# Circle is not homeomorphic

by joao_pimentel
Tags: circle, homeomorphic
 P: 68 Hi, how can I prove that a circle it is not homeomorphic to a subset of $$R^n$$ I can somehow, see that there isn't any homeomorphic application, for example between a circle in $$R^2$$ to a line, but how can I prove it? Thank you
 Sci Advisor P: 1,668 what have you tried?
 P: 68 Let a C be a circle, radius r $$C=\left\{ x \in R^2 : |x|=r \right\}$$ Let A be a subset of R between a and b, i.e. $$A=[a,b)$$ A bijective map could be $$f:A\rightarrow C$$ $$f(t)=\left(r \sin\left(\frac{2\pi(t-a)}{b-a}\right), r \cos\left(\frac{2\pi(t-a)}{b-a}\right)\right) \ t \in [a,b)$$ i can't go longer :(
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## Circle is not homeomorphic

Are you trying to show $S^{^{1}}$ is not homeomorphic to any subset of $\mathbb{R}$? If so, then think about path connectedness.
P: 68
 Quote by WannabeNewton Are you trying to show $S^{^{1}}$ is not homeomorphic to any subset of $\mathbb{R}$?
Yes, that's it!!!

Thank you
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 Quote by joao_pimentel Yes, that's it!!! Can you be more precise please in your answer? Thank you
We have to see some attempt of you first. We don't just give away the answers like that.

Think about path connectedness and what happens if you remove a point.
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 Quote by joao_pimentel Yes, that's it!!! Can you be more precise please in your answer? Thank you
Before we move any further, let's make sure you know what path connectedness is. Have you dealt with this property of topological spaces before? If you do then Micromass gave you the crucial idea so think about what he said.
P: 68
 Quote by WannabeNewton Before we move any further, let's make sure you know what path connectedness is. Have you dealt with this property of topological spaces before?
To be honest I am not that aware what path connectedness is!

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
 P: 68 Ok, using what I found in wikipedia, a continuous function f from [0,1] to the circle radius 1 could be $$f(t)=\left(\sin\left(2\pi t\right), \cos\left(2\pi t\right)\right) \ t \in [0,1]$$ 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
 PF Patron Sci Advisor Thanks P: 3,992 Yes that is what I meant by a path connected topological space. Firstly, note that since $S^{1}$ is both connected and compact, if it were to be homeomorphic to a subset of $\mathbb{R}$, 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 $[a,b]\subset \mathbb{R}$ and $S^{1}$ if I removed a point from each?
 P: 5 Here's another approach. suppose that the circle is on a plane. Naturally we will get a definition of clockwise and anti-clockwise. 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 co-image at some point on the line contaims more than one point.
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 Quote by poverlord Here's another approach. suppose that the circle is on a plane. Naturally we will get a definition of clockwise and anti-clockwise. 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 co-image at some point on the line contaims more than one point.
Not really very rigorous, unless you're going to talk about fundamental groups.
P: 68
 Quote by WannabeNewton Yes that is what I meant by a path connected topological space. Firstly, note that since $S^{1}$ is both connected and compact, if it were to be homeomorphic to a subset of $\mathbb{R}$, 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 $[a,b]\subset \mathbb{R}$ and $S^{1}$ if I removed a point from each?
Hi, sorry my delay, I've been thinking.
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?

$f:A \rightarrow B$

$f(t)=(\sin(2\pi t),\cos(2\pi t)) \ t\in [0,1)$

why this map is not hemeomorphic if I remove the last point?

Sorry and thank you very much for attention
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 Quote by joao_pimentel Hi, sorry my delay, I've been thinking. If I remove a point from each, we wouldn't have a connected path, I suppose nor it would be a closed interval...
Given $[a,b]\subset \mathbb{R}$, take some $c\in (a,b)$ and consider $[a,b] \setminus \left \{ c \right \}$. Will it always be possible to find a path between any two points in $[a,b] \setminus \left \{ c \right \}$? On the other hand, take some $p\in S^{1}$ and try to convince yourself intuitively (and explain to me your reasoning of course) that it will be possible for $S^{1}\setminus \left \{ p \right \}$.
P: 68
 Quote by WannabeNewton Given $[a,b]\subset \mathbb{R}$, take some $c\in (a,b)$ and consider $[a,b] \setminus \left \{ c \right \}$. Will it always be possible to find a path between any two points in $[a,b] \setminus \left \{ c \right \}$? On the other hand, take some $p\in S^{1}$ and try to convince yourself intuitively (and explain to me your reasoning of course) that it will be possible for $S^{1}\setminus \left \{ p \right \}$.
ok, I see that it will not be possible to find a path between any two points in $[a,b] \setminus \left \{ c \right \}$ because there will be a 'hole' in the line

I see also that it will be possible for $S^{1}\setminus \left \{ p \right \}$ 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
 PF Patron Sci Advisor Thanks P: 3,992 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.
 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
 P: 68 ok $S^1$ 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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