# 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.

$\blacksquare$

## Sources

- 2010: Raymond M. Smullyan and Melvin Fitting:
*Set Theory and the Continuum Problem*(revised ed.) ... (previous) ... (next): Chapter $3$: The Natural Numbers: $\S 3$ Derivation of the Peano postulates and other results: Theorem $3.1$