Can we halve the real number Intuitive formulation:Can we halve the real numbers in a way that changes its density by half and does not alter its analytic properties.
Formal statement:Is there a subset $A$ of $\mathbb{R}$ such that $A$ is dense and $$|A\cap[a,b]|=|(\mathbb{R}-A)\cap[a,b]|=|\mathbb{R}\cap[a,b]|$$ for all $a,b\in\mathbb{R}$
 A: Suppose $A \subset \mathbb{R}$ is dense and complete. Then its closure by assumption is $\mathbb{R}$, i.e. its limits points together with $A$ is all of $\mathbb{R}$. A limit point of $A$ is the limit of some convergent (in $\mathbb{R}$) sequence of $A$. A convergent sequence is Cauchy and so this sequence is Cauchy in $A$. $A$ is complete so the limit is in $A$. So the closure of $A$ is $A$, i.e. $A=\mathbb{R}$. 
A: Yes. Consider the Cantor set $C \subset [0, 1]$, which is in particular uncountable. The set sum
$$X := \mathbb{Q} + C = \{q + c : q \in \mathbb{Q} + c \in C\}$$
is dense in $\mathbb{R}$ and uncountable, and every intersection $X \cap [a, b]$, $a < b$ is uncountable. On the other hand, $X$ is a union the countably many nowhere dense sets $q + C$, $q \in \mathbb{Q}$, so it is meager, and in particular its complement $\mathbb{R} - X$ has uncountable intersection $(\mathbb{R} - X) \cap [a, b]$ for all $a < b$.
On the other hand $X$ has measure zero (it is a countable union of measure-zero sets), so even though this example satisfies the given condition that the cardinalities of $X \cap [a, b]$ and $(\mathbb{R} - X) \cap [a, b]$ are the same for all $a < b$, this might not be a "halving" in the sense you mean.
Perhaps a more interesting condition is that a given set $Y \subset \mathbb{R}$ (1) be measurable w.r.t. the standard measure $\mu$ on $\mathbb{R}$ and (2) satisfy $$\mu(Y \cap [a, b]) = \frac{1}{2} \mu([a, b])$$
for all $a \leq b$ (of course, $\mu([a, b]) = b - a$).
