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Hi,

I was wondering how much (if anything) $\mathcal{L}_{PA}$ can express about individual nonstandard elements in a nonstandard model of PA. For instance, presumably it can say that each has $k$-many predecessors, for each $k\in\mathbb{N}$. But:

(a) I can't see that there is any way that the type of one element in a $\mathbb{Z}$-chain differs from the type of any other in that same chain. Is this correct?

(b) Are the types of elements in separate $\mathbb{Z}$-chains also identical? I mean, clearly they won't be the same as those of elements in the initial segment $\mathbb{N}$ - these will have a finite number of predecessors - but in two of the additional chains?

To me, it looks like these questions are straightforwardly true but I could be wrong. Many thanks,

Kate

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Since the nonstandard numbers believe that every other number is even and every other number is odd, a fact that is expressible in the language you mention, it follows that the types are not the same for every two elements in a $Z$-chain. In fact, more is true: any two nonstandard natural numbers in a common $Z$-chain have distinct types, for if they differ by a finite number $n$, then they will have different residue modulo $n+1$, making their types different.

If one restricts attention only to the order, however, then any two elements in a nonstandard $Z$-chain have the same type, since there are order-automorphisms that shift within this $Z$-chain. And in a countable nonstandard model, the $Z$-chains are ordered like the rationals, and so the order automorphism group acts transitively on these elements. It follows that all the nonstandard elements have the same type in the language containing only the order.

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  • $\begingroup$ Thanks Joel, that's great. So, will we have something like the following: Let a, b be the nonstandard numbers differing by finite n. a=k(n+1) + m b=k(n+1) + m+n \Exists y[yn+1 + m = x] in tp(a) \Exists y[yn+1 + m-1 = x] in tp(a) n is finite, so definible. And so are the m, m-1, right? Thanks again $\endgroup$
    – Kate Hodesdon
    Commented Jul 1, 2010 at 10:27
  • $\begingroup$ Sorry, thats formatted horribly. I meant the lines to have linebreaks between - I was just checking that the residues are definable. $\endgroup$
    – Kate Hodesdon
    Commented Jul 1, 2010 at 10:41
  • $\begingroup$ Yes, that's the right idea. Residue mod n is definable in PA. The m here is standard finite, sine it is less than n+1, and hence definable. $\endgroup$
    – Joel David Hamkins
    Commented Jul 1, 2010 at 10:54
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To complement Joel's answer, two different $\mathbb{Z}$-chains can realize different sets of types. For example, some $Z$-chains contain prime numbers but some don't. The fact that some $\mathbb{Z}$-chains contain primes follows from the infinitude of primes. To see that some $\mathbb{Z}$-chains don't contain primes, pick a nonstandard $N$ and consider a $\mathbb{Z}$-chain contained in the infinite interval $[N!+2,N!+N]$ — since all positive integers up to $N$ divide $N!$, every element of that $\mathbb{Z}$-chain has factor which is no larger than $N$.

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    $\begingroup$ In fact, there are nonstandard models of PA in which every object is definable without parameters (start with a single nonstandard definable object, and take a Skolem Hull using the definable Skolem functions). In this case, every object has a distinct type. $\endgroup$
    – Joel David Hamkins
    Commented Jun 30, 2010 at 18:08
  • $\begingroup$ @Francois---you learn something new every day! Thanks for this answer. $\endgroup$
    – Kevin Buzzard
    Commented Jun 30, 2010 at 18:44

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