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I have four quick questions and have listed them below. I am seeking for corroboration of the first three and a bit of insight on the fourth, as I have hit a solid brick wall.

Definition. A set $A\subseteq\mathbb{R}$ is called an $F_\sigma$ set if it can be written as the countable union of closed sets. A set $B\subseteq\mathbb{R}$ is called a $G_\delta$ set if it can be written as the countable intersection of open sets.

1. Argue that a set $A$ is a $G_\delta$ set if and only if its complement is an $F_\sigma$ set.

Proof. Let $A$ be a $G_\delta$ set. Then$$A=\bigcap_{n=1}K_n,$$where each $K_n$ is an open set. By De Morgan's law, it follows that$$A^c=\left(\bigcap_{n=1}K_n\right)^c=\bigcup_{n=1}K_n^c,$$which is an $F_\sigma$ set, since it is a countable union of closed sets.

The converse statement can be proven in a similar fashion. $\square$

2. Show that a closed interval $[a,b]$ is a $G_\delta$ set.

Proof. Take$$\bigcap_{n=1}^\infty\left(a-\frac{1}{n},b+\frac{1}{n}\right)=[a,b].$$Therefore, $[a,b]$ is a $G_\delta$ set. $\square$

3. Show that the half-open interval $(a,b]$ is both a $G_\delta$ and an $F_\sigma$ set.

Proof. Take$$\bigcap_{n=1}^\infty\left(a,b+\frac{1}{n}\right)=(a,b],$$and$$\bigcup_{n=4}^\infty\left[a-\frac{a-b}{n},b\right]=(a,b].$$Therefore, $(a,b]$ is both a $G_\delta$ and an $F_\sigma$ set. $\square$

4. Show that $\mathbb{Q}$ is an $F_\sigma$ set, and the set of irrationals $\mathbb{I}$ forms a $G_\delta$ set.

I am oblivious as to how to tackle this problem. I know that I need to show that $\mathbb{Q}$ can be written as the countable union of closed sets, and that upon doing this, I can apply De Morgan's law to it to prove the second statement. However, I do not know how to begin. Do you guys have any ideas?

Thanks in advance!

Edit 1: I followed Robert's advice, and this is what I managed to weave:

Proof. Since $\mathbb{Q}$ is countable, we can write it as $\mathbb{Q}=\{r_1,r_2,r_3,\cdots\}$, where each $r_n$ is a unique rational number. It follows that we can express $\mathbb{Q}$ as the union of singletons $[r_n]$, which are closed sets, as follows:$$\mathbb{Q}=\bigcup_{n=1}^\infty[r_n].$$Therefore, $\mathbb{Q}$ is an $F_\sigma$ set, since it is a countable union of closed sets. Furthermore, it naturally follows that$$\mathbb{I}=\mathbb{Q}^c=\left(\bigcup_{n=1}^\infty[r_n]\right)^c=\bigcap_{n=1}^\infty[r_n]^c=\bigcap_{n=1}^\infty(-\infty,r_n)\cup(r_n,\infty)$$is a $G_\delta$ set, since it is a countable intersection of open sets. $\square$

Is this sound?

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Hint: a one-point set is closed. – Robert Israel Feb 23 '12 at 4:00
I think in your definition of $G_{\delta}$ set you meant "countable union of open sets", since that's what you're using later on. – Patrick Da Silva Feb 23 '12 at 4:18
another hint: $\mathbb Q$ is countable. – azarel Feb 23 '12 at 4:19
A further hint: how many rational numbers are there? (And since no one’s actually said so: your first three are fine.) – Brian M. Scott Feb 23 '12 at 4:19
I just thought we all noticed the same thing and the answer would've gotten out at some point so I answered it. – Patrick Da Silva Feb 23 '12 at 4:21
up vote 2 down vote accepted

You just need to notice that $\mathbb Q$ is countable and that singletons are closed sets, hence $$ \mathbb Q = \bigcup_{x \in \mathbb Q} \{ x \}. $$

Hope that helps,

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