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    Why are the domain and range of the continuous function in the Brouwer Fixed Point Theorem restricted to the unit ball? Why isn't any compact set valid?
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    (Original post by J.F.N)
    Why are the domain and range of the continuous function in the Brouwer Fixed Point Theorem restricted to the unit ball? Why isn't any compact set valid?
    Well, consider rotating the circle through a right angle - that has no fixed points

    The proof of the Brouwer fixed point theorem often uses algebraic topology and relies on the interior being contractible and the boundary having a fundamental group/first homology group that's infinite cyclic

    <good to hear from you again JFN - been on hols?>
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    (Original post by RichE)
    Well, consider rotating the circle through a right angle - that has no fixed points
    I still don't understand. Why not use any finite topological space? Or perhaps any compact and convex set?

    (Original post by RichE)
    The proof of the Brouwer fixed point theorem often uses algebraic topology and relies on the interior being contractible and the boundary having a fundamental group/first homology group that's infinite cyclic

    <good to hear from you again JFN - been on hols?>
    I've gathered the intuitition to see why the theorem is true for the f:[0,1]-->[0,1] case. I don't know enough topology to understand the entire proof.

    And good to be back--I've gone the way of the whigs for a while, as the say.
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    (Original post by J.F.N)
    I still don't understand. Why not use any finite topological space? Or perhaps any compact and convex set?
    Well it isn't true for all compact spaces - like the example of the circle I gave you - but the theorem is obviously a topological invariant of a space so why mention it really

    It is true for anything homeomorphic to the disc or ball (or whatever dimension you're working in)

    (Original post by J.F.N)
    Well it's essentially the IVT for the [0.1] case or any interval
    Yep

    And hope you never disappear for as long as the Whigs have :eek:
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    (Original post by RichE)
    Well, consider rotating the circle through a right angle - that has no fixed points
    I think I understand what you mean now. Are you talking about rotating the boundary of the unit circle (i.e. (x_1)^2+...(x_n)^2=1)? Clearly then the rotation has no fixed points. If thats the case, surely the set being compact and convex would take care of the problem... Unless you have an example of a compact and convex set to which the Brouwer FPT does not hold.
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    (Original post by J.F.N)
    I think I understand what you mean now. Are you talking about rotating the boundary of the unit circle (i.e. (x_1)^2+...(x_n)^2=1)? Clearly then the rotation has no fixed points. If thats the case, surely the set being compact and convex would take care of the problem... Unless you have an example of a compact and convex set to which the Brouwer FPT does not hold.
    I really don't understand your comment.

    You asked why the BFT did not apply to general compact spaces - well the example of the unit circle in 2D above x^2+y^2 = 1 is enough to show the BFT does not hold for general compact spaces.

    However the BFT has to hold for all compact convex sets as such are homeomorphic to a disc (of whatever dimension) and the BFT is a topological invariant.
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    (Original post by RichE)
    You asked why the BFT did not apply to general compact spaces - well the example of the unit circle in 2D above x^2+y^2 = 1 is enough to show the BFT does not hold for general compact spaces.
    I don't understand why it's enough. Rotating the unit circle in 2D does have a fixed point--say, the center of the circle.
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    (Original post by J.F.N)
    I don't understand why it's enough. Rotating the unit circle in 2D does have a fixed point--say, the center of the circle.
    No I mean the circle x^2+y^2 = 1 not the disc x^2+y^2 <= 1.

    The circle is a compact set (obviously not convex though)
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    (Original post by RichE)
    No I mean the circle x^2+y^2 = 1 not the disc x^2+y^2 <= 1.

    The circle is a compact set (obviously not convex though)
    Ah, ok! Then I've understood. I thought the unit circle was x^2+y^2<=1--hence, I said the boundary of the unit circle (i.e. x^2+y^2=1). Very well.

    By the way, while I'm at asking questions, do you know of any 'interesting' case where two irrational numbers sum to give a rational one? All I've been able to think of are sort-of trivial examples, i.e. Pi + (5-Pi).
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    (Original post by J.F.N)
    By the way, while I'm at asking questions, do you know of any 'interesting' case where two irrational numbers sum to give a rational one? All I've been able to think of are sort-of trivial examples, i.e. Pi + (5-Pi).
    No, I don't know of any and it would be quite a result if you could show one. All the famous irrational e, pi, rt(2), gamma are independent over Q
 
 
 
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