Theorem 1.53 (3) in page 227 of Hajek and Pudlak's book, Metamathematics of First-Order Arithmetic, says:


Theorem. If $M$ is a countable model of $I\Delta\_{0}$ such that $M$ has a proper elementary end extension, then $M\models PA$. 


Is the above theorem still true if we drop the countability assumption of $M$?



Uncountable model of bounded arithmetic with an elementary end extension
The countability assumption cannot matter. The reason is that any uncountable model $M$ is countable in a forcing extension of the set-theoretic universe. If $M$ has a proper elementary end-extension in the original universe, then this end-extension still exists in the forcing extension. So we may apply the theorem as you have stated it in the forcing extension in order to deduce that $M\models\text{PA}$. But the question whether $M$ is a model of PA or not is absolute between the set-theoretic universe $V$ and its forcing extensions $V[G]$, and so $M\models\text{PA}$ in $V$, as desired.


Although I like this kind of use of forcing, where one constructs the forcing extension in order solely to make a conclusion about what is true in the original universe, nevertheless one may omit it by taking a countable elementary substructure of a suitable $H\_\theta$, in effect replacing $M$ with a countable proxy. Applying the theorem to the proxy, you get that the proxy satisfies $\text{PA}$, and so $M\models\text{PA}$ as well.


