2
$\begingroup$

This question follows Noah Schweber's excellent answer to a corresponding question regarding second-order $ZFC$ and the continuum hypothesis: https://mathoverflow.net/a/78083/24611

Simply put, it seems that $ZFC_2$ "decides" $CH$ in a certain sense that can be made formally precise, although second-order logic is so limited as to make it impossible for us to determine which way it is decided. This seems, if my understanding is correct, to be related to $ZFC_2$ being categorical if large cardinals are excluded.

Does a similar situation exist for $ZF_2$ and $AC$?

This paper seems to show that it is, and that $ZF_2$ is likewise categorical if large cardinals are excluded. Hence, this would mean that we have the same analogous situation to with $ZFC_2$ and $CH$, so that $AC$ is likewise "decided" in $ZF_2$, but that we don't know which way it is decided.

Is this analogy correct?

Improve this question
$\endgroup$
9
  • $\begingroup$ Isn't it the same argument? The "categoricity" argument (such as it is) doesn't use AC, but rather just that the $V_\alpha$ hierarchy is "determined" by the second-order logic background in which one is undertaking the argument, since that background concept provides all the subsets of $V_\alpha$ and thus the $V_{\alpha+1}$ that arises is the same as whatever exists in the background second-order logic. $\endgroup$
    – Joel David Hamkins
    Commented Apr 15, 2018 at 18:33
  • $\begingroup$ @JoelDavidHamkins Suppose that the first failure of choice happens between the first and the second inaccessible in the von Neumann hierarchy. Won't this mean that $ZF_2$ doesn't decide AC, since there are models of $ZF_2$ it which it does or doesn't hold? (the same couldn't have happened for CH, since the failure is always determined below the first inaccessible) $\endgroup$
    – Wojowu
    Commented Apr 15, 2018 at 18:55
  • $\begingroup$ @Wojowu I thought we were assuming that there are no large cardinals. $\endgroup$
    – Joel David Hamkins
    Commented Apr 15, 2018 at 19:05
  • $\begingroup$ @JoelDavidHamkins You seem to be right, I've missed the last part of the question. $\endgroup$
    – Wojowu
    Commented Apr 15, 2018 at 19:08
  • 1
    $\begingroup$ We're talking about ZF though, not ZFC. $\endgroup$
    – Mike Battaglia
    Commented Apr 15, 2018 at 19:45

1 Answer 1

Reset to default
4
$\begingroup$

Let me start by observing that we have to be a bit careful when talking about ZF$_2$. Specifically, there is a subtle distinction between set models and class models which needs to be highlighted. In one sense, ZF proves that $V$ is a class model of ZF$_2$ - in another sense, it can't even express this claim appropriately. For this reason I'm going to take as our background theory NBG$^\circ$ = NBG without global choice (or any choice - I don't think this notation is standard but I haven't seen a notation for it), and use the word "model" to refer to class models (note that every set model is a class model).

Now NBG$^\circ$ proves the following key fact:

Every model of ZF$_2$ is isomorphic to a unique initial segment of the cumulative hierarchy (possibly the whole thing). Moreover, the set models of ZF$_2$ are exactly the levels $V_\kappa$ for $\kappa$ strongly inaccessible. Finally, there exists at least one model of ZF$_2$ (namely, $V$ itself).

Proving this is straightforward: first show that models of ZF$_2$ are well-founded, next show via second-order powerset that they "get powersets right," and finally show via second-order replacement that the height of any such model must be either an inaccessible cardinal or $Ord$ itself.

We can now make the following argument in NBG$^\circ$, as per the linked question:

  • The following are equivalent: $(i)$ CH. $(ii)$ Every model of ZF$_2$ satisfies CH. $(iii)$ Some model of ZF$_2$ satisfies CH.

The strength of $(iii)$ here comes from the fact that CH is a "bounded statement:" it only refers to objects of fixed finite order. By contrast, a failure of choice could occur for the first time very high in the cumulative hierarchy. NBG$^\circ$ can prove:

  • Suppose AC. Then every model of ZF$_2$ satisfies AC.

