Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
6  
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
show 7 more comments

1 Answer 1

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,

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.