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
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set-theory
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0Are there any assumptions of choice in the theory? – 2011-10-09
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0If $F=\emptyset$, then $T(\emptyset)=\emptyset$ for any $T$. – 2011-10-09
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0No assumption in my mind. You may assume that the cardinality of $X$ $\le$ $c$. – 2011-10-09
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0If $X$ is finite and $F=X$ then $T(F)=F$. – 2011-10-09
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0Sorry, corrected. – 2011-10-09
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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
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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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0I 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.) – 2011-10-09
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0@Asaf, it's a standard (but not trivial) consequence of AC that $\aleph_\alpha\times\aleph_\beta \simeq \aleph_{\max(\alpha,\beta)}$. – 2011-10-09
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0Come 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) – 2011-10-09
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0Great! Thank you! @Asfa, he translates every point in the interger index by $1$. – 2011-10-09
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0@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. – 2011-10-09
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0@Henning: Yes, I knew that it was this long day of working without the axiom of choice that got me confused here! :-) – 2011-10-09
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0@Asaf, yeah, that's why I assumed that it must have been the cardinal arithmetic you'd blanked out on (since that was the only use of choice in my argument). – 2011-10-09