Couching that in terms of proper classes is off. — GrandMinnow

Your use of the terminology is impressive, and I'm not able to determine whether you have advanced knowledge or not; since if you do, my own level of knowledge of these matters would not be sufficient to let me make that determination.

Nevertheless you are wrong on this particular point.

I didn't make up the claim that absent the axiom of infinity, $\omega$ (or $\mathbb N$) is a proper class. I read it somewhere a while back in a reference I can no longer find.

I did find this discussion supporting my claim. It's in a Stackexchange thread called The purpose of the Axiom of Infinity.

I will quote from the checked reply by Andrés E. Caicedo. Professor Caicedo is a well-known professional set theorist. I linked his home page so that you can determine for yourself his stature within the set theory community. (Click on the link Notes and Papers). He says:

"In particular, the axiom of infinity goes well beyond the Peano axioms (and not simply in terms of consistency strength or expressive power). The Peano axioms are provable in ZF without the axiom of infinity. In this theory, you cannot prove that $\omega$ is a set, but you can prove that as a (perhaps proper) class, it satisfies both first and second order PA."

Perhaps you can take a look at his full reply, which is of interest beyond our conversation and worth reading in general, and put it into context for me. If it means something other than what it plainly seems to say, then I'll concede the point. Else you'll need to.

ps -- In the same Stackexchange thread see Asaf Karagila's comment:

"In a model of ZFC with the negation of the axiom of infinity instead, the natural numbers are just the ordinals of the universe."

In other words, in ZFC minus infinity, the natural numbers are the proper class of ordinals. You may object to my terminology (and Professor Caicedo's), but if the extension of a predicate is not a set, what is it? It's a proper class, right?

I'm not advanced. But I do have a methodical understanding of some basics.

(1) There is a difference between ZF-Inf and (ZF-Inf)+~Inf. I'll call the later 'HF' (the theory of hereditarily finite sets).

The language of HF is the language of ZF (i.e. the language of set theory).

PA and HF can be interpreted in each other.

The usual universe for HF that we have in mind is the set of hereditarily finite sets. And of course N is also a universe for HF.

(2) Most textbooks take 'is a set' as informally primitive, but we can be precise in the language of set theory:

x is an urelement <-> (~ x=0 & ~Ey yex)

x is a class <-> ~ x is an urelement

x is a set <-> (x is a class & Ey xey)

x is a proper class <-> (x is a class & ~ x is a set)

In set theory, we can prove:

Ax x is a set (though, as mentioned, most textbooks don't bother with something so basic).

(3) The language of class theory (such as Bernays style class theory, which I'll call 'BC') has a primitive predicate 'is a set' (or a many-sorted language is used, which is essentially the same as using a primitive 1-place predicate), so in BC 'is a set' is not defined but instead certain axioms are relativized to sets.

In BC we prove:

Ex x is a proper class

(4) I explained why "N is a proper class in PA" [or whatever paraphrase] is, on its face, not coherent. But I allowed that one is welcome to adduce some particular mathematical statement instead. And I explained why it would not be a correct statement in set theory (and I would add, not even in BC). So maybe we turn to HF.

Since HF is in the language of set theory, in HF we can define any predicate of set theory, and we can define any operation of set theory for which we can prove existence and uniqueness in HF.

HF proves ~ExAy(y is a natural number -> y e x). So there is no definition of 'N' (in the sense of the set of natural numbers) in HF.

So, while HF can have predicates 'is a natural number', 'is a set', and 'is a proper class', still HF can't have the definition N = the set of natural numbers.

As far as I can tell, the best we could do in NF is this theorem:

If Ax(Ay(y is a natural number -> yex) -> x is a proper class). But that holds vacuously, since there we have ~ExAy(y is a natural number -> yex).

So, as far as I can tell, we are still thwarted from making sense of "N is a proper class in PA" or even "N is a proper class in HF".

And in set theory (and even in BC, if I'm not mistaken) the universe for a model is a set, not a proper class.

(5) Caicedo says, "in ZF without the axiom of infinity [...] you cannot prove that w is a set, but you can prove that as a (perhaps proper) class, it satisfies both first and second order PA."

I don't know why he says 'perhaps' there. And without more explanation, I don't understand what he's saying.

I do understand that, in ZF-Inf, there is not a proof that there is a set of which all natural numbers are a member (that's another way of affirming the independence of the axiom of infinity).

But when he says "you can prove", does he mean prove in ZF-Inf? Proof of satisfaction with models takes place in set theory, not in ZF-Inf nor in HF. And in set theory, universes of satisfaction are sets, not proper classes.

What is understandable to say is:

ZF-Inf does not prove there is an x such that all natural numbers are a member of x.

HF proves there is no x such that all natural numbers are a member of x.

PA and HF are mutually interpretable.

The set of natural numbers N is a universe for a model of ZF-Inf or of HF.

But saying "in Pa (or in HF), N is a proper class" makes no sense.

(6)

absent the axiom of infinity, w (or N) is a proper class. — fishfry

No, absent Inf, it is not a theorem that N is a proper class. Indeed, absent Inf, there is not even possible a definition N = the set of natural numbers. Rather, absent Inf, there is not a proof that there exists an x such that all natural numbers are in x, and there is not a proof that there is no x such that all the natural numbers are in x. In other words, "there is an x such that all natural numbers are in x" is independent of ZF-Inf. However, (ZF-Inf)+~Inf does prove "there is no x such at all natural numbers are in x", but still, it does not say anything about such a thing (which does not exist anyway in NF) being a proper class or not.

(7)

Yes, you can't define the ordinals in PA because you can't get to the first transfinite ordinal ω by successors. — fishfry

In HF, we can define the predicate 'is an ordinal' and for any finite ordinal, we can define a constant for it. But, as you mention, we can't define a set that has all the finite ordinals as members.

But even in set theory, there are specific ordinals that don't have a definition (there are more than countably many ordinals, but only countably many definitions we could form).

Perhaps one might get the notion of the universe for PA as "a proper class as far as PA is concerned" [or whatever paraphrase] from the notion regarding set theory that for every formula with n number of free variables there is the class of n-tuples for which that formula holds (where n =1, the class of individuals for which the formula holds).

