Countable union of closed sets/countable interesection of open sets

**Zalren** · October 8th 2010, 11:13 AM

A set $A$ is called an $F_\sigma$ set if it can be written as the countable union of closed sets. A set $B$ is called a $G_\delta$ set if it can be written as the countable intersection of open sets.
(a) Show that a closed interval [a, b] is a $G_\delta$ set.
(b) Show that the half open interval (a, b] is both a $G_\delta$ and an $F_\sigma$ set.
(c) Show that $Q$ is an $F_\sigma$ set, and the set of irrationals $I$ forms a $G_\delta$ set.

I have no idea where to start on this problem. Any help would be appreciated. Thank you for your time.

**TheEmptySet** · October 8th 2010, 11:21 AM

For (a) consider the family of open sets

$\displaystyle A_n=\left( a-\frac{1}{n},b+\frac{1}{n}\right)$

What happens when you take the intersection of

$\displaystyle \cap_{n=1}^{\infty}A_n$ ?

This should get you started on (a) and (b)

Also remember that in $\mathbb{R}$ points are closed and $\mathbb{Q}$ is a countable set.

Good luck.

**Plato** · October 8th 2010, 01:11 PM

I have decided to show you the irrational part of part #3.
Let $\mathbb{I}=\mathbb{R}\setminus\mathbb{Q}$ the irrational numbers.

The let $\{x\}^c=\left( { - \infty ,x} \right) \cup \left( {x,\infty } \right)$ .

Try to convince yourself that $\mathbb{I}= \bigcap\limits_{x \in \mathbb{Q}} {\left\{ x \right\}^c }$ , a $G_{\delta}$ set.

**Zalren** · October 8th 2010, 01:59 PM

Thanks all! You guys rock!

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