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In category theory, there are different ways to approach the "size issues" that crop up when we try to formalise the subject in axiomatic set theory. As far as I can tell, there are two main approaches:

  1. Work with classes. Then the categories $\mathsf{Set}$, $\mathsf{Ring}$, $\mathsf{Grp}$ etc. have a proper class-sized number of objects.
  2. Work with universes. In this formalism, we a fix a universe $U$ (say $V_{\kappa}$ where $\kappa$ is the smallest inacessible cardinal), and then work with the categories $\mathsf{Set}_U$, $\mathsf{Ring}_U$, $\mathsf{Grp}_U$, etc. of small sets/rings/groups. This approach has the advantage that we needn't worry about the set-theoretic subtleties involving classes.

From a set-theoretic perspective, the categories $\mathsf{Set}$ and $\mathsf{Set}_U$ (for example) are very different. For instance, the sets in $\mathsf{Set}_U$ have bounded rank. But my impression is that most category theorists wouldn't care for such distinctions, as concepts like rank are not invariant under isomorphism. Are there any distinctions between the two approaches that would be of interest to category theorists, and if not, is there a precise sense in which we can say that the two approaches are "the same"?

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    $\begingroup$ Taking the classes view presents naïve issues with things like functor categories between large categories etc., since you’ll want a category whose objects are proper classes which is impossible in e.g. MK class theory. $\endgroup$
    – Alec Rhea
    Commented Jun 24 at 13:08
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    $\begingroup$ Rank isn’t invariant under isomorphism, but cardinality is. $\endgroup$ Commented Jun 24 at 13:13
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    $\begingroup$ Functor categories rarely exist without restrictions on the nature of functors, if the domain of big. If you work with universes, you'll get a lot of "fake" functor categories; metaphorically speaking, you'll get something like a subring of $End(V)$ for a vector space of cardinality $a$, such that rank of the image has cardinality at most $b$. It's much more natural and less error-prone to impose those restrictions on "size of functors" explicitly. $\endgroup$
    – Denis T
    Commented Jun 24 at 13:15
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    $\begingroup$ @AlecRhea Functors not forming a category is not a bug, it is a feature. $\endgroup$
    – Denis T
    Commented Jun 24 at 13:18
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    $\begingroup$ @DenisT I'm not sure what you mean by 'rarely exist'; perhaps you mean that you don't encounter them 'in nature' in your own work, which is fine. I do, and for me functors not forming a category is pretty much always a bug, but to each their own :^). $\endgroup$
    – Alec Rhea
    Commented Jun 24 at 20:04

3 Answers 3

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From the point of view of a category theorist, I would say there are many fundamentally non-equivalent way of handling size questions - but they are in a completely different direction than what is mentioned in the question. To me, the only real distinctions between different approaches to size depend on the answer to practical questions like "what can you do with classes", "what can you do with small sets", "How many different sizes do you need", etc...

I wouldn't say that the fact (mentioned in the question) that in one case the cardinality of a set is bounded, and in the other it is not is a real difference: In both cases, the cardinality of any set is bounded by the cardinality of the class of all sets - it is just that what you call a "cardinal" in both case is not the same thing - in one case it refer to a small cardinal and not in the other. But that is not a real distinction - we are talking about the same thing but giving it different names.

But here are what I would consider three fundamentally different formalisms:

  1. If your "classes" are axiomatized within ZFC by a "formula as classes" paradigm or by NBG set theory. then you can't do much with classes, the category of classes and maps between them is pretty much just a category with finite limits (A little more than that to be honest, let's say a "Heyting pretopos" to be in line with the literature on algebraic set theory). In many cases that's enough, but not always.

  2. If you use MK set theory then you can build new class by comprehension using quantification over all class.

  3. If you use something like inaccessible cardinal or Grothendieck universes, then your category of classes is a fully formed set theory. You can form the class of functions between two classes, the class of subclass of a class, etc... Your category of classes is an elementary topos (and even more).

These corresponds to fundamentally different theory with different consistency strength : (1) is ZFC or equivalent, (2) is stronger than ZFC and (3) is even stronger. And the difference are significant in practice - I can't think of an example where (2) is significant, but if you want to talk about the category of all endofunctor of $\mathsf{Set}$, you need something like (3), while (1) only let you talk about individual functor $\mathsf{Set} \to \mathsf{Set}$.

But the differences inside a single group are just what I would call "linguistical" differences as the one I mentioned before, for example on whether the word "sets" is going to mean "small set" or "set or class", or on whether we actually have a category of all small sets or just a category equivalent to that.

Note that I'm not saying at all there are only three way of handling size issue - I'm just giving three example, but there are many other parameter you can play on:

  • You can decide the number of different sizes (just "set" and "classes", or "set/class/super class", a countable number of size, or maybe even more than that).