That is okay as an understood informality when speaking of set theory. But it does not transfer to PA or HF in the same way. The reason is that PA and HF already have a meta-theory in which everything is a set. So saying "the proper class that is the extension of the predicate" (i.e. the predicate carved out by the formula) is a way of saying that there is no set of all things that have the predicate but for convenience, and recognizing this as merely informal, we can speak of the formula itself as if it specifies a "class". However in PA or HF, any formula specifies not a proper class but a set. Even the universal predicate specifies the set that is the universe of whatever given model for the theory.

Example:

With set theory, "the proper class of ordinals". This is an informal way of saying "those x such that 'x is an ordinal' holds". It is a breezy way of speaking, when more formally we would refer only to the formula 'x is an ordinal'. In other words, saying "If x is in the proper class of ordinals" is an informal locution for "If 'x is an ordinal' holds". Or for example, "relativize F(x) to the proper class of finite sets" is a locution for "Ax(x is finite -> F(x)).

With PA, "the proper class of natural numbers". No, doesn't work, because the usual meta-theory for PA is set theory where there is the SET of all natural numbers, which encompasses those x such x is a natural number.

What I want to know is how N is defined.

Is there special use of the word 'is'? Natural numbers are N, is incomplete.

A. 1 through 9, are numbers, why?

B. Why does the number system progress, beginning from the left, proceeding to the right?

C. Is human number just a tool? — Qwex

(A)Numbers are just words and abstract concepts used to label and understand the world around us. Words are created whenever someone decides to label something around them (which hasn't been labeled as such before) and enough people agree to the convention that it sticks either because it is useful or because people just like it.
(B)By convention we think of the number system proceeding from a point of origin going to the right, but that is only when we are counting. When dealing with things such as geometry something to the right of somethings isn't necessarily more positive than it.
(C)I believe one can say that words, languages, numbers/abstract concepts, and mathematics are more or less tools that we use to define the world around us.

Further Edits:
A shadow-argument:

I understand you can count your fingers, 1 - 4, but what says a finger is a 1 and not crossed fingers? The 'whole' of the finger?

In which case it's not a single, there's an organism involved(such as under the skin of the finger), and thus, a finger is not a 1.

I understand 1 is a concept but mathmatically, 1 is a point.

Perhaps, to point at your finger you'll use the number 1 but to define it numerically it's a different number. — Qwex

Because you can only call your finger or crossed fingers '1' and not make it '1' since one is only an abstract concept and your fingers are instances of various abstract concepts. Labels and abstract concepts are created whenever you decide to call something by some name,describe some aspect or attribute of it but these labels and mental projections are not the things in and of themselves. Only the actual physical instances of things are things in and of themselves.

And on that subject numbers are never a physical instance of anything because they can only describe an attribute of a physical thing and all physical things have more than a single attribute to them.

Then I think it would be efficent to use base 8, or whatever base math is better. Definitely not base 4. Base 4 is tiny on the ratio. Like pointing a subject using selectable squares only. 08 in base 8 IS base 4. Powerful stuff.

I think the best base is containing math, the rules of containment.(or something along these lines).

I'm not advanced. But I do have a methodical understanding of some basics. — GrandMinnow

Thanks. You wrote a really interesting post and I have a lot of questions and comments. Before going into them I have to say that my stance remains unaltered: which is:

Regardless of whether it's technically accurate, it can sometimes be a useful visualization of proper classes.

So it doesn't matter if you're right on the technical part. If I myself find it a helpful visualization, that is my right. We are all entitled to our visualizations.

So at best you could possibly argue that even though I have the right to my own private visualization, I should not mention it aloud in polite company. I'll take that under advisement.

As an analogy, suppose an engineer takes calculus class and spend the rest of his professional life believing that dx is an infinitesimal. We could of course correct him mathematically, but it's a harmless belief for an engineer or physicist and a highly effective conceptual aid. Even many mathematicians who technically know better privately think of dx as an infinitesimal when analyzing a calculus problem. I hope you take my analogy as on point.


Also I did find at least one professional set theorist who was willing to make an admittedly offhand remark in agreement with my thesis. And I agree that it's not even clear what he meant. But he referenced the idea. If my intuition matches professor Caicedo's, I consider my point to have been made to my own satisfaction.

All that said, you wrote a lot of really interesting things so I'll try to comment.

(1) There is a difference between ZF-Inf and (ZF-Inf)+~Inf. I'll call the later 'HF' (the theory of hereditarily finite sets). — GrandMinnow

Oh yes right away you have actually identified a couple of points of confusion or ignorance in my mind.

In my reading the last few days I do keep seeing the notation ZF-infinity +(not infinity) and this is puzzling me greatly.

For example in discussions of the axiom of choice we see ZF or ZFC. I've never seen anyone say, "ZF + not-AC". The not-AC is emplied when you say ZF. And in a given model of set theory, there either is or isn't a choice function on every collection of nontempty sets. It has to be one way or the other, we don't have to say it twice.

But when it comes to infinity, lately I keep reading ZF-infinity + not-infinity and I have no idea why they're doing that! I'd think that likewise, in any model of ZF-infinity, there either is or isn't an infinite inductive set containing the empty set and its successors.

So why do we need to say "+ not-infinity?"

The language of HF is the language of ZF (i.e. the language of set theory). — GrandMinnow

Right, the hereditarily finite sets. Why are they considered interesting? I just thought they were the usual finite von Neumann ordinals, ie the natural numbers. Is there something else special about them?

PA and HF can be interpreted in each other. — GrandMinnow

Ok so interpretation is a technical term that I think I don't know. I know what it means to interpret an axiomatic theory, ie assigning meaning to the symbols or at least assigning elements of some model. But I'm not sure about thie bi-interpretability business.

The usual universe for HF that we have in mind is the set of hereditarily finite sets. And of course N is also a universe for HF. — GrandMinnow

Ah. How are HF and N different?

(2) Most textbooks take 'is a set' as informally primitive, but we can be precise in the language of set theory:

x is an urelement <-> (~ x=0 & ~Ey yex)

x is a class <-> ~ x is an urelement

x is a set <-> (x is a class & Ey xey)

x is a proper class <-> (x is a class & ~ x is a set)

In set theory, we can prove:

Ax x is a set (though, as mentioned, most textbooks don't bother with something so basic). — GrandMinnow

Ok, very interesting. Haven't seen those defined formally before. I learned the "informally primitive" way.