Proof. If $x$ is well-orderable, then a well-ordering of $x$ exists in the powerset of $x\times x$. Now simply use that ZF$_2$-models are closed under true powersets. $\quad\Box$

The converse, however, can fail. Let $M$ be a model of NBG$^\circ$ + "There is an inaccessible cardinal" + "Choice holds below the first inaccessible" + "$\neg$AC." (Such an $M$ exists iff NBG + "There is an inaccessible cardinal" is consistent iff ZF + "There is an inaccessible cardinal" is consistent.) Then:

  • We have $M\models$ "There are models of ZF$_2$ that disagree about AC."

Namely, $(V_\kappa)^M$ thinks AC is true while $V^M$ thinks AC is false (here $\kappa$ is the least inaccessible in $M$). Note that $V^M$ isn't quite $M$ itself, but rather the sets-part of $M$ ($M$ is a model of NBG$^\circ$, not ZF).

On the other hand, of course, the inaccessible is necessary. In NBG$^\circ$ we can prove (as an easy corollary of the bolded claim up at the top of this answer):

  • If there is no inaccessible cardinal, then there is exactly one model of ZF$_2$ - namely, $V$.

Consequently, we have:

  • If there is no inaccessible cardinal, then ZF$_2$ is categorical (in the sense of class models - it's unsatisfiable in the sense of set models) and hence decides AC.
Improve this answer
$\endgroup$
9
  • $\begingroup$ Thank you! Some quick clarification - suppose NBG° has a proper class of inaccessible cardinals in the background theory -- call it NBG°+. Then it should contain a set model of ZF2 which has no inaccessible cardinals. Is there only one such set model? Furthermore, if we add "there are no strongly inaccessible cardinals" to ZF2 as an axiom, does NBG°+ have only one model of it? $\endgroup$
    – Mike Battaglia
    Commented Apr 15, 2018 at 20:05
  • $\begingroup$ @MikeBattaglia Already NBG$^\circ$ thinks that ZF$_2$ + "There is no strongly inaccessible cardinal" has at most one model: either there are no strongly inaccessible cardinals and $V$ is the unique model (and is a proper class), or there is a strongly inaccessible cardinal and $V_\kappa$ (with $\kappa$ the least inaccessible) is the unique model (and is a set). All this is provable inside NBG$^\circ$ alone. Similarly, NBG$^\circ$ proves that the following are equivalent: (cont'd) $\endgroup$
    – Noah Schweber
    Commented Apr 15, 2018 at 20:09
  • $\begingroup$ $(i)$ ZF$_2$ has a set model. $(ii)$ ZF$_2$ + "There is no inaccessible" has a set model. $(iii)$ There is an inaccessible cardinal. The point is that if an inaccessible cardinal exists, then the level of the cumulative hierarchy corresponding to the least inaccessible cardinal satisfies ZF$_2$; conversely, in order to have a set model of ZF$_2$ you need an inaccessible cardinal. So there's really no need to add a proper class of inaccessibles: we can prove the conditional results in NBG$^\circ$ already, and they become reality once we add a single inaccessible. $\endgroup$
    – Noah Schweber
    Commented Apr 15, 2018 at 20:10
  • $\begingroup$ Thanks, that makes perfect sense. So it looks like large cardinals where things seem to get complicated then. Does this result change for other metatheories? (In general, when people talk about models of second-order theories, are they usually talking about second-order NBG as a metatheory?) $\endgroup$
    – Mike Battaglia
    Commented Apr 15, 2018 at 21:03
  • 1
    $\begingroup$ @MikeBattaglia I'm using first-order NBG (or rather NBG$^\circ$) as a metatheory here; it's not very helpful to take a second-order metatheory, since then we need to put that inside a metatheory ... As to what people generally do, or what holds for other metatheories, I think that the picture remains the same if what we're looking at is second-order set theories. However, there are other interesting second-order theories, and for these the picture can be much more complicated. $\endgroup$
    – Noah Schweber
    Commented Apr 15, 2018 at 21:06

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .