# Natural Number is Transitive Set

## Theorem

Let $n$ be a natural number.

Then $n$ is a transitive set.

## Proof

The proof proceeds by induction.

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

$n$ is a transitive set.

### Basis for the Induction

From Empty Class is Supercomplete, we have that $\O$ is supercomplete.

Thus by definition of supercomplete, $\O$ is a transitive class.

By the axiom of the empty set, $\O$ is a set.

Hence $\O$ is a transitive 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 a transitive set.

from which it is to be shown that:

$k^+$ is a transitive set.

### Induction Step

This is the induction step:

 $\ds x$ $\in$ $\ds k^+$ $\ds \leadsto \ \$ $\ds x$ $\in$ $\ds k \cup \set k$ $\ds \leadsto \ \$ $\ds x$ $=$ $\ds k$ $\, \ds \lor \,$ $\ds x$ $\in$ $\ds k$ Definition of Transitive Set: from Induction Hypothesis

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

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

In either case:

$x \subseteq k$

But we have $k \subseteq k^+$ by definition of union of class.

Hence:

$x \subseteq k^+$

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 a transitive set.

$\blacksquare$