I am trying to show that there is no homeomorphism $f:[0,1] \to [0,1]$ which is topologically mixing (a dynamical system $T:X\to X$ is called topologically mixing, provided, for every pair of open sets $U \subset X$ and $V\subset X$, there exists a natural number $m$ such that for every $n\ge m$ the intersection $T^n(U) \cap V$ is non-empty). Being topologically mixing is an invariant property: if two dynamical systems $T:X\to X$ and $F:Y\to Y$ are conjugate (i.e. one can find a homeomorphism $H:X\to Y$ for which $H \circ T=F\circ H$), then being mixing for one of $T$ or $F$ implies being mixing for the other. Even more, if the mapping $H$ is just a semi-conjugacy (i.e the mapping $H$ is just continuous and surjective), then if $T$ is mixing, so is $F$.
For my first try, my idea was to find a homeomorphism $g: S^1 \to S^1$ which is a factor of $f$ (i.e. there is a semiconjugacy $h: [0,1] \to S^1$ such that $h\circ f = g\circ h$) and then conclude that being mixing for $f$ implies being mixing for $g$ which is impossible. I tried to set $h(t) = \exp (2 \pi \mathbb{i} t) $ and then I defined $g= h^{-1} \circ f \circ h$. But since $h$ is not injective ($h(0) = h(1)$), the mapping $h^{-1}$ is defined on the interval $[0,1)$ and not on all of $[0,1]$. But if for every $(x,y)\in \mathbb R$ with $x^2 + y^2 = 1$ we set $$g(x,y) =\bigg(\cos \Big(2\pi f\big(\frac {1}{2\pi}\cos ^{-1}(x) \big) \Big), \sin \Big(2\pi f\big(\frac {1}{2\pi}\sin ^{-1}(y) \big) \Big) \bigg) $$
Then $h \circ f = g \circ h$. But I have a doubt as to whether $g$ is a homeomorphsm of $S^1$. (I think it is not injective.)
Then, I tried to show that for every homeomorphism $f:[0,1] \to [0,1]$, the mapping $f$ is strictly monotone and we must have either $f(0)=0 $ and $f(1)= 1$ or $f(0)= 1$ and $f(1) = 0$. Without loss of generality, (as professor Jason DeVito mentioned in comments) we can assume $f(0)=0 $ and $f(1)= 1$, otherwise, we work with $f^2$. Now if the only fixed points of $f$ are $0$ and $1$, depending on $f(t) < t$ or $f(t) >t$, then for any $t \in (0,1)$ we get $f^n(t) \to 0$ or $1$. Hence in each case, we can find two open sets $U$ and $V \subset [0,1]$ such that for some $n\in \mathbb{N}$ and for every $m \ge n$, the intersection $f^m (U) \cap V= \emptyset$. For example, it is enough to take two open intervals $U=(\frac{3}{10}, \frac{4}{10})$ and $V=(\frac{6}{10}, \frac{7}{10})$. Then $f^m(U) \to 0 $ or $1$ as $m\to \infty$ and so $f^m (U) \cap V= \emptyset$ for large $m$.
But what can we do when $f$ has more than $2$ fixed points. Can someone help me?
Best Answer
As I mentioned in my comment above, we may assume $f(0) = 0$ and $f(1) = 1$ and therefore, that $f$ is increasing. Now, let's show that $f$ has no other fixed points.
So, assume for a contradiction that $x\notin \{0,1\}$ is a fixed point of $f$. Because $x\notin\{0,1\}$, the sets $U = [0,x)$ and $V = (x,1]$ are open sets in $[0,1]$. We claim that these two open sets do not mix.
The key observation is the following: Let $U'\subseteq U$ be any subset. Then $f(U')\subseteq U$. Since $f(U')\subseteq f(U)$, it is sufficient to show that $f(U)\subseteq U$. First note that $f(U)$ is an interval because $U$ is and $f$ is a homeomorphism. We also know that $f(0) = 0$ so $0\in f(U)$. If there is some $u\in U$ with $f(u) > x$, then this implies that there is some $w\in U$ with $f(w) = x$. Since $f(x) = x$ and $x\notin U$, we have contradicted injectivity of $f$. Letting $U' = f^n(U)$, we deduce that the iterates $f^n(U)$ are all subsets of $U$. In particular, since $U\cap V = \emptyset$, it follows that $f^n(U) \cap V = \emptyset$ for all $n\geq 1$.