TITLE: Logic: existence of a certain type of a bijective function on an infinite set
QUESTION [4 upvotes]: Let $X$ be an  infinite set. Prove that there is a bijective function $f: X \rightarrow X$ with the property that for every $x \in X$ and all $n > 0$: $f^n(x) \neq x$.
I've tried to proved this by considering a bijective function $g: \mathbb{Z} \times X \rightarrow X$ in a certain way (by the composition of a function $f: \mathbb{Z} \times X \rightarrow \mathbb{Z} \times X$), but that's all i've got at the moment.

REPLY [1 votes]: You're on the right track. 
Consider the permutation of $\Bbb Z\times X$ defined by $(k,x)\mapsto(k+1,x)$. Now pull that to a permutation of $X$.
Do note that the axiom of choice is needed here, and in particular the fact that $\Bbb Z\times X\sim X$.

REPLY [1 votes]: You actually do have some very useful pieces; you just have to put them together properly. 

Since $X$ is infinite, there is a bijection $f:X\to X\times\Bbb Z$. 
(This actually requires some part of the the axiom of choice.)  
Find a bijection $h:X\times\Bbb Z\to X\times\Bbb Z$ that has the desired property; use the fact that $\Bbb Z$ has a bijection of the desired type.  
Consider the map $f^{-1}\circ h\circ f$.

Note: Some nonsense deleted.
{"set_name": "stack_exchange", "score": 4, "question_id": 473972}