3
$\begingroup$

Given a topological space $X$ and a dense subspace $D$, I believe it's true that for a regular open set $U$ of $X$, $U \cap D$ is regular open in $D$. Note this induces a homomorphism between the Boolean algebras of regular open sets, namely $h: \mathcal{RO}(X) \to \mathcal{RO}(D)$ by $U \mapsto U \cap D$, which is onto. When is the map $h$ also an injection?

$\endgroup$

1 Answer 1

1
$\begingroup$

It seems if $D$ is dense in $X$ then they have the same regular open algebra. Let $U, V$ be regular open sets of $X$ with $U \cap D = V \cap D$. Since $U \cap D$ is dense in $U$, $\mbox{cl}_X (U \cap D) = \mbox{cl}_X U$ (likewise for $V$, of course). Then $U = \mbox{int}\,\mbox{cl} U = \mbox{int}\,\mbox{cl}(U \cap D) = \mbox{int}\,\mbox{cl}(V \cap D) = V$, and the map is one-to-one. No?

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .