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Circle is not homeomorphic 
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#1
Feb1113, 04:12 PM

P: 68

Hi,
how can I prove that a circle it is not homeomorphic to a subset of [tex]R^n[/tex] I can somehow, see that there isn't any homeomorphic application, for example between a circle in [tex]R^2[/tex] to a line, but how can I prove it? Thank you 


#2
Feb1113, 04:23 PM

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P: 1,716

what have you tried?



#3
Feb1113, 05:58 PM

P: 68

Let a C be a circle, radius r
[tex]C=\left\{ x \in R^2 : x=r \right\}[/tex] Let A be a subset of R between a and b, i.e. [tex]A=[a,b)[/tex] A bijective map could be [tex]f:A\rightarrow C[/tex] [tex]f(t)=\left(r \sin\left(\frac{2\pi(ta)}{ba}\right), r \cos\left(\frac{2\pi(ta)}{ba}\right)\right) \ t \in [a,b)[/tex] i can't go longer :( 


#4
Feb1213, 01:39 AM

C. Spirit
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Circle is not homeomorphic
Are you trying to show [itex]S^{^{1}}[/itex] is not homeomorphic to any subset of [itex]\mathbb{R}[/itex]? If so, then think about path connectedness.



#5
Feb1213, 05:03 AM

P: 68

Can you be more precise please in your answer? Thank you 


#6
Feb1213, 06:12 AM

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P: 18,333

Think about path connectedness and what happens if you remove a point. 


#7
Feb1213, 08:37 AM

C. Spirit
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P: 5,638




#8
Feb1213, 02:56 PM

P: 68

My knowledge on topoligical spaces, is not very deep! Though, I've been reading, do you mean this http://en.wikipedia.org/wiki/Connect..._connectedness 


#9
Feb1213, 03:15 PM

P: 68

Ok, using what I found in wikipedia, a continuous function f from [0,1] to the circle radius 1 could be
[tex]f(t)=\left(\sin\left(2\pi t\right), \cos\left(2\pi t\right)\right) \ t \in [0,1][/tex] so, I know what you mean (intuitevely) that a circle is always connected, and a line is not, but I don't know how to express it mathematically Can you kindly help me? Thank you 


#10
Feb1213, 03:44 PM

C. Spirit
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Yes that is what I meant by a path connected topological space. Firstly, note that since [itex]S^{1}[/itex] is both connected and compact, if it were to be homeomorphic to a subset of [itex]\mathbb{R}[/itex], it would have to be a closed interval so we can just focus our attention on that. Can you see intuitively what happens differently, in terms of path connectedness, between [itex][a,b]\subset \mathbb{R}[/itex] and [itex]S^{1}[/itex] if I removed a point from each?



#11
Feb1213, 05:15 PM

P: 5

Here's another approach. suppose that the circle is on a plane. Naturally we will get a definition of clockwise and anticlockwise. Then imagine that there is a homeomorphism between the circle and a line. So now take a point on the circle and continuously move clockwise until you return to the same point. The image of this path should start and end at the same point. This corresponding path will have to intersect itself otherwise it would never have returned to the same point. So take one such point of intersection. This would imply that such a point has two corresponding points on the circle. Therefore the homeomorphism does not exist since its coimage at some point on the line contaims more than one point.



#12
Feb1213, 05:56 PM

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#13
Feb1413, 01:29 PM

P: 68

If I remove a point from each, we wouldn't have a connected path, I suppose nor it would be a closed interval... @poverlod Thank you very much :) I think I saw it, but still don't get it :) Can't I miss the last point, i.e. can't I make a map without the last point? [itex]f:A \rightarrow B[/itex] [itex]f(t)=(\sin(2\pi t),\cos(2\pi t)) \ t\in [0,1)[/itex] why this map is not hemeomorphic if I remove the last point? Sorry and thank you very much for attention 


#14
Feb1413, 03:52 PM

C. Spirit
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#15
Feb1413, 04:24 PM

P: 68

I see also that it will be possible for [itex]S^{1}\setminus \left \{ p \right \}[/itex] because it will be possible 'to find a path around', i.e., there will always exists a path between any two points... though, I must confess, I can't see the relation with homeomorphism, a continuous map, whose inverse is also continuous. Thank you so much 


#16
Feb1413, 04:33 PM

C. Spirit
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Yes you've got the intuition. The point is to assume there exists a homeomorphism between the unit circle and a closed subset of R and to use the aforementioned difference between the unit circle and a closed subset of R to find a contradiction.



#17
Feb1613, 07:02 AM

P: 68

ok, so we assume that there is a homeomorphism between the unit circle and a closed subset of R, and then as we remove a point, we have a connected path in the circle and we don't have it in the subset of R.
But what is the relation between path connectedness and homeomorphism? AHH, ok, the map must be continous!! Is that? Please just confirm it!! Thank you Joćo 


#18
Feb1613, 07:13 AM

P: 68

ok [itex]S^1[/itex] is compact, i.e., closed and limited and the line would have to be if there were a homeomorphism between both.
On the other hand, if I remove a point, I wouldn't have a compact set, on both, but one is still conncted and the other is not. And I suppose, that if one is path connected, and there is a homeomorphis to another set, the other set must be path connected... and then we have a contradiction.... 


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