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Circle is not homeomorphic

  1. Feb 11, 2013 #1
    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. jcsd
  3. Feb 11, 2013 #2


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    what have you tried?
  4. Feb 11, 2013 #3
    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(t-a)}{b-a}\right), r \cos\left(\frac{2\pi(t-a)}{b-a}\right)\right) \ t \in [a,b)[/tex]

    i can't go longer :(
  5. Feb 12, 2013 #4


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    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.
  6. Feb 12, 2013 #5
    Yes, that's it!!!

    Can you be more precise please in your answer?

    Thank you
  7. Feb 12, 2013 #6
    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.
  8. Feb 12, 2013 #7


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    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.
  9. Feb 12, 2013 #8
    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
  10. Feb 12, 2013 #9
    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
  11. Feb 12, 2013 #10


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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?
  12. Feb 12, 2013 #11
    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.
  13. Feb 12, 2013 #12
    Not really very rigorous, unless you're going to talk about fundamental groups.
  14. Feb 14, 2013 #13
    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...

    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
  15. Feb 14, 2013 #14


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    Given [itex][a,b]\subset \mathbb{R}[/itex], take some [itex]c\in (a,b)[/itex] and consider [itex][a,b] \setminus \left \{ c \right \}[/itex]. Will it always be possible to find a path between any two points in [itex][a,b] \setminus \left \{ c \right \}[/itex]? On the other hand, take some [itex]p\in S^{1}[/itex] and try to convince yourself intuitively (and explain to me your reasoning of course) that it will be possible for [itex]S^{1}\setminus \left \{ p \right \}[/itex].
  16. Feb 14, 2013 #15
    ok, I see that it will not be possible to find a path between any two points in [itex][a,b] \setminus \left \{ c \right \}[/itex] because there will be a 'hole' in the line

    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
  17. Feb 14, 2013 #16


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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.
  18. Feb 16, 2013 #17
    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

  19. Feb 16, 2013 #18
    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....
  20. Feb 17, 2013 #19


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    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 property--i.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

    Heine-Borel, 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.
  21. Feb 19, 2013 #20
    Thank you so very much :)
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