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Prove that if $f: \mathbb{R} \rightarrow \mathbb{R}$ is a homeomorphism, then $f$ cannot have periodic points of primitive period $3$.

The proof was given as follows:

Suppose that we have a primitive period of $3$. Then either:

1) the interval $[x \quad f(x)]$ maps to $[f(x) \quad f^2(x)]$, and $[f(x) \quad f^2(x)]$ maps to $[x \quad f^2(x)]$, or

2) $[x \quad f^2(x)]$ maps to $[f^2(x) \quad f(x)]$, and $[f^2(x) \quad f(x)]$ maps to $[x \quad f(x)]$.

Assume either 1) $x < f(x) < f^2(x)$, or 2) $x < f^2(x) < f(x)$.

Then for 1), $I_1 = [x \quad f(x)]$ and $I_2 = [f(x) \quad f^2(x)]$ and so then $f(I_1) = I_2$ and $f(I_2) = I_1 \cup I_2$. By the intermediate value theorem, there exists a $k \in I_2$ such that $f(k) = f(x)$, but since $x \neq k$ and we have that $f(k) = f(x)$, this is a contradiction.

For 2), $I_1 = [x \quad f^2(x)]$ and $I_2 = [f^2(x) \quad f(x)]$. Then $f(I_1) = I_1 \cup I_2$. Then there exists a $k \in I_1$ such that $f(k) = f^2(x)$ but $f(k) = f(f(x)) = f^2(x)$ which is a contradiction to the assumption that the primitive period was $3$. Q.E.D.

I didn't understand anything about the proof, especially this "interval" business and how the intermediate value theorem is relevant at all.

Rory Daulton
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mr eyeglasses
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1 Answers1

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This fact is quite intuitive. First of all, we can make things a little easier if we assume that $f:\mathbb R\to \mathbb R$ is an orientation preserving homeo (which means that $f$ is increasing): if $f$ was not, then $f^2$ verifies this property and the dynamics of $f^2$ is approximatively the same (the bounded orbits are the same).

Now let us suppose that there exists a point $x_0\in\mathbb R$ and $q\in \mathbb N$, $q\ge 2$ such that $f^q(x_0)=x_0$. Set $x_{k}=f^{k}(x_0)$. Since $f$ is increasing (by induction) we have $$x_0<x_1<\ldots<x_{q-1}<x_q,$$ but, oops!, $x_q=x_0$: contradiction!

Just a mathematician
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  • Sorry, I don't quite understand. If $f$ is homeo, then it can be either strictly increasing, or strictly decreasing. If it is strictly decreasing, then $f^2$ verifies what property? And what do you mean by "the dynamics...is approximately the same?" I know that an orbit is a set that lists all the elements that iterations of $f$ takes each element to, so if we have a period of $3$, then the orbit is ${x, f(x) f^2(x)}$. I don't understand what being homeomorphic has anything to do with this. – mr eyeglasses Mar 14 '15 at 14:20
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    If I'm not mistaken, we don't necessarily know that $x_0 < x_1$. But we're assuming that $x_0 \neq f(x_0)$, so either $x_0 < x_1 < \cdots < x_{q-1} < x_q$ or $x_0 > x_1 > \cdots > x_{q-1} > x_q$ holds by induction. – Ravi Fernando Mar 14 '15 at 21:10
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    @ᴇʏᴇs: if $f$ is strictly decreasing, then you can check that $f^2$ is strictly increasing. And if $x$ has period 3 under $f$, then it also has period 3 under $f^2$: its orbit is ${x, f^2(x), f^4(x)}$, where $f^4(x) = f(x)$. This allows us to assume without loss of generality that $f$ is increasing. – Ravi Fernando Mar 14 '15 at 21:14
  • @RaviFernando How come "under $f^2$" we can just ignore $f(x)$ and $f^3(x)$? You say that $f^4(x) = f(x)$ even though you exclude it. How can we be sure the value of $f(x)$ under $f$ is the same exact value of $f^4(x)$ under $f^2$? – mr eyeglasses Mar 14 '15 at 22:36
  • We're assuming that $f(f(f(x) = x$, right? Take $x$, plug it into $f^2$, get $f(f(x))$. Plug that into $f^2$, get $f(f(f(f(x))))$. But $f(f(f(x))) = x$, so $f(f(f(f(x))))$ can also be written as $f(x)$. Then plug this into $f^2$ again, and we get $f(f(f(x)))$, which happens to be $x$. So we've found the entire orbit of $x$ under the map $f^2$. – Ravi Fernando Mar 16 '15 at 03:06