The set of irrational numbers is not a $F_{\sigma}$ set. I want to proove that the set of irrational numbers is not a $F_{\sigma }$ set and also the set of rational numbers is not a $G_{\delta}$ set using Baire theorem.
I started with saying that $\mathbb{R}$\ $\mathbb{Q}$ is a $G_{\delta}$ set because $\mathbb{R}$\ $\mathbb{Q}$ = $\bigcap_{ q\in\mathbb{Q}}^{}$$\mathbb{R}$\ {q}. 
So, if $\mathbb{R}$\ $\mathbb{Q}$ were also $F_{\sigma }$ then $\mathbb{Q}$, as a complementary, would be a $G_{\delta }$ set. 
Both $\mathbb{R}$\ $\mathbb{Q}$ and $\mathbb{Q}$ are dense in $\mathbb{R}$ and ($\mathbb{R}$\ $\mathbb{Q}$) $\cup $ $\mathbb{Q}$ = $\mathbb{R}$, so the whole space can be written as a sum of two dense and $G_{\delta }$ sets. 
I've read that this is a contradiction with Baire theorem, but I can't see this. 
 A: The Baire category theorem says that if $(X,d)$ is a complete metric space and $(U_n)_{n=1}^{\infty}$ is a sequence of open, dense sets in $X$, then $\bigcap_{n=1}^{\infty} U_n$ is also dense in $X$.
If $\mathbb{Q}$ were $G_{\delta}$, say $\mathbb{Q}=\bigcap_{n=1}^{\infty} U_n$ where each $U_n$ is open, then necessarily each $U_n$ is dense in $\mathbb{R}$.  Similarly, write $\mathbb{R} \setminus \mathbb{Q}=\bigcap_{n=1}^{\infty} V_n$ where each $V_n$ is open and dense in $\mathbb{R}$.  Then we have the intersection of all the $U_n$'s and $V_n$'s being empty, which contradicts Baire as it is supposed to be dense in $\mathbb{R}$.  Thus you cannot have both $\mathbb{Q}$ and $\mathbb{R} \setminus \mathbb{Q}$ to be $G_{\delta}$.
In particular this means that $\mathbb{Q}$ is not $G_{\delta}$.
A: Essentially you have, supposing $\mathbb{Q}$ is a $G_\delta$, that
$$
\emptyset =( \mathbb{R} \setminus \mathbb{Q} )\cap \mathbb{Q} = \left( \bigcap_{q \in \mathbb{Q}} \mathbb{R} \backslash\{ q \}\right) \cap \left( \bigcap_{n=1}^\infty A_n \right)
$$
Thus, that the empty set can be expressed as a countable intersection of open dense sets of $\mathbb{R}$, however since $\mathbb{R}$ is a complete  metric space, the Baire  Category theorem implies that $\emptyset$ must be dense in $\mathbb{R}$, a contradiction!!
