6
$\begingroup$

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 $H(\kappa)$.

$\endgroup$
2
$\begingroup$

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 $H(\kappa_\omega)$.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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