Show that $S_{\mathbb{N^*}}$ isn't countable.

Question

Let $S_{\mathbb{N^*}}$ be the set of all permutations of $\mathbb{N^*}$. Show that $S_{\mathbb{N^*}}$ isn't countable.

Let $f : \mathbb{N^*} \mapsto S_{\mathbb{N^*}}$ be a bijection. I introduced $$ \sigma(k) = \begin{cases} \min \{ l \mid l\neq f(k)(k) \text{ et } l \neq \sigma(i) \, \forall i \in [1,k-1] \} ,&\text{if }k>1\\ \min \{ l \mid l\neq f(k)(k) \} &\text{if }k=1\\ \end{cases} $$ It's clear that $\sigma$ is injective, but I don't know how to show that it's surjective. Do you have any hint?

It is likely clear from context for most, but I am not familiar with the definition of $\mathbb{N}^*$, could you define it for me please? — RyanK, Oct 18 '20 at 18:58
How would this help you? To prove that a set $X$ is uncountable, you must prove that there cannot exist a bijection between $\mathbb{N}$ and $X$. — Hans Hüttel, Oct 18 '20 at 19:00
@HansHüttel If $\sigma$ is bijective, it's easy to see that $\sigma \not \in f(\mathbb{N^*})$, so $f$ isn't surjective, which is absurd. — Michelle, Oct 18 '20 at 19:01
https://math.stackexchange.com/questions/1321653/is-symmetric-group-on-natural-numbers-countable — Adriano Banchieri, Oct 18 '20 at 19:35

score 2 · Accepted Answer · answered Oct 18 '20 at 19:47

2

Unfortunately, your $\sigma$ need not be surjective without some conditions on $f$. Suppose that $f(k)(k)=1$ for each $k\in\Bbb N^*$; this is certainly possible. Then $1$ cannot be in the range of $\sigma$.

Your $\sigma$ will be surjective if $f$ has the property that $\{k\in\Bbb N^*:f(k)(k)\ne m\}$ is infinite for each $m\in\Bbb N^*$. If not, let $m=\min(\Bbb N^*\setminus\operatorname{ran}\sigma)$. If $\sigma(k)=n>m$, then either $f(k)(k)=m$, or $\sigma(i)=m$ for some $i<k$. The latter is impossible, so $f(k)(k)=m$ for all but finitely many $k\in\Bbb N^*$, is impossible if $\{k\in\Bbb N^*:f(k)(k)\ne m\}$ is infinite.

answered Oct 18 '20 at 19:47

Brian M. Scott

616,228

Thanks for your answer. By introducing $\sigma$, I wanted to use a similar argument as Cantor's diagonal argument for this question. Do you know if there's another function $\sigma$ that uses this idea (and is bijective of course)? – Michelle Oct 18 '20 at 19:58
@Michelle: You’re welcome. The accepted answer to the question to which Adriano Banchieri linked in the comments uses a diagonal argument, though it’s version of $\sigma$ is very different from yours. – Brian M. Scott Oct 18 '20 at 20:02

Show that $S_{\mathbb{N^*}}$ isn't countable.

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