TITLE: Denseness: Closed Space
QUESTION [5 upvotes]: I need this as lemma.
Topological Space
Given a topological space $\Omega$.
Consider a closed space:
$$\mathcal{S}\subseteq\Omega:\quad\mathcal{S}=\overline{\mathcal{S}}$$
Then for dense domains:
$$\mathcal{D}\subseteq\Omega:\quad\overline{\mathcal{D}}=\Omega\implies\overline{\mathcal{D}\cap\mathcal{S}}=\mathcal{S}$$
Does this really hold?
Hilbert Space
Given a Hilbert space $\mathcal{H}$.
Consider a closed space:
$$\mathcal{S}\leq\mathcal{H}:\quad\mathcal{S}=\overline{\mathcal{S}}$$
Then for dense domains:
$$\mathcal{D}\leq\mathcal{H}:\quad\overline{\mathcal{D}}=\mathcal{H}\implies\overline{\mathcal{D}\cap\mathcal{S}}=\mathcal{S}$$
Does this hold here?
Reducing Space
Given a Hilbert space $\mathcal{H}$.
Consider a closed space:
$$\mathcal{S}\leq\mathcal{H}:\quad\mathcal{S}=\overline{\mathcal{S}}$$
Denote its projection:
$$\mathcal{R}P=\mathcal{S}:\quad P^2=P=P^*$$
Regard a reducing domain:
$$P\mathcal{D}\subseteq\mathcal{D}\leq\mathcal{H}$$
Then the dense domain:
$$\overline{\mathcal{D}}=\mathcal{H}\implies\overline{\mathcal{D}\cap\mathcal{S}}=\mathcal{S}$$
Does this hold now?

REPLY [8 votes]: No. Consider $\Omega=\mathbb R$ (with the usual topology), $S=\{\pi\}$, $\mathcal D=\mathbb Q$.
EDIT: Next came the question "Ok, what about closed subsets of a Hilbert space"?
No change. Say $H=L^2([0,1])$, $\mathcal D=C([0,1])$ and let $S$ be the span of $f$, where $f$ is any discontinuous $L^2$ function. (Or rather, where $f$ is such that there does not exist a continuous $g$ with $f=g$ almost everywhere.)