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I was solving the following exercise : Is $\mathfrak{S}(\mathbb{N})$ countable ? Here $\mathfrak{S}$ stands for the set of bijections from $\mathbb{N}$ to itself. A proof could be to build an surjection from $\mathfrak{S}(\mathbb{N}) \rightarrow \mathcal{P}(\mathbb{N})$ (the power set of $\mathbb{N})$. I nonetheless had an intuition that i struggle to formalize.

It is known that for $n \in \mathbb{N}^*$, $|\mathfrak{S}(\{0,1,...,n-1\})| = n!$, and that $|\mathcal{P}(\{0,1,...,n-1\})| = 2^n$ (where |.| denotes the cardinality function).

Note that the functions used in the previous paragraph are not the same as those defined on $\mathbb{N}$, their domain is some finite segment of $\mathbb{N}$.

It is clear that $n! > 2^n$ for $n \ge 4$, and the first one grows much faster than the second one. My question is the following : is there any result stating we can "go to the limit" and thus $|\mathfrak{S}(\mathbb{N})| \ge |\mathcal{P}(\mathbb{N})|$, solving the exercise (as the second one is not countable) ?

Thank you

supermartruc
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  • An injection $\mathfrak S(\mathbb N) \to \mathcal P(\mathbb N)$ doesn't prove the former is countable. As an obvious example: there is an injection $\mathcal P(\mathbb N) \to \mathcal P(\mathbb N)$, namely the identity function, this certainly doesn't prove $\mathcal P(\mathbb N)$ to be countable! – Trebor Sep 25 '23 at 10:31
  • Not expert on the topic, but for me, it would make more sense to try to build an injection $\mathcal{P}(\Bbb N)\to \mathfrak{G}(\Bbb N)$ to show that $|\mathcal{P}(\Bbb N)|\leq| \mathfrak{G}(\Bbb N)|$ – Surb Sep 25 '23 at 10:31
  • There is continuum many bijections from $\mathbb{N}$ to $\mathbb{N}$ – Jakobian Sep 25 '23 at 10:35
  • here is another duplicate. – lulu Sep 25 '23 at 10:41
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    Yes, sorry I modified the post. I meant surjection. My question was not about finding an answer, which I did, but if the described proof attempt using the cardinality of the finite restrictions can be formalized – supermartruc Sep 25 '23 at 11:05
  • If you want a "constructive" prove, I suggest: Look at the pairs $a_n=(2n-1, 2n)\in \mathbb N\times \mathbb N$ and consider permutations which preserve those pairs. We can assemble a binary "decimal" out of that by letting the $n^{th}$ digit be $0$ if the permutation is the identity on $a_n$, and $1$ if it transposes the two elements. In that way we see that those permutations are already uncountable. – lulu Sep 25 '23 at 11:17
  • I understand the proof you give, but my initial question was rather about the argument of "inequalities between cardinals of finite sets implies inequalities about cardinals of infinite sets" – supermartruc Sep 25 '23 at 11:24
  • Ok, but that idea doesn't look terribly promising. After all, the set of finite subsets of $\mathbb N$ is countable, but that set also contains all the power sets on ${0, \cdots, n-1}$. You need the infinite subsets. – lulu Sep 25 '23 at 11:30
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    Generally speaking, anything that you can show for finite values has no guarantee of even making sense when you try to plug infinite cardinals into it, so no I don't think you can try to start with $n! \geq 2^n$ and "take the limit". – ConMan Sep 26 '23 at 02:27

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No, the behavior of the finite segments doesn't tell you anything either way. It can't distinguish the full symmetric group from the subgroup of bijections that fix all but finitely many elements of $\mathbb{N}$, which is countable; similarly it can't distinguish the full powerset from the set of finite subsets of $\mathbb{N}$, which is also countable.

To my mind it's easier to write down an injection $P(\mathbb{N}) \to \text{Aut}(\mathbb{N})$. This can be done in many ways; for example given a subset $S \subseteq \mathbb{N}$ we can construct a permutation which has cycles of length $s$ for exactly $s \in S$ (this is even an injection from $P(\mathbb{N})$ into the set of conjugacy classes in $\text{Aut}(\mathbb{N})$).

Qiaochu Yuan
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