TITLE: Locally non-enumerable dense subsets of R
QUESTION [17 upvotes]: 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

REPLY [8 votes]: $\mathbb{Q}$ 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.