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The usual way of removing large cardinals from a given model of set theory is to cut off the model below the least large cardinal of interest. But this method may have dramatic effects on the external properties of the model, the simplest of which is cardinality.

Question: Suppose there is an uncountable transitive model of ZFC. Is there also an uncountable transitive model of ZFC + "There are no inaccessibles" ?

I don't have a strong intuition about which way this should go, but I would expect the answer to be positive. Having spent some time thinking about the problem, here are some observations.

Fix an uncountable transitive model $M$ and write $M_\alpha$ for $V_\alpha^M$. We may assume that for any $\eta$ inaccessible in $M$, the initial segment $M_\eta$ is countable, for if not we can replace $M$ by the first uncountable such initial segment. Since we have excluded the possibility of cutting off the model $M$ and passing to an inner model clearly will not help with getting rid of the inaccessibles, the only alternative that seems to be left is some sort of forcing construction.

The problem now splits into a number of cases:

  • $M$ has boundedly many inaccessibles. If we let $\eta$ be the supremum of the inaccessibles of $M$ then $\eta$ (and $M_\eta$) is countable. To destroy the inaccessibles we could now try to force with a poset of size $\eta$ in $M$, for example $\mathrm{Add}(\omega,\eta)$. The issue, of course, is that, for this to have any bearing on the original question, the $M$-generic for this forcing needs to exist in $V$. This is fine if $(2^\eta)^M$ is countable, since then $M$ only has countably many dense subsets of the forcing (as seen in $V$), but if this is not the case we seem to be stuck. Alternatively, if we could at least arrange that $(2^\eta)^M<\mathfrak{c}^V$ then an appropriate fragment of MA in $V$ would still give an $M$-generic.

  • $M$ has class many inaccessibles and $M\models\mathrm{Ord\ is\ not\ Mahlo}$. In this case we can fix a definable over $M$ class club $C$ in $\mathrm{Ord}^M$ not containing any of $M$'s inaccessibles. We now force over $M$ with a class-length Easton product $\mathbb{P}$ of the posets $\mathbb{Q}_\delta=\mathrm{Add}(\delta^+,\delta^*)^M$ where $\delta\in C$ and $\delta^*$ is the next point of $C$ above $\delta$. The key point now is that $V$ has $M$-generics for $\mathbb{P}$. This is because $M$ is $\omega_1$-like, meaning that all of its elements are countable, and remains such through all of the initial stages of the forcing $\mathbb{P}$. This means that, given any $\delta\in C$, the extension $M^{\mathbb{P}_\delta}$ constructed thus far only has countably many dense subsets of $\mathbb{Q}_\delta$, which allows $V$ to build a $M^{\mathbb{P}_\delta}$-generic $G_\delta\subseteq\mathbb{Q}_\delta$. Finally, it follows that $G=\prod_\delta G_\delta\subseteq \mathbb{P}$ is $M$-generic. As it is well known that forcing with $\mathbb{P}$ preserves ZFC, the model $M[G]$ is an uncountable transitive model without inaccessibles.

  • $M\models\mathrm{Ord\ is\ Mahlo}$. In this case we would like to undertake the same construction as in the previous bullet point after first forcing over $M$ to add a class club avoiding the inaccessibles. The problem again is with the generic club existing in $V$. The model $M$ is $\omega_1$-like again, so the club shooting forcing has $\omega_1$ many definable dense classes. The forcing is $\alpha$-strategically closed in $M$ for any $\alpha$, but I don't see that this would allow $V$ to build a generic.

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    $\begingroup$ My usual solution when I'm stuck is to try and engineer a counterexample. Have you tried starting with $L$ with a single Mahlo and collapse it to be $\omega_1$? $\endgroup$ – Asaf Karagila Nov 11 '15 at 9:02
  • $\begingroup$ @AsafKaragila Well, that will give you an uncountable model with lots of inaccessibles. One would then have to argue that there are also models which are quite far from $L$, where all of the inaccessibles have been killed. This is simple enough to do if we additionally assume that in $L$ there is a transitive model containing $L_{\kappa+1}$ (where $\kappa$ is the Mahlo), but if there isn't I am not sure how to proceed. $\endgroup$ – Miha Habič Nov 11 '15 at 14:57
  • $\begingroup$ Yeah, I meant of course the latter case. How about trying to show that in that situation there is no uncountable model without an inaccessible? $\endgroup$ – Asaf Karagila Nov 11 '15 at 15:23
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I think the answer is (consistently) no.

Following Asaf's comment, let $\kappa$ be Mahlo in $L$ and let $G \subset \text{Col}(\omega,\mathord{<}\kappa)$ be an $L$-generic filter. Let $\eta$ be the least ordinal such that $\kappa < \eta$ and $L_\eta \models \mathsf{ZFC}$, and consider the model $N = L_\eta[G]$.

