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Is the standard model of arithmetic unique? If so, is there a proof of such or is it just a consequence of the definitions?

I was reading the following blog which references a standard model of Peano arithmetic: "These are often called ‘nonstandard’ models. If you take a model of Peano arithmetic—say, your favorite ‘standard’ model —" link. My question is how do we know that such is the only one?

Noah Schweber
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Aliquid Ex Nihilo
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  • Isn't a non-standard model of arithmetic any model not isomorphic to the standard model? What do you mean with unique? – Vsotvep Oct 21 '19 at 17:45
  • The standard model is simply defined as a certain structure and is, as such, unique. But I think your question is: is there a unique model for the typical axioms of arithmetic (= Peano Axioms)? To that, the answer is no. – Bram28 Oct 21 '19 at 17:46
  • I was reading the following blog which references a standard model of Peano arithmetic: "These are often called ‘nonstandard’ models. If you take a model of Peano arithmetic—say, your favorite ‘standard’ model —" [link] (https://johncarlosbaez.wordpress.com/2018/03/03/nonstandard-integers-as-complex-numbers/). My question is how do we know that such is the only one? – Aliquid Ex Nihilo Oct 21 '19 at 17:52
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    I disagree with the downvote. If one sees the phrase "the standard model" before being familiar with set theory, it's quite reasonable to worry that there's some question-begging here; and I think that this is a reasonably common occurrence based on some experiences I've had on this site. I think this is a fine question, and (per my answer) can in fact be meaningfully framed and addressed. – Noah Schweber Oct 21 '19 at 17:57
  • (That said, this question could certainly use a bit more context - I've edited in the OP's comment, which in my opinion improves the situation.) – Noah Schweber Oct 21 '19 at 18:08
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    What is a proof if not a consequence of the definitions? – Asaf Karagila Oct 21 '19 at 18:10
  • @AsafKaragila you are correct, I should have said an obvious consequence not relying on other complicated machinery – Aliquid Ex Nihilo Oct 21 '19 at 18:57
  • The following 4 wikipedia links are of interest here: $;$Second order arithmetic, $;$First order theory of arithmetic $;$Zermelo-Fraenkel set theory: Axiom of infinity and Non-standard model of arithmetic. – CopyPasteIt Oct 22 '19 at 12:15
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    https://math.stackexchange.com/questions/3230638/plurality-of-arithmetics-or-absoluteness-of-arithmetical-truths-i-e-are-all/3230706#3230706 –  Oct 27 '19 at 16:12

1 Answers1

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The standard model of arithmetic is indeed unique (up to isomorphism). Of course, to assert this we need to give the phrase above a precise definition ...

Say that a model $M$ of arithmetic is standard if for every other model $N$ of arithmetic there is a unique initial segment embedding (= injective homomorphism with image an initial segment) of $M$ into $N$. There are in fact other definitions we could use here. We can prove that $(i)$ a standard model of arithmetic exists and $(ii)$ any two standard models of arithmetic are isomorphic (indeed, uniquely isomorphic - if $M,N$ are standard models of arithmetic, then there is a unique isomorphism from $M$ to $N$). The latter justifies the use of "the" in "the standard model."

Of course, in stating the above we need to precisely define "arithmetic." But pretty much anything reasonable will work; in particular, the claim above is true for Robinson's Q and (restricting attention to addition alone) Presburger arithmetic.

The best intuition for this in my opinion is by comparison. Suppose $M,N$ are models of arithmetic; we can try to define an embedding of $M$ into $N$ in "stages" (send the least element of $M$ to the least element of $N$, the next-least element of $M$ to the next-least element of $N$, ...). This will work for a while, but possibly not "all the way up" (maybe $M$ does not in fact embed into $N$). However, we can look at the subset of $M$ consisting of all $x$ such that for every model $N$, the attempted-embedding-process works up to $x$. The set of all such $x$ can then be shown to be a standard model of arithmetic.

  • A bit more precisely: for a model $M$ of arithmetic, let $M_0$ be the set of $x\in M$ such that for every model $N$ of arithmetic there is some $y\in N$ such that $[0,x]^M$ is uniquely isomorphic to $[0,y]^N$, where "$[0,a]^B$" for $B$ a model of arithmetic and $a\in B$ is the "substructure"of $B$ consisting of all elements $\le a$ (it's not a literal substructure unless $a=0$, since if $a\not=0$ it's not closed under $+$ and $\times$, but we can resolve this either by talking about partial structures or by replacing $+$ and $\times$ with their corresponding "graph relations). It's not hard then to show that $M_0$ is a standard model of arithmetic.

