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Given these facts:

  • Countable ${AC}$ is independent of ${ZF}$
  • Godel's Completeness Theorem
  • Under ${ZF}$, Countable ${AC}$ is equivalent to "The Cartesian Product of every countable family of non-empty sets is non-empty"

There MUST be models of ${ZF}$ where ${\displaystyle\lnot{ACC}}$ (Countable $AC$), which is the same as, models of ${ZF}$ where there exists a countable family of non-empty sets, whose Cartesian Product is empty.

Can someone give a simple explanation/intuition of such a model? It seems insane to me that such a thing is conceivable.

Truth-seek
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  • Models of set theory don't necessary translate the set theory statement, $x\in X$ to mean that in the model, e object associated to $x$ is also a member of the model of $X.$ $\in$ from set theory is a first order relationship, and not a specified interpretation. – Thomas Andrews Jun 02 '25 at 19:38
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    Try explaining to a layperson an intuitive construction of a finite dimensional algebra over the reals. Sometimes things are technical and some technical preliminary knowledge is needed in order to understand their intuitive descriptions properly. – Asaf Karagila Jun 02 '25 at 19:52
  • Perhaps it is easier to imagine a situation where there is a countable sequence of two-element sets with no definable member of its cartesian product. The idea being that perhaps the elements of the pairs are difficult to systematically distinguish from one another. They are "socks" in Russell's analogy. But all of the other axioms of ZF play nicely with definability, so when we pass to a submodel consisting of only definable things, that generally still satisfies ZF. – spaceisdarkgreen Jun 02 '25 at 20:18
  • (But yes, there's a lot of technical nuance under the rug here, e.g. "definable" is not definable, so how do we define it, and the idea of how we make it so the elements are suitably hard to distinguish takes quite a while to unpack.) – spaceisdarkgreen Jun 02 '25 at 20:21
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    Would you be happy with a (boolean) topos model, rather than a model of ZF? Look at the topos of sets with a $\mathbb{Z}$-action, all of whose orbits are finite. Then each $\mathbb{Z}/n$ (with the obvious action) is a set in this topos, yet their product $\prod_{\mathbb{Z}/n}$ is empty since none of its orbits are finite! If you like I can expand on this in an answer -- especially since this model is boolean, it's very closely related to ZF... – Chris Grossack Jun 02 '25 at 20:27
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    These topoi of $G$-sets (with bonus properties, like finite orbits) are closely related to "permutation models" in classical set theory. So there's probably a similar story for ZFA, and then you can translate that (by methods I don't know but I'm told are routine) into an honest model of ZF... But I'm not enough of a set theorist to say exactly how that would work – Chris Grossack Jun 02 '25 at 20:28
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    @ChrisGrossack makes an excellent point that in the factoring 1. make a model where some things are indistinguishable 2. pass to a submodel that forgets about objects that unnaturally distinguish these things, step 1 can be drastically simplified at the expense of not really getting a model of ZF by working in more flexible backgrounds that allow for nice automorphisms of the starting model. The way to get from there to a model of ZF is to do the hard work of step 1 in a generic way, and show that we can always capture the symmetries that can be more easily implemented by a permutation model. – spaceisdarkgreen Jun 02 '25 at 20:53
  • @ChrisGrossack I would certainly be interested in hearing an expansion of your answer, it sounds quite interesting. – Carlyle Jun 02 '25 at 22:06
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    I won't close as a duplicate given there's interest in seeing a topos-flavored answer but this question could probably be justified as a duplicate (and there are links to other answers too). Note the permutation model in my answer there does violate countable choice, although a probably more intuitive model would be to start with a countable set of pairs of atoms and to take a finite support submodel under the group of permutations that swap atoms within pairs (aka the "second Fraenkel model"). – spaceisdarkgreen Jun 02 '25 at 23:03
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    I’m guessing the OP would find convincing an argument that starts with a model in which countable choice is true and constructs from that a model in which it is false. Maybe forcing would do the trick? Or maybe some more elementary argument? (of course, how comprehensible such an argument would be relative to the OP’s level of knowledge of set theory is a different matter…) – NikS Jun 03 '25 at 00:37
  • @NikS Yes, that would be very much convincing, even though I still don’t understand how that could ever be possible. – Truth-seek Jun 03 '25 at 06:30
  • @spaceisdarkgreen What’s this “second Fraenkel model” that you speak of? Anywhere that I can learn more about the same? – Truth-seek Jun 03 '25 at 08:21
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    @Truth-seek — One clarification: Are you asking “how can there possibly exist a model of ZF in which countable choice is false?” Or are you asking “how can a model of ZF in which countable choice is false possibly make sense as a foundation for the rest of mathematics?” Those are two quite different questions. The first has a clear-cut answer (it is possible). The second is more philosophical and I suspect many would say it doesn’t make sense as a foundation. – NikS Jun 03 '25 at 10:33
  • @Truth-seek I said what it was. You can find all of this stuff in Jech's book The Axiom of Choice. Note I'm not saying it's more intuitive in the sense that it's easier to understand than the model in my answer (though it's not any harder really), just that the failure of countable choice is more manifest. – spaceisdarkgreen Jun 03 '25 at 12:43
  • @NikS I fear I've talked too much in this comment thread already, but as I mentioned, one way to get models of ZF where choice fails is to take inner models of models of ZFC. The main obstacle to this is that if V=L holds in a model, there are no nontrivial inner models. Two ways around are 1) forcing 2) adding large cardinal axioms inconsistent with V=L. Both can result in interesting inner models where choice fails, the former much more flexible (and I don't know of a case of the latter where countable choice fails, but that may just be ignorance). – spaceisdarkgreen Jun 03 '25 at 14:29
  • @NikS I think I kind-of missed those two being different questions. I find ZFC axioms to be intuitive, and infinite sets to be “somewhat” intuitive, and I think I failed to understand that any model of ZF where C fails must fail to capture our “intuitive idea of sets” in some way. – Truth-seek Jun 03 '25 at 18:27

