The set of all bijections from N to N is infinite, but not countable

Question

Let $$N=\{0,1,2,3,...\}$$ be the set of all non-negative integers and $A$ the set of all bijections from $N$ to itself.

Prove that

$i)$ $A$ is an infinite set.

$ii)$ There exist no bijection from $N$ to $A$

Answer 1

Let us define the function $f_n: \mathbb{N} \to \mathbb{N}$ such that $f_n(0)=n$, $f_n(n)=0$ and $f_n(m)=m$ for all $m \in \mathbb{N} \backslash \{0,n\}$. Obviously, $f_n$ is bijective for all $n \in \mathbb{N}$ and $f_i \not = f_j$ for all $i,j \in \mathbb{N}$ with $i \not = j$. By this argument, we know that $A$ is infinite and has at least the size of $\mathbb{N}$.

Now suppose there was a bijection from $\mathbb{N} \to A$, then we could list all bijective functions from $\mathbb{N}$ to itself, i.e. $f_0$, $f_1$, $f_2$, ... are all bijections from $\mathbb{N}$ to itself and for every such bijection exists an $n \in \mathbb{N}$ sucht that $f_n$ denotes this bijection. This means that any bijection we can think of appears in our list.

Okay, now let us define a new bijection $f: \mathbb{N} \to \mathbb{N}$ as follows by induction on n. Therefore we let $X$ denote the set of even numbers.

• Choose $f(0) \in X \backslash \{f_0(0)\}$
• Choose $f(2n) \in X \backslash \{f_0(0), f_1(2), f_2(4), \dots, f_{n}(2n) \}$

Obviously, this is an injection from the even numbers to the even numbers. Then let $B$ denote the image of $f$ constructed so far, i.e. $B = \{f(0), f(2), f(4), \dots \}$. Since $Y := \mathbb{N} \backslash B$ contains at least the odd numbers but is also a subset of $\mathbb{N}$, we know that it is countable, i.e. there is a bijection $g:\mathbb{N} \to Y$. To make $f$ a total function on $\mathbb{N}$, we let $f(2n+1)=g(n)$ for all $n \in \mathbb{N}$.

So now we only have to see that $f$ is a bijection of $\mathbb{N}$ to itself, but this is clear since we first constructed an injection from the even numbers to the even numbers and then we sent the odd numbers injectively to all numbers which are left over. Moreover, we can see that $f$ does not appear in our list $f_1, f_2, f_3 \dotsc$ of all bijections from $\mathbb{N}$ to itself, since $f(2n) \not = f_n(2n)$ and hence $f \not = f_n$ for all $n \in \mathbb{N}$. But this is a contradiction, which means that there is no countable list which contains all such bijections. Hence there is no bijection from $\mathbb{N}$ to $A$.