Unique Isomorphism between Ordinal Subset and Unique Ordinal

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Theorem

Let $\operatorname{On}$ be the class of ordinals.

Let $S \subset \operatorname{On}$ where $S$ is a set.

Then there exists a unique mapping $\phi$ and a unique ordinal $x$ such that $\phi : x \to S$ is an order isomorphism.


Proof

Since $S \subset \operatorname{On}$, $\left({S, \in}\right)$ is a strict well-ordering.



The result follows directly from Strict Well-Ordering Isomorphic to Unique Ordinal under Unique Mapping.

$\blacksquare$


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