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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

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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).

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I don't see right away why this should be true for every infinite set, in fact I'm not even sure I see it for finite sets (it may just be the long day I had in thinking without the axiom of choice that is speaking, however I still can't see why all finite subsets must be moved.) –  Asaf Karagila Oct 9 '11 at 21:27
    
@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
    
Come on, I'd guess that you figured out that I'd know that by know. I don't see how this implies that there are no finite subsets which are "globally" fixed (which implies that no subset is being fixed) –  Asaf Karagila Oct 9 '11 at 21:34
    
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

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