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Let $(X,d)$ be a complete metric space with no isolated points. I want to prove that for every $(Gn)$ sequence of open and dense subsets of $X$ and for every countable set $A$$\subseteq$$X$ we have that $A$$\cap$$($$\cap_{n=1}^{\infty}$$Gn)$$\neq$$\emptyset$

According to Baire's category theorem since every $Gn$ is open and dense and since $X$ is complete, than $\cap_{n=1}^{\infty}$$Gn$ is dense in $X$. I am thinking, if i could prove that every countable set $A$$\subseteq$$X$ with no isolated points , is an open set, than my proof would be complete. Considering that if $A$ is open and $\cap_{n=1}^{\infty}$$Gn$ is dense , they must at least have a common point . Is this the right way to go ? Thank you.

Stef M
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    Something doesn't work out. In $\mathbb{R}$, take an enumeration $(r_n)$ of the rationals. Let $G_n = \mathbb{R}\setminus {r_n}$. Then $\bigcap_{n = 1}^{\infty} G_n = \mathbb{R}\setminus \mathbb{Q}$. – Daniel Fischer Dec 26 '15 at 12:06
  • @DanielFischer You are right. Something doesn't work out. I will try to find out what's wrong. Thank you for pointing this out for me. – Stef M Dec 26 '15 at 12:54
  • @DanielFischer There was an error after all . Instead of $A$$\cap$$($$\cap_{n=1}^{\infty}$$Gn)$$\neq$$\emptyset$ it is $(X-A)$$\cap$$($$\cap_{n=1}^{\infty}$$Gn)$$\neq$$\emptyset$. So I guess we need to prove that $X-A$ is open , or that $A$ is closed . And I think that is true because $A$ has no isolated points . That means it contains its own limit points . And that means its a closed set . – Stef M Dec 26 '15 at 13:44
  • $A$ need not be closed, consider $X = \mathbb{R},, A = \mathbb{Q}$ above. But the point is that $\bigcap G_n \not\subset A$, which again follows from Baire. – Daniel Fischer Dec 26 '15 at 13:48
  • @DanielFischer I understand from Baire's theorem that $\cap$$Gn$ is not countable , where as $A$ is countable . Can you help me understand how to use Baire's theorem to get to $\cap$$Gn$$\not\subset$$A$ ? – Stef M Dec 26 '15 at 17:44
  • If you know that $\bigcap G_n$ is uncountable (how do you know that, by the way?), then isn't it clear that it's not a subset of a countable set? – Daniel Fischer Dec 26 '15 at 18:18
  • @DanielFischer $\cup$$Gn$ is uncountable because we know that a) $(X,d)$ is complete and b)$Gn$ is a sequence of open and dense subsets of $X$. That is enough use Baire theorem's first form , and claim that $\cup$$Gn$ is uncountable. I understand that an uncountable set can never be a subset of an countable set. But can we say that $\cup$$Gn$$\not\subset$$A$$=>$$\cup$$Gn$$\subseteq$$X-A$. That would be really useful and solve my problem. – Stef M Dec 26 '15 at 18:42
  • You wrote $\cup$ instead of $\cap$ there, I suppose that's just a typo. I don't know what your course/book calls the first form of Baire's theorem, so I don't know the argument (it's not enough that $(X,d)$ is complete, we also need to use the assumption that $X$ has no isolated points). Okay, so you know that $\bigcap G_n$ is uncountable, and hence $\bigcap G_n \not\subset A$. That's it, that's all you get, you cannot deduce that $\bigcap G_n\subset (X - A)$, since that will in general not be the case. But that's not what you want to deduce. You want $(X-A)\cap\bigcap G_n \neq \varnothing$, – Daniel Fischer Dec 26 '15 at 18:50
  • and that is exactly $\bigcap G_n \not\subset A$. – Daniel Fischer Dec 26 '15 at 18:51
  • @DanielFischer I get it now. Thank you very much for your help and patience. – Stef M Dec 26 '15 at 19:40

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Let $A_i=X-\{a_i\}$. Then $A_i$ is open (since $\{a_i\}$ is closed as a singleton) and dense (since $a_i$ is not an isolated point). Thus $$\left(\bigcap_i A_i\right) \cap \left(\bigcap_i G_i\right)$$ is not empty by Baire, i.e., $$(X-A) \cap \left(\bigcap_i G_i\right)$$ is not empty.

andreas
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