When is it okay to intersect infinite families of proper classes?

Question

For experts who work in ZFC, it is common knowledge that one cannot in general define a countable intersection/union of proper classes. However, in my work as a ring theorist I intersect infinite collections of proper classes all the time.

Here is a simple example. Let a ring $R$ be called $n$-Dedekind-finite if $ab=1\implies ba=1$ for $a,b\in \mathbb{M}_n(R)$ (the $n\times n$ matrix ring over $R$). A ring $R$ is said to be stably finite if it is $n$-Dedekind-finite for every integer $n\geq 1$. The class of stably finite rings is the intersection (over positive integers $n$) of the classes of $n$-Dedekind-finite rings.

So my question is a straightforward one. What principle makes it okay for me to intersect classes in this manner?

Phrased a little differently my question is the following. Suppose we have an $\mathbb{N}$-indexed collection of sentences $S_n$ in the first-order language of rings. Why is it valid (in my day-to-day work) to form the class of rings satisfying $\land_{n\in \mathbb{N}}S_n$, even though the corresponding class doesn't necessarily exist if each $S_n$ is a sentence in the language of ZFC?

[Part of my motivation for this question is the idea that much of "normal mathematics" can be done in ZFC. That's how I've viewed my own work. I'm happy to think of my rings living in $V$, subject to all the constraints of ZFC. (In some cases I might go further, by invoking the existence of universes or the continuum hypothesis, but that is not the norm.) But I'm unsure how to justify my use of infinite Boolean operations on proper classes. Especially since, in some cases, my conditions on the rings are conditions on sets, from the language of ZFC!]

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Another way to think about all of this:

Looking at the (currently three) answers to this question, it appears that people took my question to be asking about definability rather than existence. Under that interpretation of my question the answers are spot on, and I appreciate what they are saying.

Moreover, those three answers helped me realize that my question was more along the lines of the MathOverflow question "The set of Godel numbers of true sentences" as interpreted by Andreas Blass, from a Platonistic viewpoint, about existence rather than definability. So, I suppose that the answer I was looking for was something like:

Pace, you were wrong to say that the intersection "doesn't necessarily exist". It exists as a subcollection $S$ of the Platonistic universe $V$. You just cannot prove (or even properly state) in ZFC the fact that the class $S$ equals the intersection you are looking at. The "principle that makes it okay for [you] to intersect classes in this manner" is your personal Platonistic view of the universe of true sets, not any sort of logical principle that can be stated in a first-order way.

Any additional thoughts people have on this are welcome.

Answer 1

You will have to come up with trickier examples.

There is a formula $\phi(R, n)$ of ZFC with parameters $R$ and $n$ whose meaning is "$R$ is an $n$-Dedekind finite ring". One might think that $n$ is a "meta-level" number, i.e., a number which is "outside" of set theory. That is not how ordinary mathematics works. We take set theory as the ambient universe in which everything happens. The natural way of reading the definition you gave is that it all happens inside set theory, and that $n$ denotes an element of a certain set $\mathbb{N}$. If one were to play tricks with logic, one could of course consider a situation in which $n$ is a meta-level number outside of set theory, but that is not how mathematics is done. Only logicians do such things.

The class of all stably finite rings is simply $\{R \mid \forall n \geq 1 \,.\, \phi(R, n)\}$. In other words, the imagined intersection of a countable collection of classes is not needed at all, because $$\bigcap_{n \geq 1} \{R \mid \phi(R, n)\} = \{R \mid \forall n \geq 1 \,.\, \phi(R, n)\}.$$ The point is that the property of being $n$-Dedekind finite is uniform in $n$ and so a single formula $\phi(n, R)$ works. The situation would be different if you proposed an infinite sequence of classes $$C_1, C_2, C_3, \ldots$$ indexed by meta-level natural numbers, where each $C_i$ is defined by some formula $\psi_i$, so that $C_i = \{x \mid \psi_i(x)\}$. Furthermore, it would have to be the case that the formulas $\psi_i$ are non-uniform, i.e., there is no single formula $\Psi$ such that $\psi_i(x) \Leftrightarrow \Psi(\overline{i},x)$. Here $i$ is a meta-level numbers and $\overline{i}$ is the element of the set $\mathbb{N}$ which corresponds to it, i.e., the $i$-fold application of successor to the element $0$. This sort of thing never happens in "ordinary" mathematics, but it happens in logic precisely because in logic we do study meta-level properties of set theory.

Lastly, let me mention that there are conservative extensions of ZFC in which classes are part of the theory proper, for example the Von Neumann-Bernays-Gödel set theory. Such theories make handling of classes quite a bit easier, so perhaps we should switch over to them, lest we get entangled in technicalities concerning meta-level and object-level numbers.