(3) The language of class theory (such as Bernays style class theory, which I'll call 'BC') has a primitive predicate 'is a set' (or a many-sorted language is used, which is essentially the same as using a primitive 1-place predicate), so in BC 'is a set' is not defined but instead certain axioms are relativized to sets.

In BC we prove:

Ex x is a proper class — GrandMinnow

Ok I never actually looked at BC but I've heard of it. Also I believe Morse-Kelley has classes.

(4) I explained why "N is a proper class in PA" [or whatever paraphrase] is, on its face, not coherent. — GrandMinnow

I have seen no such explanation; or if I have, I haven't understood it. A proper class is the extension of a predicate that's not a set. That's how I understand the term. I haven't thought about this interms of BC or Morse-Kelley so maybe you're right, I don't know.

But I allowed that one is welcome to adduce some particular mathematical statement instead. And I explained why it would not be a correct statement in set theory (and I would add, not even in BC). So maybe we turn to HF. — GrandMinnow

I admit I haven't followed the argument but that's probably my fault.

Since HF is in the language of set theory, in HF we can define any predicate of set theory, and we can define any operation of set theory for which we can prove existence and uniqueness in HF. — GrandMinnow

Ok.

HF proves ~ExAy(y is a natural number -> y e x). So there is no definition of 'N' (in the sense of the set of natural numbers) in HF. — GrandMinnow

Oh I think I see where you're going. You're saying I can't form the predicate that I think characterizes N. Am I on the right track at all?

So, while HF can have predicates 'is a natural number', 'is a set', and 'is a proper class', still HF can't have the definition N = the set of natural numbers. — GrandMinnow

In other words, if I'm understanding, then I can't form the collection of all numbers because I haven't got the language to do that. So my argument fails.

NOTE: At least the TECHNICAL argument fails; but my intuition still likes it!!

As far as I can tell, the best we could do in NF is this theorem:

If Ax(Ay(y is a natural number -> yex) -> x is a proper class). But that holds vacuously, since there we have ~ExAy(y is a natural number -> yex).

So, as far as I can tell, we are still thwarted from making sense of "N is a proper class in PA" or even "N is a proper class in HF". — GrandMinnow

I don't understand the details but perhaps this is a good argument. In which case I'm technically wrong that N is a proper class but it's still a useful intuition; and if not for everyone, at least it is for me. You can't enjoin me from thinking my thoughts, misguided though they may be.

And in set theory (and even in BC, if I'm not mistaken) the universe for a model is a set, not a proper class. — GrandMinnow

Yes, models are supposed to be sets. That's my understanding.

(5) Caicedo says, "in ZF without the axiom of infinity [...] you cannot prove that w is a set, but you can prove that as a (perhaps proper) class, it satisfies both first and second order PA."

I don't know why he says 'perhaps' there. And without more explanation, I don't understand what he's saying. — GrandMinnow

Yes. Perhaps he is speaking INFORMALLY because he has a similar intuition as mine. Perhaps he even has a context in mind where the statement could be made rigorous. Perhaps not. But we DO know that at least one professional set theoriest thinks it's something you can say in this context. That means a lot to me.

I do understand that, in ZF-Inf, there is not a proof that there is a set of which all natural numbers are a member (that's another way of affirming the independence of the axiom of infinity). — GrandMinnow

Right. Because if their were, it would witness the axiom of infinity! But I take this differently, not in terms of proofs. I imagine that in any model of ZF-inf there is not a set of all natural numbers. Because if there were it would prove the axiom of infinity, and therebey contradict ZF-inf. Am I wrong about this?

But when he says "you can prove", does he mean prove in ZF-Inf? Proof of satisfaction with models takes place in set theory, not in ZF-Inf nor in HF. And in set theory, universes of satisfaction are sets, not proper classes. — GrandMinnow

I'm afraid I have no idea what his point may have been; only that he used the magic words proper class.

What is understandable to say is:

ZF-Inf does not prove there is an x such that all natural numbers are a member of x. — GrandMinnow

Now I do not understand this. ZF-inf says: There is no x such that all natural numbers are a member of x. You said we can't prove there isn't; I'm saing that there isn't. This seems subtly different.

HF proves there is no x such that all natural numbers are a member of x. — GrandMinnow

Ok the distinction between ZF-inf and HF is totally lost on me. If I understood it perhaps I'd be enlightened.

PA and HF are mutually interpretable. — GrandMinnow

Ok. But not quite the same in some way I can't grasp.

The set of natural numbers N is a universe for a model of ZF-Inf or of HF. — GrandMinnow

Yes I believe that.

But saying "in Pa (or in HF), N is a proper class" makes no sense. — GrandMinnow

a) If you say so; and

b) But can you forbid me from thinking it? What if I promise not to tell anyone else to think it? But taken informally, it's a good visualization of what is meant by proper classes.

(6)
absent the axiom of infinity, w (or N) is a proper class.
— fishfry

No, absent Inf, it is not a theorem that N is a proper class. — GrandMinnow

Of course it's not a theorem, in ZF there is no such definition or thing as a proper class. I thought I made it clear that I understand this point. So it should be obvious that I'm speaking vaguely and metaphorically and not literally.

But in fact we're at this same "theorem" impass. It's true that there's no theorem. But morally, N is a proper class!! This is the nub of our disagreement.

Indeed, absent Inf, there is not even possible a definition N = the set of natural numbers. — GrandMinnow

Yeah yeah. You're technically right and morally wrong. When the Peano axioms say, "O is a number," what exactly is "is a number" if not a legal predicate?

Rather, absent Inf, there is not a proof that there exists an x such that all natural numbers are in x, and there is not a proof that there is no x such that all the natural numbers are in x. In other words, "there is an x such that all natural numbers are in x" is independent of ZF-Inf. However, (ZF-Inf)+~Inf does prove "there is no x such at all natural numbers are in x", but still, it does not say anything about such a thing (which does not exist anyway in NF) being a proper class or not. — GrandMinnow

I'm losing my train of concentration here, don't think I got much out of this para. My fault.

(7)
Yes, you can't define the ordinals in PA because you can't get to the first transfinite ordinal ω by successors.
— fishfry

In HF, we can define the predicate 'is an ordinal' and for any finite ordinal, we can define a constant for it. But, as you mention, we can't define a set that has all the finite ordinals as members. — GrandMinnow

Ok we found something we agree on!

