Can two different models of arithmetic have non-comparable views of peano arithmetic? 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 $V(M') \nsubseteq V(M)$.

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.
