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I am working on Exercise 1.1 and I think the way to do this would be to show that open sets are homeomorphic to $\mathbb{R}^n$ or open balls in $\mathbb{R}^n$. Is this even true? I'm not sure how to go about proving it.

BTW: The exercise is from Lee's Introduction to Smooth Manifolds.

Luigi Traino
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iYOA
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    Here's a hint: An open set $U$ in $\mathbb R^n$ has the property that for any $p\in U$, there is a neighborhood of $p$ contained in $U$ which is homeomorphic to $\mathbb R^n$ (a ball suffices). – Milo Brandt Mar 31 '15 at 03:13
  • Not all open subsets of $\mathbb{R}^n$ are homeomorphic to $\mathbb{R}^n$. For example, $(0, 1)\cup(2, 3)$ is open in $\mathbb{R}$ but it certainly isn't homeomorphic to $\mathbb{R}$ as it is not connected. – Michael Albanese Mar 31 '15 at 03:13
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    To answer your question in a nutshell. Yes, open balls are homeomorphic to $\mathbb R^n$. First show $(0,1)$ is homeomorophic to $\mathbb R$ and you'll see why. – Gregory Grant Mar 31 '15 at 03:16
  • A more interesting question might be whether or not any contractible open subset of $\mathbb{R}^n$ is homeomorphic to $\mathbb{R}^n$? I think it's true for convex open sets, no? – shalop Mar 31 '15 at 03:18
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    @Shalop there's a famous counterexample for $n=3$ called the Whitehead manifold. – Kevin Carlson Mar 31 '15 at 03:33
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    Oh wow! Now that's really awesome imo. – shalop Mar 31 '15 at 03:36
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    @Shalop: In dimension 2, essentially by the uniformization theorem of Riemann surfaces, every noncompact contractible surface is homeomorphic to $\Bbb R^2$. In addition, every compact contractible surface is homeomorphic to the closed disc $D^2$; and every compact contractible 3-manifold is homeomorphic to $D^3$. This all falls apart in dimension 4, where there are loads of compact contractible 4-manifolds. –  Mar 31 '15 at 05:45

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Open balls in $\Bbb R^n$ are homeomorphic to $\Bbb R^n$, but it’s not true in general that (non-empty) open sets in $\Bbb R^n$ are homeomorphic to $\Bbb R^n$: $\Bbb R^n$ and its open balls are connected, but there are lots of open sets in $\Bbb R^n$ that are not connected. However, if $U$ is an open nbhd of $x$ in $\Bbb R^n$, then there is an open ball $B$ such that $x\in B\subseteq U$, so if every point $M$ has a nbhd homeomorphic to some open $U\subseteq\Bbb R^n$, then it automatically has one homeomorphic to an open ball in $\Bbb R^n$. The other direction is trivial, since every open ball in $\Bbb R^n$ is an open set in $\Bbb R^n$.

Finally, to prove that an open ball in $\Bbb R^n$ is homeomorphic to $\Bbb R^n$ itself, it suffices to prove it for the open unit ball centred at the origin. Consider the map from the open unit ball to $\Bbb R^n$ that sends $x$ to $\left(\tan\frac{\pi|x|}2\right)x$.

Brian M. Scott
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  • Okay thanks! So it seems my idea does not work. How would I go about solving the exercise then? – iYOA Mar 31 '15 at 04:49
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    @iYOA: You're welcome! That's what I was getting at with the However sentence: the moment you have a nbhd $N$ of $x$ that's homeomorphic to some open $U\subseteq\Bbb R^n$ by a homeomorphism $h$, find an open ball $B$ such that $h(x)\in B\subseteq U$, and look at $h^{-1}[B]$: it's a nbhd of $x$ homeomorphic to an open ball. – Brian M. Scott Mar 31 '15 at 05:22
  • ohh okay I see now. – iYOA Apr 01 '15 at 00:18
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    Hi @BrianM.Scott , is every open connected subspace of an euclidean space homeomorphic to the whole space? – Anguepa May 01 '17 at 15:56
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    @Anguepa, Since a proper subspace is not a open set of the whole space, and it's well known that $\mathbb{R}^m$ and $\mathbb{R}^n$ not homeomorphic if $m\ne n$, I guess that you mean open connected subset (not subspace), then open connected subsets with one or many holes are not homeomorphic to the whole space. – ydhhat Feb 25 '20 at 16:36