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June 5, 2019

Nonstandard Models of Arithmetic

Posted by John Baez

A nice quote:

There seems to be a murky abyss lurking at the bottom of mathematics. While in many ways we cannot hope to reach solid ground, mathematicians have built impressive ladders that let us explore the depths of this abyss and marvel at the limits and at the power of mathematical reasoning at the same time.

This is from Matthew Katz and Jan Reimann’s nice little book An Introduction to Ramsey Theory: Fast Functions, Infinity, and Metamathematics. I’ve been been talking to my old friend Michael Weiss about nonstandard models of Peano arithmetic on his blog. We just got into a bit of Ramsey theory. But you might like the whole series of conversations.

  • Part 1: I say I’m trying to understand “recursively saturated” models of Peano arithmetic, and Michael dumps a lot of information on me. The posts get easier to read after this one!

  • Part 2: I explain my dream: to show that the concept of “standard model” of Peano arithmetic is more nebulous than many seem to think. We agree to go through Ali Enayat’s paper Standard models of arithmetic.

  • Part 3: We talk about the concept of “standard model”, and the ideas of some ultrafinitists.

  • Part 4: Michael mentions “the theory of true arithmetic”, and I ask what that means. We decide that a short dive into the philosophy of mathematics may be required.

  • Part 5: Michael explains his philosophies (plural!) of mathematics, and how they affect his attitude toward the natural numbers and the universe of sets.

  • Part 6: After explaining my distaste for the Punch-and-Judy approach to the philosophy of mathematics (of which Michael is not guilty), I point out a strange fact: our views on the infinite cast shadows on our study of the natural numbers. For example: large cardinal axioms help us name larger finite numbers.

  • Part 7: We discuss Enayat’s concept of “a T-standard model of PA”, where T is some set of axioms extending ZF. We conclude with a brief digression into Hermetic philosophy: “as above, so below”.

  • Part 8: We discuss the tight relation between PA and ZFC with the axiom of infinity replaced by its negation. We then chat about Ramsey theory as a warmup for the Paris–Harrington Theorem.

  • Part 9: Michael sketches the proof of the Paris–Harrington Theorem, which says that a certain rather simple theorem about combinatorics can be stated in PA, and proved in ZFC, but not proved in PA. The proof he sketches builds a nonstandard model in which this theorem does not hold!

Posted at June 5, 2019 6:41 PM UTC

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Re: Nonstandard Models of Arithmetic

A joke:

If we add up all of a nonstandard set of natural numbers, do we still get 112-\frac{1}{12}?

Posted by: Blake Stacey on June 6, 2019 12:24 PM | Permalink | Reply to this

Re: Nonstandard Models of Arithmetic

Taking the joke seriously, the answer has to be yes, right? Suppose we take one of the various theorems making rigorous the statement that nn=1/12\sum_{n \in \mathbb{N}} n = -1/12. It’s still a theorem for any nonstandard \mathbb{N}, isn’t it?

I admit, I don’t know the mathematical definition of “nonstandard model of the natural numbers”. (Or of “standard model”.) I skimmed through almost all of John and Michael’s conversations, rather fast, and I understood this to be actually one of the central points. Is that right?

Posted by: Tom Leinster on June 6, 2019 10:21 PM | Permalink | Reply to this

Re: Nonstandard Models of Arithmetic

I’m pretty sure “It’s still a theorem for nonstandard \mathbb{N}” is a sentence that would make most logicians squirm… though it’s truer than is false. I feel the pedantic need to say it better. Pardon me if you know all this:

Theorems are things you prove starting from a theory—a bunch of axioms. Peano arithmetic is a theory. You can ask if some sentence is a theorem of Peano arithmetic. It either is or isn’t. Nothing about models here—not directly, anyway.

Models of a theory are structures in which the theory is “valid”. In other words, all the axioms, and all theorems following from them, are “satisfied”.

Gödel’s completeness theorem ties these concepts together. If a sentence is satisfied in every model of your theory, it must be a theorem!

So, you really meant to say something like:

Suppose we take one of the various theorems making rigorous the statement that nn=1/12\sum_{n \in \mathbb{N}} n = -1/12. These sentences are still valid in nonstandard models of Peano arithmetic, because they are theorems in Peano arithmetic.

