# Zermelo-Frankel set theory for algebraists

$$\DeclareMathOperator\Var{Var}\DeclareMathOperator\CRings{CRings}\DeclareMathOperator\Grp{Grp}\DeclareMathOperator\Sets{Sets}$$I'm not a logician/set theorist, and I have some questions on set theory and references that may seem "trivial" for experts. Still I ask the question – if you have references this would be interesting.

In algebraic geometry (See Hartshorne's book, Appendix A) the following theorem is proved:

Let $$\Var(k)$$ be the "category of non-singular quasi-projective varieties over an algebraically closed field $$k$$ and morphisms of varieties over $$k$$. This category is defined in Hartshorne's book.

Theorem 1.1. There is a unique intersection theory $$A^*(X)$$ for algebraic cycles on $$X\in \Var(k)$$ modulo rational equivalence satisfying the axioms A1–A7.

The axioms A1–A7 are listed on page 426-427 in the book. For a variety $$X\in \Var(k)$$ one defines a commutative unital ring $$A^*(X)$$ – the Chow ring – and this construction is unique. There is only one way to do this, meaning there is a unique functor

$$A^*(-) : \Var(k) \rightarrow \CRings$$

such that axioms A1–A7 hold. Here $$\CRings$$ is the category of commutative unital rings and maps of unital rings.

In algebra one defines a group $$(G, \bullet)$$ as a set $$G$$ with an operation $$\bullet: G\times G \rightarrow G$$ satisfying $$3$$ axioms: G1 Associativity, G2 existence of identity and G3 existence of inverse. One defines a morphism of groups and the "category of groups" $$\Grp$$. Clearly the category of groups $$\Grp$$ contains non-isomorphic groups, hence the axioms G1–G3 does not uniquely determine one group. There are many different groups satisfying G1–G3.

In ZF set theory set theorists write down 9 axioms ZF1-ZF9, and these axioms determine $$\Sets$$ – the "category of sets". $$\Sets$$ is a category with "sets" as objects and "maps between sets" as morphisms. We would like the category $$\Sets$$ to be uniquely determined by the axioms ZF1–ZF9 similarly to what happens for the Chow ring. Is it? Is there a unique category $$\Sets$$ fulfilling the axioms ZF1–ZF9? If yes I ask for a reference.

For reference, Wikipedia has the page ZF set theory.

• "unital rings and maps of rings": you probably mean "[homomorphisms] of unital rings"? – YCor Dec 15 '20 at 17:21
• I'm puzzled by the comparison with groups since in groups you're discussing uniqueness of groups, not of the category of groups. So if you consider the category of sets, the analogous fact is just that a set is not unique up to set isomorphism... that is, there exist two sets that are not in bijection... – YCor Dec 15 '20 at 17:28
• I don't understand what's so hard about this. Given any mathematical object, we can ask if it can be characterized uniquely by axioms, and given any list of axioms, we can ask if they characterize a unique object. The axioms of a group do not characterize a unique object, since there are lots of groups, and similarly nor do the axioms of a category or the axioms of ZF. But there are individual groups that can be characterized uniquely by axioms, at least up to isomorphism, and similarly individual categories that can be characterized uniquely by axioms, at least up to equivalence. – Mike Shulman Dec 15 '20 at 18:34
• Of course one can make plenty of analogies between Set and Grp, but as far as I can see the question was not about any of those, but rather about whether Set can be characterized uniquely by axioms, in the way that a particular group might be characterized by axioms (though not by "the axioms of a group"). (Of course there was some confusion between the category Set and the corresponding model of ZF, but that's beside this point.) – Mike Shulman Dec 15 '20 at 18:37
• @MikeShulman I agree the question isn't hard to understand, but isn't the OP asking if Set is the unique category satisfying the 9 axioms of ZF, and if so wouldn't this be false since any autological topos has the full strength of ZF? – Alec Rhea Dec 15 '20 at 20:08

## 5 Answers

We say that a mathematical theory is categorical if it has exactly one model, up to isomorphism.

We intend some theories to be categorical, for instance the Peano axioms for natural numbers, Euclid's planar geometry, and set theory. Other theories are designed not to be categorical, i.e., the theory of a group, the theory of a ring, etc.

You are asking whether there are general theorems about categoricity, and whether in particular the Zermelo-Fraenkel set theory is categorical. First we have:

Theorem: If a theory expressed in first-order logic is categorical, then it axiomatizes a unique (up to isomorphism) finite structure.

Thus Zermelo-Fraenkel set theory and Peano arithmetic are not categorical. In fact, they both have many models:

Theorem: (Löwenheim-Skolem theorem) If a theory expressed in first-order logic has an infinite model, then it has an infinite model of every cardinality.

