2
$\begingroup$

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 $\prec$.

$\endgroup$
2
$\begingroup$

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 $\in$.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.