partitioning $\mathbb{R}$ into two dense sets of equal cardinality Just out of curiosity, is it possible to partition $\mathbb{R}$ into two dense sets of equal cardinality?
I was thinking something like this: Let $S$ be the basis of $\mathbb{R}$ over $\mathbb{Q}$. Then $S$ is of equal cardinality as $\mathbb{R}$. Let $A$ be an uncountable subset of $S$ such that $S\setminus A$ is also uncountable. Such a set exists, for example if $f$ is a bijection from $S$ to $\mathbb{R}$, put $a$ in $A$ if and only if $f(a)>0$. Now let $T$ be the subset of $\mathbb{R}$ such that $t\in T$ if and only if $tq\in A$ for some rational number $q$. Then $T$ and $\mathbb{R}\setminus T$ gives the desired partition.
But, as far as I know, the existence of a basis of $\mathbb{R}$ over $\mathbb{Q}$ depends on the Axiom of Choice. Is it possible to prove the above statement without assuming the Axiom of Choice?
 A: First set: irrational numbers outside $[0,1]$ and rational numbers in $[0,1]$.
Second set (the complement): rational numbers outside $[0,1]$ and irrational numbers in $[0,1]$.
A: Let $D$ be the set of all dyadic rationals (rationals of the form $\frac{a}{2^i}$ with $a$ an integer, $i$ a nonnegative integer). Let $E=\mathbb{Q}-D$. Both $D$ and $E$ are countable dense subsets of $\mathbb{R}$.
Now let $A=D\cup\Bigl( (-\infty,0)-\mathbb{Q}\Bigr)$, and $B=E\cup\Bigl( (0,\infty)-\mathbb{Q}\Bigr)$. That is, $A$ is the set of all dyadic rationals and all negative irrationals; $E$ is the set of all non-dyadic rationals and all positive irrationals. Then $A\cap B=\emptyset$, $A\cup B=\mathbb{R}$, and since both $A$ and $B$ contain dense subsets of $\mathbb{R}$, they are dense.
A: How about $[(\mathbb{R}\setminus \mathbb{Q})\cap(-\infty,0)]\cup (\mathbb{Q}\cap[0,\infty))$ and its complement?
A: I'll add a few solutions for a stronger problem: generating an infinite number of such subsets of $\mathbb{R}$ such that their union is $\mathbb{R}$, and such that each one is dense and uncountable in every closed interval.  Obviously, these solutions can be used to easily generate solutions for a finite number of such sets.
First, consider equivalence classes generated $\mathbb{R}/\mathbb{Q}$  Specifically, start with the number 0, and define $S_1$ to be a maximal set of reals such that no linear combination gives a non-0 rational.  By establishing a counting order for the rationals and letting $S_n = {r_n + x : x \in S_0}$ where $r_n$ is your $n$th rational, you establish an infinite number of such sets. 
The second and third solutions require writing your decimal expansion in binary.  Ignore expansions ending in an infinite string of 1s (to avoid the $.999 \dots = 1$ problem).  For the second, define $r = \lim_{n \to \infty}\frac{k}{n}$ where $k$ denotes the number of 1s in the first $n$ digits of the binary expansion.  Then, define an extra set for numbers for which that limit does not converge.  
My third solution defines $S_n$ to be the set of all numbers whose decimal expansions contain an infinite number of repetitions of strings of $n$ 1s, but a finite number of repetitions of strings of more than $n$ 1s.  Then, define $S_0$ to be the set of all numbers that do not fit in the other sets (0, or have an infinite amount of arbitrarily long strings of 1s).  
A: Does $(0,1)\cup\mathbb{Q}$ work?
