# Order Types and Replacement Schema

Using Replacement Schema we can prove any well-ordering is isomorphic to an ordinal number.

Q: Is the following consistent?

$ZFC-Rep+\neg Rep+\text{Any well-ordering is isomorphic to an ordinal number}$

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It seems like a very Cantorian set theory. – Asaf Karagila Jan 8 '14 at 14:16
Your theory makes no sense. What can $\lnot\mathrm{Rep}$ possibly mean? – Andrés Caicedo Jan 8 '14 at 15:26
The question is trivial: $\mathsf{ZFC}$ is not finitely axiomatizable over $\mathsf{ZC}$ but adding "any well-ordering..." to $\mathsf{ZC}$ obviously is finitely axiomatizable over $\mathsf{ZC}$. So there are models of $\mathsf{ZC}+$"Any well-ordering..." that do not satisfy $\mathsf{ZFC}$. – Andrés Caicedo Jan 8 '14 at 15:38
@AndresCaicedo: It might be worth posting that as an answer. It's not easy to understand the difference between axioms and axiom schemes and this is a nice example where the difference is visible. – François G. Dorais Jan 8 '14 at 18:37

Let $\kappa$ be a $\beth$-fixed point -- that is, let $\kappa = \beth_\kappa$. Then $V_\kappa$ models ZFC - Rep + "every well-order is isomorphic to an ordinal". But ZFC proves the existence of $\beth$-fixed points. Thus, when $\kappa$ is the least such fixed point, $V_\kappa$ models ZFC - Rep + "every well-order is isomorphic to an ordinal" but fails to model ZFC (since being a $\beth$-fixed point is absolute for $V_\kappa$).
It isn't true that replacement must fail in every such $V_\kappa$, for there is no reason to expect that the cofinal $\omega$ sequence is definable from parameters inside $V_\kappa$, which is what it would take to violate replacement, and one can show that $V_\kappa\models\text{ZFC}$ is possible even when $\kappa$ has cofinality $\omega$. Nevertheless, as you suggest one can use a more specific $\kappa$, such as the first $\beth$-fixed point (or the next one after any given ordinal, or the next limit of fixed points, etc., as long as it is sufficiently absolutely definable). – Joel David Hamkins Jan 8 '14 at 13:03