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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?

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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.)

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    $\begingroup$ $H=L^2([0,1])$, $D=C([0,1])$, $S=$ the span of $f$ where $f\in H\setminus D$. $\endgroup$ – David C. Ullrich Jun 26 '15 at 18:53
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    $\begingroup$ @Freeze_S that would not work either. Let $e^x$ be an element in $L^2[0,1]$ and suppose $S= span\{ e^x\}$. This is a finite dimensional subspace and therefore it is closed. If we let $D$ consist of all polynomials, then $\overline D = L^2[0,1]$. However, $D\cap S = \emptyset$. $\endgroup$ – Joel Jun 26 '15 at 18:55
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    $\begingroup$ @Joel. Right, except your $D\cap S$ is not quite empty... $\endgroup$ – David C. Ullrich Jun 26 '15 at 18:59
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    $\begingroup$ Right of course. Typo. I meant $D \cap S = \{0\}$. About as close to empty as you can get for a subspace. $\endgroup$ – Joel Jun 26 '15 at 19:00
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    $\begingroup$ @Freeze_S You mean an example of $f$ so that there does not exist continuous $g$ with $f=g$ ae? No, such examples are phenomenally easy. Take $f=\chi_{[0,1/2]}$ for example. $\endgroup$ – David C. Ullrich Jun 26 '15 at 19:38
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$\overline{D\cap S}$ is the smallest closed subset containing $D \cap S$. If $S \subset D$ then $D \cap S = S$ and since $\overline{S} = S$ we have $\overline{D \cap S} = \overline{S} = S$.

However, if $S \not \subset D$, the answer is less clear. Certainly $\overline{D\cap S} \subset S$ in that case. The answer of equality depends on the particular topology.

Take for instance $\Omega = [0,1]$ with the topology given by the ordering of the real numbers. If $D=[0,1)$ and $S = \{1\}$ we see that $\overline D = [0,1]$ but $D \cap S = \emptyset$.

However, if we have the set $\Omega=\{1,2,3,4\}$ with the topology $\tau =\{ \emptyset, \{1,2\}, \{3,4\}, \Omega\}$ we can have a different answer. Suppose $S = \{1,2\}$. This is an open set, and its compliment is $\{3,4\}$ which is also open. Thus $S$ is closed. If we let $D=\{1,3\}$, then $\overline{D}=\Omega$ since $\Omega$ is the smallest closed set containing $D$. Now we also have $D \cap S = \{1\}$ and $\overline{ D\cap S} = \{1,2\} = S$ since $S$ is the smallest closed subset containing $\{1\}$.


(Appended after the question was edited)

As for the Hilbert space question. The statement does not hold there either.

Let $e^x$ be an element in $L^2[0,1]$ and suppose $S=span\{e^x\}$. This is a finite dimensional subspace and therefore it is closed. If we let $D$ consist of all polynomials, then $\overline{D}=L^2[0,1]$. However, $D\cap S=\{0\}$ and $\overline{D \cap S}=\{0\}$.

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