And that would be true if these really are theorems in Peano arithmetic. But I’ve never seen someone try to develop complex analysis using just Peano arithmetic. After all, Peano arithmetic is just about the natural numbers! But you can do a surprisingly large amount using surprisingly little, using various clever coding tricks. So it’s an interesting question: can you develop complex analysis to the point of studying the Riemann zeta function, using just Peano arithmetic?

Usually people, even logicians trying to develop math on bare-bones axioms, would use a bit more. An expert on reverse mathematics could tell us.

If you can’t even say ζ(1)=1/12\zeta(-1) = -1/12 in Peano arithmetic, then it doesn’t make sense to ask if it’s valid in some models and not others.

In short: now you see, Blake: when talking to logicians, no joke will go unpunished!

Posted by: John Baez on June 7, 2019 12:58 AM | Permalink | Reply to this

Re: Nonstandard Models of Arithmetic

Thanks, John, and I probably made you squirm too. I was indeed being careless.

However, I’d like to argue that what I wrote was less squirmworthy than it might appear. My argument is to do with the way in which categorical logic blends (a critic might say “muddles up”) syntax and semantics in a way that logicians of a more traditional mindset might find disturbing. I dimly remember some quote by Tarski or some such figure about Lawvere, along the lines of “that guy doesn’t know the difference between syntax and semantics”. It was, I believe, prompted by Lawvere theories.

More specifically: as I said, I never understood quite what was meant by “(non)standard model”. But I had in my head that it would be something like this.

We have “the” category of sets, which is really just a topos satisfying a couple of extra axioms. It has a unique-up-to-iso natural numbers object. No ambiguity there… except that there isn’t really such a thing as “the” category of sets. There are very many inequivalent categories satisfying the topos-plus-a-bit-more axioms, and they all have a uniquely-determined natural numbers object.

Perhaps we wave our hands a bit and distinguish one category satisfying these axioms as “the” category of sets — or as you might call it, the “standard” category of sets, Set\mathbf{Set} — while calling the other ones “nonstandard”. This is certainly handwavy, but it looks to me to be in a similar spirit to what Michael was saying, and no vaguer.

It’s then natural to refer to the NNO of Set\mathbf{Set} as the “standard \mathbb{N}” and the NNO of the nonstandard categories of sets as “nonstandard \mathbb{N}s”.

The point is this. You’ve got all these different, inequivalent, toposes. There are statements that are true in some of them and not in others. In particular, there are statements about the NNO that are true in some but not others. So you really can talk about a statement being “true for any nonstandard \mathbb{N}”.

I shouldn’t have carelessly used the word “theorem” in a discussion about logic, but I was using it in the mathematicians’ rather than logicians’ sense, like if I was to ask you:

is the classification theorem for finite abelian groups true in an arbitrary topos?

Anyway, that’s how I was thinking. I realize your discussion with Michael wasn’t at all categorical. But when you think about logic categorically, the boundary between syntax and semantics is different and blurrier than in the classical treatment.

Posted by: Tom Leinster on June 7, 2019 10:21 PM | Permalink | Reply to this

Re: Nonstandard Models of Arithmetic

Tom wrote:

More specifically: as I said, I never understood quite what was meant by “(non)standard model”. But I had in my head that it would be something like this.

We have “the” category of sets, which is really just a topos satisfying a couple of extra axioms. It has a unique-up-to-iso natural numbers object. No ambiguity there… except that there isn’t really such a thing as “the” category of sets. There are very many inequivalent categories satisfying the topos-plus-a-bit-more axioms, and they all have a uniquely-determined natural numbers object.

Perhaps we wave our hands a bit and distinguish one category satisfying these axioms as “the” category of sets — or as you might call it, the “standard” category of sets, Set\mathbf{Set} — while calling the other ones “nonstandard”. This is certainly handwavy, but it looks to me to be in a similar spirit to what Michael was saying, and no vaguer. It’s then natural to refer to the NNO of Set\mathbf{Set} as the “standard \mathbb{N}” and the NNO of the nonstandard categories of sets as “nonstandard \mathbb{N}s”.

This is all very much worth thinking about, but this is not what classical logicians mean by “nonstandard models of the natural numbers”. So, when I’m talking to Michael Weiss, and I mention nonstandard models of the natural numbers, that’s not what I mean.

