Regular open sets in topology generated by regular open sets

Question

Let $Ro( \tau )$ denote the set of regular open sets of topology $\tau$. Is it possible for a set to be regular open in the topology generated by $Ro( \tau)$, but not in $\tau$? Obviously $\tau$ cannot be a regular topological space.

Answer 1

Let $\tau^\prime$ denote the topology generated by $\mathrm{Ro}(\tau)$. Note that $\tau^\prime$ is coarser than $\tau$.

I claim that $\tau$ and $\tau^\prime$ have the same regular open sets. To begin I will establish the following.

Fact. If $U \subseteq X$ is $\tau^\prime$-open, then $\operatorname{cl}_{\tau} ( U ) = \operatorname{cl}_{\tau^\prime} ( U )$.

• proof of Fact. Since $\tau^\prime$ is coarser than $\tau$ we have that $\operatorname{cl}_{\tau} ( U ) \subseteq \operatorname{cl}_{\tau^\prime} ( U )$.

  Given $x \in X \setminus \operatorname{cl}_{\tau} ( U )$, then $V := X \setminus \operatorname{cl}_{\tau} ( U )$ is a $\tau$-open neighborhood of $x$ disjoint from $U$, and since $U$ is $\tau$-open it follows that $\operatorname{cl}_{\tau} ( V ) \cap U = \emptyset$, and therefore $\operatorname{Int}_\tau ( \operatorname{cl}_{\tau} ( V ) ) \cap U = \emptyset$. Note that $\operatorname{Int}_\tau ( \operatorname{cl}_{\tau} ( V ) )$ is regular open in $\tau$, and so it is $\tau^\prime$-open, and clearly $x \in \operatorname{Int}_\tau ( \operatorname{cl}_{\tau} ( V ) )$. Therefore $x \notin \operatorname{cl}_{\tau^\prime} ( U )$.

Using de Morgan's laws from this Fact it follows that for any $\tau^\prime$-closed $F \subseteq X$ we have $\operatorname{Int}_{\tau} ( F ) = \operatorname{Int}_{\tau^\prime} ( F )$.

We know prove that every regular open set in $\tau$ is regular open in $\tau^\prime$, and vice versa.

• If $U \subseteq X$ is regular open in $\tau$, then $U$ is $\tau^\prime$-open, and by the Fact and its corollary we have $$ U = \operatorname{Int}_{\tau} ( \operatorname{cl}_{\tau} ( U ) ) = \operatorname{Int}_{\tau} ( \operatorname{cl}_{\tau^\prime} ( U ) ) = \operatorname{Int}_{\tau^\prime} ( \operatorname{cl}_{\tau^\prime} ( U ) ). $$ Thus $U$ is regular open in $\tau^\prime$.

• If $U \subseteq X$ is regular open in $\tau^\prime$, then using the Fact and its corollary we have $$U = \operatorname{Int}_{\tau^\prime} ( \operatorname{cl}_{\tau^\prime} ( U ) ) = \operatorname{Int}_{\tau} ( \operatorname{cl}_{\tau^\prime} ( U ) ) = \operatorname{Int}_{\tau} ( \operatorname{cl}_{\tau} ( U ) )$$ and so $U$ is regular open in $\tau$.