# Permutation without fixed finite set

Let $X$ be an arbitrary infinite set, can we always find a bijective map $T: X\rightarrow X$ such that for any finite (nonempty) subset $F\subset X$, $T(F)\neq F$ ? This question is related to another post.

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Are there any assumptions of choice in the theory? – Asaf Karagila Oct 9 '11 at 21:23
If $F=\emptyset$, then $T(\emptyset)=\emptyset$ for any $T$. – SL2 Oct 9 '11 at 21:26
No assumption in my mind. You may assume that the cardinality of $X$ $\le$ $c$. – Syang Chen Oct 9 '11 at 21:26
If $X$ is finite and $F=X$ then $T(F)=F$. – George Lowther Oct 9 '11 at 21:31
Sorry, corrected. – Syang Chen Oct 9 '11 at 21:32

This is easy enough when $X$ is finite or countably infinite. In the general case I think it requires the Axiom of Choice. Given AC we know $X\simeq X\times \mathbb Z$ for any infinite $X$, and $X\times\mathbb Z$ can be made to satisfy your property by letting $T$ shift each copy of $\mathbb Z$ one position to the right (or left).
@Asaf, it's a standard (but not trivial) consequence of AC that $\aleph_\alpha\times\aleph_\beta \simeq \aleph_{\max(\alpha,\beta)}$. – Henning Makholm Oct 9 '11 at 21:32
Great! Thank you! @Asfa, he translates every point in the interger index by $1$. – Syang Chen Oct 9 '11 at 21:35
@Asaf, if you have a finite nonempty $F\subset X\times\mathbb Z$, then there must be a least $n$ such that $(x,n)\in F$ for some $x$. But this $(x,n)$ cannot be a member of $T(F)$ because $T$ adds one to all of the $n$'s. – Henning Makholm Oct 9 '11 at 21:38