Finite Ordinal is not Subset of one of its Elements

Theorem
Let $n$ be a finite ordinal.

Then:


 * $\nexists x \in n: n \subseteq x$

that is, $n$ is not a subset of one of its elements.

Proof
Let $S$ be the set of all those natural numbers $n$ which are not a subset of any of its elements.

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

We know that $0 = \varnothing$ is not a subset of any of its elements, as $\varnothing$ by definition has no elements.

So $0 \in S$.

Now suppose $n \in S$.

From Set is Subset of Itself, $n \subseteq n$.

But as $n \in S$ it follows by definition of $S$ that $n \notin n$.

By definition of the successor of $n$, it follows that $n^+ \nsubseteq n$.

Now $n^+ \subseteq x \implies n \subseteq x$ from Subset Relation is Transitive.

But since $n \in S$ it follows that $x \notin n$.

So $n^+ \nsubseteq n$ and $\forall x \in n: n^+ \nsubseteq x$.

So $n^+$ is not a subset of any of its elements.

That is, $n^+ \in S$.

So by the Principle of Finite Induction, $S = \omega$.

Hence the result.