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Thread: [SOLVED] Topology

  1. #1
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    [SOLVED] Topology

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

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

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

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

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

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

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

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

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

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