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Math Help - [SOLVED] Topology

  1. #1
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    [SOLVED] Topology

    Let (A_\lambda)_{\lambda\in\mathbb{N}} be a collection of open sets of real numbers. Suppose F\subset\mathbb{R} is such that \overline{F\cap A_\lambda}=F \forall \lambda\in\mathbb{N}.

    Show that \overline{F\cap\bigcap_{\lambda\in\mathbb{N}}A_\la  mbda}=F

    I've managed to prove that \overline{F\cap\bigcap_{\lambda\in\mathbb{N}}A_\la  mbda}\subset F.

    How can i prove the other inclusion?

    Thanks in advance.
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  2. #2
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    Quote Originally Posted by JoachimAgrell View Post
    Let (A_\lambda)_{\lambda\in\mathbb{N}} be a collection of open sets of real numbers. Suppose F\subset\mathbb{R} is such that \overline{F\cap A_\lambda}=F \forall \lambda\in\mathbb{N}.

    Show that \overline{F\cap\bigcap_{\lambda\in\mathbb{N}}A_\la  mbda}=F

    I've managed to prove that \overline{F\cap\bigcap_{\lambda\in\mathbb{N}}A_\la  mbda}\subset F.

    How can i prove the other inclusion?
    This looks suspiciously like the Baire category theorem (see the section headed "Proof" in that link).
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  3. #3
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    I actually used Baire category theorem. Here goes the solution:

    Let  A:=\bigcap_{\lambda\in\mathbb{N}}A_\lambda.

    Since F=\overline{A_\lambda\cap F}\text{ }\forall\lambda\in\mathbb{N}, it follows that A_\lambda\cap F is a dense subset of F for all lambda. Hence each A_\lambda is a dense subset of F.

    Suppose x\in F but x\notin \overline{F\cap A}. That means x is not an adherent point of F\cap A. Therefore \exists \delta such that I_{\delta}(x)\cap (F\cap A)=\emptyset. But x\in F, so it must be true that I_\delta(x)\cap A=\emptyset.

    On the other hand, since each A_\lambda is a dense subset of F, the category theorem guarantees that the countable intersection of these sets is a dense set in F. Thus I_\delta(x)\cap A\neq\emptyset, a contradiction.
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