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Math Help - Fixed point property.

  1. #1
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    Fixed point property.

    I'm trying to find if the following has the fixed point property:

    S= \{ (x,y) \, : \, x^2+y^2=1 \}


    Is it enough to show:

    Let f \, : S \rightarrow S

    0^2 + (-1)^2=1 \, so (0,-1) belongs to S but f(0,-1) = (0,1)

    Thus, S does not have the fixed point property.
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  2. #2
    A Plied Mathematician
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    Not sure I understand. Typically, you think of a function as having the fixed point property. What is your function f from S to S?
    Last edited by Ackbeet; May 2nd 2011 at 12:37 PM. Reason: Fixed grammar.
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  3. #3
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by MathSucker View Post
    I'm trying to find if the following has the fixed point property:

    S= \{ (x,y) \, : \, x^2+y^2=1 \}


    Is it enough to show:

    Let f \, : S \rightarrow S

    0^2 + (-1)^2=1 \, so (0,-1) belongs to S but f(0,-1) = (0,1)

    Thus, S does not have the fixed point property.
    What's your mapping? What about e^{i\theta}\mapsto e^{i\theta+\frac{\pi}{2}} this is evidently continuous and doesn't have a fixed point.
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Ackbeet View Post
    Not sure I understand. Typically, you think of a function as having the fixed point property. What is your function f from S to S?
    Ackbeet, generally a structure X is said to have the 'fixed point property' if every morphism X\to X has a fixed point.
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  5. #5
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    Quote Originally Posted by Drexel28 View Post
    What's your mapping? What about e^{i\theta}\mapsto e^{i\theta+\frac{\pi}{2}} this is evidently continuous and doesn't have a fixed point.
    Okay, forget the second part. I was just trying to find out if and why/why not S had the fixed point property...
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  6. #6
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    Quote Originally Posted by Drexel28 View Post
    Ackbeet, generally a structure X is said to have the 'fixed point property' if every morphism X\to X has a fixed point.
    then this would depend on the structure imposed on S. for example, if one is regarding S as a topological group, then sure, because then (1,0) has to be a fixed point. but if one is just regarding S as a topological space, then no, we can pick any rotation by an angle that is not an integral multiple of 2\pi
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  7. #7
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Deveno View Post
    then this would depend on the structure imposed on S. for example, if one is regarding S as a topological group, then sure, because then (1,0) has to be a fixed point. but if one is just regarding S as a topological space, then no, we can pick any rotation by an angle that is not an integral multiple of 2\pi
    Thank you, I am aware of this. But I think it's clear from the OPs question that we are talking about top. spaces and not top. groups.
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  8. #8
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    Now that I look at it again, the question has it as S^1

    Does that change things?
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  9. #9
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by MathSucker View Post
    Now that I look at it again, the question has it as S^1

    Does that change things?
    Now, \mathbb{S}^1 is just the unit circle and any rotation by 0<\theta<2\pi is a continuous map without a fixed point.
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