1
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

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).

$\endgroup$
  • $\begingroup$ Thank you very much for your clear answer. Just to verify that I understand correctly: You actually prove more, you prove that every open subset $W$ is an increasing union $\bigcup_{i=1}^{\infty} U_i$ of regular open subsets in $X$, that are $\underline{\text{also}}$ regular open in $W$ (since by construction, their closures are contained in $W$, and so $U_i=\mathrm{int}_X(\mathrm{cl}_X(U_i))=\mathrm{int}_W(\mathrm{cl}_W(U_i))$ for each $i$. $\endgroup$ – Kiko Aug 9 at 15:15
  • $\begingroup$ @Kiko That is true, but it is not more since any regular open set in $X$ that is a subset of an open set $W$ is also regular open in $W$. $\endgroup$ – user87690 Aug 9 at 15:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.