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 am trying to prove that the set of irrational numbers $\mathbb{R} \setminus \mathbb{Q}$ is not an $F_{\sigma}$ set. Here's my attempt:

Assume that indeed $\mathbb{R} \setminus \mathbb{Q}$ is an $F_{\sigma}$ set. Then we may write it as a countable union of closed subsets $C_i$: $$ \mathbb{R} \setminus \mathbb{Q} = \bigcup_{i=1}^{\infty} \ C_i $$ But $\text{int} ( \mathbb{R} \setminus \mathbb{Q}) = \emptyset$, so in fact each $C_i$ has empty interior as well. But then each $C_i$ is nowhere dense, hence $ \mathbb{R} \setminus \mathbb{Q} = \bigcup_{i=1}^{\infty} \ C_i$ is thin. But we know $\mathbb{R} \setminus \mathbb{Q}$ is thick, a contradiction.

This seems a bit too simple. I looked this up online, and although I haven't found the solution anywhere, many times there is a hint: Use Baire's Theorem. Have I skipped an important step I should explain further or is Baire's Theorem used implicitly in my proof? Or is my proof wrong? Thanks.

EDIT: Thin and thick might not be the most standard terms so:

Thin = meager = 1st Baire category

share|improve this question
    
I have added (baire-category) tag, since the proof outlined in the post is about meager sets and it uses Baire category theorem. –  Martin Sleziak Jan 10 at 7:44
add comment

2 Answers 2

up vote 8 down vote accepted

Your solution is correct. You could also argue that $\mathbb{R} = \bigcup_{i =1}^{\infty} C_{i} \cup \bigcup_{q \in \mathbb{Q}} \{q\}$, so by Baire one of the $C_{i}$ must have non-empty interior, contradicting the fact that $\mathbb{R} \smallsetminus \mathbb{Q}$ has empty interior.

share|improve this answer
    
@milcak: I forgot to mention that you assume Baire implicitly when asserting that $\mathbb{R}\smallsetminus\mathbb{Q}$ is thick –  t.b. Mar 12 '11 at 2:59
add comment

Suppose $\mathbb{R} \setminus \mathbb{Q}$ is an $F_{\sigma}$ set. Then $\mathbb{Q}$ is a $G_{\delta}$ set which is a contradiction. In other words, suppose $$ \mathbb{R} \setminus \mathbb{Q} = \bigcup_{i=1}^{\infty} \ C_i $$ Then by Demorgan's Laws $$ \mathbb{Q} = \bigcap_{i=1}^{\infty} \ C_i^{c} $$ which is a contradiction since $\mathbb{R} \setminus \mathbb{Q}$ is a countable intersection of open sets and we know that the intersection between the rational is irrationals is $\emptyset$. So $\emptyset$ is a countable intersection of open dense subsets which contradicts Baire's category theorem.

share|improve this answer
1  
What does this line mean: " we know that the intersection between the rational is irrationals is ∅"? –  Soarer Mar 11 '11 at 7:41
    
isn't it obvous? $(\mathbb{R} \setminus \mathbb{Q}) \cap \mathbb{Q}= \emptyset$.. –  Shir Sivroni Dec 29 '13 at 16:22
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.