**MW:** Indicators: we don’t *need* to discuss these, to prove the Paris-Harrington theorem. But I think they offer valuable insight.

Let’s start with what an indicator indicates. Suppose *N* is a nonstandard model of PA, and we want to know if *N* has an initial segment *I* with some property P, that also “lies between” two given elements *a* and *b*. “Lies between” means that *a*∈*I*<*b*, and “*I*<*b*” means that *b* is a strict upper bound for *I*.

The intuition: the larger the “gap” between *a* and *b*, the more initial segments we have with *a*∈*I*<*b*. That makes it easier to find one enjoying property P. OK: let’s say a function *Y*(*a*,*b*) is an *indicator* for P, if *Y*(*a*,*b*) is nonstandard when and only when there exists an initial segment *I* with *a*∈*I*<*b* possessing property P. (Standing assumption: if *b<**a*, then *Y*(*a*,*b*)=0. In mathspeak, by the way, if you possess something, you automatically enjoy it. Or at least this is true for mathematical objects.)

Example: *Y*(*a*,*b*)=*b−a* is an indicator for “is a cut”, i.e., is closed under successor. Clearly we have a cut *I* with *a*∈*I*<*b* if and only if *a*+*n*<*b* for all standard *n*.

(In keeping with the standing assumption, it’s implicit that *Y*(*a*,*b*)=0 when *b*<*a*; strictly speaking, the indicator is “if *b<a* then 0 else *b−a”*. This goes without saying in the future.)

Second example: *Y*(*a*,*b*)=⌊log* _{a} b*⌋ is an indicator for “is an initial substructure”, i.e., is closed under plus and times, provided

*a*>1. Clearly we have an initial substructure

*I*with

*a*∈

*I*<

*b*if and only if

*a*<

^{n}*b*for all standard

*n*. In other words, iff ⌊log

*⌋≥*

_{a}b*n*for all standard

*n*. (To eliminate the proviso

*a*>1, use

*Y*(

*a*,

*b*)=⌊log

_{a+2}

*b*⌋ instead.)

So *Y*(*a*,*b*) measures the “size of the gap”. If *Y*(*a*,*b*) is highly “compressive”, then *Y*(*a*,*b*) being nonstandard indicates a “large” gap. Certainly ⌊log* _{a} b*⌋ is more “compressive” than

*b−a*.

Incidentally, people usually write *Y*(*a*,*b*)>ℕ to mean *Y*(*a*,*b*) is nonstandard.

**BS:** That is all very interesting, and understandable.

**MW:** ‘Well-behaved’ indicators (as Kaye calls them) obey some other conditions; we’ll talk about them later. Right now, to get a feel for the indicator concept, let’s temporarily set *a*=0, writing *Y*(*b*).

In this special case, *Y*(*b*)>ℕ means that “there’s room enough” below *b*, to fit an initial segment I with property P. Conversely, *Y*(*b*)∈ℕ means there *isn’t* enough room.

The growth rate of *Y* matters. If *Y* is slow-growing (like *Y*(*b*)=⌊log_{2}* b*⌋, compared to *Y*(*b*)=*b*), you’ll need to make *b* pretty big to get enough “headroom”. The slower *Y* grows, the bigger *b* needs to be.

**BS:** But isn’t ⌊log_{2}* b*⌋ nonstandard iff *b* is nonstandard?

**MW:** Oops! You’re right. Let me see where I went astray…

I wanted to emphasize, for now, how *Y*(*a*,*b*) depends on *b*. I forgot, though, that there’s *always* an initial segment *I *with 0∈*I*<*b* and with property P, for any nonstandard *b* and for the indicators we’ll be looking at. Namely ℕ! So *Y*(0,*b*) will be nonstandard for *b*>ℕ.

Let’s just assume *a* is some fixed value; I’ll write *Y*(*b*) sometimes for *Y*(*a*,*b*).

**BS:** Ok—so far, so good.

**MW:** How far out do you have to go for *Y* to “reach or crash through” a ceiling of height *h*? In other words, let’s define

*F*(*a*,*h*) = μ*b *(*Y*(*a*,*b*)*≥h*),

sometimes writing just *F*(*h*). Here μ*b* φ(*b*) stands for the least *b* such that φ(*b*) holds, and is undefined if there is no such *b*. (Standard notation in recursion theory.)

**BS:** So the slower *Y* grows, the faster the corresponding *F* grows?

I’ve broken the vertical axis into the standard and nonstandard ranges. That’s because we’re asking “Is *Y*(*b*)>ℕ?” The horizontal axis goes from *a *to *b*, since *Y*(*a*,*b*)=0 for *b*<*a* (our standing assumption).

