On page 164 of his book **Models of Peano Arithmetic**, Kaye states Friedman's Theorem:

Let $M{\vDash}PA$ be nonstandard and countable, let $a\in M$ and let $n\in {\mathbb N}$. Then there is a proper initial segment $I\subseteq_c M$ ($\subseteq_c$ means "cofinal in") containing $a$ such that $I\cong M$ and $I<_{\Sigma_n} M$.

However, on the next page he writes "Nor can we expect in general to get initial segments $I$ with $M\cong I < M$ and $M\neq I$, i.e., elementary for all formulas. For example if $M=K_T$ (where $T\neq Th({\mathbb N})$ is a complete extension of PA) then $M$ has no proper elementary substructures, and so certainly has no proper elementary initial segments!"

I am confused. If two models in the same language are isomorphic, are they not elementarily equivalent? Of course the converse need not be true.

An isomorphism of models is a bijective homomorphism of the language's algebraic portion (constants and functions) which preserves **and reflects** all relations of the language (Hodges, **Model Theory**, p5). Therefore by induction on the structure of any ${\mathcal L}_{\omega,\omega}$ formula, the isomorphism will both preserve and reflect it. So an isomorphism preserves **all** formulas (Hodges, Theorem 2.4.3(c)).

If the proper initial segment is isomorphic as a model to the entire model, how could any first-order sentence possibly be true in one and not in the other?

Thanks,

- a

initialorend. (The typo is in the OP's question, not in Kaye's book). $\endgroup$ – jeq Dec 3 '15 at 15:41