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Note: "Update" at the end of this question hopefully summarizes/clarifies the original language (original text left in place for context).

Philosophical Preface: For the purposes of this question, I am assuming that the notion of "standard $\mathbb{N}$" is not an illusion. In other words, that there is a unique (up to isomorphism) model of $\mathsf{PA}$ which is standard in an "absolute" sense (namely, the numbers we all learned to count with in elementary school, as laid out in the first section of these notes for example). And it is in relation to this "absolutely standard" $\mathbb{N}$ that we can meaningfully speak of whether the $\omega$ of a given set universe is actually standard (i.e. notwithstanding that every universe "thinks" its $\omega$ is standard, when viewed "from the outside" this $\omega$ may or may not be "actually" standard, depending on whether it's order-isomorphic to our "absolutely standard" $\mathbb{N}$). In what follows, this is what I mean when speaking of "standard $\mathbb{N}$" (hopefully this is all relatively conventional and uncontroversial so far?)

Some Background:

At one time, I used to be under the impression that:

  • The first-order Peano axioms have many models (both standard $\mathbb{N}$, and also various systems of non-standard natural numbers).
  • But Peano axiomatization with second-order induction and "full" second-order semantics ($\mathsf{PA2}$) has a unique (up to isomorphism) model, namely the standard natural numbers (the above-described "absolutely standard" $\mathbb{N}$).

But it now seems to me that second bullet point is probably not quite correct. Instead (as I understand it) any given "domain of discourse" does indeed have a unique (up to isomorphism) model of $\mathsf{PA2}$, but depending on the assumptions you make about the domain you're working in, its model of $\mathsf{PA2}$ may or may not be standard $\mathbb{N}$. For example, if you assume you are working in a $\mathsf{ZFC}$ universe with non-standard $\omega$ (and successor function defined in the usual way as $S(x) = x \cup \{x\}$), then your unique model of $\mathsf{PA2}$ is the $\omega$ of this universe, which by hypothesis is not order-isomorphic to standard $\mathbb{N}$.

The Actual Question:

If one's goal is some formalism whose only possible model (regardless of any additional assumptions about the "domain of discourse") is (up to isomorphism) the above-described "absolutely standard" $\mathbb{N}$, then (if my above-described understanding is correct) $\mathsf{PA2}$ doesn't fully accomplish this goal.

So, is there any formalism (whether an axiomatization in second-order logic, or something else) which does accomplish this goal?

  • If you're assuming $\mathsf{ZFC}$ as your foundation, then augmenting $\mathsf{PA2}$ with an additional stipulation that "the $\omega$ of my $\mathsf{ZFC}$ universe is standard" might seem to be a solution. But that's unhelpfully circular, because how do you formalize the "standardness" of $\omega$? That's exactly the problem you're trying to solve in the first place.

Update -- A couple additional points:

  1. Viewed from the perspective of set-theoretic ($\mathsf{ZFC}$) multiversism, I think my question can be equivalently recast as follows: If I philosophically accept the multiverse, I nonetheless may wish speak of distinctions between one category of universes versus others, in particular to distinguish universes with "standard $\omega$" from others with "non-standard $\omega$". When attempting to articulate this distinction, I certainly have a fairly concrete but somewhat informal concept in mind of what I mean by "standard $\omega$" (see "Philosophical Preface" above). But can this concept of standardness of $\omega$ be formalized? In other words, is there some formal criterion (whether formulated in second-order logic, or some other way) that effectively distinguishes the universes with standard $\omega$ from the other (non-standard $\omega$) universes?

  2. The notion of "standardness" of $\mathbb{N}$ becomes, I suppose, somewhat more formal if we think of it as: $X$ is a "standard" natural number iff it equates to any term in first-order logic in which the last symbol is $\mathbf{0}$ and every other symbol is the successor function symbol $\mathbf{S}$ (hence: $\mathbf{0}$, $\mathbf{S0}$, $\mathbf{SS0}$, etc.). However this falls somewhat short of what I have in mind as really "formalizing" standardness. Specifically, it seems like it's kind of "cheating" by using the grammar rules of the language (first-order logic) to inform the semantic meaning of "standardness" in the theory (arithmetic). But maybe this is the best we can do towards formalizing the notion of "standardness" of $\mathbb{N}$?

