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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

Martin Sleziak
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milcak
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  • I have added ([tag:baire-category]) tag, since the proof outlined in the post is about meager sets and it uses Baire category theorem. – Martin Sleziak Jan 10 '14 at 07:44

2 Answers2

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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.

t.b.
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  • @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 02:59
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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.

NebulousReveal
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    What does this line mean: " we know that the intersection between the rational is irrationals is ∅"? –  Mar 11 '11 at 07:41
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    isn't it obvous? $(\mathbb{R} \setminus \mathbb{Q}) \cap \mathbb{Q}= \emptyset$.. – topsi Dec 29 '13 at 16:22