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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 $R^n$ or open balls in $R^n$. Is this even true? I'm not sure how to go about proving it.

btw, the exercise is from Lee's Smooth Manifolds book.

  • $\begingroup$ 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). $\endgroup$ – Milo Brandt Mar 31 '15 at 3:13
  • $\begingroup$ 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. $\endgroup$ – Michael Albanese Mar 31 '15 at 3:13
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    $\begingroup$ @Shalop there's a famous counterexample for $n=3$ called the Whitehead manifold. $\endgroup$ – Kevin Carlson Mar 31 '15 at 3:33
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    $\begingroup$ Oh wow! Now that's really awesome imo. $\endgroup$ – Shalop Mar 31 '15 at 3:36
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    $\begingroup$ @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. $\endgroup$ – user98602 Mar 31 '15 at 5:45

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 $\left(\tan\frac{\pi|x|}2\right)x$.

  • $\begingroup$ Okay thanks! So it seems my idea does not work. How would I go about solving the exercise then? $\endgroup$ – iYOA Mar 31 '15 at 4:49
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    $\begingroup$ @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. $\endgroup$ – Brian M. Scott Mar 31 '15 at 5:22
  • $\begingroup$ ohh okay I see now. $\endgroup$ – iYOA Apr 1 '15 at 0:18
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    $\begingroup$ Hi @BrianM.Scott , is every open connected subspace of an euclidean space homeomorphic to the whole space? $\endgroup$ – Anguepa May 1 '17 at 15:56

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