Register to reply 
Are open sets in R^n always homeomorphic to R^n? 
Share this thread: 
#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

Sci Advisor
P: 1,170

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

Sci Advisor
P: 1,807

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

Sci Advisor
P: 1,170

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

Sci Advisor
P: 1,807




#6
Jan2213, 07:13 PM

Mentor
P: 18,290




#7
Jan2313, 04:31 PM

Sci Advisor
HW Helper
P: 9,484

notice R^n is contractible. But even that is not enough. Look at this:
http://math.stackexchange.com/questi...ctomathbbrn 


Register to reply 
Related Discussions  
Homeomorphism between the open sets of the circle and the open sets of real line  Calculus & Beyond Homework  1  
3 questions concerning open and closed sets for sets having to do with sequence space  Calculus & Beyond Homework  0  
SIMPLE QUESTION! Open interval homeomorphic to R?  Topology and Analysis  7  
Showing two sets are not homeomorphic in subspace topology.  Calculus & Beyond Homework  0  
Can all open sets in R^n be expressed as countable union of open cubes?  Set Theory, Logic, Probability, Statistics  5 