Properties of basis of regular open sets in separable spaces

Question

I'm trying to prove that in any regular Hausdorff separable space, given a dense countable set D, for any open set $U$ such that $U=\text{int Cl}(U)$ it happens that $U=\text{int Cl}(U\cap D)$. I've tried to prove this by double inclusion and also by assuming that it is false to achieve a contradiction to the fact that $U=\text{int Cl}(U\cap D)$, but I haven't succeed, can someone provide me any hint?

Answer 1

This is actually a straightforward consequence of the following much more general fact.

Fact. Given any topological space $X$, any $A \subseteq X$ and any open $V \subseteq X$, $\overline{ V \cap \overline{A} } = \overline{ V \cap A }$.

To prove this fact we demonstrate the two inclusions:

• As $V \cap A \subseteq V \cap \overline{A}$, then $\overline{ V \cap A } \subseteq \overline{ V \cap \overline{A} }$.
• Note that $W = X \setminus \overline{ V \cap A }$ is open, and $(W \cap V ) \cap A = W \cap ( V \cap A ) = \emptyset$. Since $W \cap V$ is open set disjoint from $A$, it follows that it is also disjoint from $\overline{A}$, meaning that $W \cap ( V \cap \overline{A} ) = ( W \cap V ) \cap \overline{A} = \emptyset$. Therefore $X \setminus \overline{ V \cap A } = W \subseteq X \setminus \overline{ V \cap \overline{A} }$, or $\overline{ V \cap \overline{A} } \subseteq \overline{ V \cap A }$.

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Using the notation from the question, this fact allows us to calculate as follows: $$ \operatorname{Int} ( \overline{U \cap D} ) = \operatorname{Int} ( \overline{U \cap \overline{D} } ) = \operatorname{Int} ( \overline{U \cap X} ) = \operatorname{Int} ( \overline{U} ) = U. $$