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

Claim: If $\langle X, \prec_X \rangle$ and $\langle Y, \prec_Y \rangle$ are isomorphic strict partial orders then they have the same Mostowski collapse.

Proof: Let $f: X \to Y$ be an order isomorphism. Let $F: X \to \alpha$ and $G : Y \to \beta$ be the respective collapsing functions. Then $G \circ f : X \to \beta$ is also a collapsing function. By the uniqueness of the Mostowski collapse it follows that $\alpha = \beta$.

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  • $\begingroup$ I'm not an expert but if it's unique, then your proof is obviously correct. $\endgroup$ – yo' Dec 5 '12 at 21:30
  • $\begingroup$ @tohecz Thank you. Nothing is ever "obvious" to me, unfortunately. $\endgroup$ – Rudy the Reindeer Dec 5 '12 at 21:39
  • $\begingroup$ @tohecz: It is unique. Furthermore on transitive sets the collapse is the identity (i.e. nothing is changed). $\endgroup$ – Asaf Karagila Dec 5 '12 at 23:59
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Since the algebras $\langle X,\preceq_X\rangle$ and $\langle Y,\preceq_Y\rangle$ differ by an isomorphism, anything that relies only on them being these algebras must at most differ by an isomorphism as well.

This is the idea you use in the proof. I'm, however, not entirely sure that $\alpha=\beta$, but according to this wiki site it seems to (the uniqueness seem to be global, so isomorphisms play no role).

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