User:Dfeuer/Natural Number is Transitive

Theorem
Let $n$ be a natural number.

Then $n$ is transitive.

Proof
Proof proceeds by induction:

Let $a$ be the class of all natural numbers $n$ such that $n$ is transitive.

$\varnothing \in a$ by User:Dfeuer/Empty Class is Supercomplete.

Suppose $n$ is transitive.

Let $m \in n^+$.

Let $x \in m$.

By the definition of successor, $m \in n \cup \{n\}$.

By the definition of class union and the definition of singleton, $m \in n$ or $m = n$.

If $m = n$, then $x \in n$.

If $m \in n$, then since $n$ is transitive, $x \in n$.

Thus $x \in n$.

Therefore $x \in n^+$.

As this holds for all such $m$ and $x$, $n^+$ is transitive, so $n^+ \in a$.

By the principle of mathematical induction, $a = \omega$.

Thus every natural number is transitive.