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Today, there is a problem with a bijective function of Charles C. Pinter's book.

Let $A$ be any set with more than one element, prove that there exists a bijective function $f\colon A\to A$ such that $f(x)\neq x$ for all $x\in A$.

Please guide me with a proof. Thank you for your kindness.

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    In a book: Set Theory of Charles C. Pinter, page 120.2011-09-23

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If $A$ is finite with $|A|=n\geq 2$, write $A=\{a_0,a_1,\ldots,a_{n-1}\}$. Then the function $f:A\to A$ with $f(a_k)=a_{k+1}$ for $k and $f(a_{n-1})=a_0$ will be a bijection without fixed points.

If $A$ is infinite then by the axiom of choice there is a bijection $g$ between $A$ and $A\times\{0,1\}$. Define $h:A\times\{0,1\}\to A\times\{0,1\}$ with $h(a,0)=(a,1)$ and $h(a,1)=(a,0)$ for any $a\in A$. Then $f=g^{-1}\circ h\circ g:A\to A$ is a bijection without fixed points.

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    @LostInMath: Interesting question. I will have to think about it.2011-09-23