Is this possible?

For each cardinal $\kappa$, we can define $H(\kappa) = \{ x \mid |trclx| < \kappa\}$, where trcl is the transitive closure, and $|A|$ is the cardinality of $A$. For every regular cardinal, and any axiom $\phi$ of ZFC aside from the axiom of infinity and the power set axiom, we have $H(\kappa) \models \phi$.

Additionally, we have a Levy Reflection Principle to the effect that for any formula expressable in ZFC $\phi$, we have that:

$\forall \alpha \in ON \ \exists \beta \in ON [\beta \geq \alpha \text{ and } \phi \text{ is absolute for } R(\beta)]$

Where $R(\beta)$ is the cumulative hierarchy.

As a homework exercise, I am to state a corresponding principle for $H(\beta)$, and to address the following question: is it possible that when we apply the principle to the power set axiom, that we have a model of ZFC $H(\kappa)$ for a $\kappa$ that is not inaccessible?

Inaccessibility here is taken to mean regular plus unreachable via power sets (if $\alpha < \kappa$, then $|P(\alpha)| < \kappa$).

My thoughts: since $H(\kappa)$ is a model of the power set axiom, it will be closed under the power set operation and so we have $P(\alpha) < \kappa$ for every $\alpha < \kappa$. As for regularity, if we assume it's not regular then we have a cofinal subset of cardinality less than $\kappa$. Each element will be in $H(\kappa)$, so we have a set of subsets of $H(\kappa)$ indexed by a set in $H(\kappa)$. So I want to say that the union should be in $H(\kappa)$, which would be a contradiction (if it's in $H(\kappa)$ it must have cardinality less than $\kappa$, but the union of this set should have cardinality $\kappa$. But I'm not sure if it can be shown that the union really is in $H(\kappa)$.


Question: Is it consistent that $H(\kappa)\vDash ZFC$ even though $\kappa$ is not inaccessible?

Answer: yes. Let $\kappa$ be inaccessible. Within $V_\kappa$ we can adapt the usual argument for reflection to show:

Theorem: There is a club $C\subseteq \kappa$ such that $\forall\alpha\in C(H(\alpha)\prec V_\kappa)$ (that is, $H(\alpha)$ is an elementary substructure of $V_\kappa$).

Clearly, for each such $\alpha$, $H(\alpha)\vDash ZFC$. But since the ordinals below $\kappa$ with co-finality $\omega$ are club, there will be club $\alpha$ such that $H(\alpha)\vDash ZFC$ which are not inaccessible.

Question: Are there $H(\kappa)$ which model $ZFC - Replacement + \neg Replacement$?

Answer: yes. We can define the beth function $\beth_\alpha$ such that $\beth_0 = 0$, $\beth_{\alpha+1} = 2^{\beth_\alpha}$, and $\beth_\lambda = \bigcup_{\alpha<\lambda}\beth_\alpha$. By recursion, we can define $\kappa_0 = \omega$, $\kappa_{n+1}$ to be the least ordinal such that for all $\beta<\kappa_\alpha$, $\beth_\beta< \kappa_{n+1}$, and $\kappa_\omega = \bigcup\kappa_n$. Then, $\kappa_\omega = \beth_{\kappa_\omega}$, and so $H(\kappa_\omega) \vDash ZFC - Replacement$. But since the $\kappa_n$ are definable, replacement will fail in $H(\kappa_\omega)$.

  • $\begingroup$ So if I understand you correctly, this $H(\kappa)$ with cofinality $\omega$ will not model ZFC. But this does not rule out the possibility of other non-inaccessible $\kappa$ such that $H(\kappa)$ does model ZFC. $\endgroup$ – Saigyouji Dec 10 '14 at 14:29
  • $\begingroup$ I think I was confused about what your question was. One is: if we reflect on the Powerset axiom to some $H(\kappa)$, will it always model ZFC. I answered "no, since $\kappa$ won't always be regular". Another is: is it consistent to have $H(\kappa)\vDash ZFC$ but $\kappa$ not be inaccessible (equivalently, not be regular). I will answer that now. (Are there other questions I've missed?) $\endgroup$ – GME Dec 10 '14 at 14:52
  • $\begingroup$ No, though it would be helpful to explain why what you wrote implies that $H(\kappa)$ does not model ZFC. It seems that you have countably many things in $H(\kappa)$, the set of which is not in $H(\kappa)$ and this should be impossible, but I don't immediately see why (what axiom(s) does this violate?) Alternatively, by using which axioms can we take unions of things in $H(\kappa)$ indexed by a set in $H(\kappa)$. We can definitely do this in ZFC (we do it all the time in standard mathematics) but the justification escapes me. $\endgroup$ – Saigyouji Dec 10 '14 at 14:57
  • $\begingroup$ It might well model ZFC (see my edit). Even though there is a countable co-final sequence, the model might not "see" that there is such a sequence. In other words, it might not be definable in the model. This is similar to the case with non-well founded models. From the outside, we know that they are non-well founded. But the models themselves might not know this (as it were). $\endgroup$ – GME Dec 10 '14 at 15:06
  • $\begingroup$ I appear to be confused. You answered "if we reflect on the Powerset axiom to some H(κ), will it always model ZFC. I answered "no, since κ won't always be regular"", which seems to suggest that $\kappa$ has to be regular for $H(\kappa)$ to model ZFC. But since $H(\kappa)$ always models the power set axiom (when using reflection on the power set axiom), doesn't this imply $\kappa$ inaccessible if $H(\kappa)$ models ZFC? $\endgroup$ – Saigyouji Dec 10 '14 at 15:16

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