But even in set theory, there are specific ordinals that don't have a definition (there are more than countably many ordinals, but only countably many definitions we could form). — GrandMinnow

Ok. I concede all your points even though there are some I don't understand, and among those I understand, some I disagree with.

But what of it? It's obvious that since there are no proper classes in ZF, when I speak of proper classes in ZF I mean metaphorically. You can go into any set theory book and they'll tell you that the set of all sets is a proper class, even as they say it's not technically so because we're working in a set theory that doesn't have proper classes!

Ok my head is officially confused. There is much I don't understand about these matters. But I retain my private intuition that N is morally a proper class with respect to PA; and that if I were to share my intuition in print, I would be helping more people than I'd be hurting.

Ok for that!! More words than this subject is worth. Not a hill I need to die on. I'll consider myself suitably chastized for my belief that N is a proper class with respect to PA. As Galileo whispered as they forced him to recant his belief that the earth moves around the sun: And yet it moves.

ZF-inf says: There is no x such that all natural numbers are a member of x. — fishfry

That is not correct.

ZF-Inf is ZF but without the axiom of infinity. (The '-' here means 'without'; it doesn't mean 'the negation of'.)

(ZF-Inf)+~Inf is ZF but with the axiom of infinity replaced by the negation of the axiom of infinity.

the hereditarily finite sets. [...] I just thought they were the usual finite von Neumann ordinals, ie the natural numbers. — fishfry

No, the finite ordinals are a proper subset of the set hereditarily finite sets. For example, {0 2} is an hereditarily finite set but it's not an ordinal.

Is there something else special about them? — fishfry

They may be of interest for many reasons, but for starters, they are the usual universe for a model of "finite set theory" = (ZF-Inf)+~Inf = HF.

so interpretation is a technical term that I think I don't know. I know what it means to interpret an axiomatic theory, ie assigning meaning to the symbols or at least assigning elements of some model. — fishfry

This is a different sense of 'interpretation' (but closely related). Simplifying here: We interpret a theory T into a theory T' by defining the symbols of T in the language of T' so that every every theorem of T is a theorem of T' plus the added definitions. And we say the theories are equivalent when there is such an interpretation from T into T' and vice versa. (This deserves a sharper statement, but it's too many technical details for a post.)

How are HF and N different? — fishfry

HF is a theory. N is a set.

You're saying [in HF] I can't form the predicate that I think characterizes N. — fishfry

Defining a predicate symbol is not a problem. But there is no definition of a constant symbol (such as 'N') such that N = {y | y is a natural number}, since HF does not prove that there exists an object that has as members all the natural numbers.

In any language, in any theory, we can define whatever predicate symbols we want. It's only function symbols (including constant symbols, where a constant symbol is a 0-place function symbol) that require the supporting existence and uniqueness theorem

I can't form the collection of all numbers because I haven't got the language to do that. — fishfry

In HF, you have the language, but you don't have the existence theorem ExAy(y is a natural number -> y e x).

in any model of ZF-inf there is not a set of all natural numbers — fishfry

No, the sentence ExAy(y is a natural number -> y e x) is a theorem, but that doesn't preclude what the members of the universe for the model may be.

For every infinite cardinality, there is a model of ZF-Inf with a universe of that cardinality. And that universe can have as members any sets whatsoever. Same for (ZF-Inf)+~Inf. Same for PA.

For example, we can have a model of PA whose universe is {w, w+1, w+2} and each of those members of the universe is infinite.

But wait, (ZF-Inf)+~Inf has a theorem ~Ex Ix [where 'I' is a defined 1-place predicate symble we are read in English as "is infinite"], so how can the universe of a model have a member that is infinite? Well, because for such a model, the predicate symbol 'e' is interpreted not as the ordinary membership relation but rather as some other "bizarre" relation and so also my 'I' be interpreted differently from "is infinite". When we talk about models in general, we can't presume that any given model of a theory "captures" the way we ordinarily "read off" the theorems of the theory. If we want to narrow the discussion to only models that adhere to the way we "read off' the theorems, then we should confine to talking about standard models.

in ZF there is no such definition or thing as a proper class. — fishfry

In ZF, we may define:

x is a proper class <-> Ey y e x & ~Ez x e z

And we may prove:

~Ex x is a proper class.

When the Peano axioms say, "O is a number," — fishfry

First order PA doesn't have a primitive 'is a natural number'.

Peano's historical own formulation should not be conflated with first order PA.

the set of all sets is a proper class — fishfry

There is not a set of all sets, not even in class theory. There is the class of all sets, and it is a proper class. And I explained why referring to proper classes in discussion about set theory can be understood as an informal rendering for an actual formal notion in the background, but that is lacking here in saying N is a proper class in discussion about PA.

And N is a set, which is not needing exceptions in view of the fact that in PA there can be no definition N = {x | x is a natural number}. If one wishes to say "N is a proper class with respect to PA" but not formulate the exact mathematical meaning of "with respect to" or even to a clearly articulate an intuitive/heuristic notion that is still consistent with the ordinary mathematical result that N is a set, and hopefully has value as a metaphor rather than confusing the subject with impressionistic use of terms, then, of course, I cannot opine whether or not in one's own mind it somehow makes sense nevertheless. But I do say, and have explained, that it makes no sense to me.

PA and HF can be interpreted in each other. — GrandMinnow

I just wrote a post about the bi-interpretability of PA and HF. The worst problem I had in investigating this idea was to find examples of how to translate standard set builder notation into arithmetic-set notation.

For example, { 1, 4, 7 } in set builder notation is equivalent with φ(x):= 1 - sgn((x-1) (x-4) (x-7))² in arithmetic-set notation.

Now if you want to find even one publication where they give such example, good luck, l because I could not find even one!

Below is a link to an article about HF. (They do call it ZF-Inf, contrary to some others, including me, who call it (ZF-Inf)+~Inf):

https://projecteuclid.org/download/pdfview_1/euclid.ndjfl/1193667707

[If the link doesn't work, do Internet search for 'richard kaye hereditarily finite' and it will be the first result.]

Note that the article warns against overlooking certain technicalities, as doing so leads to a merely "folktale" understanding. Indeed, the subject of interpretation of theories into another, and especially the exact sense of "equivalence", especially with PA and HF, is a lot of technical details that are doomed to be mangled in the confines of short posts.

