# Thread: Homeomorphism of the circle

1. ## Homeomorphism of the circle

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

Anybody got an idea?

2. 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.

3. ok thanks. Can we somehow expand this proof to prove the following fact?
If $\displaystyle f: \mathbb{S}^1 \rightarrow \mathbb{S}^1$ is an orientation preserving homeomorphism with at least one of minimal period MATH]n > 1[/tex], dhow that then every periodic point of $\displaystyle f$ has minimal period n.

4. 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.