Is the number of classes countable in axiomatic set theory?

Question

I'm recently learning mathematical logic and axiomatic set theory. In Takeuti and Zaring's Introduction to Axiomatic Set Theory Chapter 4, it says

For each wff (well formed formula) $\varphi(a, a_1, ... , a_n)$ we will introduce a class symbol $\{x| \varphi(x, a_1, ... , a_n)\}$...

This seems to imply that a class is just another expression of a well-formed formula. This answer also says a class is a collection of "all sets which have a property which we can describe in the given language".

My question is, since in the formal language we used to describe set theory there is only countable symbols (finite quantifiers, finite connectives, finite relationships, finite constants, and countable variables), it is obvious that the number of all well formed formulas can only be countable, and since each class is just another expression of wffs, does that mean there are only countable classes in axiomatic set theory?

This seems strange to me because people say "every set is a class" and the number of sets is uncountable. Is there an explaination of this apparent contradiction?

My guess is my understanding of classes is wrong. If so, could anyone give me a precise definition of a class?

update:

To be more precise, consider these two questions:

Is it true that for every class $A$, there exists a well formed formula $\varphi(x)$ such that $A = \{x| \varphi(x)\}$?
Is it true that every subset of natural number set $\mathbb N$ can be described with a finite string (since every wff is at least a finite string of symbols)?

If the answers to both questions above are Yes, then it looks strange to me. I want to know which statement above is wrong.

You're conflating theory and meta-theory. What constitutes 'countable' in the meta-theory is not what constitutes 'countable' in the theory. If this puzzles you, you'll be shocked to learn that if ZFC is consistent, it has countable models. — Chad K, Apr 19 '23 at 19:54
The parameters $a_1,\ldots, a_n$ range over all sets. So “every set is a class” in the sense that ${x:x\in a}$ is a class symbol that for each set $a$ gives a class coextensive with it. — spaceisdarkgreen, Apr 19 '23 at 19:59
@JohnDoe Thanks for the comment John. I've learned from my mathematical logic textbook that every consistent formula set which contains all axioms of logic has a countable model. I'm not too shocked with this conclusion because having a countable model does not mean only have a countable model, and the set theory that we are using everyday is obviously not a countable model of ZFC (that is because, there are other axioms added which are not descibed in first-order language, for example the Dedekind axiom, which leads to an uncountable model of real numbers). — Yongyi Yang, Apr 19 '23 at 20:13
@JohnDoe However, I still don't understand what does it mean by "conflating theory and meta-theory". Could you explain this in more details? — Yongyi Yang, Apr 19 '23 at 20:14
@YongyiYang: I think spaceisdarkgreen gave the answer you looking for - which is that parameters guarantee you have at least as many classes as the size of you model. However, on the subject you're asking - how do you know the set theory we're using everyday is not a countable model? Every model of ZFC (if exist) believes it has countable and uncountable sets in it. It has $\Bbb{N}$ and $P(\Bbb{N})$ of different cardinality. It can't tell how big its universe is in the meta-theory. The different cardinality just means there is no bijection in the model between $\Bbb{N}$ and $P(\Bbb{N})$ — Chad K, Apr 19 '23 at 20:28
@spaceisdarkgreen Thanks for the comment but I don't quite understand what does it mean by "$a$ ranges over all sets"? Since $x \in a$ is simply just a formula where $x$ is a variable and $a$ another variable? Say consider a formula $Q = x \in a$. Does ${x | Q(x)}$ define a set? If so? which set it does it define? (I guess my understanding of set is not quite right?) — Yongyi Yang, Apr 19 '23 at 20:36
@JohnDoe I guess I'm kind of confused with the definition of sets. I kind of don't want to think of sets as a model of ZFC, since we used the language of set theory to define the concept of model (like, we say a model maps every constant in the language to a set $M$, but what does mapping even mean if we havn't developed set theory?) (and moreover, the proof of every consistent formula set which consists axioms of logic having a countable model even used axiom of choice....). I would more prefer to understand a set as a formula $P$ such we can derive $\exists z \forall x, x \in z \iff P(x)$.. — Yongyi Yang, Apr 19 '23 at 20:45
@JohnDoe ...but that again leads to the same contradiction as described in the question. So I guess we can not understand sets as in this way.... I'm kind of confused about all these things now.... — Yongyi Yang, Apr 19 '23 at 20:47
There's probably some other book about set theory that should be studied before studying Takeuti and Zaring, given that question. The book could be just as rigorous as Takeuti and Zaring, but should help make some fundamental concepts clear before trying to deal with Takeuti and Zaring. — Ren Eh Daycart, Apr 20 '23 at 02:56
This reminds me of a question about the least upper bounds. (Unfortunately, I cannot post it as a separate question until I first "improve" my existing questions that nobody is interested in.) Does any theorem rely upon the full generality of the least upper bound property, or is it good enough to be able to apply it to every subset of the reals of the form {x∈$\mathbb{R}$: formula(x)}? Is it standard practice to introduce uncountably many claims where countably many would be good enough, just to be able to avoid using ordinary language and avoid formulating a least upper bound schema? — Ren Eh Daycart, Apr 20 '23 at 03:12
You're wrong about the model existence theorem. Any consistent countable FOL theory has a countable model, and this fact can be proven within ACA0 (which is conservative over PA), so any actual practical FOL theory satisfies the model existence theorem with absolutely no reliance on ZFC, not to say AC. — user21820, Apr 21 '23 at 07:20