Of course this depends on a background theory within which this proof takes place (this is distinct from the theory we choose to call "arithmetic" - we're proving something about arithmetic, in the context of a different, "larger" theory). Usually when we don't state that theory it's taken to be ZFC, and that certainly suffices here. However, in general ZFC is massive overkill, and this is such a case: the existence of the standard model of arithmetic can be appropriately stated and proved in the much weaker theory ACA$_0$, and in fact this is optimal in a precise sense (I don't know a citation for this fact but this is not hard to prove).

  • Incidentally, the general study of what axioms are needed to prove particular theorems is indeed a topic of much interest, most of it falling under the umbrella of reverse mathematics

  • Also, it's worth pointing out that both ACA$_0$ and ZFC have a "distinguished" model of arithmetic, and so in each context we generally prove the stronger fact that that distinguished model is standard; indeed, in the ACA$_0$ context I think that's how we have to proceed.

On the other hand, we could always consider frameworks which actively refute (rather than simply are too weak to prove) the theorem above. See e.g. this article of Yessenin-Volpin (an ultrafinitist who interestingly believed that ZFC, and indeed much more, is consistent!) or more recently some approaches to a multiverse framework of set theory (see the discussion before the "well-foundedness mirage" bit on page $25$ - this would be an extreme version of that principle). Whether one finds these alternate theories compelling is of course subjective; personally, I happen to be at least somewhat sympathetic to them.

Noah Schweber
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  • Incidentally it's worth pointing out that ACA$_0$ is drastically less well-known than ZFC; indeed, I'd be extremely surprised to find more than a few people outside logic who are familiar with it! This is because its primary interest is within logic. Outside logic the general interest in something like ZFC is instrumental: that it's an apparently-consistent theory within which basically all of math can be developed. So there's generally not much call for something like the (reverse-math-motivated) ACA$_0$ outside of logic. I mention it here, though, since I do think it's relevant to the issue. – Noah Schweber Oct 21 '19 at 18:11
  • Noah, to be fair, your answers are great. But do you ever spend a few minutes checking if something is a duplicate? – Asaf Karagila Oct 21 '19 at 18:12
  • @AsafKaragila The closest thing I found was this, which I don't really think is an exact duplicate (although it's really close) since it's taking as given the existence of a thing we're considering the "standard model." Is there a closer duplicate I'm missing? – Noah Schweber Oct 21 '19 at 18:13
  • (To be fair I do take a somewhat restrictive view towards what constitutes an "exact duplicate," and I often do forget to check for duplicates. So you're not wrong.) – Noah Schweber Oct 21 '19 at 18:15
  • @NoahSchweber Regarding your point about reverse mathematics, is there a particular axiom (or combination of axioms) of ACA0 that makes the standard model minimal model? – Aliquid Ex Nihilo Oct 21 '19 at 19:17
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    @AliquidExNihilo Well, ACA$_0$ is RCA$_0$ (which we consider to correspond to "computable" mathematics) together with a single new axiom; there are various ways to formulate this axiom, the most natural of which is probably "Every function has a range." To see why this is a nontrivial principle note that the range of a simple function can be quite complicated since a priori being certain that a given number is not in the range of a function requires searching over all possible inputs. Even if a function is computable, its range might not be: every c.e. set, like the halting problem, is such. – Noah Schweber Oct 21 '19 at 20:40
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    So this additional axiom is the culprit since it's the only additional axiom. (The way this additional axiom is usually phrased, though, is "For every set $X$, the Turing jump of $X$ exists." While more technical, this is better fits the unifying computability-theoretic theme of the subject.) Note that ACA$_0$ cannot actually prove that the "maximal common part" of any two models of arithmetic exists; in some sense, it's "barely" able to prove that arithmetic has a standard model. – Noah Schweber Oct 21 '19 at 20:43
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    It's also worth noting that in the ACA$_0$ context we need to be more careful than usual about what we mean by "arithmetic" - PA would be an inappropriate choice here, since in fact ACA$_0$ cannot prove that PA has a model, let alone that PA is true of the "canonical" natural numbers! See the first part of this Mathoverflow answer of mine. – Noah Schweber Oct 21 '19 at 20:43