## Is it possible to prove that $\{\mathbb{N}\} = \{\}$ ?

A standard definition of the naturals is considered once again: $$\begin{array}{l} 0 = \{\} \\ 1 = \{0\} = \{\{\}\} \\ 2 = \{0,1\} = \{ \{\} , \{\{\}\} \} \\ 3 = \{0,1,2\} = \{ \{\} , \{\{\}\} , \{ \{\} , \{\{\}\} \} \} \\ \cdots \\ n+ = \{0,1,2,3,4, .. ,n\} = [ \cdots \mbox{ oh, well } \cdots ] \\ \cdots \end{array}$$ The axiom of Infinity in ZFC is as follows. There exists a set which contains with each member $n$ also its successor $n+$, starting with $0$. So the natural numbers are defined by a successor function, named $s$, where: $$s(n) = n+ = n + 1$$ We subsequently find: $$\begin{array}{l} 0\\ 1 = s(0) = \{0\} \\ 2 = s(1) = s(s(0)) = \{0,1\} \\ 3 = s(2) = s(s(1)) = s(s(s(0))) = \{0,1,2\} \\ \cdots \\ n+ = s(n) = s(s(s(s( .. s(s(0)) .. ))) = \{0,1,2,3,4, \cdots ,n\} \\ \cdots \end{array}$$ Notation for multiple function composition. Let $m,n$ be any naturals: $$S_n(m) = \stackrel{n}{\overbrace{s \cdot s \cdot s \cdot s \cdots s \cdot s}}(m) = s(s(s(s( .. s(s(m)) .. ))) \quad [ : n \mbox{ parentheses pairs} ]$$ It is easily shown that $S_n(m) = S_{n+m}(0)$ . And: $$S_\infty(m) = \lim_{n\to\infty} S_n(m) = \lim_{n\to\infty} \{0,1,2,3,4, \cdots ,n+m\} = \mathbb{N}$$ Definition: $$T_n(\mathbb{N}) = \{S_n(0),S_n(1),S_n(2),S_n(3),\cdots,S_n(m),\cdots \}$$ Then it is easily shown that: $$T_n(\mathbb{N}) = \{\{0,1,2,3,4, \cdots ,n\},\{0,1,2,3,4, \cdots ,n+1\}, \cdots ,\{0,1,2,3,4, \cdots ,n+m\}, \cdots \}$$ So, with help of the above: $$T_\infty(\mathbb{N}) = \lim_{n\to\infty} T_n(\mathbb{N}) = \{S_\infty(0),S_\infty(1),S_\infty(2),\cdots,S_\infty(m),\cdots \} = \{\mathbb{N},\mathbb{N},\cdots,\mathbb{N},\cdots\} = \{\mathbb{N}\}$$ Mind the subtlety: the infinite composition $T_\infty$ of successors is not the naturals, but the singleton with the naturals as the only element in it.
However, we also have the following sequences: $$\begin{array}{l} \mathbb{N} = \{0,1,2,3,4,5,6,7,8,9, .. ,m, .. \} \\ T_1(\mathbb{N}) = \{1,2,3,4,5,6,7,8,9, .. ,m+1, .. \} \\ T_2(\mathbb{N}) = \{2,3,4,5,6,7,8,9, .. ,m+2, .. \} \\ T_3(\mathbb{N}) = \{3,4,5,6,7,8,9, .. ,m+3, .. \} \\ \cdots \\ T_n(\mathbb{N}) = \{n,n+1, .. ,m+n, .. \} \\ \cdots \end{array}$$ Therefore, all naturals from $0$ up and including $(n-1)$ are not present in the range of the composite function $T_n$.
Therefore the range of the mapping $T_\infty(\mathbb{N}) = \lim_{n\to\infty} T_n(\mathbb{N})$ is a set of naturals with not a single natural left in it; therefore it is the empty set $=\{\}$ .
But we have also shown that $T_\infty(\mathbb{N}) = \{\mathbb{N}\}$. Consequently: $\{\mathbb{N}\} = \{\}$ .
Take a good look at the latter formula. It says exactly this: the set containing only the set of all naturals is empty. This can mean nothing else than: the set of all naturals $\mathbb{N}$ does not exist. Which is clearly absurd.

What is right and what is wrong? That's the question.