Is a dense and co-dense subset $G_\delta$ or co-$G_\delta$ Let $A \subset \mathbb{R}$ such that $A$ and $A^C$ are both dense.
By Baire's Theorem at most one of $A$ and $A^C$ is $G_\delta$ (i.e. a countable intersection of open sets)
I couldn't think of an easy example of such A, where neither $A$ nor $A^C$ is $G_\delta$.
So is there a dense and co-dense subset of $\mathbb{R}$, which is neither $G_\delta$ nor co-$G_\delta$?
 A: Yes. In fact, by a counting argument, most dense co-dense sets are neither $G_\delta$ nor $F_\sigma$. The point is that there are exactly as many $G_\delta$ or $F_\sigma$ as there are real numbers, but there are as many dense co-dense sets as there are sets of real numbers.
In somewhat more detail: There are only countably many rationals, so there are countably many open intervals with rational endpoints. Any open set is the union of a countable sequence of such intervals, so there are at most as many open sets as countable sequences of pairs of rationals, $|2^{\mathbb N}|$, but this is precisely the same size as $\mathbb R$. Any $G_\delta$ set is an intersection of a countable sequence of open sets, so there are at most as many $G_\delta$ sets as countable sequences of reals, $|\mathbb R^{\mathbb N}|=|\mathbb R|$. On the other hand, any singleton is $G_\delta$. This shows that there are exactly as many $G_\delta$ sets as there are reals. Since a set is $F_\sigma$ iff its complement is $G_\delta$, there are also as many $F_\sigma$ sets as there are reals.
On the other hand, given any subset $A$ of $\mathbb R\setminus\mathbb Q$ there is a dense codense subset of $\mathbb R$ whose intersection with the irrationals is $A$: Split $\mathbb Q$ into two dense sets $C,D$, and consider $C\cup A$. This shows that there are at least as many dense codense sets as there are subsets of the irrationals. But there are as many irrationals as there are reals.  
Finally, we can also find (arguing in $\mathsf{ZF}$) examples of dense codense Borel sets that are neither $F_\sigma$ nor $G_\delta$. We can even do this at the next level of complexity, the $G_{\delta\sigma}$ sets, those sets that are a countable union of $G_\delta$ sets. One way of arguing this is to note that for any set $C$ of rationals, if $E$ is any subset of the irrationals that is $G_{\delta\sigma}$ in the subspace topology, then $C\cup E$ is $G_{\delta\sigma}$ as a subset of $\mathbb R$. It suffices to exhibit such a set $E$ that is neither $F_\sigma$ nor $G_\delta$. For an example, see Theorem 2.1 of 

Arnold Miller. Long Borel hierarchies, Math Logic Quarterly, 54, (2008), 301-316.

Arnie's example is a subset of Cantor space $2^\omega$. To turn this into a set of irrationals it is enough to note that $2^\omega$ is homeomorphic to the set of all reals whose continued fraction expansion is 
 $$ a_0 + \frac 1{a_1+\frac 1{a_2+\dots}} $$
where each $a_i$ is either $1$ or $2$.
