Are any two open, connected subsets of $\Bbb{R}^n$ homeomorphic? This seems intuitively true.
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1See: necessary and sufficient conditions to be homeomorphic to $\mathbb R^n$. – Feb 25 '13 at 18:04
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1Think about donuts and disks. – Neal Feb 25 '13 at 18:53
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1 $\neq$ 0 $\neq$ 8
(Think of these as being open subsets of the plane. The set consists of the area that is colored black. Remove the infinitely thin boundary)
Rasmus
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I am not sure how to think of these as open subsets of the plane? Certainly not 1... – Zev Chonoles Feb 25 '13 at 17:58
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I think the idea is that you should draw the symbols in the plane, and then take small open neighbourhoods of them. – mdp Feb 25 '13 at 17:59
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2The set consists of the area that is colored black. Remove the infinitely thin boundary. – Rasmus Feb 25 '13 at 18:00
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@julien: I don't think so. On the board, you would of course make a nicer drawing. – Rasmus Feb 25 '13 at 18:38
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When $n>1$, an open annulus $$\{x\in\mathbb{R}^n\mid a<|x|<b\},\quad 0<a<b$$ and an open ball $$\{x\in\mathbb{R}^n\mid |x|<r\},\quad 0<r$$ are open connected subsets of $\mathbb{R}^n$ that are not homeomorphic.
When $n=1$ or $0$, it is true that any open connected subsets of $\mathbb{R}^n$ are homeomorphic. This is because any open connected subset of $\mathbb{R}$ is an open interval, and $\mathbb{R}^0$ is just a point.
Zev Chonoles
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Nope ! For example $\rm B(0, 2)$ is not homeomorphic to $\rm B(0,2) \backslash \rm \overline{B(0,1)}$.
Damien L
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If the statement was refined to say: any two open, simply-connected subsets of $\mathbb{R}^n$, would the statement then be true? – Feb 25 '13 at 18:00
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5@Lev: It certainly would not be true for $n>2$ because these spaces are simply-connected in that case. – Zev Chonoles Feb 25 '13 at 18:01
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2@LevLivnev: in higher dimensions one should then replace simply connected with "n-connected." – Rasmus Feb 25 '13 at 18:40
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3@LevLivnev the idea holds when you define higher dimensional analogues to "simply connected". The idea behind these examples is the fact that the fundamental groups of these spaces are non-isomorphic, the fundamental groups $\pi_n$ detect $n$-dimensional holes. I have no idea to justify this without using these concepts. Has anyone other explanation? – Vinicius M. Feb 25 '13 at 18:45
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2@ViniciusM. There are contractible open manifolds that are not homeomorphic to Euclidean space, for example: http://en.wikipedia.org/wiki/Whitehead_manifold Does this answer your question? – MartianInvader Feb 25 '13 at 23:30