1 Answers1

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A boolean topos is a model of a kind of set theory, whose relationship with ZF is well understood (see here, for instance). In this answer we'll build a topos which has countably many nonempty sets $C_n$ with the strange property that $\prod_n C_n$ is empty, and hopefully we'll make this computation seem "obvious", or at least very doable. For this you might have to take some basic topos theory on faith, but you can learn more in the excellent book Sheaves in Geometry and Logic. In this answer we'll only work with boolean topoi (those which satisfy the law of the excluded middle), so we won't need to worry about the usual intuitionistic subtleties that can come up when doing topos theory.

First, for any group $G$ the category of $G$-sets is a topos. Indeed, given two $G$-sets $X$ and $Y$ we know how to take their product and disjoint union, the "powerset" of $X$ is the usual powerset where the $G$ action sends $A \subseteq X$ to $g \cdot A = \{g \cdot a \mid a \in A\} \subseteq X$, and the axiom of infinity comes from $\mathbb{N}$ with the trivial $G$-action.

Next we'll consider the subcategory of all those $G$-sets whose orbits are finite. Note that the inclusion into all $G$-sets is left exact and admits a right adjoint. Without the jargon, these (respectively, morally) say that

  • If you compute finite products in the subcategory, that's the same as computing finite products among all $G$-sets, and every subset of something in this subcategory is already in the subcategory.
  • If you have any $G$-set, you can turn it into a $G$-set with finite orbits by just throwing out any infinite orbits. This is a kind of "inverse" to the inclusion.

Now it's a theorem (Thm 8.4 on pg251 of Sheaves in Geometry and Logic) that in any such situation, the subcategory is again a topos, and the second condition tells us how to compute with it! Given some set-theoretic construction we want to do to $G$-sets with finite orbits, we can just do the construction worry free (using the fact that anything we do to a bunch of $G$-sets will give us a $G$-set), and then at the end of the day we throw out any infinite orbits to make sure we end up back in the "subuniverse" of $G$-sets with finite orbits.

Finally, I claim the topos of $\mathbb{Z}$-sets with finite orbits violates choice in the way you're looking for!

Indeed, consider the $\mathbb{Z}$-sets $\mathbb{Z}/n$, all of whose orbits are finite (there's only one orbit, of size $n$). So these sets all live in the topos of $\mathbb{Z}$-sets with finite orbits, and they're all nonempty. What happens if we take their product?

Well, to compute the product of elements we just compute the usual product as $G$-sets, and then throw out any infinite orbits! But we see that every orbit in $\prod_n \mathbb{Z}/n$ is infinite, so that when we throw out infinite orbits we're left with the empty set. That is, this topos thinks that

$$ \prod_n \mathbb{Z}/n = \emptyset $$

even though all the things we're producting together are nonempty!


I definitely read this argument in a paper at one point, but for the life of me I can't find which paper it was... While looking I was able to find it as an exercise (Chapter 5, Exercise 5 on pg 163 of Johnstone's Topos Theory). There's also a subtlety that I'm not going into about the "external" vs "internal" axioms of choice in a topos, and the way you actually build something that looks like the cumulative hierarchy to get a model of ZF out of this topos. For instance, Freyd's famous The Axiom of Choice mentions that while the model I gave here violates the internal axiom of choice, if you use Fourman's construction of a cumulative hierarchy the resulting model of ZF will (counterintuitively) model choice (since the well founded objects in this topos are just those with the trivial $G$-action).

Of course, I think this is still a very instructive example to see how you might take a product of nonempty things and end up with something empty.


I hope this helps ^_^