Non-standard Models of Arithmetic 12

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JB: It’s been a long time since Part 11, so let me remind myself what we’re talking about in Enayat’s paper Standard models of arithmetic.

We’ve got a theory T that’s a recursively axiomatizable extension of ZF. We can define the ‘standard model’ of PA in any model of T, and we call this a ‘T-standard model’ of PA. Then, we let PA^T to be all the closed formulas in the language of Peano arithmetic that hold in all T-standard models.

This is what Enayat wants to study: the stuff about arithmetic that’s true in all T-standard models of the natural numbers. So what does he do first?

MW: First Enayat offers a brief interlude, chatting about the strength of PA^T for various T’s. PA^T is stronger than PA, because it can prove Con(PA) and PA can’t. That’s because T extends ZF, and ZF can prove that its is a model of PA. ZF would be a pretty poor excuse for a set theory if it couldn’t prove that!

Next: PA^ZF = PA^ZFC+GCH, where GCH is the generalized continuum hypothesis. Instead of looking at ZFC+GCH, let’s consider ZFL, i.e., ZF+“every set belongs to Gödel’s constructible universe L”. As Gödel showed, ZFL implies ZFC+GCH, and also Con(ZF) implies Con(ZFL); in fact any model M of ZF contains a so-called inner model L^M of ZFL. Now, ordinals are absolute between transitive models of ZF; not only that, but ω is absolute, and the ordinal operations are absolute. (“We don’t have to search all over the kingdom to do arithmetic with finite ordinals.”) So M and L^M have the same “” as their “standard model of PA”.

Since M and L^M have the same “standard model of PA”, the same arithmetic statements hold in both of them. To see if some φ belongs to PA^ZF, we have to verify the truth of in all models of ZF. But in the process, we also verify that truth in all inner models L^M of ZFL. And when we’ve finished that task, we’ve actually verified that truth in all models of ZFL, just because every model of ZFL is the inner model of at least one model of ZF—namely itself! So PA^ZF = PA^ZFL.

Contrast this with ZFI, i.e., ZF+“there is an inaccessible cardinal”. PA^ZFI is strictly stronger than PA^ZF. Here’s why: ZFI proves Con(ZF), since if we gather all the sets of rank below the first inaccessible cardinal and toss them in a bag, we’ve got a model of ZF! ZF does not prove Con(ZF), by Gödel’s second incompeteness theorem. So Con(ZF) is our example of a statement in L(PA) that PA^ZFI delivers and PA^ZF doesn’t.

JB: By the way, just as a stupid little side note, you’re assuming here that ZF is consistent. But I’m willing to keep that as a background assumption in all these deliberations. (Maybe you already mentioned this and I forgot!)

MW: Yup. Kind of hard to talk about the omegas of models of ZF, if there are no models of ZF!

You might wonder what’s the chief difference between the two cases, ZFI and ZFL. Models of ZF contain inner models of ZFL, and models of ZFI contain inner models of ZF. Ah, but there are models of ZF not contained in any model of ZFI (at least if ZF is consistent). Namely, if ZF is consistent, then ZF+¬Con(ZF) is also consistent, as we’ve just seen. So it has a model. I leave it as an exercise that a model of ZF+¬Con(ZF) can’t possibly be contained in a model of ZFI.

JB: Nice!

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