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I know that open balls are homeomorphic to the entire Euclidean space, and any convex open set can be proved to be homeomorphic to the entire Euclidean space. So I was wondering if all the open sets in $\mathbb{R}^n$ are homeomorphic?

xzhu
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    No ... there are for example disconnected open sets ... – martini Dec 07 '12 at 17:25
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    Even with connected sets, they might have different kinds of "holes" in them. – Jonas Meyer Dec 07 '12 at 17:26
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    This is not the only obstruction. See https://math.stackexchange.com/questions/427787/characterization-of-the-subsets-of-euclidean-space-which-are-homeomorphic-to-the?rq=1 for the exact obstructions for open subsets to be homeomorphic to the space itself. – Christian May 03 '18 at 10:02

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No. For example, the open annulus $$\{(x, y) \in \Bbb R^2 : 1 < x^2 + y^2 < 2\}$$ is not simply connected, but the open ball $$\{(x, y) \in \Bbb R^2 : x^2 + y^2 < 1\}$$ is.

Henry T. Horton
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