1
$\begingroup$

I'm trying to show that for any topological space $X$ and any $ O\subseteq X$ with $O$ an open set, there is an regular open set $W$ such that $ O \subseteq W$ and $O$ is dense in $W$.

I'm not quite sure what "$O$ is dense in $W$" means. Does this typically mean that $O$ is dense in the subspace topology induced by $W$ or for any nonempty open set contained in $W$, $O$ meets it?

I appreciate any clarification

Edit: I believe that I can just let $W = $ Int(Cl($O$)) as the regular open set, but I'm still trying to show the density part.

$\endgroup$
3
$\begingroup$

$O$ is dense in $W$ means that $O$ is dense in the subspace topology induced by $W$

$\endgroup$
1
$\begingroup$

Your interpretation of denseness is quite correct.

To show it for your suggestion $W= \operatorname{Int}(\operatorname{Cl}(O))$ : suppose $U$ is a non-empty open subset of $W$.

Suppose that $O \cap U=\emptyset$, it follows that $U \cap \operatorname{Cl}(O)=\emptyset$ too ($ O \subseteq U^\complement$, so $\operatorname{Cl}(O) \subseteq U^\complement$, as the right hand set is closed, so $U \cap \operatorname{Cl}(O) = \emptyset$), and so $U \cap \operatorname{Int}(\operatorname{Cl}(O)) = \emptyset$ as well, but this just says $U \cap W= \emptyset$, which cannot be as $\emptyset \neq U \subseteq W$ by assumption. So contradiction and thus $U \cap W \neq \emptyset$, QED.

$\endgroup$

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.