TITLE: $A \subset \mathbb{R}$ such that $A$ is homeomorphic to $\mathbb{R} \setminus A$
QUESTION [6 upvotes]: For 2 topological spaces $(X,T_X)$ and $(Y,T_Y)$, I write $X \simeq Y$ if $X$ and $Y$ are homeomorphic. If $A \subset X$, I always endow $A$ with the subspace topology.
I wonder if it is true that : 

$\exists A \subset \mathbb{R}$ such that $A \simeq \mathbb{R} \setminus A$.

It doesn't hold for connected sets (i.e intervals, either because $\mathbb{R} \setminus A$ would not be connected or one of $A$ or $\mathbb{R} \setminus A$ would not be connected after removing a point whereas the other one may still be), nor for compact sets (as $\mathbb{R}$ would be the union of two compact sets, so compact), nor for countable or co-countable sets (as $A$ and $\mathbb{R} \setminus A$ would have different cardinalities).
If this is true, I wonder if moreover : 

$\exists A \subset \mathbb{R}$ open such that $A \simeq \mathbb{R} \setminus A$.

Thanks.

REPLY [4 votes]: Noah answered the first question.
For the second question, the answer is no.
Because if $A$ were homeomorphic to $\mathbb{R} \backslash A$ with homeomorphism being $f$, it would follow that $A \to \mathbb{R} \backslash A \hookrightarrow \mathbb{R}$ is an open map (invariance of domain), hence $\mathbb{R} \backslash A$ would be open in $\mathbb{R}$. Hence, it should be $\mathbb{R}$ (since it is also closed, being the complement of an open set, and clearly cannot be the empty set). It follows that $A$ is empty, an absurd.