Natural Number is Ordinary Set

Theorem
Let $n$ be a natural number.

Then $n$ is an ordinary set.

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.

By definition, $\O$ has no elements.

Hence:
 * $\O \notin \O$

Thus by definition $\O$ is an ordinary set.

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:


 * $k \notin k$

$k^+ \in k^+$.

If $k^+ \in k$ then, because $k$ is transitive, $k^+ \subseteq k$.

If $k^+ = k$, then, because $k^+ = k \cup \set k$, it follows that $k^+ \subseteq k$ by definition of union of class.

In either case:
 * $k^+ \subseteq k$

But we have:
 * $k \in k^+$

and because $k^+ \subseteq k$ it follows that:
 * $$ \in $$

But this contradicts our supposition that $k \notin k$.

Hence by Proof by Counterexample:
 * $k^+ \notin k^+$

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.