How many connected components are left after removing a line from the plane? Let $A \subseteq \mathbb{R}^2$ be a subset of the plane which is homeomorphic to $\mathbb{R}$. How many connected components does $\mathbb{R}^2 \setminus A$ have?
My conjecture is that only one or two components are possible. Is it true? How can I prove this?
(To see these options are in fact possible, consider $A=(0,1) \times \{ 0\}$ and $A=\mathbb{R} \times \{0\}$).
Update:
It says here* that the Jordan Brouwer separation theorem implies that a closed subset of $\mathbb{R}^n$ homeomorphic to $\mathbb{R}^{n-1}$ disconnects the space into two connected components. 
So, we need only to consider the case where $A$ is not closed in $\mathbb{R}^2$.

*Here is an explanation, following Stefan Hamcke's comment: 
Consider the homeomorphism $h:\mathbb{R}^{n-1} \to A$. We can view $h$ as a map $\mathbb{R}^{n-1} \to \mathbb{R}^{n}$. Since $A$ is closed (by assumption) $h$ is a closed map. A closed map having the property that the pre-image of every point is compact is proper (Proof). The requirement on pre-images is satisfied here trivially, since $h$ is injective. 
As a proper map, $h$ extends to the one-point compactifications, so we can view it as an embedding $h:\mathbb{S}^{n-1} \to \mathbb{S}^{n}$. Thus, theorem 2B.1 (Hatcher, Algebraic Topology) implies $\mathbb{S}^{n} \setminus h(\mathbb{S}^{n-1})$ has two connected components, which in turn implies $\mathbb{R}^n \setminus A$ have two components.
 A: It is proven in 
S. Eilenberg, An invariance theorem for subsets of $S^n$. 
Bull. Amer. Math. Soc. 47, (1941). 73–75.
that if $A, B\subset S^n$ are homeomorphic subsets of $S^n$ (which need not be closed) then the number of connected components of $S^n-A$ and $S^n -B$ is the same. 
It follows from this theorem, at least, that a bounded subset of $R^2$ homeomorphic to $R$ cannot separate $R^2$. If $R$ is properly embedded in $R^2$ then it separates $R$ in $2$ components (since it is closed). The remaining case, which I do not quite see how to deal with is when $A\subset R^2$ is homeomorphic to $R$, is unbounded but is not closed. I will come back to this when I have more time. 
There are relevant papers by Sitnikov from 1950s, where he proves duality theorems for general subsets of $S^n$, but I do not have access to them. 
Edit. Everything indeed works, one obtains, in fact a more general result:
Theorem. If $B\subset R^n$ is homeomorphic to $R^{n-1}$, $n\ge 2$, 
then either $B$ is closed and, hence separates $R^n$ in two components or $B$ does not separate. 
To prove this one considers the subset $B'=B\cup \{\infty\}$ in $S^n= R^n \cup \{\infty\}$. The separation properties of $B$ and $B'$ (in $R^n$ and $S^n$ resp.) are, of course, equivalent. There are two cases to consider: $B'$ is compact. Then everything follows from Jordan separation theorem in $S^n$. Assume, therefore, that $B'$ is not compact. One proves the following
Lemma. $H_{n-1}^c(B', {\mathbb Q})=0$, where $H_k^c$ denotes the Chech homology with compact support, i.e.
$$
\lim_K H_k(K, {\mathbb Q}) =0,
$$ 
where the direct limit is taken over all compacts $K\subset B'$ and $H_k$ denotes Chech homology (with rational coefficients).  
Given this lemma, one then uses Theorem 8 in 
S. Kaplan, Homology properties of arbitrary subsets of Euclidean spaces. 
Trans. Amer. Math. Soc. 62, (1947) 248–271.   
which is a more elaborate version of Eilenberg's paper. Here $B'$ as above is $S^n - A$ in Kaplan's notation. When you unravel what this theorem 8 says in our setting, it shows that if $B'$ is not connected, then $H^c_{n-1}(B')\ne 0$, which contradicts the Lemma. 
