I've some trouble calculating the cardinality of the set of the permutations of a given set $ A $. For notational purpose let $ k = |A|$ and define $ P_A = \{ f : A \to A | f \text{ is a bijective function } \} $

Clearly if $ A $ is a finite set, the cardinality of $ P_A $ is $ k! $, but the interesting case is when $ A $ is infinite. Again, it is obvious that $|P_A| \leq 2^k $. But i can't find neither a lower upper bound or the lower bound to conclude by Schroeder-Berstein. Any hints or corrections? Thanks in advance!

Following the hint given by Asaf Karagila, I worked out this answer:

I'll put [AC] where i used AC to prove the following fact

So starting from [AC] $ k \cdot k = k$ [AC] there exists a family of $ k $ pairwise disjoint subset of $ A $ each of them of cardinality $k$. Let's call them $\{ B_i \}_{i \in k} $. Let $ F := { f \colon k \to \{0, 1} | \text{ f is a function } \} $. Define $ F^* := F \setminus \text{ costant functions }$. Note that $|F^*|= 2^k $. For each $ f \in F^*$ let $ J_f :=\{i \in k | f(i) =1\} $ and let $ U_f := \{ \bigcup_{j \in J_f} B_j \}$. It is immediate to prove that for each $ f $ we have $|U_f|=|U_f^c| =k $. So we have the $2^k $ partitions of $A $. With a little abuse of notations I denote with $2^k $ an index set of this size. For each $ i \in 2^k $ let $ (P1_i, P2_i) $ a partition of $ A $. [AC] for each partition choose one bijection $g_i: P1_i \to P2_i $. Now we can define $ 2^k $ permutations in this way. For each $ i\in 2^k $ let $ h_i : A \to A $ with $ h_i(a)=g_i(a) $ if $ a \in P1_i $ otherwise $ h_i = g^{-1} $. is is easy to see that it is a permutation.

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    $\begingroup$ Sorry to be slow, but what's your obvious injection of $P_A$ into the powerset of $A$ getting $|P_A|\leq 2^k$? $\endgroup$ – Kevin Carlson Feb 3 '14 at 20:38
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    $\begingroup$ @Kevin, $P_A\subseteq\mathcal P(A\times A)$. $\endgroup$ – Asaf Karagila Feb 3 '14 at 20:40
  • $\begingroup$ @Asaf Karagila i was ispired by one of your (very good) answer (math.stackexchange.com/questions/191006/… ) when you say "there are $2^k $ way of enumerating a set in $[ A]^k $" so i supposed that if two functions "differ" by a permutation of the domain they enumerate the same subset $\endgroup$ – Riccardo Feb 3 '14 at 20:58
  • $\begingroup$ And so tried to calculate the size of the set $\endgroup$ – Riccardo Feb 3 '14 at 20:59
  • $\begingroup$ A (probably) silly question: by constant functions in $F$ do you mean $f_0(i) = 0, f_1(i) = 1, i \in k$? So $\text{card}~F^* = \text{card}~F - 2 = 2^k - 1 = 2^k$? $\endgroup$ – Andrey Surovtsev Aug 17 at 14:17

HINT: Prove that there are $2^k$ partitions of $A$ into equipotent two parts. Given such partition $\{A_1,A_2\}$ choose a bijection $f\colon A_1\to A_2$ and use $f$ to define a permutation of $A$.

Note the heavy use of the axiom of choice. It is needed.

Based on the suggested solution, here's a much better outline for a solution:

Fix a bijection $f\colon A\to A\times A$, and fix a permutation of $A$ which is not the identity, $\pi$. Now given $A'\subseteq A$ define a permutation of $A\times A$, $$\pi_A(a,b)=\begin{cases}(a,\pi b) & a\in A'\\ (a,b) & a\notin A'\end{cases}$$

This defines $2^{|A|}$ permutations of $A\times A$, and therefore of $A$.

  • $\begingroup$ Ok, now I try to work it out the hints $\endgroup$ – Riccardo Feb 3 '14 at 21:43
  • $\begingroup$ Mmmh it doesn't allow me to post long comments, I'll edit my question, can you please check the reasoning? I'm interested in quoting correctly when AC is needed $\endgroup$ – Riccardo Feb 3 '14 at 22:28
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    $\begingroup$ @Steven: I think that it's provably never $|A|$. Also note that every set has a cardinality, it just doesn't have to be an $\aleph$ number. Some sets would have strictly more than $2^{|A|}$ permutations, but I'm not sure how well we can fine tunes these inequalities. Controlling power sets cardinality is one of the greatest challenges when creating models of set theory. $\endgroup$ – Asaf Karagila Feb 3 '14 at 22:48
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    $\begingroup$ @Ric: No. Just any non-trivial permutation works. Try to see why! $\endgroup$ – Asaf Karagila Feb 4 '14 at 9:57
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    $\begingroup$ @AndreySurovtsev: Yes, that's the idea. You're right about the subscript, but I think it's best to not make trivial edits like that on posts which are this old, so I'm just going to keep it this way for now. $\endgroup$ – Asaf Karagila Aug 18 at 8:31

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