# [SOLVED] Topology

• Feb 8th 2010, 01:53 AM
JoachimAgrell
[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?

• Feb 8th 2010, 06:35 AM
Opalg
Quote:

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).
• Feb 8th 2010, 06:36 PM
JoachimAgrell
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.