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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.

Riccardo
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    Sorry to be slow, but what's your obvious injection of $P_A$ into the powerset of $A$ getting $|P_A|\leq 2^k$? – Kevin Carlson Feb 03 '14 at 20:38
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    @Kevin, $P_A\subseteq\mathcal P(A\times A)$. – Asaf Karagila Feb 03 '14 at 20:40
  • @Asaf Karagila i was ispired by one of your (very good) answer (http://math.stackexchange.com/questions/191006/number-of-countable-subsets-of-mathbbr ) 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 – Riccardo Feb 03 '14 at 20:58
  • And so tried to calculate the size of the set – Riccardo Feb 03 '14 at 20:59
  • A (probably) silly question: by constant functions in $F$ do you mean $f_0(i) = 0, f_1(i) = 1, i \in k$? So $\text{card}~F^* = \text{card}~F - 2 = 2^k - 1 = 2^k$? – Andrey Surovtsev Aug 17 '19 at 14:17
  • @Riccardo May I also ask a question, what justifies the existence of a family ${B_i}_{i \in k}$ in the first place? As I understand it, for any two sets of the cardinalities $k$ and $k$ the cardinality of their Cartesian product should be $k$ due to $k \cdot k = k$. How can I be assured that for a particular set $A$ there exists the said partition? – Andrey Surovtsev Aug 18 '19 at 02:54

1 Answers1

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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 $A$.

Asaf Karagila
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  • Ok, now I try to work it out the hints – Riccardo Feb 03 '14 at 21:43
  • 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 – Riccardo Feb 03 '14 at 22:28
  • @AsafKaragila Without AC, I presume it's possible to go all the way down to $|P_A| = |A|$? Or is it just that $P_A$ doesn't necessarily have a cardinality, but if it does then that cardinality has to be $2^{|A|}$? – Steven Stadnicki Feb 03 '14 at 22:42
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    @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. – Asaf Karagila Feb 03 '14 at 22:48
  • @Ric: Actually, your interpretation of my hint made me realize that there is a much simpler way to do so. I've edited to improve my hint. – Asaf Karagila Feb 03 '14 at 22:50
  • Ok I'll take a look! (My -heavy- solution is correct? ) – Riccardo Feb 03 '14 at 23:03
  • @Ric: Your solution is correct, although you seem to go a bit too far in the proof that there are $2^k$ partitions; instead of talking about functions into ${0,1}$, just talk about subsets. It's clearer and easier. – Asaf Karagila Feb 03 '14 at 23:08
  • W-O-W very very good reasoning – Riccardo Feb 03 '14 at 23:08
  • @Ric: I wouldn't have done that without your suggested solution. :-) – Asaf Karagila Feb 03 '14 at 23:09
  • @Asaf just an observation, you need that $\pi $ is a permutation without any fixed point, to have $ 2^{|A|} $ bijections right? – Riccardo Feb 04 '14 at 09:51
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    @Ric: No. Just any non-trivial permutation works. Try to see why! – Asaf Karagila Feb 04 '14 at 09:57
  • @AsafKaragila Thank you, I think your updated proof is the only one that I could completely understand :) So I hope I done it right: it's easy to see that $\pi_A(a, b)$ is a permutation on $A$ (by the way, didn't you mean to index it $\pi_{A'}$? – Andrey Surovtsev Aug 18 '19 at 04:30
  • @AsafKaragila Then, for any $A_1' \subseteq A, A_2' \subseteq A, A_1' \neq A_2'$, there exist $\alpha \in A_1', \alpha \notin A_2'$ (of course the proof is similar if the belonging is reversed). Any non-trivial permutation will have some $\beta \in A$ s.t. $\pi \beta = \beta', \beta' \neq \beta$. Then, $\pi_{A_1}(\alpha, \beta) = (\alpha, \beta')$ whereas $\pi_{A_2}(\alpha, \beta) = (\alpha, \beta)$, and those values are not equal, so the procedure is "injective" with respect to subsets and permutations. – Andrey Surovtsev Aug 18 '19 at 04:31
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    @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. – Asaf Karagila Aug 18 '19 at 08:31