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Math Help - Homeomorphism of the circle

  1. #1
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    Homeomorphism of the circle

    Let f: \mathbb{S}^1 \rightarrow \mathbb{S}^1 be an orientation preserving homeomorphism with at least one fixed point. Show that every periodic point of f has period 1, i.e. is a fixed point.

    Anybody got an idea?
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  2. #2
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    Let p0 be a fixed point and p be a periodic point with period n, that is, f^n(p) = f(f(...f(p))) = p while f^k(p) not equals p if k < n.
    Let A be the shorter arc between p0 and p, since f is a homeomorphism, f(A) is also an arc with f(p) being the second end point. ( p0 is the first one).
    Suppose the length of f(A) is greater than A. Denote this fact simply as f(A) > A.
    Since f(A) contains A, f(f(A)) contains f(A). And since f(f(p)) does not equal to f(p), we get f(f(A)) > f(A).
    With this process we get a series of arcs, f(A) < f(f(A)) < ... f^n(A). Now f^n(p) = p, so f^n(A) must cover the circle S^1 one or more times, ending at p. So A is covered at least twice.
    While f^n is also a homemorphism, a contradiction.
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  3. #3
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    ok thanks. Can we somehow expand this proof to prove the following fact?
    If f: \mathbb{S}^1 \rightarrow \mathbb{S}^1 is an orientation preserving homeomorphism with at least one of minimal period MATH]n > 1[/tex], dhow that then every periodic point of f has minimal period n.
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  4. #4
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    g=f^n is an orientation preserving homemorphism having at least one fixed point. According to your first question, every periodic point of g is a fixed point, that is, a periodic point of period n for f. DONE.
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