Showing models of T are isomorphic to disjoint unions

Question

Let L = {S,0}, where S is a unary function symbol and 0 is a constant symbol. Let T be the L-theory consisting of the following L-sentences:

• { axioms of equality },
• ∀∀(()=()→=),
• ∀∃(()=),
• ∀¬(=()),
• ∀¬(=(())),
• ∀¬(=((()))),
• and countably infinite number of similar axioms each stating that applications of don't cycle.

Let N = ($\mathbb N$,0,S) and Z = ($\mathbb Z,$ 0, S) where S is the successor function.

I am asked to prove that every model of T is isomorphic to a model which is the disjoint union of one copy of N and some (possibly infinite) number of copies of Z. For a specific example clearly M = ($\mathbb R,$0,S) satisfies T but I don't see what its isomorphism would be. Any help would be appreciated.

Answer 1

I assume that the third axiom writes $\forall x (\exists y S(y) = x \leftrightarrow x \neq 0)$ (otherwise $\mathbb{N}$ isn't a model).

Here's how one can prove the claim :

Let $M \models T$. For each $x \in M$, define the closure of $x$ as follow :

$$ cl(x) := \{y \in M \ \big| \ \exists n \in \mathbb{N}, S^n(x) = y \vee S^n(y) = x \}$$

The relation $x \operatorname{E} y : \Leftrightarrow cl(x) = cl(y)$ is an equivalence relation. Furthermore, we have $x \operatorname{E} y \Leftrightarrow y \in cl(x)$. There are two possibilities for each $x \in M$ :

• either $0 \in cl(x)$, and then $cl(x) \cong \mathbb{N}$
• or $0 \notin cl(x)$ and then $cl(x) \cong \mathbb{Z}$

The case $M = (\mathbb{R} \setminus \{-n \ \big| \ n \in \mathbb{N}^*\}, 0, S)$ (I had to change this a bit so that it fits the third axiom) :

Here, you have one copy of $\mathbb{N}$ plus $\beth_1$ copies of $\mathbb{Z}$ : one for each $x$ in the interval $]0;1[$.