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I have been thinking about the following problem:

Let $A$ be a set of cardinality $k$ and denote $\sum_A$ the set of all bijection from $A$ to $A$.

Also denote $k! = \mathrm{card}\left(\sum_A \right)$. Prove that $k!=2^k$.

My proof consists of finding a bijection $F:\sum_A\to P(A)$ which associates each bijection from the left to the set of its fixed points. Then the result would follow. ($P(A)$=the power set of $A$).

Since this proof seems quite easy I am afraid it is wrong. Can someone enlighten me? Thank you very much!

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Whats My Name
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    Your $F$ isn’t a bijection: many different permutations can have the same set of fixed points. – Brian M. Scott May 11 '13 at 20:01
  • F isn't a bijection. (This is obvious if you look at some explicit elements of $\Sigma_A$ when A has size, say, 3.) – Billy May 11 '13 at 20:02
  • Ok, now I see. But if I take $F:P(A)->\sum_A$ as before I would get an injective function so I have $2^k\leq K!$ – Whats My Name May 11 '13 at 20:04
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    @Cubic: There isn't such an $F$: if $x \in A$, then $A \setminus { x }$ cannot be the set of fixed points of a permutation! That flat can be patched up, though. It is probably worth making a precise statement of what you are assuming to make your argument work. (and then prove the precise statement) –  May 11 '13 at 20:06
  • Notationally, I would write $\Sigma_A$ and not $\sum_A$. That is, \Sigma not \sum. – GEdgar Dec 05 '17 at 14:22

2 Answers2

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First of all, unless you limit yourself to infinite sets this statement is wrong. It is wrong because for $k=3$ we have $3!=6<8=2^3$.

For infinite sets, assuming the axiom of choice, this is true. To see why note that $f\colon A\to A$ means that $f\subseteq A\times A$, so $\Sigma_A\subseteq\mathcal P(A\times A)$.

Assuming the axiom of choice $|A|=|A\times A|$ and therefore $|\Sigma_A|\leq 2^k$. You still have to show the other direction holds as well. In order to show that, you need to find $2^k$ distinct bijections from $A$ to itself.

Hint: There are $2^k$ different pairs $A_1,A_2\subseteq A$ such that $\{A_1,A_2\}$ is a partition of $A$ and $|A_1|=|A_2|$. For every such pair we define a unique $f\colon A\to A$ which sends $A_1$ to $A_2$ and vice versa, conclude the wanted equality of cardinals.

Daniel Fischer
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Asaf Karagila
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  • Two questions: 1) To show there are $2^k$ different partitions of equal sizes, it is enough to show ${B\subset A;|;|B| = |A\backslash B|}\sim P(A)$. An injection is clear, but surjection I could not find. 2) To conclude that for each $A_1,A_2$ we have such a $f:A\to A$ with $A_1$ bijectively sent to $A_2$ and vice versa, are we using AC? – mez May 15 '13 at 22:59
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    Hint: Partition $A$ into three parts of the same size as $A$, $A_1,A_2,A_3$. Then take any $B\subseteq A_1$ and consider $A_2\cup B$ and $A_3\cup(A_1\setminus B)$ as a partition. How many subsets of $A_1$ are there? – Asaf Karagila May 15 '13 at 23:01
  • Also, we have to use the axiom of choice. Otherwise one can come up with counterexamples. – Asaf Karagila May 15 '13 at 23:01
  • Your new hint is a great argument, never would thought about that! – mez May 15 '13 at 23:06
  • How to argue that I can have $A_1\sqcup A_2 \sqcup A_3 = A$ all of same size as $A$? Are you using Dedekind infinite or just infinite? – mez May 15 '13 at 23:11
  • @mezhang: You have to use Dedekind-infiniteness, sets which are Dedekind-finite have no such partitions. But you have to use more than that. You have to use the fact that the axiom of choice implies that $|A|=|A|+|A|$, and therefore $|A|+|A|+|A|$. – Asaf Karagila May 15 '13 at 23:13
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Hint:

  • Every on $A$ bijection is a function from $A$ to $A$.
  • Every function from $A$ to $A$ is a relation on $A\times A$.
  • A relation on $A\times A$ is just a subset of $A\times A$.
  • What's the cardinality of $A\times A$ when $A$ is infinite?
Petr
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