Element of Finite Ordinal iff Subset

Theorem

Let $m, n$ be distinct finite ordinals.

Then:

$m \in n \iff m \subseteq n$

Proof

Let $m \in n$.

Since $n$ is an ordinal, it is transitive.

Therefore, it follows directly that $m \subseteq n$.

Now let $m \subseteq n$.

We have by hypothesis that $m \ne n$.

From Natural Numbers are Comparable it follows that either $m \in n$ or $n \in m$.

Suppose $n \in m$.

Then $\exists x \in m: m \subseteq x$ which contradicts Finite Ordinal is not Subset of one of its Elements.

The only option left is that $m \in n$.

Hence the result.

$\blacksquare$

Sources

• 1960: Paul R. Halmos: Naive Set Theory ... (previous) ... (next): $\S 13$: Arithmetic