The blue indicator, *Y*, is faster-growing, while the red *Y*′ is slower-growing. As a result, *Y*(*b*)>ℕ while *Y*′(*b*)∈ℕ. For comparison’s sake, I’ve shown a ceiling at a *standard* height *n*.

**BS:** This is all understandable, but I’m not sure where you’re going with it…

**MW:** Let’s go through some examples of kinds of initial segments we might be interested in, and figure out how to define indicators for each one. How about an *I* which is a cut, but isn’t an initial substructure, so we can compare the indicators *b−a* and ⌊log* _{a} b*⌋? Remember we want

*a*∈

*I*<

*b*.

**BS:** I think we can just include every element up to *a*+*n* for arbitrary standard *n*. I.e., include all of *a’*s ℤ-copy and below. This is a cut, but not closed under either plus or times. I think this corresponds to *Y*(*a*,*b*)=*b−a*.

**MW:** Right. A little more formally, *I* is the downward closure of {*a*+*n* : *n*∈ℕ}.

**BS:** Let me see if I can come up with an initial segment closed under plus but not times, and containing *a*… I *think* we just want to include everything reachable by *an* for standard *n*. That is, the downward closure of {*a**n* : *n*∈ℕ}.

**MW:** Good. Can you think of an indicator for this property, “closed under plus”?

**BS:** Let’s see… I think ⌊*b/a*⌋ works. (I guessed this by following this pattern: if you want *I* to be closed under some operation, it only has to be closed under iterating that operation a standard number of times, and then the indicator can just compute that number of iterations, by inverting that operation. I’m not explaining this very clearly.)

**MW:** That’s right. Your explanation seems clear enough to me. So we have three indicators: *b−a*, ⌊*b/a*⌋, and ⌊log* _{a} b*⌋. What are the corresponding

*F*’s?

**BS:** *F* is just “solves for *b*” in *Y*(*a*,*b*) = *h*, so (ignoring the effects of inverting ⌊⌋): *a*+*h*, *ah*, *a ^{h}*. They look a lot more straightforward in that form!

**MW:** Yup. OK, now for the indicators we really care about. They’re for the property “is a model of PA”. Given the progression “*a*+*h*, *ah*, *a ^{h}*…”, we expect

*F*(

*h*) to grow even faster. I’ll talk more about that in a moment. But first, it’s time to list the conditions Kaye gives for a

*well-behaved*indicator:

a. *Y*(*a*,*b*)=*h* is a Σ_{1} formula with only the free variables shown.

b. PA ⊢ ∀*a*∀*b*∃!*h **Y*(*a*,*b*)=*h.*

c. PA ⊢ ∀*a*∀*b** **Y*(*a*,*b*)≤*b.*

d. PA ⊢ ∀*a*∀*b*∀*c*∀*d* (*a*≤*b* ∧ *c*≤*d* → *Y*(*b*,*c*)≤*Y*(*a*,*d*)).

(b) says that the formula for *Y*(*a*,*b*) defines a function. The notation makes this look inevitable, but if you wrote out ⌊log* _{a}b*⌋=

*h*explicitly in L(PA), you’d have a pretty elaborate formula; proving (b) for it would take some doing.

**BS:** I understand. I suspect that could even be a Δ_{0} formula (not that I could personally construct it), but I know that before long, we’ll need the unbounded quantifier in Σ_{1} . But I’m willing to grant you that that will be sufficient for any function we’d consider computable (not that I can personally prove that, either).

**MW:** In ℕ, (a) implies that *Y*(*a*,*b*) is recursive. In a nonstandard model *N*, the meaning of ‘recursive’ gets a little fuzzy, so Kaye adopts (a) and (b) together as the definition of *provably recursive*.

**BS:** Don’t you need both Σ_{1} and Π_{1} (i.e. Δ_{1}) for that, with Σ_{1} alone implying only r.e?

**MW:** For relations, that’s exactly right. But functions are special. ¬*Y*(*a*,*b*)=*h* is equivalent to ∃*k*(*k≠h *∧ *Y*(*a*,*b*)=*k*), so if *Y*(*a*,*b*)=*h* is Σ_{1} then its negation is also Σ_{1}. In other words, if *Y*(*a*,*b*)=*h* is Σ_{1}, then it’s Δ_{1}.

(c) is a technicality. Finally, (d) says that *Y*(*a*,*b*) gets bigger (or at least not smaller) if we move *a* to the left and *b* to the right. As a trivial consequence, *Y*(*a*,*b*) is monotonically increasing as a function of *b* when we hold *a* fixed. So our picture isn’t misleading.

