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It is a standard result that the open ball $$B^2=\{(x,y)\in\mathbb{R}^2:x^2+y^2<1\}$$ is homeomorphic to $\mathbb{R}^2$ itself. Also, distorting $B^2$ by any continuous bijective transformation will also give a path-connected open set homeomorphic to $\mathbb{R^2}$. So, I was wondering the following:

Question: Are all path-connected open subsets of $\mathbb{R}^2$ homeomorphic to $\mathbb{R}^2$?

I state the question in $\mathbb{R}^2$ for simplicity, but of course we could ask the same question in $\mathbb{R}^n$ which I suspect will have the same answer.

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Take $\Bbb R^2-0$. A rather weird result is that every nonempty proper simply connected subset of $\Bbb R^2$ is homeomorphic, actually biholomorphic, to $B(0,1)$. This is the Riemann mapping theorem. It is rather weird because a biholomorphic mapping preserves angles and orientation, and one can produce rather erratic simply connected subspaces of $\Bbb R^2$ which can be mapped "nicely" into $B(0,1)$. As an example, take $(0,1)\times (0,1)$, and from the point $(0,0)$ draw a line of length say $1/n$ and angle of $2\pi /n$ radians for each $n$. This looks like a square with a "duster deleted in the corner" and is simply connected, but it is hard to imagine a conformal mapping of this onto the unit disk.

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  • $\begingroup$ Right, indeed. Thanks for the quick answer. $\endgroup$ – Chern Dec 11 '14 at 22:45
  • $\begingroup$ Are all simply-connected open subsets of $\mathbb{R}^2$ homeomorphic to $\mathbb{R}^2$? $\endgroup$ – Chern Dec 11 '14 at 22:46
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    $\begingroup$ @Chern Yes! The Riemann mapping theorem states that there is a holomorphic map from any such region to the unit disk. $\endgroup$ – D Wiggles Dec 11 '14 at 22:47
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    $\begingroup$ @Spenser It does not generalize to $\mathbb{R}^n$ for $n\geq 3$ because, for example, $\mathbb{R}^n-\{0\}$ is simply connected and open. But there is maybe a condition on the homology of the set that would be enough though. I don't know. $\endgroup$ – Chern Dec 11 '14 at 23:14
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    $\begingroup$ @Chern The Whitehead manifold is an open contractible subset of $\Bbb R^3$ that's not homeomorphic to $\Bbb R^3$. So if you want to generalize it, you need to use things that aren't homotopy invariant. I wouldn't hold out much hope. $\endgroup$ – user98602 Dec 11 '14 at 23:39
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Take any small enough open neighborhood of the $n$-sphere $S^n$, with $n\ge 1$, in $\mathbb R^{n+1}$. That is a path-connected open subset of the ambient space $\mathbb R^{n+1}$ which is not homeomorphic to $\mathbb R^{n+1}$.

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