spaceisdarkgreen · Accepted Answer · 2023-04-20T03:00:09.740

4

Informally, a class is "a collection of sets". There are three ways this is typically treated in axiomatic set theory:

Every mathematical object is a set and so variables in our formal theory range over sets. Classes are just a convenient mode of expression for what are really first-order statements about sets.
Similar the previous, but we're a little more deliberate about it. We add class comprehension terms $\{x:\varphi(x,\vec a)\}$ to the formal language and then define class membership by $y\in \{x:\varphi(x,\vec a)\}\iff \varphi(y,\vec a).$ Our variables still always range over sets.
Classes are bona-fide mathematical objects, and our formal theory quantifies over classes and sets, not just sets.

Your question concerns the first and second approach, which are only different in the formal details... it's similar to how you can work in the language where $\in$ is the only non-logical symbol, or can work in a definitional extension, including constant/relation/function symbols for empty power set, union, etc. We generally add these things mentally, anyway, and it's just a matter of whether for formal purposes you conceive these things to be abbreviations or as additions to the formal language.

Let's take approach 2 explicitly. You are correct, of course, that there are countably many formulas, so only countably many class comprehension terms, but that's not quite the same thing as saying there are only countably many classes. The class comprehension terms have parameters, so don't describe one notion of class membership, but a whole family of notions of class membership indexed by the parameter variables.

For instance, we can write down the term $\{x: x\in a_1\land x\in a_2\}$ (which is another way of writing $a_1\cap a_2$), or for a proper example, we can write $\{x: a_1\in x\}.$ When we actually put together a formal argument involving these class symbols, we will wind up instantiating the variables $a_1$ and $a_2,$ and we'll be talking about a "specific" class at that point (quotation marks since $a_1$ or $a_2$ may well be instantiated as arbitrary sets).

In this set-up, we can make sense of the idea that every set is a class, since we can write down the class term $C:=\{x:x\in a\}.$ Then we have by definition $x\in C\iff x\in a.$ So any time we reason about a set, we can assign it the variable $a$ and also talk about the corresponding class $C,$ which has the exact same members.

Now, we'll only ever instantiate finitely many variables in a given argument, or in our lifetime, and there are only countably many possible formal proofs anyway, but that's not generally taken as cause to say there's only countably many sets. So nor should the fact that there are only countably many class comprehension terms be cause to say there are only countably many classes.

I didn't want to bring model theory into this since you seemed resistant to it in the comments, but the idea of "what a class is" becomes much snappier in this arena. A class (in the sense of approaches 1 and 2) in a model of set theory is nothing more or less than a definable-with-parameters subset of the model.

As for your summary questions, hopefully the first one is superseded by the explanation above. For the second one, no there's no reason that we should expect every subset of the natural numbers to be definable. But one thing to appreciate is that in order to even make precise what we mean by this, we need to talk about models... (global) definability is not a property that we can define in set theory.

edited Apr 20 '23 at 03:00

answered Apr 20 '23 at 02:37

spaceisdarkgreen

58,508

Thank you for the answer! Can we say like this (under your second approach): if we have a model of ZFC, then a class can be precisely defined as a formula $\varphi(x,a_1, a_2 \cdots a_n)$, together with the mapping of $a_1, a_2 \cdots a_n$? – Yongyi Yang Apr 20 '23 at 05:10
1

