Tell me more ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.
up vote 2 down vote favorite

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.

share|improve this question
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
show 1 more comment

1 Answer

up vote 4 down vote accepted

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).

share|improve this answer
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
show 2 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.