If *Y*(*b*) is monotonically increasing, then *F*(*h*) = μ*b*(*Y*(*b*)*≥h*) is strictly increasing (so far as it’s defined). Also we have (restoring the implicit argument *a*)

*F*(*a*,*h*)≤*b* ⇔ *h**≤**Y*(*a*,*b*).

**BS:** Ok, this is all making perfect sense.

**MW:** Here are two theorems about well-behaved indicators for “is a model of PA”:

- For each
*n*∈ℕ, PA ⊢ ∀*a*∃!*b**F*(*a*,*n*)=*b.* - Not [ PA ⊢ ∀
*h*∀*a*∃!*b**F*(*a*,*h*)=*b*].

Loosely speaking, *F*(*a*,*h*) increases so fast (as a function of *h*) that PA can’t prove it’s defined for all *h*. In ℕ, *F* is recursive but not *provably* recursive.

**BS:** Ok—it’s at least plausible to me that a property like that is very related to what you’d need to prove your cut *I* “extends far enough” to be a model of PA—since certainly it has to be closed under all *provably* recursive functions.

**MW:** Yes. That’s the key idea to proving (2). Or rather the contrapositive. If we can find a model of PA with elements *a* and *h* such that *F*(*a*,*h*) is undefined, then we can’t have PA⊢∀*h*∀*a*∃!*b**F*(*a*,*h*)=*b*.

So suppose *N* is a nonstandard model of PA, and *a* is a nonstandard element of *N*. There’s one trick: use *a* for *h*. That is, we’ll look at *F*(*a*,*a*). It’s a diagonal argument!

If *F*(*a*,*a*) is undefined in *N*, we’re done. If not, let *F*(*a*,*a*)=*b*. Do you want to see if you can finish the argument?

**BS:** Let’s see: we have *Y*(*a*,*b*)≥*a*, because *F*(*a*,*a*)=*b*. But *a* is nonstandard, so *Y*(*a*,*b*) is nonstandard, so that is saying there is a cut *I*, which is a model of PA, which includes *a* and not *b*. Now, reasoning within that model *I*, *F*(*a*,*a*) is undefined, since *b* is not in that model. (I am assuming *F* is “sufficiently absolute” for this reasoning to work—since *Y* is Δ_{1}, I am guessing so is *F*, since the same formula defines them, except for the technicality of which variable is the output which needs to be “rounded to an integer”.) Anyway, since *F*(*a*,*a*) is undefined in that model *I*, PA can’t prove it’s defined in every model, which completes the proof of (2).

**MW:** Good! Let’s look at the absoluteness issue. If we had *F*(*a*,*a*)=*c* for some *c*∈*I*, then we’d have *Y*(*a*,*c*)≥*a* for some *c*∈*I.* That’s *inside I*; that is, *I*⊧*Y(a,c)≥a.* But *Y*(*a*,*c*)*≥a* is a Σ_{1} formula (see below), and these are persistent upwards for end extensions. So we’d have *N*⊧*Y*(*a*,*c*)*≥a* with *c*∈*I*. Which means that *b* wouldn’t be the *least x* with *N*⊧*Y*(*a*,*x*)*≥a*, contradicting *N*⊧*F*(*a*,*a*)=*b*. So indeed, *F*(*a*,*a*) is undefined in *I*.

*Y*(*a*,*c*)*≥a* is shorthand for ∃*h*(*Y*(*a*,*c*)*=h* ∧ *h*≥*a*). And *Y*(*a*,*c*)*=h* is Σ_{1} (in fact Δ_{1}, but we don’t need that here). So *Y*(*a*,*c*)*≥a* is Σ_{1}.

**BS:** I notice I didn’t explicitly use conditions (c) or (d). Are they used in the technicality I glossed over, about how the formulas for *F* and *Y* are related?

**MW:** Well, (d) gave us the implication *F*(*a*,*h*)≤*b* ⇒ *Y*(*a*,*b*)≥*h*, which you used. I don’t think we’ll use (c) for anything, but I wanted to list all of Kaye’s conditions.

Now for the role indicators play in the Paris-Harrington theorem. The proof sketched in post 9 featured this setup, reprised at the end of post 21: a nonstandard model *N* of PA, and an initial submodel *M*, with the Paris-Harrington principle (PHP) holding in *N* but not in *M*. To state the PHP, we start with a combinatorial object sitting on top of an interval [*a*,*b*]. I’ll call the object a ‘colored hypergraph’, because that’s what it is, but we don’t need the details right now. We want to know if there’s a subset *H* of [*a*,*b*] of size *h*, with a certain property involving the colored hypergraph. The PHP says that for all *a* and *h*, there is *b* such that the desired *H* exists. In other words, just having *b* big enough will guarantee us the *H*. In fact, *b* will typically have to be *much much* bigger than *a*. So the PHP takes the form,

∀*a*∀*h* ∃*b* φ(*a*,*b*,*h*)

If *F*(*a*,*h*) = μ*b* φ(*a*,*b*,*h*), then the PHP says that *F*(*a*,*h*) is well-defined for all *a* and *h*. Where there’s an *F*, there’s a *Y*: one can define a well-behaved indicator *Y*_{PH} for the PHP.

