An alternative proof of set of all permutations on naturals is uncountable by Cantor's diagonalization.

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I was trying to show that the set of all permutations on $\mathbb N$ is uncountable.It is quite easy to show it directly using Riemann Rearrangement Theorem of Series.Take any real $r$ and choose a permutation of $\sum\limits_{n=1}^{\infty} (-1)^{n+1}\frac{1}{n}$ which converges to $r$ and mark it by $p_r$.Define a map $p : \mathbb R \to Sym(\mathbb N)$ by $p(r)=p_r$ and that is an injection and hence $|Sym(\mathbb N)|\geq c$ and hence $Sym(\mathbb N)$ is uncountable.

But now I want to prove it directly from Cantor's diagonalization argument:

Suppose $Sym(\mathbb N)$ is countable,then the permutations can be named as $x_1,x_2,.....$.

Suppose $x_n=x_{n1} x_{n2} x_{n3}...x_{nn}...$ in sequence form.$[i.e. x_n(i)=x_{ni}]$

Now if we define a new permutation on $\mathbb N$,namely $\sigma$ as follows:

Choose $\sigma(1)=\min(\mathbb N-\{x_{11}\})$

and $\sigma(n)=\min(\mathbb N-\{\sigma(1),...,\sigma(n-1),x_{nn}\})$ for each $n\geq 2$

does $\sigma \in Sym(\mathbb N)$ ,it is clearly injective but is it surjective too?

If it is,the $\sigma$ is a permutation on $\mathbb N$ which is not in the countable list.which is a contradiction.

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Your function $\sigma$ is surjective, too. Assume there is an $a \in \mathbb N$ such that $\sigma(n)\ne a, \forall n \in \mathbb N$ and assume that $a$ is the smallest number in $\mathbb N$ with this property. Then there exist numbers $n_1,\ldots,n_{a-1}$ such that $\sigma(n_i)=i, i \in \{1,\ldots,a-1\}$. Let $b=\max\{n_1,\ldots,n_{a-1}\}$. Then $x_{k,k}=a, \forall k>b$. So we have $x_{k,1}\ne a,\forall k>b$. So there are at most $b$ rows in this matrix that have $x_{k,1}=a$. But there are infinitely many rows in this matrix with $x_{k,1}=a$, if this list would contain all permutations.


Here ist is another notation for a permutation $\sigma$ of $\mathbb N$.

the sequence $\sigma=(n_1,n_2,n_3,\ldots)$ is the following function $$\sigma(1)=n_1\\ \sigma(k)=n_k\text{-th smallest element of } \mathbb{N} \setminus \{\sigma(1),...,\sigma(k-1)\}$$

Here you can use the diagonalization argument without any problems