5

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?

C-star-W-star
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2 Answers2

8

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.)

C-star-W-star
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David C. Ullrich
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  • Haha ya just had the same idea with exactly the same sets. ^^ – C-star-W-star Jun 26 '15 at 18:47
  • What about closed subspaces $\mathcal{S}\leq\mathcal{H}$ within a Hilbert space $\mathcal{H}$ and a $\mathcal{D}\leq\mathcal{H}$. – C-star-W-star Jun 26 '15 at 18:49
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    $H=L^2([0,1])$, $D=C([0,1])$, $S=$ the span of $f$ where $f\in H\setminus D$. – David C. Ullrich Jun 26 '15 at 18:53
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    @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, $D\cap S = \emptyset$. – Joel Jun 26 '15 at 18:55
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    @Joel. Right, except your $D\cap S$ is not quite empty... – David C. Ullrich Jun 26 '15 at 18:59
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    Right of course. Typo. I meant $D \cap S = {0}$. About as close to empty as you can get for a subspace. – Joel Jun 26 '15 at 19:00
  • @DavidC.Ullrich: Do you mind adding that counterexample to your answer? – C-star-W-star Jun 26 '15 at 19:15
  • @DavidC.Ullrich: Thanks alot!!! :) – C-star-W-star Jun 26 '15 at 19:28
  • @DavidC.Ullrich: Aren't such examples extremely hard to construct explicitely? (The Dirichlet function won't work.) – C-star-W-star Jun 26 '15 at 19:35
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    @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. – David C. Ullrich Jun 26 '15 at 19:38
  • @DavidC.Ullrich: Aha right thanks again!!!! :D – C-star-W-star Jun 26 '15 at 19:43
  • @DavidC.Ullrich: I added another case that is very important for my concerns. Do you mind adding that one. Suggestion: $P\mathcal{D}\subseteq\mathcal{D}\iff\mathcal{D}=\mathcal{S}\cap\mathcal{D} \oplus\mathcal{S}^\perp\cap\mathcal{D}$, $\mathcal{H}=\overline{\mathcal{D}}=\overline{\mathcal{S}\cap\mathcal{D}}\oplus \overline{\mathcal{S}^\perp\cap\mathcal{D}}$ – C-star-W-star Jul 01 '15 at 09:02
  • That has nothing to do with the original question - seems to me it belongs elsewhere. (Yes, it's related to you because this is all part of that stuff you're thinking about.) – David C. Ullrich Jul 01 '15 at 15:44
  • @DavidC.Ullrich: Besides, I won't get notified if you don't ping me by '@Freeze_S'. Would you mind doing so in future, please? – C-star-W-star Jul 01 '15 at 16:51
  • @DavidC.Ullrich: Well I didn't see this from the beginning. However opening dozen of threads for every little aspect seems overrun for MSE, IMHO. But I think it is ok for now as I remarked this in the comment. ;) (Remind T.A.E.'s slogan!) – C-star-W-star Jul 01 '15 at 16:54
  • @DavidC.Ullrich: I tried to leave it this way but it bothers me so much :( Would you mind adding that last case, pleease? (Hidden in the comments.) – C-star-W-star Jul 14 '15 at 16:44
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    @Freeze_S Why are you always asking me to rewrite my posts? Yes, that's better than just rewriting them for me. But if there's something you feel needs posted why not just post it yourself? We know you know you can answer your own question... – David C. Ullrich Jul 14 '15 at 16:47
  • @DavidC.Ullrich: I would but the point is a kind of like your answer and keeping the cases together in one asnswer would be nice. Of course, if you want to it's perfectly fine in your own words. ...because I often edit so long until I'm happy with it. (Supposedely that time comes...) – C-star-W-star Jul 14 '15 at 16:51
4

$\overline{D\cap S}$ 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 $\overline{D \cap S}=\{0\}$.

Joel
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