How should one react to these results? Perhaps we need not worry about it. So what if there are many models of Peano arithmetic and set theory? If we can accept the fact that there are many different groups, why not accept the fact that there are many different set theories? The mathematical universe just gets richer this way (but the search for "absolute truth" has to shift focus).

We could also "blame" first-order logic for these undesirable phenomena. For instance, whereas Peano axioms do not "pin down" the natural numbers, the category-theoretic notion of the natural numbers object does: all natural number objects in a category are isomorphic (because they are all the initial algebras for the functor $$X \mapsto 1 + X$$). This is possible because the category-theoretic description speaks about the entire category, not just the object of natural numbers.

We can in fact do the same for set theory: assuming a suitable "category of classes", the Zermelo-Fraenkel universe of sets can be characterized (uniquely up to isomorphism) as a certain initial "ZF-algebra" (this point of view has been studied in algebraic set theory). Note however that from a foundational point of view we have not achieved much, as we just shifted the problem from sets to classes.

As an algebraist, should you be worried that the category of sets is not "uniquely determined" by the Zermelo-Fraenkel axioms? I don't think so. Algebra is generally quite robust, and works equally well in all models of set theory. Of course, there are also parts of algebra that depend on the set-theoretic ambient, but is that not a source of interesting mathematics?

• +1. To clarify for the OP, the theorem implies that categorical theories like "$\mathsf{PA}$ with true induction" and "geometry with the completeness axiom" are not in fact first-order theories - usually second-order logic is secretly at play. Such theories generally do have first-order analogues, e.g. first-order $\mathsf{PA}$ (which modern sources tend to just call "$\mathsf{PA}$" but older sources may use that for second-order $\mathsf{PA}$) is gotten by replacing the second-order induction axiom with a bunch of first-order induction "special cases," but those analogues are vastly weaker. – Noah Schweber Dec 15 '20 at 19:31
• Also note that LS is often unnecessary: e.g. for $\mathsf{ZFC}$ and (first-order) $\mathsf{PA}$ we can also use a combination of Godel's incompleteness and completeness theorems. And more specialized techniques can be applied to particular theories (e.g. forcing shows that $\mathsf{ZFC}$ has some models where $\mathsf{CH}$ holds and others where it fails). – Noah Schweber Dec 15 '20 at 19:35
• @NoahSchweber: I was not sure how to explain the significance of second-order theories. These can be categorical, but bring in their own bag of problems. But how to explain in plain words why second-order logics aren't as good as first-order logics? (In fact, I don't actually believe this to be the case!) – Andrej Bauer Dec 15 '20 at 19:37
• First of all, I hope my previous comments didn't come off as criticism - that's definitely not how they were intended, but I'm not always good at tone in text. As to the question of relative value of FOL and SOL, I think - as with a lot of foundational questions - this really reveals that the question is conflating multiple issues at once. In this case I'd say that the primary conflation is between whether we want to express things or prove things. The latter points strongly away from SOL (although FOL has strenthenings which are still effective); the former is more subtle IMO. (contd) – Noah Schweber Dec 15 '20 at 19:49
• Expressive power and actual usability are in general in tension. FOL suffers on the former front but is pretty decent on the latter (it's not decidable but it is c.e.) while SOL is awful on the latter but works amazingly on the former. Since one of the goals of a foundational theory is to provide a common system which can be commonly used by mathematicians of wildly differing philosophies (even if they don't think it's 'true'), there's a definite place for FOL which SOL can't occupy here (without equipping it with an incomplete proof system anyways, at which point it's really FOL in disguise). – Noah Schweber Dec 15 '20 at 20:01

The axioms of ZF are not axioms which are expressible in categorical terms - they involve not just objects and morphisms in $$Sets$$, but also the membership relation $$\in$$. Therefore it doesn't even make sense to ask if some category satisfies those axioms.

Instead, those axioms are satisfied by certain structures called models of ZF. A model is just some set (or class) $$M$$ together with a relation $$\in_M$$, which may or may not coincide with the usual membership relation, such that all of the axioms ZF1-ZF9 are satisfied when we restrict the quantifiers to range over elements of $$M$$ and we replace $$\in$$ with $$\in_M$$ in those formulas. Under the assumption of consistency of ZF, there are many models like that, of various cardinalities.

To relate back to categories, to any model $$(M,\in_M)$$ of ZF we can associate a category, let me denote it $$Sets_M$$, whose objects are elements of $$M$$ and whose morphisms are functions in $$M$$ (which, in the set-theoretic definition of a "function", are themselves some sets which in a suitable sense relate the domain and codomain). The category $$Sets$$ itself is what happens if you take $$M$$ to be $$V$$, the class of all sets, and $$\in_M=\in$$. Therefore, while the question as stated doesn't exactly make sense, hopefully this illustrates why the axioms of ZF do not determine $$Sets$$ uniquely in any sense.

