Are the complements of homeomorphic compact connected subsets homeomorphic?

[AKG]

Let A and B be compact connected subsets of the plane, homeomorphic to one another. Are their complements homeomorphic?

I think they are. If A and B are empty, there's nothing to prove. Otherwise, let C denote the complement of A, let D denote the complement of B. We look at the connected components of C and D. Each will have one unbounded component with one hole in it. The unbounded component of C should be homeomorphic to the unbounded component of D. The rest of the components of C are bounded and, I believe, contractible and/or simply connected (are the two equivalent in the plane?). I want to argue that since A and B are homeomorphic, C and D have an equal number of these bounded components. I also want to argue that any two bounded, open, connected, contractible and/or simply connected subsets of the plane are homeomorphic. Actually, it might even be that any two open contractible subsets of the plane are homeomorphic. If this is so, then in technical terms, I believe we can find a bijection f from the collection of connected components of C to the collection of connected components of D such that X is homeomorphic to f(X). Since each X is open, I would like to argue that this induces a homeomorphism from all of C to all of D.

Is the above good?

 

[Hurkyl]

Is this a homework question, or something you want to use to prove a homework question?

 

[AKG]

Neither...

 

[matt grime]

Then why is it in the homework section?

 

[AKG]

Oops, force of habit I guess. It should be moved, but it doesn't matter to me.

 

[StatusX]

I believe this theorem is true, although it does not generalize to R³, with alexander's horned sphere as a counterexample. Your argument isn't quite rigorous, and it will take some work for it to handle pathological examples similar to the horned sphere that may arise in 2D.

 

[Palindrom]

Any two open, simply connected subsets of the plane are homeomorphic. We know that thanks to the Riemann mapping theorem in complex analysis.

 

[AKG]

It seems the hard part to prove is the following:

Suppose A and B are connected compacts in the plane, homeomorphic to one another. If A^C has a connected component that is simply connected, so does B^C.

It seems obvious, but how would one go about proving it?