What do the traditional logicians mean? Fix your favorite version of the category of sets. Call it SetSet. There are many nonisomorphic structures in SetSet obeying the axioms in first-order classical logic called PA. There are lots of different countable ones, lots of different uncountable ones, etc. These are called “models of PA” or more loosely “models of the natural numbers” (which covers situations where we add or remove some axioms from PA to get some other closely related theory in first-order classical logic).

There’s a whole category of models of PA in our chosen category SetSet. The initial objects in this category are all uniquely isomorphic to each other. People call any one of them “the standard model”. The rest are called “nonstandard models”. Each one has a copy of the standard model sitting inside it.

I’m also very interested in varying the category we call SetSet, and Enayat’s paper Standard models of arithmetic does a bit of that too.

But it’s good to remember that while the second-order Peano axioms have a unique model up to isomorphism, the first-order system called PA does not. It can’t, even if we add finitely many more axioms expressed in first-order logic, or even a recursively enumerable infinite list of such axioms, thanks to Gödel’s incompleteness theorem.

The point is that in the first-order version, we only have mathematical induction for predicates that can be expressed in the language of arithmetic, while in the second-order version we have mathematical induction for all predicates — which is only possible because in second-order logic we can quantify over predicates.

Topos theory is a form of higher-order logic, and the definition of “natural numbers object” accomplishes something equivalent to mathematical induction for all predicates.

Posted by: John Baez on June 8, 2019 12:32 AM | Permalink | Reply to this

Re: Nonstandard Models of Arithmetic

Thanks; that’s a much-needed reminder. I used to know this stuff! I did the relevant course as an undergraduate and knew it like the back of my hand. But I’ve never needed this stuff since then, so it’s faded away in the way things do.

The statement:

There’s a whole category of models of PA in our chosen category SetSet. The initial objects in this category are all uniquely isomorphic to each other. People call any one of them “the standard model”. The rest are called “nonstandard models”.

is very crisp and clear. I hadn’t picked that up from your conversation with Michael.

Posted by: Tom Leinster on June 8, 2019 9:01 AM | Permalink | Reply to this

Re: Nonstandard Models of Arithmetic

Tom wrote:

The statement:

There’s a whole category of models of PA in our chosen category SetSet. The initial objects in this category are all uniquely isomorphic to each other. People call any one of them “the standard model”. The rest are called “nonstandard models”.

is very crisp and clear.

Thanks! Traditional model theorists never say this, perhaps because they’re not comfortable enough with category theory to say “initial object”. But they know the relevant facts:

The standard model \mathbb{N} is initial. It maps via a monomorphism to every model NN. The elements in the range of this inclusion are called “standard” elements of NN, while the rest are called “nonstandard”.

You can define a linear ordering on any model, and the standard elements always form an initial segment. The rest, NN - \mathbb{N}, is isomorphic as an ordered set to a bunch of copies of \mathbb{Z}. There are infinitely many of these copies, and between any two copies there are infinitely many more.

This is sort of easy to see: for example, if NN is nonstandard, there’s a nonstandard number N/2\lfloor N/2 \rfloor with

0<N/2<N 0\; &lt; \; \lfloor N/2 \rfloor \; &lt; \; N

and the difference between NN and N/2\lfloor N/2 \rfloor is too big to be standard, so NN and N/2\lfloor N/2 \rfloor live in different copies of \mathbb{Z}.

By the Löwenheim–Skolem theorem, the cardinality of NN can be any infinite cardinal. So, the number of these copies of \mathbb{Z} in NN can be any infinite cardinal.

All countable nonstandard models of PA have the same order type. However, they are not isomorphic in other ways! Indeed there are uncountably many nonisomorphic countable models of PA. To cook them up, just create a set SS of infinitely many sentences that are independent of PA, and each other. For any subset XSX \subseteq S, there’s a countable model of PA where the sentences in XX are valid and those in SXS - X are not.

I hadn’t picked that up from your conversation with Michael.

Sorry: we never said any of the above stuff, since we’re just chatting and we both know this stuff.

