Is there a bijection $\phi : [0,1]\to [0,1]$ with no finite orbits?

Question

Is there any bijection $\phi : [0,1]\to [0,1]$ such that for every $x\in [0,1]$ and every $n\in\mathbb{N}$ we have $\phi ^n (x)\ne x$?

I don't even know how to approach this other than the bijection can't be continuous.

Thanks!

Answer 1

Fix a bijection of $[0,1]$ with $[0,1]\times\Bbb Z$, now consider the permutation of $[0,1]\times\Bbb Z$ given by $(x,k)\mapsto(x,k+1)$. Now apply the inverse of the bijection you started with.

Answer 2

You can get something like that for the interior points very easily, say $\phi_1(x)=x^2$. However this maps $0$ to $0$ and $1$ to $1$. This can be fixed by mapping these two points to the interior. Let $\psi(0)=1/3$, $\psi(1/3)=0$, $\psi(2/3)=1$, $\psi(1)=2/3$ and $\psi(x)=x$ otherwise. Now setting $\phi(x):= \phi_1(\psi(x))$ should give you a bijection with no finite orbits.

Answer 3

The mapping

$$\phi(x)=\begin{cases}x+1-a, & x< a, \\ x-a, & x\ge a,\end{cases}$$

with any $a\in (0,1)\setminus\Bbb Q$ should do the job, since all it does is translate each point by an irrational value.

EDIT: Sorry, this only works for $[0,1]$ replaced by $[0,1)$. But maybe someone sees how this can be repaired?