4
$\begingroup$

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

Q: Is the following consistent?

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

$\endgroup$
4
  • $\begingroup$ It seems like a very Cantorian set theory. $\endgroup$
    – Asaf Karagila
    Jan 8, 2014 at 14:16
  • 3
    $\begingroup$ Your theory makes no sense. What can $\lnot\mathrm{Rep}$ possibly mean? $\endgroup$ Jan 8, 2014 at 15:26
  • 3
    $\begingroup$ 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 $\mathsf{ZFC}$. $\endgroup$ Jan 8, 2014 at 15:38
  • $\begingroup$ @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. $\endgroup$ Jan 8, 2014 at 18:37

1 Answer 1

6
$\begingroup$

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 $V_\kappa$).

$\endgroup$
2
  • 3
    $\begingroup$ 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). $\endgroup$ Jan 8, 2014 at 13:03
  • $\begingroup$ You're right, of course; I was unclear. I've updated my answer. Thanks! $\endgroup$ Jan 8, 2014 at 15:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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