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Consider the theory $\text{Th}(\mathbb{N}, S, 0)$, which we know to be axiomatized by the following axioms:

  1. $\forall x (S(x) \neq 0)$
  2. $\forall x \forall y (S(x) = S(y) \rightarrow x=y)$
  3. $\forall x (x \neq 0 \rightarrow \exists y (x=S(y))$

plus the axiom schema:

  1. $\forall x (S^n(x) \neq x)$ for each $n \in \mathbb{N}$ (where $S^n(x)$ is the aplication of $S$ $n$ times to $x$).

Now, it is well known that models of these theory are exactly those that have the form $\mathbb{N} + \lambda \cdot \mathbb{Z}$, where $\lambda$ is a finite or infinite cardinal. My question is about how to prove this. It is not difficult to show that any structure of this form is a model for the theory, but what about the converse, i.e. that any model of the theory has this form?

Now, it is clear that any model of this theory will consist of an initial segment which is a copy of $\mathbb{N}$. My thought was then to define an equivalence relation $x \equiv y$ iff there is $n$ such that $S^n(x)=y$, and then show that the equivalence classes will all be of the form $\mathbb{N}$ or $\mathbb{Z}$. Is this in the right direction?

Also, as a side note, just to check, but the $1$-types of this theory are all either isolated by a formula of the form $S^n(0)=x$, or else the unique non-isolated type $\{S^n(0) \neq x \; | \; n \in \mathbb{N}\}$, correct?

Vivaan Daga
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Nagase
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    Your proposed equivalence relation has the right idea but doesn't actually work: it's not symmetric. Once you fix that, though, you've got the right idea. – Noah Schweber Mar 08 '22 at 17:58
  • @NoahSchweber - Right, of course! You mean I should put in the right hand side "there is $n$ such that either $S^n(x) = y$ or $S^n(y) = x$, right? Can you post your comment as an answer? – Nagase Mar 08 '22 at 18:05
  • You say: models of these theory are exactly those that have the form ℕ+⋅ℤ. What about models of the form ℕ+ℚ⋅ℤ ? – Primo Petri Mar 08 '22 at 19:13
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    @PrimoPetri - Since $|\mathbb{Q}|=\aleph_0$, they are of type $\mathbb{N} + \aleph_0 \cdot \mathbb{Z}$. I don't think successor arithmetic can distinguish between the order type of the copies of $\mathbb{Z}$ (that is, the way the $\mathbb{Z}$s are ordered among them). – Nagase Mar 08 '22 at 20:07
  • Isn't (4) redundant? Isn't that "looping back" disallowed by (1) and (2). – Dan Christensen Mar 09 '22 at 13:50
  • @DanChristensen - No, they are not redundant. Consider a model $\mathcal{M}$ consisting of $\mathbb{N}$ and a single element, say $a$, not in $\mathbb{N}$. Define $S$ as the successor as usual over $\mathbb{N}$ and set $S(a)=a$. Then $\mathcal{M}$ satisfies 1-3, but does not satisfy 4 for $n=1$. – Nagase Mar 09 '22 at 15:08
  • Thanks. Why not then just use the more natural principle of induction and avoid constructing the $S^n(x)$ function? – Dan Christensen Mar 09 '22 at 17:44
  • @DanChristensen - You can use an induction schema, similar to what you have with first-order PA (but restricted to formulas containing only the successor function, of course). This will give another infinite set of axioms for the theory. But I find it easier to analyze the models of the theory using this particular schema. – Nagase Mar 09 '22 at 17:45
  • I think you would need induction to prove the existence of the $S^n(x)$ function. Just to clarify, is it the case that $S^0(x)=x, ~S^1(x) = S(x), ~S^2(x)=S(S(x))$, and so on? Then $S^n(x)=x+n$. – Dan Christensen Mar 09 '22 at 18:25
  • @DanChristensen - There is no "$S^n$ function. The official language contains only $S$. I only use $S^n$ as an abbreviation for applying the function $n$ times to a term. A1$_1$ is $\forall x (S(x) \neq x)$, A2$_2$ is $\forall x (S(S(x)) \neq x)$, etc., so that none of the axioms actually contains $S^n$. – Nagase Mar 09 '22 at 20:59
  • In 2nd order PA (in the language of set theory), you can formally prove the existence of such a function starting from Peano's Axioms. I understand you can't do that in 1st order PA. – Dan Christensen Mar 10 '22 at 00:38

1 Answers1

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You've got the right idea, but it needs a tweak: the "finite distance" relation is what you want, but it's not what you've written (what you've written isn't symmetric). Set $x\sim y$ iff either there is some finite $n$ such that $S^n(x)=y$ or there is some finite $n$ such that $S^n(y)=x$ (note that this is just the usual "graph metric").

Now the key observations are that in any model of successor arithmetic there is exactly one element without a predecessor and the successor function is injective. These two facts quickly imply that any model of successor arithmetic consists of exactly one $\sim$-class of "type $\mathbb{N}$" and all other $\sim$-classes (if any) have "type $\mathbb{Z}$."

(It may help to think about what successor arithmetic proves about the "distance one" relation $x=S(y)\vee y=S(x)$; in my experience this is a bit easier to conceptualize, although I'm not sure why.)

Noah Schweber
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