## 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.

(set-theory)(paradoxes)

### 1 Answer

The answer is, of course, that the conclusion may be only absurd in the paradise of Cantorian mathematics.

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 a Cantorian mathematician, would conceive a natural as the set of all
naturals preceding it.