Disjoint Exhaustive Subsets with Alternate Elements and equal Cardinality Suppose  $U$ is an ordered infinite set, I want to construct subsets $A$ and $B$ such that:
(1) (Disjoint) $A \cap B = \phi $ 
(2) (Exhaustive) $A \cup B = U$ 
(3) (Alternate elements) $\forall x,y \in A, x<y, \exists z \in B $ such that $x < z<y$ and vice versa for B. 
(4) (Same cardinality) $|A|=|B|$ (i.e. There exists a bijection $h : A\rightarrow B$).
Can we construct $A$ and $B$ for $U = \mathbb{R}$ (set of real numbers)

For $U = \mathbb{Z}$ (integers), we can take $A=\{2z\}, B = \{2z+1\}, z \in \mathbb{Z}$
For ${U} = \mathbb{Q}$ (set of rationals), we can define $A = \{(2m)/n\}, B = \{(2m+1)/n \}$ (where $m$ and $n$ have no common factor)
My main question is what if $U=\mathbb{R}$ (set of real numbers)? Can we construct $A$ and $B$ satisfying all 4 properties? And further on any arbitrary interval?
 A: We define the relation $\sim$ on $\mathbb{R}$ like so:
$$
x\sim y\,\leftrightarrow\, x-y\in\mathbb{Q}
$$
One can easily confirm that $\sim$ is an equivalence relation. one can also easily confirm:
$$
[x]_\sim =\{x+q\,:\,q\in\mathbb{Q}\}=f_x(\mathbb{Q})
$$
Where $f_x(y):=x+y.$ Using the axiom of choice, let us select a single representative from each equivalence class in $\mathbb{R}\,/\sim\,$. Let us denote the set of these representatives by $R$. 
We already have a partition $\{A,B\}$ of $\mathbb{Q}$ satisfying (1)-(4). Using this partition, define: 
$$
\textbf{A}:=\bigcup_{x\in R} f_x(A) \;\;\;\;,\;\;\;
\textbf{B}:=\bigcup_{x\in R} f_x(B)
$$
One can easily confirm that $\{\textbf{A}, \textbf{B}\}$ is a partition of $\mathbb{R}$ satisfying (1)-(4). In fact, $\{\textbf{A}, \textbf{B}\}$ satisfies (1)-(4) on any interval in $\mathbb{R}$.
A: How about letting $A$ be the set of nonnegative rationals and negative irrationals, while letting $B$ be the set of negative rationals and nonnegative irrationals?
