Locally non-enumerable dense subsets of R Today after lunch I was hungry for math problems so I started begging for some at the department and finally someone threw me this: Can $\mathbb{R}$ be partitioned into two non-countable dense subsets? It was a good starter, after a few minutes I got: Take the irrationals less than $0$ and the rationals greater than $0$, this is one subset, the complement of course works. Then the following question came into my mind: Can $\mathbb{R}$ be partitioned into two locally-non-countable dense subsets?
I'm still hungry 
 A: $\mathbb{Q}$ has $2^{\aleph_0}$ many cosets in the additive group $\mathbb{R}$.  Therefore $\mathbb{R}$ can be partitioned into 2 sets each of which is a union of $2^{\aleph_0}$ many cosets of $\mathbb{Q}$.  Each coset is dense, so each of these sets is locally uncountable.
A: Updated in light of the comment:  How about we express all numbers in base 3.  For set $A$ take all the numbers that eventually have $1$'s in all the even places behind the base 3 point-things like $0.x1x1x1x1x1\ldots$ (where the $x$'s can be any digit and need not be the same), $0.xxx1x1x1x1x1x1\ldots$, $0.xxxxx1x1x1x1x1\ldots$ and for $B$ take everything else.
A: The answer is yes. I think that one way to do it is to iteratively construct Cantor sets in each interval $[n,n+1]$, $n\in\mathbb{Z}$, i.e., for a given interval, construct a Cantor set, then construct a Cantor set in each of the (countably many) disjoint segments whose union is the complement of that Cantor set intersected with the original interval, and repeat this process. In the end you would consider the union of those Cantor sets, and the complement of that union.
