TITLE: Uncountable dense subset whose complement is also uncountable and dense
QUESTION [7 upvotes]: Possible Duplicate:
Locally non-enumerable dense subsets of R 

This may be a standard result... I don't know... anyway...
Does there exists an uncountable dense subset of $\mathbb{R}$ whose complement is also uncountable and dense?

REPLY [14 votes]: Funny, I asked myself the same question at one point! This should work:
$$A:=((-\infty,0]\cap\mathbb{Q})\cup ((0,+\infty)\cap(\mathbb{R}\setminus\mathbb{Q})),$$
$$B:=((-\infty,0]\cap(\mathbb{R}\setminus\mathbb{Q}))\cup ((0,+\infty)\cap\mathbb{Q}) $$
They are both uncountable and dense, and form a partition of $\mathbb{R}$.

REPLY [6 votes]: Take any uncountable set $A_0$ with measure $0$, e.g., the Cantor set. Let $$A = \bigcup_{q \in \mathbb{Q}} (A_0 + q) = \{a + q : a \in A_0, q \in \mathbb{Q}\}.$$
Then $A$ is uncountable, dense and it still has measure $0$ since it is the union of countably many translates of $A_0$. Since the complement of a measure $0$ set is always uncountable and dense, this set $A$ fits the bill.