  • you can change what the category of small set need to satisfy: For example if "small sets" just needs to form an elementary topos with NNO (so only satisfies bounded replacement) then even if you ask your classes to be a full model of ZFC, you have a theory not logically stronger than ZFC. If you are asking even less properties of your small set them taking "small" to mean $\kappa$-small for some regular cardinal $\kappa$ might be enough and you have a proper class of different size at your disposition just within ZFC, and more freedom to apply your results.

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  • $\begingroup$ Thanks for this answer. Would you be clarify the statement "In both cases, the cardinality of any set is bounded by the cardinality of the class of all sets - it is just that what you call a "cardinal" in both case is not the same thing - in one case it refer to a small cardinal and not in the other." $\endgroup$
    – Joe Lamond
    Commented Jun 25 at 10:18
  • $\begingroup$ If I am reading this correctly, your point is related to how the ZFC-statement corresponding to "for every cardinal $\lambda$, the universe is larger than $\lambda$" is true both in the class of all sets $V$, but also internally true in all models of ZFC, including (uncountable) Grothendieck universes. Is this part of the reason you consider this to not be a real distinction? (Apologies if I am misrepresenting your view – I'm not that well-versed in models of set theory and logic.) $\endgroup$
    – Joe Lamond
    Commented Jun 25 at 10:19
  • $\begingroup$ Yes that is exactly it. If you working using a Grothendieck Universe $U$ to model your large vs small distinction and says that the cardinality of (small) set has an upper bound, this upper bound is the cardinality of the universe. If you are working in plain ZFC or in NGB and uses class to model large set (and set for small set) then it is still true that every set is small than the class of all sets - it is just that now you are not considering this upper bound as being a cardinal. $\endgroup$ Commented Jun 25 at 10:35
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    $\begingroup$ I really agree with Simon’s point regarding cardinality; in particular, when we translate from ZFC with inaccessibles to higher order class theory we see that the ‘class of all ordinals’ at each stage canonically corresponds to the appropriate inaccessible under the translation. $\endgroup$
    – Alec Rhea
    Commented Jun 25 at 14:45
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Although people often talk as though it just doesn't matter which approach you use — perhaps all universes are alike? Let me prove that in a strict sense this is not true. The nature of the mathematical setting that arises is necessarily different in the various accounts, different in regards to some core foundational issues in category theory. All universes are not alike, and perhaps none of them is like the proper class category, even with respect to first-order internally expressible features relevant for category theory. It is never the case that all universes have the same internal theory as the full universe $V$ or each other, although it is possible that some of them do. In this sense, yes, there are substantive differences; and no, they are not "the same."

Suppose we place ourselves at first in an ambient set-theoretic background of ZFC with unbounded many inaccessible cardinals, that is, where the universe axiom holds. In this case, the category of sets considered as a definable class will have the property that it thinks there are many Grothendieck-Zermelo universes inside it, and, in particular, every set exists in such a universe. In short, the existence of all those inaccessible cardinals is observable inside the category of sets.

But this is a feature that is not true in every Grothendieck-Zermelo universe itself. For example, if $\kappa$ is the smallest inaccessible cardinal, then the category of $\kappa$-small sets will not have any GZ-universes inside it, a strong failure of the universe axiom relativized to that universe. The next inaccessible will have exactly one universe inside it, and it will think that not every set is in a universe. The category of sets made with the $\omega$th inaccessible cardinal will think there are infinitely many GZ universes, but still bounded in the cardinals, since the $\omega$th inaccessible, being regular, is not the limit of the first $\omega$ many inaccessible cardinals.

If the ambient universe has a proper class of inaccessible cardinals, but no inaccessible limits of inaccessible cardinals, then the background universe (category of sets as a proper class) will think that every set can be placed inside a universe, but none of the universes themselves will think this is true. In other words, in this background, the universe axiom is true inside the category of sets using the class approach, but not when using any particular universe.

In general, knowing the universe axiom is true doesn't entitle you to know that it is true inside any particular universe.

In my paper with Robin Solberg, we investigate the extent to which the various Grothendieck-Zermelo universes admit categorical characterizations — these are the universes that can be characterized by an internal feature, expressed either by a first-order sentence, a first-order theory, a second-order sentence, or a second-order theory.

If there are more than continuum many inaccessible cardinals, then many of them must have the same first-order theory, since there aren't that many theories, but never is it the case (assuming there are at least two) that all universes have the same theory, since the smallest one thinks there are no universes but all others think there are some.

It is possible that some universes have the same first-order theory as the full universe $V$, and this occurs when $V_\kappa\prec V$, that is, some universe is an elementary substructure of the full universe. This is a large cardinal axiom with the same consistency strength as ZFC+"Ord is Mahlo", which is strictly weaker than the existence of a Mahlo cardinal, which is itself a rather mild large cardinal axiom.

