# Finite Ordinal is not Subset of one of its Elements

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## 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 finite ordinals $n$ 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 from Subset Relation is Transitive:

- $n^+ \subseteq x \implies n \subseteq x$

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 Mathematical Induction:

- $S = \omega$

Hence the result.

$\blacksquare$

## Sources

- 1960: Paul R. Halmos:
*Naive Set Theory*... (previous) ... (next): $\S 12$: The Peano Axioms