Here’s how the PHP shows up in the proof of the Paris-Harrington theorem. We need to find a nonstandard model where the PHP fails. Let *N* be a nonstandard model of PA. If *N* does not satisfy the PHP, then done. Otherwise, apply the PHP with a nonstandard *h*, getting an *H* of nonstandard size. Extract the set *B* of diagonal indiscernibles from this *H*. (How? Topic for later.) Define *M* as the downward closure of *B*, and we’re off to the races: the properties enjoyed by *B* ensure that *M* is a model of PA.

How do we show that *M* doesn’t satisfy the PHP? Well, both *a* and *h* turn out to belong to *M*, but *b* doesn’t. So you see it’s just like the proof of Theorem 2: *F*(*a*,*h*) is defined in *N* but not in *M*, and *M* is the initial submodel of *N* whose existence is guaranteed by the indicator.

**BS:** I see—that definitely helps motivate what we’ve been doing.

**MW:** We don’t need Theorem 1, but do you want to take a crack at it? I’ll start you off. By the completeness theorem, it’s enough to show that for all *n*∈ℕ and all models *M* of PA, *M*⊧∀*a*∃!*b**F*(*a*,*n*)=*b.* Rephrase as an assertion about *Y*. Now we quote the MacDowell-Specker theorem: every model of PA has a proper elementary end extension, so let *M*⊂_{e}*N*.

**BS:** Ok, let’s see. Rephrasing, *M*⊧∀*a*∃*b*(*Y(a,b*)*≥n*). Suppose for contradiction that, for some *a* we’ll hold fixed, there is no such *b*. Now you wouldn’t have mentioned *N* unless I should use it, but all I know about it is it has elements bigger than anything in *M—*and it has an initial segment *M* which is a model of PA. So pick an arbitrary *c*∈*N*\*M*, and consider *Y*(*a,c*). Since there is a model of PA containing *a* and not *c*, namely *M*, we better have *Y*(*a,c*) > ℕ. ** **

Now let’s visualize the situation, to help find the contradiction. As a function of increasing *x*, *Y*(*a,x*) gets stuck at some maximal standard value (call it *h*) for an end-segment of *M*, but eventually reaches a nonstandard value for *x* = *c*. But consider the set of *x* for which *Y*(*a*,*x*)* = h*. That includes all elements of *M* from some point onwards. By overspill relative to the cut *M*, this set has to extend beyond the end of *M*, say to some *d* in *N*\*M*. But then *Y*(*a*,*d*) = *h*, which is standard, contradicting the existence of *M* as a model of PA between *a* and *d* (i.e., *a*∈*M*<*d*). QED.

(I guess I never did use *c*, only *d*. But I’ll leave it in, since it shows how I figured this out.)

**MW:** Very nice! I didn’t realize you could use overspill to prove this. The argument I had in mind is direct, and goes like this: since *N*⊧*Y(a,c*)*≥n* (because *Y*(*a,c*)>ℕ), *N*⊧∃*x*(*Y(a,x*)*≥n*). Now *M* is an elementary submodel of *N*, so *M*⊧∃*x*(*Y(a,x*)*≥n*), which is what we wanted to prove.

**BS:** That is indeed much more concise, and it uses the fact that *N* is an *elementary* extension—which I never used, or thought of trying to use. I need to get in the habit of paying more attention to that kind of assumption!

**MW:** Before we go, a couple more remarks. I’ve stated Theorems 1 and 2 for indicators for PA. Everything still works if you replace PA with an extension theory, say *T*. But if *T* is weaker than PA, no such luck. However, Kaye has versions of the theorems for any theory *T *in the language of arithmetic (§14.1). Things get considerably more complicated.

Wong’s Consistency notes (lecture 22) treat a special case with a weaker theory, the one known as IΔ_{0}(exp). This allows for induction only over Δ_{0} predicates, and has additional axioms saying (basically) that exponentials exist and have the expected properties. Here, the function *F* is an iterated exponential. Wong uses this case as a paradigm for the Paris-Harrington Theorem. Very nicely done; well worth a look.

**BS:** Thanks, I will definitely check that out!