if two models of ZFC have their ordinals isomorphic then there is a isomorphism between their constructibles?

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    $\begingroup$ I don't understand the downvote and vote to close - this seems like a reasonable question (if a bit poorly asked). $\endgroup$ Jan 28, 2017 at 22:51

1 Answer 1


The answer is no.

It's not hard to show - similarly to the situation with respect to PA - that if $M$ is a countable model of ZFC such that $\omega^M$ is non-well-founded, then $ON^M$ has ordertype $(\omega+\zeta\cdot\eta)\cdot(1+\eta)$, where $\zeta$ and $\eta$ represent the ordertypes of $\mathbb{Z}$ and $\mathbb{Q}$ respectively.

In particular, this means:

If $M, N$ are countable models of ZFC with illfounded natural numbers, then $ON^M$ and $ON^N$ are isomorphic.

So now let $p$ be any sentence independent of ZFC+V=L (which exists by Goedel), and let $M, N$ be countable models of ZFC+V=L with illfounded natural numbers such that $M\models p$ and $N\models\neg p$ (which exist by Compactness + Lowenheim-Skolem). Then $ON^M\cong ON^N$ but $L^M=M\not\cong N=L^N$.

Even if we restrict attention to $\omega$-models the answer is no, as long as we allow ill-founded $\omega$-models. This is a bit more complicated - there are lots of ordertypes of ordinals in countable $\omega$-models of ZFC - but can still be handled similarly to the above.

Namely, any theory extending $ZFC$ has a model whose ordinals have ordertype $$\omega_1^{CK}+\omega_1^{CK}\cdot\eta,$$ where $\omega_1^{CK}$ is the first nonrecursive ordinal; the ordertype above is called the Harrison order and is a standard counterexample in computable structure theory. So, again, pick a sentence $p$ which is true in some $\omega$-models of ZFC+V=L and false in others (e.g. "There is an inaccessible cardinal"), and let $M, N$ be countable models of $ZFC+V=L$ whose ordinals each have ordertype the Harrison order, and $M\models p$ but $N\models\neg p$.

Finally, if we restrict attention to well-founded models, the answer is yes - if $M, N$ are transitive models with $ON^M=ON^N$, then by induction on $\alpha$ we have $L_\alpha^{M}=L_\alpha^N$ for every $\alpha\in ON^M$.

  • $\begingroup$ thank you.And if both models are omega models? $\endgroup$
    – luis
    Jan 28, 2017 at 23:05
  • $\begingroup$ @luis See my edit. $\endgroup$ Jan 28, 2017 at 23:12
  • $\begingroup$ @luis I fixed a silly ordertype mistake. $\endgroup$ Jan 29, 2017 at 1:29
  • $\begingroup$ I mean, I shouldn't be writing comments around 4 am in Lapland. $\endgroup$
    – Asaf Karagila
    Jan 29, 2017 at 7:19
  • $\begingroup$ Nice; in each case you prove the stronger there are $M$, $N$ with $M\not\equiv N$. This makes me think of the same question with the additional hypothesis of $M\equiv N$. $\endgroup$ Jan 29, 2017 at 13:53

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