TITLE: Elementary end extensions of models of Peano Arithmetic in uncountable languages
QUESTION [9 upvotes]: A well-known theorem of Mills asserts that there is a model of Peano Arithmetic $M$ in an uncountable language such that $M$ has no elementary end extension (e.e.e.). I ask whether every complete extension of $PA$ in an uncountable language can have models of every cardinality with arbitrary large e.e.e.'s. To put it in more precise form, suppose $L$ is an uncountable language expanding the language of arithmetic $\{+,.,0,1,<\}$. By $PA(L)$ we mean $PA^{-}$ plus the induction axiom for all $L$-formulas $\phi(x,\bar{y})$. Let $T$ be a complete $L$-theory extending $PA(L)$ and let $\kappa$ be a cardinal $\geq |L|$.
Question. Does $T$ have a model $N$ such that $|N|=\kappa$ and $N$ has e.e.e.'s of every cardinality $\geq\kappa$?

REPLY [6 votes]: It may be of interest to note that Jerabek and Dorais's positive solution to the above question and their use of the MacDowell-Specker theorem motivated me to take a closer look at the following paper of myself:
Sh.Mohsenipour, On Keisler Singular-Like Models, Math. Logic Quart. 54 (2008), 324-330,  
in order to obtain another proof and then I realized that the following model theoretic result can be easily derived from the main result of the paper:
Theorem. Suppose $L=\{<,\dots\}$ is a first-order language where $<$ is a linear order and $T$ is an
$L$-theory such that for any finite $T^{'}\subset T$, there is a strong limit cardinal $\theta$ so that $T^{'}$ has a 
$\theta-$like model $M$ and also $M$ can be represented as the union of an e.e.e. chain of its submodels. Let $\lambda$ be a cardinal $\geq|L|$, then:
(i) $T$ has a model $N$ of cardinality $\lambda$ such that $N$ has e.e.e.'s of every cardinality $\geq\lambda$.
(ii) If $\lambda$ is a singular cardinal, then the above $N$ can be a $\lambda-$like model with the additional property that for any singular cardinal $\lambda^{'}$ with $cf(\lambda^{'})<\lambda$ and $\lambda^{'}>\lambda$, $N$ has a $\lambda^{'}-$like e.e.e.
Now turning to the above question it is easily seen that the MacDowelـ‎Specker theorem implies that every completion $T$ of $PA(L)$ satisfies the hypothesis of the above theorem and therefore by part (i) we get another solution to the question. 
This is a magic property of MO that makes you understand even your own work better!