metrizable is fine).