# Homeomorphism in the plane (Elementary?)

Let $U$ be an open and connected subspace of the Euclidean plane $\mathbb{R}^2$, $A$ be a union of $n$ disjoint subspaces in $U$ each of which is homeomorphic to the closed unit interval. Also, let $P$ be a union of $n$ disjoint subspace in $U$ each of which is closed, connected and contractible.

Then, are the two spaces $U \setminus A$ and $U \setminus P$ always homeomorphic?

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Are $A$ and $P$ both sets consisting of $n$ points? This is true for any two sets of $n$ points $F_1,F_2$: we have $U\setminus F_1\cong U\setminus F_2$. – Grumpy Parsnip Sep 24 '13 at 19:32
@GrumpyParsnip Thank you for the comment. I changed the question a bit. I hope this is more clear. – JW. Sep 24 '13 at 19:46
What if $n=1$ and $P$ consists of a single closed loop? Then $U\setminus A$ is connected and $U\setminus P$ is not. Or have I misunderstood? – GCD Sep 24 '13 at 19:49
@GCD Right.. I should have written that each of subspace in P is contractible (if this makes sense..) – JW. Sep 24 '13 at 19:52

This is true, but probably difficult to prove. It should follow from the classification of non compact surfaces. See this question. If you assume that the contractible subsets $P$ are the insides of simple closed curves, then the result will be easier to prove. One can use the Jordan Curve theorem to find a self homeomorphism of the plane where all of the subsets $P$ are turned into round circles, and it is not hard to prove that the complement of a round circle locally is homeomorphic to the complement of a point.