
#19
Feb1713, 12:46 AM

Sci Advisor
P: 1,168

Assume there is a homeomorphism h between S^1 and E , where E is a subset
of the real line. Since connectedness is a topological propertyi.e., connectedness is preserved by homeomorphisms ( continuous maps will do) h(S^1) is a connected subset of the real line. Then h(S^1) is an interval ; by compactness of S^1 (which must be preserved by h), h(S^1) is also compact in the real line, it then follows by HeineBorel, h(S^1) is closed and bounded in the real line , so h(S^1)=[a,b]. Now, if h is a homeomorphism, the restriction of h to any subset of S^1 is a homeomorphism (into its image). Consider x in S^1 with h(x) not an endpoint , i.e., h(x)≠ a,b; say h(x)=c . Now consider the restriction of h to S^1 {x}. This is a homeomorphism from the connected space S^1{x} to the disconnected space [a,b]{c} . This is not possible , so no such h can exist. Moral of the story/ general point: disconnection number is a homeomorphism invariant. 



#20
Feb1913, 06:42 AM

P: 68

Thank you so very much :)




#21
Feb1913, 08:58 PM

Sci Advisor
P: 1,716

It might be fun to try to prove this by approximating a homeomorphism of the circle in R^n by a sequence of smooth maps to see how far off from a diffeomorphism the homeomorphism can get. the image of a smooth approximation must have measure zero so it can not produce a homeomorphism between the circle and euclidean space.



Register to reply 
Related Discussions  
Are Quotients by Homeomorphic Subspaces Homeomorphic?  Topology and Analysis  11  
Prove that the cone over the unit circle is homeomorphic to the closed unit disc.  Calculus & Beyond Homework  4  
Why the circle can't be homeomorphic to a real interval  Topology and Analysis  17  
Can say me why annulus and circle are not homeomorphic?  Topology and Analysis  2  
The set {A in GL(n;R)  det(A)>0} is homeomorphic to the set {A in GL(n;R)  det(A)<0  Differential Geometry  4 