[SOLVED] Topology

**JoachimAgrell** · February 8th 2010, 01:53 AM

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.

**Opalg** · February 8th 2010, 06:35 AM

Originally Posted by JoachimAgrell

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).

**JoachimAgrell** · February 8th 2010, 06:36 PM

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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