TITLE: Uncountable model of bounded arithmetic with an elementary end extension
QUESTION [9 upvotes]: 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 $M$?

REPLY [10 votes]: Having seen the other answers, let me point out yet another quick argument using just Lowenheim-Skolem.
Let $M$ be an uncountable model and $N$ a proper elementary end extension of it. Expand $N$ by adding an extra unary predicate naming the subset $M$. Call the resulting structure $(N,M)$. Take a countable elementary substructure of it, say $(N',M')$. Then $N'$ is a proper elementary end extension of $M'$ and $M'$ is countable, so the theorem applies. Hence $M'$ is a model of PA and so is $M$ since they are elementarily equivalent.