@YongyiYang Yes. – spaceisdarkgreen Apr 20 '23 at 05:12
Thanks! That solves my problem. – Yongyi Yang Apr 20 '23 at 05:15
@YongyiYang: To add on one more point, if you only want to work within ZFC, then you can indeed treat classes as nothing more than syntactic sugar as per (1) and (2). This is really no different from having on-the-fly definitorial expansion for new predicate-symbols. Writing "E∈C" where C is a class is just convenient syntactic sugar for writing "C(E)" where C is a newly defined predicate-symbol. On the other hand, if you want every set in the intended model of your set theory to be a class, then you must go to (3)... – user21820 Apr 20 '23 at 09:27
But I should say that ZFC has a very dubious ontology. An intended model of ZFC is a structure that you ought to be able to imagine satisfying the axioms of ZFC, but there really is no way you can imagine such a structure without making strong assumptions that essentially are equivalent to assuming existence of such a structure. And so far, nobody has given any non-circular justification for these strong impredicative assumptions. Going to (3) would be extremely much worse since it is way more impredicative. – user21820 Apr 20 '23 at 09:40
@user21820 I have the same feeling. To me it is kind of strange to talk about the model of ZFC, while the definition of a model actually uses the language of set theory (as far as I learned). I mean, I can accept that there should be a meta-theory to develop set theory, and FOL is one of them. I can accept FOL since it is simple enough (as long as we don't talk about the properties of FOL). But when talking about the model of ZFC, as you said, we must imagine a relatively complex structure, and this makes me unsettled. Do you know any ways to solve this issue? – Yongyi Yang Apr 21 '23 at 07:15
1

@YongyiYang: You do not need ZFC in order to meaningfully reason about models of ZFC. As I mentioned in my other comment (which I just fixed to add in the word "consistent" that I accidentally omitted), you only need ACA0. The catch is that the intended model of ACA0 is only ℕ and its arithmetically definable subsets! So philosophically ACA0 can only be justified to be able to talk about countable models. After all, if we cannot justify the existence of any uncountable structure in the real world, on what basis can we even assume that any uncountable structure exists? – user21820 Apr 21 '23 at 07:22
Of course, this idea may not sit well with the common interpretation of Cantor's theorem, but actually you need to realize that there is also a syntactic version of Cantor's theorem. There are clearly countably many 1-parameter sentences over PA, but for any 2-parameter sentence F over PA we can construct a 1-parameter sentence Q and prove that ¬∃x∈ℕ ∀y∈ℕ ( Q(y) ⇔ F(x)(y) ). Thus no 2-parameter sentence over PA can 'reflect' the countability of all 1-parameter sentences! This doesn't imply size disparity, but merely complexity disparity. Same with set-theoretic Cantor... – user21820 Apr 21 '23 at 07:32
Using ACA0 to talk about models of ZFC of course doesn't tell you that any model of ZFC exists, but at least you can talk about them without any big assumptions. If you want a model of ZFC to exist then you are out of luck. ACA0+Con(ZFC) proves that, but it's circular. – user21820 Apr 21 '23 at 07:36
@user21820 Thanks! I'm not familiar with the theory of ACA0, but do you mean that ACA0 can be used to reason about the model of ZFC (instead of using ZFC itself), but can only be applied to a countable model? – Yongyi Yang Apr 22 '23 at 01:54
@YongyiYang: Yes. ACA0 is briefly sketched in the PDF by Henry Towsner linked from this post (which you may be interested in as well). In short, ACA0 is a 2-sorted FOL theory where one sort is intended for ℕ and another sort is intended for arithmetical subsets of ℕ, and you have PA for ℕ and arithmetical comprehension (i.e. you can construct { x : x∈ℕ ∧ Q(x) } where Q is any property that quantifies only over ℕ). With this, you can easily reason about functions from ℕ→ℕ (encoded as subsets of ℕ×ℕ) and hence about countable FOL theories. – user21820 Apr 22 '23 at 05:39
@user21820 Thanks for providing the reference! I will look at it. – Yongyi Yang Apr 26 '23 at 00:31

score 3 · Answer 2 · answered Apr 20 '23 at 04:34

To answer your questions explicitly, your first question depends on what you mean by a "class." There are certainly collections of sets in any model of $\mathsf {ZFC}$ that are not definable by a well-formed formula, simply because there are (in the metatheory) uncountably many collections of sets and there are only countably many formulas.

The second question is a little more complicated. It depends on whether you're talking about subsets of $\Bbb N$ in the theory or in the metatheory, and precisely what you mean by a "subset." Since there are countable models of $\mathsf{ZFC}$, it's conceivable that every subset of $\Bbb N$ that the theory recognizes as a subset is definable in the metatheory. That's because the model is countable (in the metatheory), so it can only recognize countably many (in the metatheory) subsets of $\Bbb N$ as members of the model and it's conceivable that for each of the "recognizable" (in the theory) subsets there's a formula that defines it.

But if you're talking about subsets of $\Bbb N$ in the metatheory, then since there are uncountably many of those, there aren't enough formulas to define them all, but it's possible that none of the undefinable subsets (in the metatheory) are recognized (in the theory) as elements of the model.

Is the number of classes countable in axiomatic set theory?

2 Answers2