• This is a repeat "The axioms of ZF are not axioms which are expressible in categorical terms - they involve not just objects and morphisms in Sets, but also the membership relation ∈. Therefore it doesn't even make sense to ask if some category satisfies those axioms." The axioms of ZF set theory were formulated in order to "formulate a theory of sets free of paradoxes". In order to have a well defined and consistent "theory of sets", would want a "construction" of the "category of sets" and the "category of finite sets". Does such a construction exist? – hm2020 Dec 15 '20 at 19:50
• We can define an internal membership relation in a category with finite limits using power objects (ncatlab.org/nlab/show/power+object) and ask if the category we're in satisfies internal versions of the axioms of ZFC, like the internal axiom of choice -- it holds in some toposes and fails in others even when the background universe has choice. If a category has finite limits and power objects for all its objects it is a topos, which is where we can internally formulate enough logic to ask about all the axioms of ZFC internally. – Alec Rhea Dec 15 '20 at 19:52
• For more on this stuff, Mike Shulman has an excellent paper arxiv.org/abs/1004.3802. – Alec Rhea Dec 15 '20 at 19:59

You have to specify what you mean by a category 'fulfilling a list of axioms of set theory', because axioms of set theory are usually statements about sets in some logical language.

One reasonable interpretation would be to use the internal logic of a category to form an internal logical language, then formulate the axioms of set theory in the internal language and ask if they're true in the ambient category. Set can trivially internally formulate and satisfy all of the set-theoretic axioms of whatever background set-theoretic universe we're in, so in particular if we're working in $$ZF$$ then Set will satisfy ZF1-ZF9.

It turns out that other categories do this as well, so Set is not unique in this regard as other answers state. Specifically, Mike Shulman wrote a nice paper exploring how to extend the usual internal logic of a pretopos to a more general interpretation called the stack semantics, which allows us to formulate axioms involving unbounded quantifiers. Using the stack semantics of a topos, we can formulate all the axioms of $$ZF$$ internally and ask if they're true in the ambient topos.

Collection and replacement are satisfied in the stack semantics of any topos, but the separation axiom requires the introduction of a new topos-theoretic axiom schema -- any topos satisfying this additional axiom schema is called an autological topos, and these are precisely the categories which internally have the full strength of $$ZF$$ set theory.

So it seems like the answer to your question is:

No, Set is not the only category satisfying ZF1-ZF9 in the above sense, and the (bi)category of all categories satisfying ZF1-ZF9 would be precisely the (bi)category of autological toposes.

If you are looking for a unique characterization of Set, it is the terminal object in the category of Grothendieck toposes and geometric morphisms.

As mentioned in Andrej Bauer's answer it is also the initial 'ZF-algebra'. For a summary on this view with definitions you can take a look at these notes on Algebraic Set Theory by Steve Awodey, or take a look at the classical reference on Algebraic Set Theory by Joyal and Moerdijk for a full explanation.

Linked paper: arXiv:1004.3802 [math.CT]

There is such a thing as Algebraic Set Theory where:

models of set theory are simply algebras for a suitably presented algebraic theory and then many familiar set theoretic conditions (such as well-foundedness) are thereby related to familiar algebrauc ones (such as freeness) ... it was developed by Joyal & Moerdjik in 1988 and was first presented in detail in a book published in 1995 by them.

To the extent I can understand your question, I think I could probably say it's a no.

Because, for example, suppose there is a worldly cardinal (a cardinal $$\kappa$$ such that $$V_{\kappa}$$ is a model of first-order ZFC), which is an assumption which would usually be considered reasonable although it is not provable in ZF. Then there are transitive sets which satisfy axioms ZF1-ZF9 but are not equal to the entire universe of sets, and in fact $$V_{\kappa}$$ is one example, but there are countable examples as well. And there would be other ways to argue the same point, necessarily using assumptions which go a little bit beyond ZF.

So that is one way to answer your question. Assuming I've got a clear handle on what it is you're asking.

• You don't need a worldly cardinal for this: as long as $\mathsf{ZFC}$ is consistent, it has nonisomorphic models. If for some reason we want to restrict attention to well-founded models, it's still overkill: as long as $\mathsf{ZFC}$ has well-founded model, it has a countable well-founded model, and we can force over that. – Noah Schweber Dec 15 '20 at 20:09
• Consistency is definitely enough for existence of a model, yes, although you need a bit more if you want a well-founded model. And yes, even then, a worldly cardinal is overkill. – Rupert Dec 18 '20 at 21:04