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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.2012-02-23
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    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.2012-02-23
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    another hint: $\mathbb Q$ is countable.2012-02-23
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    A further hint: how many rational numbers are there? (And since no one’s actually said so: your first three are fine.)2012-02-23
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    I just thought we all noticed the same thing and the answer would've gotten out at some point so I answered it.2012-02-23
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    @PatrickDaSilva I agree that something is amiss. "[Countable intersection of open sets](http://en.wikipedia.org/wiki/G-delta_set)" is probably what is meant. My French is sufficiently crummy that I can't remember the word for closed begins with an "F", but it is pretty easy to remember that $\sigma$ stands for _somme_.2012-02-23
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    @Dylan : Lol, you wrote closed instead of open, you too. I speak French daily ; the word for "closed" is "Fermé". But why are you speaking about French? I don't understand your point.2012-02-23
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    Indeed. There aren't so many ways to get this wrong and I think we've collectively managed to hit all of them. The wiki, though, is always right. [To respond to your other comment, I mentioned French because that's where the $F$ and the $s/\sigma$ appear to come from and I find that it's a good (not good enough, apparently) way to remember what's what. Confusingly, the $G$ and the $d/\delta$ come from German, but I usually don't have to read papers written in German, so that isn't so helpful for me.]2012-02-23
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    @Dylan : You mean that the $F$ comes from ferme (closed)? Then what would the $G$ stand for? "Open set" in german doesn't begin with a $G$.2012-02-23
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    @PatrickDaSilva According to the Wiki (not me), _gebiet_ means "area" or "neighborhood".2012-02-23
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    @Dylan : Ohhhh. That makes sense. Hehe, a +1 in my mathematical culture!2012-02-24
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    Concerning the notational-discussion, the $\delta$ comes from $Durchschnitt$ which is a German word for intersection in set theory, to which this notation is accredited to Hausdorff -- $Real~And~Complex~Analysis$, 3rd. Ed., W. Rudin, PG. 12. The letters $F$ and $G$ were used for closed and open sets, respectively, at the time.2017-10-01
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    Also, shouldn't the index for the union be 1 in part 3, where you are showing $(a,b]$ is an $F_{\sigma}$-set?2017-10-01

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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,