ZF-inf says: There is no x such that all natural numbers are a member of x.
— fishfry

That is not correct. — GrandMinnow

Ok. In general terms, I noticed you didn't engage with my point that "N is a proper class in PA" is a personal visualization that I find helpful; even though it is not literally true and, according to your thinking, is so flagrantly false that I shouldn't even think it.

If you didn't engage with this point I assume you accept it and are just explaining some of the technical points I brought up, which I appreciate.

ZF-Inf is ZF but without the axiom of infinity. (The '-' here means 'without'; it doesn't mean 'the negation of'.)

(ZF-Inf)+~Inf is ZF but with the axiom of infinity replaced by the negation of the axiom of infinity. — GrandMinnow

Right, perfectly well understood. But why don't they do the same thing with ZF-infinity +(not-infinity)? Nobody every does that. Rather, ZF-infinity means ZF plus the negation of the axiom of infinity by default. Why is that?

No, the finite ordinals are a proper subset of the set hereditarily finite sets. For example, {0 2} is an hereditarily finite set but it's not an ordinal. — GrandMinnow

Oh ok it's all the finite sets. I think I knew that then forgot. All good.

Is there something else special about them?
— fishfry

They may be of interest for many reasons, but for starters, they are the usual universe for a model of "finite set theory" = (ZF-Inf)+~Inf = HF. — GrandMinnow

Right.

so interpretation is a technical term that I think I don't know. I know what it means to interpret an axiomatic theory, ie assigning meaning to the symbols or at least assigning elements of some model.
— fishfry

This is a different sense of 'interpretation' (but closely related). Simplifying here: We interpret a theory T into a theory T' by defining the symbols of T in the language of T' so that every every theorem of T is a theorem of T' plus the added definitions. And we say the theories are equivalent when there is such an interpretation from T into T' and vice versa. (This deserves a sharper statement, but it's too many technical details for a post.) — GrandMinnow

I didn't follow all that but it's ok, I'll check out the def one of these days. Especially since bi-interpretability came up the other day. PA is bi-interpretable with ZF-infinity + (not-infinity) but apparently I'm not getting the subtleties of that statement.

How are HF and N different?
— fishfry

HF is a theory. N is a set. — GrandMinnow

Aiiiyyyy now I'm confused. You just explained to me that HF are all the finite sets in ZF. So if I'm understanding you, HF is the inductive definition of the collection of all finite sets in ZF; and that collection of finite sets is a model of HF. But it's wrong to also call that HF?

You're saying [in HF] I can't form the predicate that I think characterizes N.
— fishfry

Defining a predicate symbol is not a problem. But there is no definition of a constant symbol (such as 'N') such that N = {y | y is a natural number}, since HF does not prove that there exists an object that has as members all the natural numbers. — GrandMinnow

So your argument comes down to saying that since N is not a definable symbol in PA, I can't say "N is a proper class" because I have no idea what N is. Is that right?

In any language, in any theory, we can define whatever predicate symbols we want. It's only function symbols (including constant symbols, where a constant symbol is a 0-place function symbol) that require the supporting existence and uniqueness theorem

I can't form the collection of all numbers because I haven't got the language to do that.
— fishfry

In HF, you have the language, but you don't have the existence theorem ExAy(y is a natural number -> y e x). — GrandMinnow

Didn't track that sorry. But I'll agree that N's not a definable symbol.

in any model of ZF-inf there is not a set of all natural numbers
— fishfry

No, the sentence ExAy(y is a natural number -> y e x) is a theorem, but that doesn't preclude what the members of the universe for the model may be.

For every infinite cardinality, there is a model of ZF-Inf with a universe of that cardinality. And that universe can have as members any sets whatsoever. Same for (ZF-Inf)+~Inf. Same for PA. — GrandMinnow

Yes ok ... not sure where this is going ...

For example, we can have a model of PA whose universe is {w, w+1, w+2} and each of those members of the universe is infinite. — GrandMinnow

That's not a model of PA. w+2 has no successor. What am I missing here?

But wait, (ZF-Inf)+~Inf has a theorem ~Ex Ix [where 'I' is a defined 1-place predicate symble we are read in English as "is infinite"], so how can the universe of a model have a member that is infinite? Well, because for such a model, the predicate symbol 'e' is interpreted not as the ordinary membership relation but rather as some other "bizarre" relation and so also my 'I' be interpreted differently from "is infinite". When we talk about models in general, we can't presume that any given model of a theory "captures" the way we ordinarily "read off" the theorems of the theory. If we want to narrow the discussion to only models that adhere to the way we "read off' the theorems, then we should confine to talking about standard models. — GrandMinnow

"Read off"? Is that a technical term? Lost. I do understand that set theorists prefer models in which $\in$ is set membership.

in ZF there is no such definition or thing as a proper class.
— fishfry

In ZF, we may define:

x is a proper class <-> Ey y e x & ~Ez x e z

And we may prove:

~Ex x is a proper class. — GrandMinnow

That's very interesting. Makes perfect sense. I've never seen this done. They always say that ZF doesn't talk about proper classes so whenever we say proper class, we are being informal. In fact I've heard that so many times that it's probably why I do use "proper class" informally. But your definition makes perfect sense. I do wonder why I haven't seen it.

When the Peano axioms say, "O is a number,"
— fishfry

First order PA doesn't have a primitive 'is a natural number'. — GrandMinnow

Grrrrrr. The first thing it says is that "0 is a number." Is that not true? You know, everything you say directly contradicts Wiki. Is that what you meant earlier when you talked about Peano's original formulation? Is what everyone thinks of as PA not what you mean by PA?

Here's Wikipedia on the subject:

The Peano axioms define the arithmetical properties of natural numbers, usually represented as a set N or $\mathbb N$. The non-logical symbols for the axioms consist of a constant symbol 0 and a unary function symbol S.

The first axiom states that the constant 0 is a natural number:

0 is a natural number.

Now if by PA we mean what is described by Wikipedia, then I am frankly right and you are wrong. So it must be that by PA you mean something else. Is that the source of our misunderstanding? I will stipulate that they shouldn't have used the word set, but rather collection. Or (ahem) proper class!

Peano's historical own formulation should not be conflated with first order PA. — GrandMinnow

Ah this is what I meant a moment ago. You don't think PA is what is described by Wiki, and I and probably millions of other people are confused about this. Is that what you are saying?

the set of all sets is a proper class
— fishfry

There is not a set of all sets, not even in class theory. There is the class of all sets, and it is a proper class. — GrandMinnow

I'm terribly sorry, of course I know better. See how you have me confused!! I meant that the collection of all sets is a proper class. Just misspoke myself.