Indeed, one can arrange a proper class of inaccessible cardinals $\kappa$ with $V_\kappa\prec V$, and all such universes will have the same first-order theory as $V$ itself, even with parameters. However, it doesn't follow from this that these $V_\kappa$ are themselves models of Ord is Mahlo, since if there is one then there is a least one and perhaps we were already inside that least one, in which case none of the universes below model that theory.

There could be universes $V_\kappa$ that have the same second-order theory as the full universe $V$, simply on cardinality grounds, since there are only continuum many possible second-order theories, and so if there are more than that many inaccessible cardinals then two of them will have the same theory and so we can cut off at the second one to realize that situation. (And these second-order assertions are relevant for category-theoretic assertions such as the existence of functors of certain kinds.) This is a kind of argument that Robin and I use a lot in our paper.

I won't speak about whether these differences will be of interest to category theorists, and perhaps they don't matter much in the practice of category theory. To my set-theorist way of thinking, however, the differences I mentioned revolve around the foundational uses of universes in category theory, and in particular, the subtle differences in how the universes themselves view the nature of the universe axiom and its failures, and so the properties strike me as fundamental, even if most applications are not affected.

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    $\begingroup$ This answer has a lot of very nice observations, but I think it misses the point of the question. You’re comparing the logical theories of different universes; the question is about the practical effects of different approaches to handling size issues. $\endgroup$ Commented Jun 24 at 22:08
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    $\begingroup$ As a push-back against the pushback: most category theorists would probably view the differences described in this way similar to how the independence of CH usually impacts their actual theorems ... pretty much never. It may be true that there are real differences in difference approaches, but if they are orthogonal to the statements of the theorems that category theorists are interested in, then one can usually ignore these. Much like the existence of different models of arithmetic in different set theories really doesn't impact theorems characterising natural number objects in a topos... $\endgroup$
    – David Roberts
    Commented Jun 25 at 13:19
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    $\begingroup$ Who are these category theorists who naively think that "all universes are the same"? Are they made of straw? But let me ask a mathematical question: if we allow non-well founded sets, can we arrange for a $\mathbb{Z}$-indexed sequence of universes that are "all the same"? $\endgroup$ Commented Jun 25 at 21:04
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    $\begingroup$ I agree with Andrej that probably category theorists don't think all universes are the same, but a lot of CT is indeed done "universe polymorphically", and we tend to only care about relative sizes. Concretely, apart from Adamek-Rosicky's use of Vopenka's principle mentioned above, I am only aware of two results in which "absolute size" plays a role: 1. Isbell's result that N is codense ("right adequate" in his terminology) in Set iff no measurable cardinals exist maths.ed.ac.uk/~tl/docs/Isbell_Adequate_subcategories.pdf $\endgroup$
    – Jonas Frey
    Commented Jun 26 at 1:40
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    $\begingroup$ @JonasFrey Your remark about "universe polymorphically" is exactly the kind of thing that I had intended to be talking about. Evidently I overstated things in a way that annoyed people, and for that I am sorry. $\endgroup$ Commented Jun 26 at 9:44
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One naïve difference is the handling of things like functor categories between large categories. Let $\mathcal{C}$ and $\mathcal{D}$ be large categories. If we want to consider something like $\mathcal{C}^\mathcal{D}$ the objects are functors $F:\mathcal{D}\to\mathcal{C}$, which are in turn large since they’re pairs of subsets of ${\bf Hom}_\mathcal{D}\times{\bf Hom}_\mathcal{C}$ and ${\bf Ob}_\mathcal{D}\times{\bf Ob}_\mathcal{C}$ whose first factor contains all of the relevant piece of $\mathcal{D}$. This makes $\mathcal{C}^\mathcal{D}$ ‘super large’, since it’s objects are all large objects.

  1. Using universes, $\mathcal{C}$ being “large” in a given universe just means that its hom/object sets are subsets of the universe instead of objects in it. We can always jump a universe upwards to make subsets objects in the new universe, and we can go transfinitely ‘large’ using something like Grothendieck’s universe axiom.
  1. Using classes naïvely, we’re boned as soon as we want something like $\mathcal{C}^\mathcal{D}$ since (in class theories like MK) one class being a member of another forces it to be a set. We can get around this by working in a higher order class theory, e.g. as proposed in this note I wrote a few years back.

To answer your final question, over ZFC the universe axiom up to some transfinite stage is equiconsistent with that many inaccessible cardinals existing, which is in turn equiconsistent with that many levels of higher order class existing (modulo the exact phrasing of the axioms). In this sense, all approaches are equivalent and it’s all just a matter of preference and taste.

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