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The standard definition of the naturals is considered once again:
$$
\begin{array}{l}
0 = \{\} \\
1 = \{0\} = \{\{\}\} \\
2 = \{0,1\} = \{ \{\} , \{\{\}\} \} \\
3 = \{0,1,2\} = \{ \{\} , \{\{\}\} , \{ \{\} , \{\{\}\} \} \} \\
\cdots \\
n+ = \{0,1,2,3,4, .. ,n\} = [ \cdots \mbox{ oh, well } \cdots ] \\
\cdots \end{array}
$$
It should be noticed that facts in Implementable Set theory are quite different
from this: $$\{0,1,2,3,4, .. ,n\} = 2^{n+1}-1$$ But let's ignore and proceed.
The axiom of Infinity in ZF 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: $$\Large \{\mathbb{N}\} = \{\}$$
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.
This is clearly absurd in the paradise of Cantorian mathematics. But the simple truth is that
the completed infinite set of all naturals is an absurdity in itself.
It is important
to notice, though, that the above proof critically depends upon the construction of the naturals as von Neumann
ordinals. Actually, it's an "unwanted" side-effect of that construction. Which is not unexpectedly so, because
not a sensible person in the world, except some mathematician, would conceive a natural as the set of all naturals
preceding it.