And I explained why referring to proper classes in discussion about set theory can be understood as an informal rendering for an actual formal notion in the background, but that is lacking here in saying N is a proper class in discussion about PA. — GrandMinnow

It's still a good conceptual metaphor IF not literally true. But again, according to Wiki your point is not so clear. So you must be objecting to the common Wiki understanding of PA.

And N is a set, which is not needing exceptions in view of the fact that in PA there can be no definition N = {x | x is a natural number}. — GrandMinnow

There is no SET such as that because first, PA doesn't have sets, and second, even if it did that would not be a valid specification of a set since it violates the axiom schema of specification.

But "is a natural number" must be a predicate, what else can it be? It's a thing that classifies all the objects in the universe into "yes it's one of these" and "no it's not." So the number 3 is a number and a tunafish sandwich is not a number.

Therefore I can form the COLLECTION, or "predicate satisfier," or as it's officially called the extension of the predicate, N = {x | x is a natural number}. N is a class and it's not a set. So it's a proper class.

If one wishes to say "N is a proper class with respect to PA" but not formulate the exact mathematical meaning of "with respect to" or even to a clearly articulate an intuitive/heuristic notion that is still consistent with the ordinary mathematical result that N is a set, and hopefully has value as a metaphor rather than confusing the subject with impressionistic use of terms, then, of course, I cannot opine whether or not in one's own mind it somehow makes sense nevertheless. But I do say, and have explained, that it makes no sense to me. — GrandMinnow

But it makes perfect sense to me! And you haven't explained what "is a number" could be, if not a predicate. And for any predicate whatever you can always form its extension consisting of all and only those things that satisfy it. Which, if it's a set, is a set; and if it isn't, is a proper class.

tl;dr: By PA do you mean something other than what's called PA on Wikipedia? That's the only way your post makes sense.

impressionistic use of terms — GrandMinnow

You say that like it's a bad thing! LOL.

Here's an attempt at making this terminological situation more precise. Perhaps we should distinguish two senses of proper class. In the absolute sense, a proper class is a collection that is not a set according to your favorite conception of what a set is (for a recent survey of such conceptions, cf. Luca Incurvati's Conceptions of Set and the Foundation of Mathematics). For instance, suppose you adhere to something like the limitation of size conception, according to which sets are collections that are not too big (say, are not the size of the universe). Then a proper class is a collection which is too big (i.e. in bijection with the universe of sets). In this sense, the natural numbers are not usually considered a proper class, unless, of course, one is a strict finitist.

But there is also a relative sense. Perhaps we can say that a collection is a proper class relative to some theory T if: (i) there is a predicate P such that x belongs to the class iff P(x), (ii) T proves that there is an x such that P(x) but (iii) T proves that there is no y such that x belongs to y iff P(x). For example, in ZFC, the class of all sets is a proper class, since x=x is a predicate satisfied by every set, but ZFC proves that there is no y such that x belongs to y iff x=x. Whether a collection is a proper class, then, would be relative to the strength of the theory. Perhaps a weaker theory (say, Kripke-Platek) would consider a proper class something that some other theory (say, ZFC) would consider a set.

Unfortunately, it seems that even in this relative sense, the natural numbers are not a proper class relative to PA. Now, there seems to be some confusion here about whether to take first-order or second-order PA. Notice that, when people take about PA, they generally mean first-order PA. In any case, that's the theory that is bi-interpretable with ZF-Inf (to save me from typing ZF-Inf+~Inf all the time, let's just adopt the convention that ZF-Inf means the latter. I'll rarely make mention of this theory anyway, so no confusion will ensue). Moreover, second-order PA can't be at issue, because second-order PA proves that there is an infinite set, namely the set of all natural numbers, so the natural numbers cannot be a proper class relative to this theory. What about first-order PA?

Here, the problem is that there is no predicate N such that N(x) iff x is a natural number expressible in PA. For suppose there is. Add a constant c to the language and consider the set of all formulas An:= n<c . Let PA* be PA + {An} (n in N) + Nc. This theory is obviously finitely satisfiable. So, by compactness, it is satisfiable, whence there is a model M in which c is a natural number greater than every natural number! This is a contradiction. So there can be no such predicate.

P = NP is false because solving a hard problem is a harder process than NP, so 2NP is regarded.

The computer has to solve logic, this logic, N, is opposite to P which is the objective. If P is placed before N, we commit a fallacy and our stance is as logic is already countable; we have frozen time, we expect that P = NP when NP is 2NP in regards N requiring it's own formation.

We assume N has no answer, to say P = NP. What is N in NP if not 2NP?

More properly phrased P = N(P)? Yes it does, P is the solving of a problem, via 'calculator' N, but whether the problem can be solved 'quickly' is innacurate, it's more fluency, it could take days, it depends on N. P = NP is a fallacy.

Read further down in the Wikipedia article, and you will see the axioms for first order PA. There is no predicate 'is a number'.

/

Regarding your notion about improper sets relative to PA as personal visualization, I didn't ignore it - you even quoted me remarking on it. I said I don't opine as to what does or does not make sense in your mind. But I said your notion makes no sense to me. And I would add that I think it does muddle discussion. But I didn't say you shouldn't think it.

ZF-infinity means ZF plus the negation of the axiom of infinity by default. — fishfry

Some people use it that way. And I have seen it lead to misunderstandings as casual readers fail to note that, in this context, we need the negation of the axiom of infinity and not just leaving out the axiom of infinity.

it's all the finite sets — fishfry

No, not all the finite sets. Only the hereditarily finite sets.

You just explained to me that HF are all the finite sets in ZF — fishfry

I defined 'HF' to stand for the theory (ZF~Inf)+~Inf.

