analogue of the Jordan curve theorem for closed curve I wonder whether there are some generalization of the Jordan curve theorem :  Can the theorem be generalized into closed curve?
$C$ is a closed curve , then $\Bbb R^2\setminus C$ consists of several connected components. One of these components is unbounded  and the rest  is bounded，and  the boundary of each component is but a small  part of the curve $C$.
$\Bbb R^2\setminus C$ is a non-empty open set. As such,  it is the disjoint union of domains, the components of $\Bbb R^2\setminus C$
However I am not quite sure  whether it is right or not. If it is wrong, can you give  an counterexample? 
Thanks a lot!
 A: Can you specify what the components you're talking about are?
I am pretty sure there is no such neat generalization. I know this from ODE where in 2-dimensions the dynamical systems are so easily classified (Poincare-Bendixon Theorem), but there arises chaos in 3-dimensions and above. 
A: Let $C\subset {\mathbb R}^2$ be a compact subset; then  $A= {\mathbb R}^2 -C$  is an open subset of the plane. Then each connected component of $A$ is connected (see this question) and open. The latter is true for connected components $A_o$ of any open subset $A$ of any locally connected topological space $X$. Indeed, for each $x\in A_o$ pick a connected open neighborhood $U\subset A$ of $x$ in $X$. Then, since $x\in U$ and $U$ is connected, it follows from the definition of connected components that $U\subset A_o$.
This answers your second question.  
Now, for each component $A_o\subset A$, $\partial A_o= cl(A_o)\cap cl({\mathbb R}^2 - A_o)=  cl(A_o) \cap ({\mathbb R}^2 - A_o)$. I claim that $\partial A_o$ is contained in $C$. Indeed, 
$$
{\mathbb R}^2 - A_o= C \sqcup \coprod_{i\in I} A_i
$$
where $A_i, i\in I$, are the components of $A$ different from $A_o$. Since each $A_i$ as above is open, its intersection with $cl(A_o)$ is empty. Therefore, 
$$
cl(A_o)\cap \amalg_{i\in I} A_i=\emptyset. 
$$
Hence, $\partial A_o\subset C$. This answers your first question. 
A: No there need not be several components, take for example something like the Peano curve, while that specific one is not closed one could use a similar approach to construct a closed square filling curve. That one would definitely show that $\mathbb R\setminus C$ need not have more than one component.
The rest of it is not especially controversial and follows from Weierstrass' extrem value theorem. To see that precisely one of the components of the complement is unbounded you use the fact that the curve is bounded (since there is a point on it with maximal modulus). And that the complement is open is also shown with WEVT since for a point outside the curve there must be a nearest point on the curve and hence a positive distance to the curve.
Finally that the boundary of the components are part of the curve follows from that the components are open. So if a point $x$ on the boundary of of one of the sets $A$ would be within another component $B$ then there would be a neighbourhood $N$ of $x$ such that $N\subseteq B$, but since $x$ is a limit point of $A$ we would have $N\cap A\ne\emptyset$. This would mean that $A$ and $B$ intersects.
