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Are open sets in R^n always homeomorphic to R^n? 
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#1
Jan1913, 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 4ball is it homeomorphic to R^4?)
My book doesn't talk about anything general like that and only gives examples from R^2. 


#2
Jan1913, 02:47 PM

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P: 1,169

No for general open sets; look at, e.g., an open annulus, or any disconnected open set. But an open ndisk D:={x in R^n : x<1 } (or any translation of it) is homeomorphic to R^n.



#3
Jan2213, 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.



#4
Jan2213, 08:05 AM

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P: 1,169

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 simplyconnected.



#5
Jan2213, 06:14 PM

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#6
Jan2213, 07:13 PM

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#7
Jan2313, 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...ctomathbbrn 


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