But now I realize that writers often use 'HF' to stand for a class. So my choice to use 'HF' as the abbreviation was not good. From now on, I won't use it to stand for the theory (ZF~Inf)+~Inf. Instead I'll use:

TF = (ZF~Inf)+~Inf

since N is not a definable symbol in PA, I can't say "N is a proper class" because I have no idea what N is. Is that right? — fishfry

To be more precise, whatever symbol 's' we pick, TF does not support a definition:

s = {x | x is a natural number}

because the theory does not prove that there is a such an object.

your definition [of 'proper class' makes perfect sense. I do wonder why I haven't seen it. — fishfry

In class theory, it is well understood that a proper class is a class that is not a member of any class. All I'm doing is pointing out that we can also say that in set theory and conclude in set theory that there are no proper classes. It might be annoying, because it's not a very useful series of formulations. But it its technically correct, and I find that it sharpens the picture. Especially it goes against a common misconception that we can define a predicate symbol only to stand for a relation (sets are 1-place relations) that has members. No, we can always define an empty predicate. For example:

dfn: Jx <-> (x is odd and x is even)

is allowable, even if rather pointless.

That's not a model of PA. w+2 has no successor. — fishfry

Was a typo of omission; I meant {w, w+1, w+2 ...}

"is a natural number" must be a predicate [...] Therefore I can form the COLLECTION, or "predicate satisfier," or as it's officially called the extension of the predicate, N = {x | x is a natural number}. N is a class and it's not a set. So it's a proper class. — fishfry

Yes, we can have a predicate 'is a natural number' in TF. And upon an interpretation of the language, it has an extension (a subset of the universe for the model) and that extension is a set, not a proper class.

PA doesn't have sets, and second, even if it did that would not be a valid specification of a set since it violates the axiom schema of specification. — fishfry

Hard to discuss a counterfactual here.

So let's turn to TF.

It's not a matter of being consistent with the axiom schema of specification.

Instead, in the absence of the axiom of infinity, we do not have a supporting existence theorem for a definition:

N = {x | x is a natural number}

suppose you adhere to something like the limitation of size conception, according to which sets are collections that are not too big (say, are not the size of the universe) — Nagase

I don't begrudge anyone from that notion, but, for me, it's too vague. What is "too big"? And which universe?

Perhaps we can say that a collection is a proper class relative to some theory T if: (i) there is a predicate P such that x belongs to the class iff P(x), (ii) T proves that there is an x such that P(x) but (iii) T proves that there is no y such that x belongs to y iff P(x). — Nagase

I might suggest saying (only a subtle difference with yours):

For a theory T with 'e' in the language, we say a formula F(x) in the language for T "invokes" a proper class relative to T
iff
(i) T proves Ex F(x). (ii) T proves ~EyAx(F(x) -> xey).

So there is no proper class mentioned there. Instead there is a two place relation between a theory T and a formula F. Okay, fair enough, as maybe that's what fishfry has in mind. And it does work where T is set theory itself, indeed as it is common even in informally stating such things as the axiom schema of replacement where we refer to a 'function class'.

But I warn against thinking conflating this with the one place predicate 'is a proper class', except quite informally.

But worse, this manifestly clashes with ordinary terminology.

Let T = NBG (I call it 'BC' for Bernays class theory). For example:

BC proves Ex x is an ordinal. But BC does not prove ~EyAx(x is an ordinal -> xey), indeed BC proves the negation of that. But we do say, with BC, that the class of ordinals is a proper class.

Moreover, second-order PA can't be at issue, because second-order PA proves that there is an infinite set, namely the set of all natural numbers, so the natural numbers cannot be a proper class relative to this theory. — Nagase

Yep.

"Read off"? Is that a technical term? — fishfry

No. I was giving practical advice to not overlook that when we read natural language renderings of formulas, then we can't expect that how we naturally take such locutions in English is preserved with every interpretation (model) for the formal language.

No. I was giving practical advice to not overlook that when we read natural language renderings of formulas, then we can't expect that how we naturally take such locutions in English is preserved with every interpretation (model) for the formal language. — GrandMinnow

Yes ok. Agreed.

PA doesn't have sets, and second, even if it did that would not be a valid specification of a set since it violates the axiom schema of specification.
— fishfry

Hard to discuss a counterfactual here. — GrandMinnow

That made me chuckle.

So let's turn to TF.

It's not a matter of being consistent with the axiom schema of specification.

Instead, in the absence of the axiom of infinity, we do not have a supporting existence theorem for a definition:

N = {x | x is a natural number} — GrandMinnow

Yes ok I get that you assert that.

But when Wikipedia says that the first axiom of PA is, "0 is a natural number," are you asserting that "is a number" is something other than a predicate?

My chain of reasoning is:

P1: I take Wiki's account of PA as reasonably accurate (whereas I suspect that perhaps you don't);

P2: Wiki says that the first axiom of PA is "0 is a natural number";

C: If "is a number" is not a predicate, what the heck is it? In which case N = { x : x is a natural number} is the extension, or "predicate satisfier" as I think of it, of the predicate "is a natural number." It's not a set. What is a predicate satisfier that is not a set? It's a proper class.

Can you please tell me where you disagree?

It goes back to Frege. A predicate defines a class. Of course he thought that a predicate defined a set, which was falsified by Russell. But a predicate still defines a class.

This is how I understand what's going on.

That’s not first order PA. Look further down in the article where the axioms of first order PA are listed.

Having spent an entire math career not concerned with these things, I now find there is literally no escape: https://en.wikipedia.org/wiki/Reverse_mathematics

I wasn't even aware of this before opening the brochure I got in the mail today from the Princeton Press.

:worry:

For more on the limitation of size idea, I strongly recommend reading the summary in Incurvati's book. Still, here are two ways of making the idea more precise: (i) A collection A is too big iff there is an injection from the class of all ordinals into A (Cantor); (ii) A collection A is too big iff it is in bijection with the collection determined by the formula x=x (von Neumann).

On NBG, I thought it proved that there is no set of all ordinals? If so, then it does not prove the negation of your formula, since typically we would employ (à la Bernays) two sorts of variables, one for classes, one for sets, and lower case letters would correspond to sets... or else we could explicitly have two predicates, say M(x) for x is a set and C(x) for x is a class, and then adapt the formula.

We should not overlook that ‘class’ does not mean just ‘proper class’. Some classes are sets and other classes are proper classes. Everything in NBG is a class.

