Let A and B be compact connected subsets of the plane, homeomorphic to one another. Are their complements homeomorphic?(adsbygoogle = window.adsbygoogle || []).push({});

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?

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# Homeomorphic Complements

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