Uncountable dense subset whose complement is also uncountable and dense 
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?
 A: 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.
A: 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 $\mathbb{R}$.
A: Probably not the best approach, but you can also try this: Let $B$ be a basis of $\mathbb R$ over $\mathbb Q$. Write $B=B_1 \cup B_2$ with $B_1 \cap B_2 =\emptyset$ and $B_1, B_2$ of the same cardinality.
Then $V= span_{\mathbb Q} B_1$ will do it. It is uncountable and dense ($B_1$ in uncountable, thus it contains two elements with irrational ration) and $\mathbb R \backslash V$ contains $W= span_{\mathbb Q} B_2$.

Added a Second example
Here is another example:
$$A:= \{ x=m.x_1x_2...x_n...| 0.x_2x_4x_6... {\rm \, is \, periodic \,} \}\,.$$
More exactly, $A$ consists of all those real numbers for which, the digits after $.$ of even order form a periodic sequence.... 