I am also starting a conversation with him on category-theoretic approaches to the foundations of first-order logic and arithmetic, and I’ll announce it here after it’s gone on for a while. You just raised a fun question: does every nonstandard model of Peano arithmetic in SetSet become a natural numbers object in some suitable topos? There’s some wiggle room in making this question precise.

Posted by: John Baez on June 8, 2019 7:29 PM | Permalink | Reply to this

Re: Nonstandard Models of Arithmetic

Analytic continuation – i.e.,

“For any analytic f:f : \mathbb{C} \to \mathbb{C} on some open region UU \subseteq \mathbb{C}, there is a unique analytic continuation f +:f^{+}: \mathbb{C} \to \mathbb{C}

will be a theorem of second-order arithmetic Z 2Z_2 using coding tricks. I suspect it’s a theorem of ACA 0\mathsf{ACA}_0, since it doesn’t seem to be impredicative, or to involve any fancy recursion. I looked this up the Reverse Mathematics literature, but couldn’t find that exact theorem.

Letting f(s)f(s) be the usual sum n=1 n s\sum_{n=1}^{\infty} n^{-s}, then this function converges when Re(s)Re(s) > 11. The 1+2+3…-theorem required is then “f +f^{+} exists and is unique and f +(1)=1/12f^{+}(-1) = -1/12”.

This (I suspect) is a theorem of ACA 0\mathsf{ACA}_0.

Posted by: Jeff Ketland on June 8, 2019 9:01 PM | Permalink | Reply to this

Re: Nonstandard Models of Arithmetic

Jeff wrote:

Analytic continuation – i.e.,

“For any analytic f:f : \mathbb{C} \to \mathbb{C} on some open region UU \subseteq \mathbb{C}, there is a unique analytic continuation f +:f^{+}: \mathbb{C} \to \mathbb{C}

will be a theorem of second-order arithmetic Z 2Z_2 using coding tricks. I suspect it’s a theorem of ACA 0\mathsf{ACA}_0, since it doesn’t seem to be impredicative, or to involve any fancy recursion.

Nice! Of course the Riemann zeta function is not analytic on \mathbb{C}, but we can probably work around that since we know exactly the domain on which it is analytic.

I’m actually hoping we need even less, though. To analytically continue the Riemann zeta function, we don’t need a general theorem on analytic continuation. I think we can write down a few explicit power series that prove we can analytically continue this function from the region Re(z)>1\mathrm{Re}(z) &gt; 1 to latexz=1latex z = -1. We should be able to show by explicit computations that these power series agree on the regions where they’re both defined. I haven’t done this, but I think I could if required. All the arguments should be on the same order of logical complexity as proving that

n=0 1n 2=π 26 \sum_{n = 0}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}

Trig identities, convergence of series where you have explicit formulas for how fast they converge, etc. I don’t think you really need anything as abstract as, say, the intermediate value theorem… which is essential when you’re trying to prove something about all continuous functions.

This is why I suggested that PA may be enough.

Posted by: John Baez on June 8, 2019 9:25 PM | Permalink | Reply to this

Re: Nonstandard Models of Arithmetic

John, I’m not sure your response hits at the crux of what Jeff was saying. Sure, there are any number of ways of showing there exists an analytic continuation of ζ(s)= n01n s\zeta(s) = \sum_{n \geq 0} \frac1{n^s} to larger regions, and one of those might be codable in PA. The question though is how you know there’s only one – and that, it seems to me, is the crucial thing one wants to know to justify the summing of the divergent series.

For what it’s worth, the nicest proof I know of that ζ(s)\zeta(s), or more precisely Ψ(s)=ζ(s)1s1\Psi(s) = \zeta(s) - \frac1{s-1}, analytically continues the region Re(s)>0Re(s) \gt 0 uses some integrals. Here is what I wrote into the nLab some while ago:

Let us bound the summands in the definition of Ψ(s)\Psi(s):

| n n+11n s1x sdx| maxnxn+1|1n s1x s| maxnxn+1maxntx|st s+1(xn)| maxntn+1|st s+11| |s|n Re(s)+1\array{ \left| \int_n^{n+1} \frac1{n^s} - \frac1{x^s}\; d x \right| & \leq & \underset{n \leq x \leq n+1}{\max}\; \left|\frac1{n^s} - \frac1{x^s} \right|\\ & \leq & \underset{n \leq x \leq n+1}{\max}\; \underset{n \leq t \leq x}{\max}\; \left| \frac{s}{t^{s+1}} \cdot (x-n)\right| \\ & \leq & \underset{n \leq t \leq n+1}{\max}\; \left|\frac{s}{t^{s+1}}\cdot 1\right| \\ & \leq & \frac{{|s|}}{n^{Re(s) + 1}} }

where the second inequality follows from the mean value theorem. Since

n=1 1n Re(s)+1\sum_{n=1}^\infty \frac1{n^{Re(s) + 1}}

converges over the region Re(s)>0Re(s) \gt 0, it follows that the series for Φ(s)\Phi(s) converges absolutely, and Ψ(s)\Psi(s) is a holomorphic function, in that region.

Over the region Re(s)>1Re(s) \gt 1, we have

Ψ(s) = n=1 1n s n n+11x sdx = n=1 1n s n=1 n n+11x sdx = ζ(s) 1 1x sdx = ζ(s)1s1\array{ \Psi(s) & = & \sum_{n=1}^\infty \frac1{n^s} - \int_n^{n+1} \frac1{x^s}\; d x \\ & = & \sum_{n=1}^\infty \frac1{n^s} - \sum_{n=1}^\infty \int_n^{n+1} \frac1{x^s}\; d x \\ & = & \zeta(s) - \int_1^\infty \frac1{x^s}\; d x \\ & = & \zeta(s) - \frac1{s-1} }

as asserted.

Granting the uniqueness clause of analytic continuation, this would allow us to extend the equation

ζ(s)=η(s)12 s\zeta(s) = \frac{\eta(s)}{1 - 2^{-s}}

into the region Re(s)>0Re(s) \gt 0 (where η(s)\eta(s) is the eta function), and then, provided with the knowledge that one can analytically continue still further to the left (integral representations, etc.), evaluate the formally divergent series η(1)\eta(-1) by any of a number of methods such as series acceleration (mentioned in the eta function article), or generalized Cesàro summations, etc. But I’m still puzzled about how you would address the uniqueness aspect.

Posted by: Todd Trimble on June 9, 2019 3:28 PM | Permalink | Reply to this

Re: Nonstandard Models of Arithmetic

Sorry for this additional comment. The nLab page I meant to link to was this, and the proof I quoted may look a little garbled because I should have added a bit more context (namely, how the proposition was stated). Let me fix that now.

Proposition: The function Ψ\Psi defined by the expression

Ψ(s)= n=1 n n+11n s1x sdx\Psi(s) = \sum_{n=1}^\infty \int_n^{n+1} \frac1{n^s} - \frac1{x^s}\; d x

converges absolutely for Re(s)>0Re(s) \gt 0, giving a holomorphic function in that region, and

Ψ(s)=ζ(s)1s1\Psi(s) = \zeta(s) - \frac1{s-1}

over the region Re(s)>1Re(s) \gt 1.

Proof: Let us bound the summands in the definition of Ψ(s)\Psi(s):

| n n+11n s1x sdx| maxnxn+1|1n s1x s| maxnxn+1maxntx|st s+1(xn)| maxntn+1|st s+11| |s|n Re(s)+1\array{ \left| \int_n^{n+1} \frac1{n^s} - \frac1{x^s}\; d x \right| & \leq & \underset{n \leq x \leq n+1}{\max}\; \left|\frac1{n^s} - \frac1{x^s} \right|\\ & \leq & \underset{n \leq x \leq n+1}{\max}\; \underset{n \leq t \leq x}{\max}\; \left| \frac{s}{t^{s+1}} \cdot (x-n)\right| \\ & \leq & \underset{n \leq t \leq n+1}{\max}\; \left|\frac{s}{t^{s+1}}\cdot 1\right| \\ & \leq & \frac{{|s|}}{n^{Re(s) + 1}} }

where the second inequality follows from the mean value theorem. Since

n=1 1n Re(s)+1\sum_{n=1}^\infty \frac1{n^{Re(s) + 1}}

converges over the region Re(s)>0Re(s) \gt 0, it follows that the series for Φ(s)\Phi(s) converges absolutely, and Ψ(s)\Psi(s) is a holomorphic function, in that region.

