Element of Minimally Inductive Set is Transitive Set

Theorem
Let $\omega$ be the minimal infinite successor set.

Let $n \in \omega$.

Then $x \in n \implies x \subseteq n$.

That is, every element of $n$ is also a subset of it.

In other words, each element of $\omega$ is a transitive set.

Proof
Let $S \subseteq$ be the set of all transitive elements of $\omega$.

That is:
 * $n \in S \iff n \in \omega \land \forall x \in n: x \subseteq n$

It is vacuously true that $0 \in S$, as there are no $x \in 0$.

Now suppose $n \in S$.

If $x \in n^+$ then either $x \in n$ or $x = n$.

In the first case, $x \subseteq n$ as $n \in S$, and so $x \subseteq n^+$

In the second case, $x \subseteq n^+$ by definition of successor set.

By the Principle of Finite Induction it follows that $S = \omega$.

Hence the result.