Answer 2

If in defining the classes, you’ve worked within ZFC in the usual way, then you can always take their intersection/union; that’s just universal/existential quantification over the parameter indexing the family. Unions/intersections are only problematic if, at some step in your reasoning, you’ve already done something which steps outside ZFC, and reasons about it from the outside, and hence have ended up with a family of classes indexed by some “meta-level” number or similar.

So the core of this question isn’t really about intersections/unions; it’s about when, in normal mathematical reasoning, does one step outside ZFC and work on the meta-level? And the answer to that is essentially never, unless you’re explicitly talking about logic, satisfaction of formulas, and the like.

It’s hard to accidentally step outside the foundational system you’re working in, because the foundational system is designed explicitly to include all the standard techniques of mathematical reasoning. And if you’re stepping outside it deliberately, e.g. by talking about formulas and their satisfaction like in user57888’s answer, then if you’re being as rigorous about that as you are about the rest of your mathematics, you should understand the logical issues involved well enough to understand which constructions are happening internally and which are happening at the meta-level.

The only way I’ve ever seen people accidentally step outside ZFC — that is, do something which actually requires meta-level reasoning, without realising it — is in examples like the one in user57888’s answer. It’s easy to mistakenly think that you can, within ZFC, talk about whether formulae of the language of set theory are satisfied by actual objects (that is, objects of the ZFC you’re working in). This is a lapse of rigour — if you try to fill in the details of the definition of that satisfaction relation, you’ll find you need to step outside outside your normal foundation at some point, e.g. by wanting to quantify over “all classes”. It’s an easy error to make, but once you know about it, you can recognise any talk of “satisfaction of formulas” or “definability” as a red flag. (To see why this is impossible, look at Berry’s paradox and Tarski’s undefinability of truth theorem.)

If however you’re talking about some more restricted language, e.g. the first-order language of rings, then yes — as you have presumably seen, one can (within ZFC) define satisfaction of such formulae in an arbitrary ring. So that’s why you can take rings satisfying $S_n$ for all $n$, where $S_n$ is a sequence of sentences in the language of rings, as you suggested.

In summary: If you are working rigorously, like a mathematician in principle should, then you can never step outside ZFC accidentally. Even if you’re a little less than perfectly rigorous, it’s hard to step outside it by mistake once you know what mistakes to look out for.

Answer 3

Here is an example of a family of classes of rings which you couldn't intersect: The class $A_n$ consists of all rings of power greater than n and the rings of cardinality $k≤n$ such that $ϕ_k$, where $ϕ_k$ is your favorite enumeration of sentences of ZFC. So for example, if $ϕ_0$ is "there exists an infinite set" and $ϕ_3$ is "$\mathbb{N}$ has at least 17 elements", $\phi_2$ is "there are uncountable sets of cardinality less than continuum" and CH happens to hold and $ϕ_1$ is "ZFC is inconsistent" and ZFC happens to be consistent, then $A_3$ is the class consisting of rings of cardinality 0,3 or greater. This is something you cannot write internally.

EDIT: To stress my point: if you could intersect the family of classes described above, this would simply result in an inconsistency.

Answer 4

Your edit doesn't make sense to me. Just treat each family of classes as a parametrized formula over ZFC, where the last parameter is for the input and the other parameters are for parametrizing the family. Then it's obvious that the intersection of an $ω$-indexed family of classes is a class, because $ω$ is definable over ZFC. To be explicit, for convenience use set-notation for classes. Then if the $k$-th class in the family is $\{ x : φ(k,x) \}$ then the intersection is simply $\{ x : ∀k∈ω\ ( φ(k,x) ) \}$.

So there is absolutely no need to accept a "platonic" set-theoretic universe for it to be acceptable to manipulate classes in this manner, since "stepping outside" here is merely stepping outside to a syntactic meta-system that merely knows how to manipulate finite symbol strings. All you need in order to accept that permitting such intersection of classes does not go beyond ZFC is to accept the basic properties of finite strings, because there is a concrete translation from a proof that use classes in this manner to a plain ZFC proof that uses no classes.

Are you able to come up with a family that you cannot actually express as a parametrized formula over ZFC? If so, then yes you will have a problem. If not, then you're almost surely nowhere near the limit of ZFC, since you probably never use replacement, and even if you do it is really unlikely that you use unbounded replacement.

Even if you escape the reach of ZFC, it is still some way before you can escape MK. The difference is that MK classes are permitted to quantify over classes (but still only have sets for members). With such quantification we can no longer treat a class as a formula over ZFC, because we essentially have a full second-order world.