NikS
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  • What does PA, first or second order, have to do with "the universe"? The universe is governed by ZFC. PA is merely one its residents. – Asaf Karagila Dec 22 '23 at 09:52
  • Well, if I assume some universe of $\mathsf{ZFC}$ as my foundation, then presumably the “objects” (natural numbers) described by the $\mathsf{PA2}$ axioms are the elements of some set, and also the (graph of) the successor function is some set. So if I assume I am building the model of $\mathsf{PA2}$ inside this universe, then it seems like differing assumptions about the universe will have differing implications as to what model of $\mathsf{PA2}$ can exist in it (standard vs nonstandard) – NikS Dec 22 '23 at 10:40
  • Like, when the second-order induction axiom speaks of “every subset” of $\mathbb{N}$, surely our assumptions about the universe will impact the precise meaning of “every subset” (i.e. which subsets actually exist in $\mathcal{P}(\mathbb{N})$)? – NikS Dec 22 '23 at 10:48
  • Yeah, but then the discussion about $\Bbb N$ is a red herring. You're asking about the universe of sets not being "a concrete and singular object", which is a different question, and when phrased like that, strips away a lot of the distractions of focusing on $\Bbb N$. The short answer is an easy one, though. Everything that is, exists; everything that exists, is. But truth and existence are relative to a universe. So different universes have differing opinions. You can either make peace with the multiverse, choose to believe there is just one "true" universe, or decide none of it matters. – Asaf Karagila Dec 22 '23 at 11:12
  • If I interpret your question purely formally, as "is every statement in the (1st order) language of $\mathsf{PA}$ decidable in the theory $\mathsf{PA2}$", then the answer is no, e.g. the statement "$\mathsf{PA2}$ is consistent" is not decidable. – user8268 Dec 22 '23 at 11:19
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    @user8268: You need to be careful when talking about decidability in SOL since it's not very conducive to a good proof theory, so in what sense is something decidable in second-order logic? – Asaf Karagila Dec 22 '23 at 13:14
  • @AsafKaragila I got the impression that $\mathsf{PA2}$ is actually a 1st order theory with 2 types of objects (elements and subsets of $\mathbb{N}$); if that is not true then my comment was wrong (or meaningless). – user8268 Dec 22 '23 at 14:21
  • @user8268: No. It is a second-order theory. If it were first-order, it would have had arbitrarily large models and would not prove that all of its models are isomorphic. You can look at the first-order theory of the structure $\Bbb N\cup\mathcal P(\Bbb N)$ with $<$ and $\in$, and so on. But it will be a different theory than $\sf PA_2$, so to an extent you will have a theory that isn't strong enough to capture that structure or one that is not recursively enumerable (or both!), so it gets complicated again to the point of "it's not at all clear what should be the 'obvious' interpretation". – Asaf Karagila Dec 22 '23 at 14:44
  • @AsafKaragila thanks! – user8268 Dec 22 '23 at 14:47
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    I think you're argumentation is somewhat of a fallacy. If you really believe in an absolute standard $\mathbb{N}$ in the real world (such as in elementary schools) then the place to interpret $PA2$ is in the real world and it does uniquely determine it. It's like saying we can't pin down exactly the Eiffel Tower because maybe there is a parralel universe in which there is a slightly different Eiffel Tower. – Jonathan Schilhan Dec 24 '23 at 15:40
  • @AsafKaragila : From the multiverse perspective, my question becomes (I think) equivalent to: is there a *formalizable* criterion for distinguishing between standard-$\omega$ universes versus nonstandard-$\omega$ universes? (additional detail on this, beyond the character-count limit for comments, is in the "update" section I added to the original question) – NikS Dec 26 '23 at 08:09
  • @NikS in this last case, I believe Łoś's theorem to the effect that ultrapowers are elementarily isomorphic to their 'base' models answers it in the negative – ac15 Dec 26 '23 at 14:52
  • @ac15 : Do you mean Los’s theorem proves the impossibility of formalizing the notion “standardness of $\omega$” as a criterion for categorizing universes? – NikS Dec 28 '23 at 05:35
  • Los’s theorem (to my understanding) says the truth value of all *first-order* sentences is the same in the “$\omega$-standard” base model and in the “$\omega$-nonstandard” ultrapower. But that doesn’t (I think?) preclude some other mechanism (perhaps formulated in second-order logic) from differentiating the two models. – NikS Dec 29 '23 at 05:35
  • @NikS oh yeah, my bad, for a moment I thought only of first order set theory(ies), and forgot you are willing to consider extensions/stronger logics – ac15 Dec 29 '23 at 14:03

