A derangement of a set $A$ is a bijection from $A$ to itself with no fixed points. Is it the case that every infinite set has a derangement?
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1If the set is finite, any cycle is a derangement. Approximately $1/e$ of all permutations is a derangement. – lhf Nov 06 '14 at 00:02
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1And if $A$ is infinite, I suspect this is equivalent to the set having a well ordering, which is equivalent to the axiom of choice. – Simon S Nov 06 '14 at 00:04
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If the set is countable, fix an enumeration of it and swap pairs of adjacent elements, as in $2n-1\leftrightarrow 2n$. – lhf Nov 06 '14 at 00:08
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^ right ... also equivalent to Zorn's Lemma. – Simon S Nov 06 '14 at 00:23
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Sets of cardinality $1$ have no derangements. All other finite sets, including the empty set, have at least one derangement. – Michael Hardy Nov 06 '14 at 00:52
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@lfh: Cycles of length $1$ are not derangements. – Michael Hardy Nov 06 '14 at 00:54
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If $A$ is infinite, then it is in bijection with $\lbrace 0,1\rbrace\times A$. Swapping $0$ with $1$ is a derangement of $\lbrace 0,1\rbrace\times A$, and hence, using the mentioned bijection, a derangement of $A$.
Olivier Bégassat
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Assuming all the elements are different, the bijections of a set of size $n$ are equivalent to permutations: label the elements $1,2,\ldots,n$. So there are as many derangements for a set of size $n$ as there are for the numbers $1,2\ldots,n$
Alex R.
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Obviously it's true for the finite case. I'm mainly asking about the infinite case. – Nishant Nov 06 '14 at 00:09
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