Then in $N$ there is an uncountable transitive model of $\mathsf{ZFC}$, namely $L_\kappa$. It suffices to show that every such model has an inaccessible cardinal.

Let $M$ be an uncountable transitive model of $\mathsf{ZFC}$ in $N$. Note that $\text{Ord}^M \ge \kappa$ by uncountability and in fact $\text{Ord}^M = \kappa$ because otherwise $L^M$ would violate the minimality of $\eta$.

The set of $L$-inaccessibles below $\kappa$ is stationary in $L$ by definition, and this stationarity is preserved by the Levy collapse forcing that adds $G$ to obtain $N$. So the model $M$ satisfies "the class of $L$-inaccessibles is definably stationary."

Therefore (because $M \models \mathsf{ZFC}$) there is an $L$-inaccessible $\alpha < \kappa$ that is a strong limit cardinal in $M$. Then $\alpha$ is inaccessible in $M$ by Jensen's covering theorem (note that $0^\sharp$ could not have been added by forcing over $L$.)

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  • $\begingroup$ Wooo! My idea was correct! Also, it's good seeing you around Trevor! :) $\endgroup$ – Asaf Karagila Nov 17 '15 at 3:48
  • $\begingroup$ @Asaf Thanks! And I should edit my answer to mention that the idea came from your comment; not sure how I forgot that. $\endgroup$ – Trevor Wilson Nov 17 '15 at 4:11
  • $\begingroup$ Great argument! The Jensen covering trick at the end was the key piece that I was missing. $\endgroup$ – Miha Habič Nov 17 '15 at 21:00
  • $\begingroup$ @Miha Thanks. I was surprised to see covering get involved; I wouldn't have guessed it initially. $\endgroup$ – Trevor Wilson Nov 17 '15 at 21:36
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i added an answer and then realized it was a complete nonsense, i was getting at the following question.

Suppose kappa is an inaccessible cardinal. Is there a transitive model M of ZFC such that

  1. M has height kappa
  2. for some stationary in kappa set S, for every alpha in S, M computes alpha^+ correctly,
  3. for some stationary set D, for every alpha in D, alpha is inaccessible in M
  4. there are no inacessibles below kappa.

    One can show that if the answer is yes then there is an inner model with a proper class of measurables.

Assume M is as in the statement. Assume there is no inner model with a proper class of measurables. Then clause 2 implies that K^M and K must coiterate (K is the core model). This means that K has stationary set of inaccessibles, so by Trevor's trick V must also have inaccessibles. So we must have covering fails in V which then implies that K doesn't exist.

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  • $\begingroup$ Can't we use Radin forcing to produce such a model M. Assume $GCH+\kappa$ is $(\kappa+2)-$strong. Let $M=V_\kappa$. Force with Radin forcing, using a measure sequence of length $\kappa^+,$ to add a club of former regulars into $\kappa,$ and preserve $\kappa$ inaccessible. All limit points in the club are singular in the extension. Let $C$ be the set of limit points of the club, and force with Easton forcing to make $2^{\alpha^+}=\alpha_*^+,$ where $\alpha < \alpha_*$ are successive points in $C$.. $\endgroup$ – Mohammad Golshani Dec 13 '15 at 11:58
  • $\begingroup$ In the final extension, there are no inaccessibles below $\kappa,$ but the elements of $C$ (except its first element that we can assume is below the least inaccessible) were inaccessible in $V$, hence in $M$> $\endgroup$ – Mohammad Golshani Dec 13 '15 at 11:58
  • $\begingroup$ are you sure about the argument? i know i did the K stuff fast and there are details there to check, but kappa is kappa+2 strong sounds something that shouldn't be enough. $\endgroup$ – Grigor Dec 13 '15 at 17:29
  • $\begingroup$ You have K^M and K that coiterate, if K^M only has measurable with a few measures then any non-measurable inaccessible of K^M is inaccessible in K. So K has a stationary set of inacessibles. Then because K has covering, you get that this inaccessibles are inacessibles in V. Of course one has to check the details and i haven't done it, but on the face of it, you def need at least a strong...a woodin may be a stretch, as i know one has to be careful... $\endgroup$ – Grigor Dec 13 '15 at 17:29
  • $\begingroup$ ok, the problem is in my use of covering, i forgot that it doesn't quite hold when there are measurables around...so the argument from my note would only give you a proper class of measurables...this matches well with your example, i guess, i wonder if there is exact equiconsistency. $\endgroup$ – Grigor Dec 13 '15 at 18:29

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