1 Answers1

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Is "standard $\mathbb{N}$" in fact not "fully formalizable"?

If we identify "standard $\mathbb{N}$" with the/a theory of true arithmetic, then of course, by incompleteness, the answer is no: being consistent and negation complete, it cannot be 'effective'

One may initially object that this seems too tied to (a) specific (choice of) formalisms; Boolos, et al (2007), for example, may perhaps give such an impression

17.7 Theorem (Gödel’s first incompleteness theorem). There is no consistent, complete, axiomatizable extension of Q. [Robinson arithmetic]

[...]

The import of Gödel’s first incompleteness theorem is sometimes expressed in the words ‘any sufficiently strong formal system of arithmetic (or mathematics) is incomplete, unless it is inconsistent’. Here by ‘formal system’ is meant a theory whose theorems are derivable by the rules of logical derivation from a set of axioms that is effectively decidable, and hence (assuming Church’s thesis) recursive. So ‘formal system’ amounts to ‘axiomatizable theory’, and ‘formal system of arithmetic’ to ‘axiomatizable theory in the language of arithmetic’.

But when one remembers that one can do a reasonable amount of set theory inside finite-type arithmetic, and type theory inside higher-order logic, and higher-order logic inside second-order logic (Hintikka, Montague) $-$ plus Church-Turing, of course $-$, then it seems pretty much safe to answer your request

So, is there any formalism (whether an axiomatization in second-order logic, or something else) which does accomplish this goal?

for an 'absolute pinning' or 'absolute description' of "(absolutely) standard $\mathbb{N}$" with "No"

ac15
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  • Is the decidability of arithmetic really the decisive factor though? I could equally well frame this question as pertaining to the "standard" model of Presburger arithmetic (which *is* decidable) – NikS Dec 26 '23 at 08:04
  • @NikS there are two senses of 'decidable' going on here: the sense in Boolos's quote is that there is an algorithm for, given a statement, checking if it is an (instance of some the) axiom(s) (remember sometimes there are axiom schemas to deal with), or, given a list of statements, checking if it's a proof (that is, checking if each statement is either an axiom, or if it's obtained from previous statements in the list by mean of the allowed inference rules); the sense in your comment is that there is an algorithm for, given a statement, finding a proof of it, if there is one, or [continues] – ac15 Dec 26 '23 at 14:41
  • finding a proof of its negation, otherwise. Terminology in logic is sometimes confusing like this, unfortunately. But besides that, knowing that addition and multiplication (plus full induction) suffice for fully defining all other recursive functions, and that addition alone doesn't, I believe framing the question as being about Presburger arithmetic would not be equaly good – ac15 Dec 26 '23 at 14:48
  • Based on responses (and links) here and in linked question, it seems clear that the short answer to this question is "no". Therefore marking this answer as accepted. – NikS Feb 04 '24 at 09:00