Existence of a bijective function with no fixed points

Question

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.

Since $f(x)\in A$ and $f(x)\neq x$: What value would you assign to $f(f(x))$? — Dirk, Sep 23 '11 at 11:16
It could be useful if you give the explicit name of the book and the page. — Asaf Karagila, Sep 23 '11 at 11:49

score 5 · Accepted Answer · answered Sep 23 '11 at 11:42

5

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 $k<n-1$ 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.

answered Sep 23 '11 at 11:42

LostInMath

4,528

4

You don't need the whole axiom of choice for $|A|=|A|+|A|$. – Asaf Karagila Sep 23 '11 at 11:50
@Asaf: Indeed, it's a nice piece of additional information. Do you happen to know what is the relationship between $ZF$+"$|A|=|A|+|A|$" and $ZF$+"for infinite $A$ there is a derangement of $A$"? The latter is strictly weaker that the former, I suppose. – LostInMath Sep 23 '11 at 12:21
@AsafKaragila: I know, but I do not know how to prove. – Aj I Sep 23 '11 at 12:24
@LostInMath: Thanks for the proof. I understand that for these problems. – Aj I Sep 23 '11 at 13:13
@LostInMath: Interesting question. I will have to think about it. – Asaf Karagila Sep 23 '11 at 13:24

Existence of a bijective function with no fixed points

1 Answers1

Linked