Can two different models of arithmetic have non-comparable views of peano arithmetic?

Question

For a given model of arithmetic $M$, we say that models view of peano arithmetic, $V(M)$, is $\{\phi : M \models (PA \vdash \phi) \}$.

For example the view of the standard model is $\{\phi : PA \vdash \phi \}$. On the other hand, for any model $X$ of $PA + \lnot Con(PA)$, $V(X)$ is the set of all statements in arithmetic. For a model $Y$, of $PA + Con(PA) + \lnot Con(ZFC)$, $(ZFC \vdash 0=1) \in V(Y)$ but $0=1 \notin V(Y)$. So $V(\text{standard model}) \subset V(Y) \subset V(X)$.

Can we have two model, $M$ and $M'$, of arithmetic (which are models of peano arithmetic) such that $V(M) \nsubseteq V(M')$ and $V(M') \nsubseteq V(M)$.

Answer 1

First of all, I believe you would be interested in a paper by Kikuchi and Kurahashi, "Illusory Models of Peano Arithmetic". They explore related questions in detail in that paper. In their notation, what you call "V(M)" they call $\mathrm{Thm}_{\mathsf{PA}}(M)$.

Let's first note that if $M \models \lnot \mathrm{Con}(\mathsf{PA})$, then $\mathrm{Thm}_{\mathsf{PA}}(M)$ will contain all sentences in the language. They refer to such models as "insane", and models of $\mathrm{Con}(\mathsf{PA})$ as "sane". So the real question here is whether, given sane models $M$ and $N$, the sets $\mathrm{Thm}_{\mathsf{PA}}(M)$ and $\mathrm{Thm}_{\mathsf{PA}}(N)$ are necessarily linearly ordered. The answer is no; in fact the main result regarding this in their paper actually shows that the family $\mathcal{T} = \{ \mathrm{Thm}_{\mathsf{PA}}(M) : M$ is sane $ \}$ has cardinality $2^{\aleph_0}$.

The main crux behind getting these independence results is that $\mathrm{Thm}_{\mathsf{PA}}(M)$ is determined entirely by the $\Sigma_1$ theory of $M$. Further, given any recursively enumerable theory $T$, there is a $\Sigma_1$ statement $\phi$ such that $T + \phi$ and $T + \lnot \phi$ are both consistent (this is a usual incompleteness argument, though they use a stronger version in order to get that $|\mathcal{T}| = 2^{\aleph_0}$). so let $T$ be the theory $\mathsf{PA} + \mathrm{Con}(\mathsf{PA})$, and then you can find two models of PA which have different $\Sigma_1$ theories, and therefore have incompatible "theorems of PA".

I strongly recommend reading that paper as they ask and answer many related questions that you might be interested in.

Answer 2

First, note that if $\phi \equiv \exists x\psi(x)$ is $\Sigma_1$ and $M$ is a model of $PA$, then $M \models \phi$ implies $\phi \in V(M)$. This is because if $M \models \psi(n)$, then $M$ is able to write $n=1+\cdots+1$ and then prove $PA \vdash \psi(1+\cdots+1)$.

Assume PA is consistent, and let $A,B$ be a recursively inseparable pair of disjoint $\Sigma_1$ subsets of $\mathbb{N}$. Build this disjointness explicitly into their defining formulas, so that you can prove they're disjoint. We claim that for some $n$, there is a pair of models $M_1 \models( n \in A)$ and $M_2 \models (n \in B)$. Since the sets are $\Sigma_1$ and provably disjoint, we get $n\in A$ in $V(M_1)$ but not $V(M_2)$, and the other way round.

For a contradiction, assume that for all $n$ there is no such pair. In particular, either there are no models with $n \in A$, or there are none with $n \in B$. By the Completeness Theorem, the corresponding formula is a theorem of PA. Define a recursive set $C$ by: If $PA \vdash n \not \in B$, put $n$ into $C$; and if $PA \vdash n \not \in A$, put $n$ into the complement $\bar C$. But then $C$ is a recursive separator for $A$ and $B$, contradicting the choice of $A$ and $B$. The proof is over.