Yes, NBG does not prove that there is a set of all the ordinals. To be clear, with NBG we don’t have variables for just proper classes. With the multi-sorted version we have general variables for all objects - all classes, i.e sets and proper classes. And we have another special variable for just sets. Where ‘x’ and ‘y’ are general variables (class variables) we have the theorem

EyAx(x is an ordinal -> x e y)

And where ‘s’ is a set variable, we have the theorem

~EsAx(x is an ordinal -> x e s)

As you mentioned, having two kinds of variables is equivalent to having a primitive predicate M for ‘is a set’ and relativizing. (But a primitive predicate C for ‘is a class’ turns out to be the universal predicate: Cx <-> x=x.) So the second theorem is

~Ey(My & Ax(x is an ordinal -> x e y))

But the first theorem itself shows that your formulation doesn’t work unless you revise it in some way, perhaps with such relativization, and we would have to see exactly what that formulation would be.

By the way, since historically different writers formulate class theory differently, for sake of definiteness, I choose one in particular: ‘Model Theory’ by Chang and Keisler.

Numbers are names for quantities. — creativesoul

I would add a quantity precisely defined to every other quantity (with that we exclude the problem of quantity defined being a "heap" of something, for example).

That would be a great definition, but it ain't.

We don't allow the infinitesimal or infinity to be numbers. :roll:

Numbers are not the only names for quantities. They are names for quantities nonetheless. So... that criticism misses the mark.

If you consider NBG as a two-sorted theory, and take the lower case variables to range over sets, then there's no need to revise my statement, since it employed only lower case variables...

I'll use M, because it stands out better.

So:

For a theory T with 'e' in the language, a formula F(x) in the language for T "invokes" a proper class relative to T
iff
(i) T proves Ex F(x). (ii) T proves ~Ey(Mx & Ax(F(x) -> xey)).

But that depends on the theory having M as primitive or defined (or being in a 2-sorted logic) And M not to be just an arbitrary predicate, I guess the theory has to prove Ax(Mx <-> Ey xey).

So:

For a theory T with 'e' in the language, and having a 1-place predicate symbol 'M' as mentioned below, a formula F(x) in the language for T "invokes" a proper class relative to T
iff
(i) T proves Ex F(x). (ii) T proves ~Ey(Mx & Ax(F(x) -> xey)). (iii) T proves Ax(Mx <-> Ey xey).

This is not a problem for any language with 'e', as such an 'M' can be defined if not primitive.

So 'x is an ordinal' invokes a proper class relative to NBG.

But PA is not eligible for applying this definition, because PA does not have 'e'. So what about (ZF-Inf)+~Inf as a surrogate for PA?

We want to see whether 'x is a natural number' invokes a proper class relative to (ZF-Inf)+~Inf.

I guess it works. (?)

Read further down in the Wikipedia article, and you will see the axioms for first order PA. There is no predicate 'is a number'. — GrandMinnow

If you mean the part where it lists the axioms only referring to the successor function $S$ and the symbol $\in$, why can't I define

$\mathbb N = \{x : x = 0 \lor \exists y (x = S(y))\}$?

Regarding your notion about improper sets relative to PA as personal visualization, I didn't ignore it - you even quoted me remarking on it. I said I don't opine as to what does or does not make sense in your mind. But I said your notion makes no sense to me. And I would add that I think it does muddle discussion. But I didn't say you shouldn't think it. — GrandMinnow

Yes I think that appeared in your post after I wrote that. But I really don't get why you think the idea is so terribly wrong.

But now I realize that writers often use 'HF' to stand for a class. So my choice to use 'HF' as the abbreviation was not good. From now on, I won't use it to stand for the theory (ZF~Inf)+~Inf. Instead I'll use:

TF = (ZF~Inf)+~Inf — GrandMinnow

I'm still confused on this point, since nobody ever describes ZF as ZF-C + (not-C). I sort of see your point that "ZF" is being noncommittal on the point, but generally from context you can tell when they explicitly mean ZF-C + (not-C). So perhaps people should be more formal about that.

To be more precise, whatever symbol 's' we pick, TF does not support a definition:

s = {x | x is a natural number}

because the theory does not prove that there is a such an object. — GrandMinnow

As above, what of

$\mathbb N = \{x : x = 0 \lor \exists y (x = S(y))\}$?

Surely I can form that class directly from the symbols 0 and S referenced in the PA axioms.

In class theory, it is well understood that a proper class is a class that is not a member of any class. All I'm doing is pointing out that we can also say that in set theory and conclude in set theory that there are no proper classes. It might be annoying, because it's not a very useful series of formulations. But it its technically correct, and I find that it sharpens the picture. Especially it goes against a common misconception that we can define a predicate symbol only to stand for a relation (sets are 1-place relations) that has members. No, we can always define an empty predicate. For example:

dfn: Jx <-> (x is odd and x is even)

is allowable, even if rather pointless. — GrandMinnow

We're agreed on all this but I don't see how it invalidates my point. But I'll concede that if I understood you better you might well be right.

That's not a model of PA. w+2 has no successor.
— fishfry

Was a typo of omission; I meant {w, w+1, w+2 ...} — GrandMinnow

Yes sorry, obvious in retrospect but now I don't remember what point was being made ...

Yes, we can have a predicate 'is a natural number' in TF. And upon an interpretation of the language, it has an extension (a subset of the universe for the model) and that extension is a set, not a proper class. — GrandMinnow

You're saying that my definition of $\mathbb N$ is a set? I confess I don't follow your reasoning but I admit I don't necessarily disagree with it. I just don't follow it. Maybe I haven't the background to follow your argument. I don't see now this could be a set in TF.

Instead, in the absence of the axiom of infinity, we do not have a supporting existence theorem for a definition:

N = {x | x is a natural number} — GrandMinnow

Right, but I gave the correct formula based on the formal axioms you pointed me to in the Wiki piece. We can't form a set with that definition but it's certainly the extension of a predicate. And if it's that, and it's not a set ... that's what I call a proper class. Maybe the people who told me that were speaking more loosely than I realized, that would be your point.

If you mean the part where it lists the axioms only referring to the successor function S and the symbol e — fishfry

No the part that has the headline:

First-order theory of arithmetic

That is first order PA. (In this context, By ‘PA’ we mean first order PA.)

And PA does not have set abstraction notation.

(ZF-C)+~C is just ZF+~C.

That is not ZF.

I’ve never seen that theory discussed (though maybe it comes up somewhere.)

I take ‘-‘ to mean ‘without’ and ‘+’ to mean ‘also with’.

So ZF-Inf is ZF but without the axiom of infinity.

And (ZF-Inf)+~Inf is ZF without the axiom infinity and also with the negation of the axiom of infinity.

Two very different theories.