# ZF + DC + LM + multiple inaccessible cardinals

Solovay proved that from ZFC + existence of an inaccessible cardinal there is a model of ZF + DC + LM, where LM is "every set of reals is Lebesgue measurable."

If we instead start from ZFC + existence of $n$ inaccessible cardinals, does a construction like Solovay's give a model of ZF + DC + LM + existence of at least $n-1$ inaccessible cardinals?

I believe the answer is yes, but I am a complete novice in this area, so I am unsure. The sketch of an argument I have in mind is the following:

1. Solovay starts by taking the Levy collapse $V[G]$ for a generic $G$ that collapses all cardinals less than the first inaccessible $\kappa$ to $\omega$. I believe this preserves the other inaccessible cardinals. (Actually, I guess it even preserves $\kappa$?)

2. One can then pass to the inner model $L(R)$, where the desired result holds. (Kanamori points out that this is different than Solovay's original argument, but that it also works). But inaccessible cardinals are downward absolute so we still have at least $(n-1)$ inaccessible cardinals in $L(R)$. (Right?)

I understand that there are different notions of inaccessibility in a setting without choice, so I guess I should clarify that what I'm really after is something that would give me (several) Grothendieck universe(s). The reason is that it seems that a significant amount of measure theory and analysis can be done with ZF + DC (e.g. as in Fremlin's book), but in some other fields it is a technical convenience to be able to have Grothendieck universes.

Sorry if the question is rather naive/confused. I mainly want to use this result as a black box given the considerable prerequisites (which are admittedly rather fascinating!).

This question seems to be partially addressed by Asaf Karagila's answer to Implications of existence of two inaccessible cardinals? but I'm not completely certain.

Yes - Solovay's construction (at least the initial forcing- your concern about what "inaccessible" means without choice is a good one) only kills the first inaccessible, so the other $n-1$ survive.
Incidentally, this is a situation where we have to be careful about the word "preserves." You're right that the initial forcing preserves $\kappa$, in the sense that $\kappa$ remains a cardinal. However, we're interested in preserving more than cardinal-ness - we want to think about preserving inaccessibility! In this sense $\kappa$ is not preserved: it becomes the new $\omega_1$, which is of course not a limit cardinal. The forcing, however, does preserve the inaccessibility of the inaccessibles $>\kappa$. Of course the previous sentence requires proof, but it's not too bad and makes a good exercise.
If $\kappa$ is inaccessible in $V$ (where $V\models$ ZFC), then $L(\mathbb{R})^V\models$ "$(V_\kappa)^{L(\mathbb{R})^V}$ is a Grothendieck universe." (Note that the axioms for Grothendieck universes don't require choice!)
• Thanks! So, just to make sure I understand the second half of your answer: since $V[G] \vDash ZFC$, the upshot is that from a model of ZFC + $n$ inaccessible cardinals, then there is a model of ZF + DC + LM + a hierarchy of $(n-1)$ Grothendieck universes, $U_1, ..., U_{n-1}$, with the reals in $U_1$? Does the fact you quote in the second half have a reference or is it just sufficiently obvious to someone who knows about this stuff? (Solovay seems to say something similar here, #3 under "why v-inaccessible...") – user43585 Apr 4 '18 at 16:01
• I don't know of a reference; I would consider it a standard exercise. Note that in fact Grothendieck universes have to contain every real number: if $U$ is a universe, we have $\omega\in U$; now for $r\subseteq\omega$ consider the function $f$ sending $n\in\omega$ to $\langle n, 0\rangle$ if $n\in r$ and to $\langle n, 1\rangle$ if $n\not\in r$, where $\langle\cdot,\cdot\rangle$ is your favorite pairing function on $\omega$. Thinking along these lines lets you prove more, see en.wikipedia.org/wiki/Grothendieck_universe#Properties. – Noah Schweber Apr 4 '18 at 16:05
• @user43585: The easy way of doing that is to use the equivalence between $\kappa$ is inaccessible (in the universe definition) if and only if there is no $\alpha<\kappa$ and a surjection from $V_\alpha$ onto $\kappa$. Then it's easy to prove with absoluteness arguments. – Asaf Karagila Apr 4 '18 at 17:02