Natural Number is Ordinary Set

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Theorem

Let $n$ be a natural number.

Then $n$ is an ordinary set.


That is:

$n \notin n$


Proof

The proof proceeds by induction.

For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:

$n$ is an ordinary set.


Basis for the Induction

By the Axiom of the Empty Set, $\O$ is a set.

From Empty Set is Ordinary:

$\O \notin \O$

Thus $\map P 0$ is seen to hold.


This is the basis for the induction.


Induction Hypothesis

Now it needs to be shown that if $\map P k$ is true, where $k \ge 0$, then it logically follows that $\map P {k + 1}$ is true.


So this is the induction hypothesis:

$k$ is an ordinary set.


from which it is to be shown that:

$k^+$ is an ordinary set.


Induction Step

This is the induction step:

By the Induction Hypothesis:

$k \notin k$

From Successor Set of Ordinary Transitive Set is Ordinary:

$k^+ \notin k^+$

So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.


Therefore:

$\forall n \in \N: n$ is an ordinary set.

$\blacksquare$


Sources

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