# Element of Finite Ordinal iff Subset

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## 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: Proof using Minimal Infinite Successor Set 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