Over the region Re(s)>1Re(s) \gt 1, we have

Ψ(s) = n=1 1n s n n+11x sdx = n=1 1n s n=1 n n+11x sdx = ζ(s) 1 1x sdx = ζ(s)1s1\array{ \Psi(s) & = & \sum_{n=1}^\infty \frac1{n^s} - \int_n^{n+1} \frac1{x^s}\; d x \\ & = & \sum_{n=1}^\infty \frac1{n^s} - \sum_{n=1}^\infty \int_n^{n+1} \frac1{x^s}\; d x \\ & = & \zeta(s) - \int_1^\infty \frac1{x^s}\; d x \\ & = & \zeta(s) - \frac1{s-1} }

as asserted.

Posted by: Todd Trimble on June 9, 2019 4:08 PM | Permalink | Reply to this

Re: Nonstandard Models of Arithmetic

Todd wrote:

Sure, there are any number of ways of showing there exists an analytic continuation of ζ(s)= n01n s\zeta(s) = \sum_{n \geq 0} \frac1{n^s} to larger regions, and one of those might be codable in PA. The question though is how you know there’s only one – and that, it seems to me, is the crucial thing one wants to know to justify the summing of the divergent series.

Oh, I wasn’t thinking of proving any result in PA about uniqueness of analytic continuations. I was just thinking about writing down some explicit formula for the analytic continuation of ζ(s)\zeta(s) to the region around s=1s = -1, and showing that it equals 1/12-1/12 at s=1s = -1. What you want to do sounds like it would require more general results — and thus, perhaps, a stronger theory.

Posted by: John Baez on June 16, 2019 1:20 AM | Permalink | Reply to this

Re: Nonstandard Models of Arithmetic

Reverse mathematics uses systems of second-order arithmetic, so not PA in the usual first-order sense, and even then, all sorts of weird restriction on induction are made to get the systems intermediate between the base system (RCA 0RCA_0) and full second-order arithmetic (Z 2Z_2), the latter of which corresponds to a good old NNO in a topos.

Posted by: David Roberts on June 7, 2019 1:44 AM | Permalink | Reply to this

Re: Nonstandard Models of Arithmetic

I was just thinking: RCA0, which is the least powerful of the ‘big five’ systems of second-order arithmetic, is already able to state and prove very general theorems like the intermediate value theorem or the Baire category theorem for separable complete metric spaces. (I’m getting this from Wikipedia; I keep rereading this article and forgetting the details.) But an equation like ζ(1)=1/12\zeta(-1) = -1/12 is very specific, basically just the result of a calculation: we’re not quantifying over all continuous functions f:[0,1]f \colon [0,1] \to \mathbb{R}, or all separable metric spaces. To state this equation we don’t really need to talk about real numbers in general, just certain computable ones. So it might be true that this equation, or some other easier identities, can be encoded into Peano arithmetic somehow, and proved there.

Posted by: John Baez on June 7, 2019 2:03 AM | Permalink | Reply to this

Re: Nonstandard Models of Arithmetic

Well, I guess since there’s an elementary statement equivalent to RH (pdf), maybe there’s an elementary statement equivalent to the evaluation of the analytic continuation of 12 n=1 1n s12\sum_{n=1}^\infty \frac{1}{n^s} at s=1s=-1 being 1-1, via some kind of estimates etc.

Posted by: David Roberts on June 7, 2019 12:27 PM | Permalink | Reply to this

Re: Nonstandard Models of Arithmetic

Clearly, I need to make jokes more often.

Posted by: Blake Stacey on June 14, 2019 12:51 AM | Permalink | Reply to this

Re: Nonstandard Models of Arithmetic

It seems that two slightly different versions of this post were accidentally posted; see also.

Posted by: L Spice on June 6, 2019 6:10 PM | Permalink | Reply to this

Re: Nonstandard Models of Arithmetic

Yeah, John posted it a nonstandard number of times.

Posted by: Tom Leinster on June 6, 2019 10:14 PM | Permalink | Reply to this

Re: Nonstandard Models of Arithmetic

Whoops! I don’t know how that happened. Fixed.

Posted by: John Baez on June 7, 2019 12:38 AM | Permalink | Reply to this

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