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Are open sets in R^n always homeomorphic to R^n?

by iLoveTopology
Tags: homeomorphic, sets
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iLoveTopology
#1
Jan19-13, 01:47 PM
P: 11
I know that open intervals in R are homeomorphic to R. But does this extend to any dimension of Euclidean space? (Like an open 4-ball is it homeomorphic to R^4?)

My book doesn't talk about anything general like that and only gives examples from R^2.
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Bacle2
#2
Jan19-13, 02:47 PM
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No for general open sets; look at, e.g., an open annulus, or any disconnected open set. But an open n-disk D:={x in R^n : ||x||<1 } (or any translation of it) is homeomorphic to R^n.
disregardthat
#3
Jan22-13, 05:37 AM
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Any connected open set in R^n is homeomorphic to R^n, for any n. An open set in R^n is homeomorphic to the disjoint union of equally many R^n's as connected parts of your open set.

Bacle2
#4
Jan22-13, 08:05 AM
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Are open sets in R^n always homeomorphic to R^n?

Actually, an open annulus is open and connected, but not homeomorphic to R^n, since it is not simply-connected.
disregardthat
#5
Jan22-13, 06:14 PM
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Quote Quote by Bacle2 View Post
Actually, an open annulus is open and connected, but not homeomorphic to R^n, since it is not simply-connected.
thanks for the correction, i meant simply connected :)
micromass
#6
Jan22-13, 07:13 PM
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Quote Quote by disregardthat View Post
thanks for the correction, i meant simply connected :)
Also not true, then open set [itex]B(0,1)\setminus\{0\}[/itex] of [itex]\mathbb{R}^3[/itex] is simply connected but not homeomorphic to [itex]\mathbb{R}^3[/itex].
mathwonk
#7
Jan23-13, 04:31 PM
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notice R^n is contractible. But even that is not enough. Look at this:

http://math.stackexchange.com/questi...c-to-mathbb-rn


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