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.
Permutation without fixed finite set
2
$\begingroup$
set-theory
-
0The statement that $T(F)\not=F$ for any finite $F$ is equivalent to saying that the orbits of $T$ are infinite. Or, equivalently, $T^m(x)\not=T^n(x)$ for $m\not=n$ and $x\in X$. – 2011-10-09
1 Answers
4
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).
-
0@As$a$f, ye$a$h, th$a$t's why I $a$ssumed th$a$t it must h$a$ve been the c$a$rdin$a$l $a$rithmetic you'd blanked out on (since that was the only use of choice in my argument). – 2011-10-09