# Proof of “ordinal if and only if is Mostowski collapse”

I tried to prove the following, can you tell me please if my proof is correct? Thank you.

Claim: A set $\alpha$ is an ordinal iff $\alpha$ is the Mostowski collapse of a strict well-order $\langle X, \prec \rangle$. The Mostowski collapse of a strict well-order $\langle X, \prec \rangle$ is a unique transitive set $\alpha$ together with a collapsing function $F: X \to \alpha$ such that $F$ is a bijection and $\forall x,x' \in X ( x \prec x' \leftrightarrow F(x) \in F(x'))$, that is, $F$ is an order isomorphism.

Proof: $\implies$ Let $\alpha$ be an ordinal. Then by definition, $\alpha$ is transitive and strictly well-ordered with respect to $\in$. Hence $\mathrm{id}: \alpha \to \alpha$ is a collapsing function and every ordinal is its own Mostowski collapse.

$\Longleftarrow$: Let $\alpha$ be the Mostowski collapse of a strict well-order $\langle X, \prec \rangle$. Then $\alpha$ is transitive. It remains to be verified that $\alpha$ is strictly well-ordered with respect to $\in$. Since we have $\forall x,x' \in X ( x \prec x' \leftrightarrow F(x) \in F(x'))$, $F$ is a bijection and $X$ is a strict total order, it follows that $\alpha$ is a strict total order. Assume $\alpha$ was not well-founded and let $\beta \subset \alpha$ be a set containing an infinite descending chain. Then $F^{-1}\beta$ would be a subset of $X$ containing an infinite descending chain hence contradicting well-foundedness of $\prec$.

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It looks okay, but there is a shorter argument in the $\Longleftarrow$ direction: $\alpha$ is isomorphic to $\langle X,\prec\rangle$. Therefore $\alpha$ is transitive and well-ordered by $\in$.