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Let $A$ a set such that $|A|>1$. Iwant to prove that there exists a bijection $f:A\rightarrow A$ such that $f(x)\neq x$.

First, by AC, I can obtain a well order for $A$, and w.l.o.g I suppose that it has order type $\alpha$. So $A=\{x_\xi\}_{\xi<\alpha}$. Now, define the set $C_\xi=A\setminus\{x_\xi\}$. By transfinite recursion I can define the element $y_\xi=\min C_\xi\setminus\{y_\beta:\beta<\xi\}$.

Then, the function $f:A\rightarrow A$ such that $f(x_\xi)=y_\xi$ is the desire function.

Is it fine my approach?

Federico
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YCB
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  • How do you prove that $C_\xi \setminus {y_\beta : \beta<\xi} \neq \emptyset$? – Federico Dec 03 '18 at 17:36
  • @Federico Fine, this construction works if $\alpha$ is a limit ordinal, right? – YCB Dec 03 '18 at 17:38
  • Well, intuitively I would say that it always works, but I don't see immediately why the set is non-empty – Federico Dec 03 '18 at 17:42
  • The axiom of choice tag is meant for questions where you're asking specifically about the necessity of choice in the proof. Not for questions whose answers use choice (since that would encompass a significant portion of this site). – Asaf Karagila Dec 03 '18 at 17:44
  • @AsafKaragila Well maybe it's interesting to know if this problem requires AC – Federico Dec 03 '18 at 17:44
  • @Federico: Of course it does, but that would be a duplicate of at least two questions. – Asaf Karagila Dec 03 '18 at 17:45
  • @AsafKaragila Which ones? I cannot find them – Federico Dec 03 '18 at 17:47
  • You should mention "bijection" in the body of the question, not just in the title (without that it's a triviality). – Robert Israel Dec 03 '18 at 17:57
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    @Federico: For example https://math.stackexchange.com/questions/134152/prove-that-every-set-with-more-than-one-element-has-a-permutation-without-fixed – Asaf Karagila Dec 04 '18 at 08:08

2 Answers2

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Alternatively, you can do the following.

If $A$ is finite but not a singleton, then it's simple: just cycle the elements.

If $A$ is infinite, show that $A=A_1 \sqcup A_2$ with $|A_1|=|A_2|$ and take any bijection $\phi:A_1\to A_2$. Then $f|_{A_1}=\phi$ and $f|_{A_2}=\phi^{-1}$ is a solution.

Federico
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You can also use the fact that every infinite set $X$ can be partitioned into blocks of size 2 using ordinal theory; see this. So we partition $X$

$\tag 1 X = \cup \; F_\lambda \text{ with #}(F_\lambda) = 2$

Once you have that you are in business. Set

$\tag 2 F_{(\lambda, \rho)} = (F_\lambda \times F_\lambda) \setminus \{(x,x) \, | \, x \in F_\lambda\}$

The union of the $F_{(\lambda, \rho)}$ is a bijective correspondence $f$ with $f(x) \ne x$. It can be viewed as a union of transpositions.

CopyPasteIt
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