Assuming axiom of choice , for any set $S$ with more than one point , there exist a bijection $f:S \to S$ such that $f(s) \ne s , \forall s \in S$ . Is the converse true , i.e. Does the statement " For every set $S$ with more than one point , there exist a bijection $f:S \to S$ such that $f(s) \ne s , \forall s \in S$ " implies Axiom of Choice ? Or at least can we prove that any set with more than point has a permutation with no fixed point , without axiom of choice ? Please help. Thanks in advance
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My answer in the duplicate shows that it is much weaker than choice. It is not provable without choice, and there are threads here with such example, but I have to go and it will have to wait (or you can search for it yourself). – Asaf Karagila Apr 13 '15 at 15:26
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@AsafKaragila : I do not see any reference there of why "it is not provable without choice " – Apr 13 '15 at 15:29
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As I said, this is in a different question, and you can look for it yourself or wait until I am by a keyboard to do that for you. – Asaf Karagila Apr 13 '15 at 15:33
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http://math.stackexchange.com/questions/103161/is-there-an-element-with-no-fixed-point-and-of-infinite-order-in-operatorname/ – Asaf Karagila Apr 13 '15 at 15:36
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This statement allready follows from the addition theorem, which states that $A \times \{0,1\}$ is equipotent with $A$ for every infinite $A$ (as on $A \times \{0,1\}$, $(a,x) \mapsto (a, 1+x)$ is fixed-point-free). The addition theorem is known to be strictly weaker than the axiom of choice (a result by Sageev).
martini
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True it follows , but I didn't know that this is strictly weaker than axiom of choice , could you please give a reference ? – Apr 13 '15 at 15:26
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The paper was fully published, in 180 painful pages of forcing in pre-Shoenfield times. – Asaf Karagila Apr 13 '15 at 15:34