Is it possible to divide the real line into two disjoint totally disconnected spaces of equal cardinality? Is it explicitly possible to take $ \mathbb R$ and to divide it into two sets (say $\mathbb A$ and $\mathbb B$), which are : disjoint ($\mathbb A \cap \mathbb B = \emptyset$), totally disconnected (i.e. contain no open intervals) (EDIT:) and for all intervals $(a,b)$, $|\mathbb A \cap (a,b)| = |\mathbb B \cap (a,b)| = \mathfrak c$
I asked my measure theory professor, and he said he couldn't think of why such a construction would be impossible, but whenever I try to create such two sets I always end up with undefinable points, nor did an internet search reveal anything about this.
 A: Consider the set
$$\big( (-\infty, 0) \cap \Bbb Q \big) \cup \big([0, \infty) \cap (\Bbb R - \Bbb Q) \big)$$
---that is, the union of the negative rationals and the nonnegative irrationals---and its complement.
A: Yes: 
$$A= \{ x=b.b_1b_2..b_n... | 0.b_2b_4b_6... \in \mathbb Q \} \\
B= \{ x=b.b_1b_2..b_n... | 0.b_2b_4b_6... \notin \mathbb Q \} $$
For any interval $(a,b)$ it is easy to construct onto functions from $A \cap(a,b) , B \cap(a,b) $ to $(0,1)$.
Indeed, pick some $a < \frac{k}{10^{2n}} <  \frac{k+1}{10^{2n}} < b$ and show that 
$$ x=b.b_1b_2..b_n... \to 0.b_{2n+1}b_{2n+3}...$$
is onto function from $A \cap(a,b)$ and $B \cap(a,b) $ to $(0,1)$.
This example has the roots in the example of a discontinuous function with the IVP that appears in Sierpiński (I think).
A: Let
$$\mathbb A=C+\mathbb Q=\{x+y:x\in C,\ y\in\mathbb Q\}$$
where $C$ is the Cantor set, and let $\mathbb B=\mathbb R\setminus\mathbb A.$
Since $\mathbb A$ and $\mathbb B$ are Borel sets ($F_\sigma$ and $G_\delta$ respectively), it will suffice to show that $\mathbb A\cap(a,b)$ and $\mathbb B\cap(a,b)$ are uncountable.
Since $\mathbb A$ has measure zero (as the union of countably many translates of $C$), $\mathbb B\cap(a,b)$ has positive measure, so it's uncountable.
Choose $q\in\mathbb Q\cap(a,b).$ Since every neighborhood of $0$ contains uncountably many points of $C,$ every neighborhood of $q$ contains uncountably many points of the set $C+q;$ in particular, the interval $(a,b)$ contains uncountably many points of the set $C+q\subseteq C+\mathbb Q=\mathbb A.$
