My question follows from the 1958 result of MacDowell–Specker (located originally in Modelle der Arithmetik, J. Symbolic Logic Volume 38, Issue 4 (1973), 651-652) of the proof of the following theorem:

Theorem 1. Every model $M$ of $PA$ has a proper elementary end extension $E_M\subset M$, where we take $\subset$ to mean $E_M\subseteq M$ and $\neg (E_M=M$).

The question I ask is broken into two parts which I believe are answerable in one post. Intuitively it feels that they are related as questions by the Macdowell-Specker as stated below:

Theorem 2. Every nonstandard model $N$ of $PA$ has a proper elementary end extension with $E_N\subseteq N$.

Question 1) Following from the existence of the proper end extension $E_M$ by the Macdowell-Specker theorem, can one assume that $E_N\models\mathrm{PA}$ if each $E_i\subset N$ is nonstandard? Perhaps more briefly, I ask if $E_N\models\mathrm{PA}$ still holds when $E_N$ is not a proper elementary extension of $N$.

Question 2) By the assumption of the "nonstandardess" of $E$, is it a sufficient condition for [Theorem 2] that if $E_N\models\mathrm{PA}$ then $N$ has its proper elementary end extensions $|E_N|\leq \omega$?

Thank you for your help, and I apologize if only one question should be asked in this context. The only similar question on the website I can find possibly related to the above is https://mathoverflow.net/questions/141845/elementary-end-extensions-of-models-of-peano-arithmetic-in-uncountable-languages, but this question concerns uncountability, which I do not feel is straightforwardly related to my question.

  • $\begingroup$ I'm having trouble following your notation. Why doesn't Theorem 2 follow immediately from Theorem 1: if N is a nonstandard model of PA it is, a fortiori, a model of PA, and so has a proper elementary end extension? $\endgroup$ – Henry Dec 22 '14 at 13:50
  • $\begingroup$ @Henry Yes it does follow from theorem 1, and is in fact the same theorem but just with the existence of the nonstandard model stated explicitly for convience. I see now your point of how it is redudant in view of theorem 1. My two questions are about the nonstandard model of PA below so I felt I was being more direct with my two questions; I am new to this website so I apologize for that, and I do not mind editing it out. $\endgroup$ – cmn1 Dec 22 '14 at 15:57
  • $\begingroup$ That clarifies the role of Thm 2, but I still don't understand your questions. In both questions, what is $E_N$? If it's the model promised by MS, it obviously satisfies PA; if it's some other model, what model is it? In question 2, does $|E_N|\leq\omega$ mean countable cardinality? Are you assuming that $N$ has countable cardinality? What does it mean to have a sufficient condition for a Theorem in the first place? $\endgroup$ – Henry Dec 22 '14 at 16:49
  • $\begingroup$ @Henry I denote by $$E_N$ the nonstandard proper elementary end extension such that $E_N \subseteq N$ and $\subseteq$ is defined as in theorem 1. By question 1, I ask if $E_N$ must be a proper elementary end extension to prove that every nonstandard model of PA has an end extension (not necessarily proper). In question two, I mean by "sufficient condition" if it is necessary to assume that $|E_N|\leq \omega$ in order to prove that $E_N$ is a proper nonstandard elementary end extension of $PA$. Thanks again, and I hope that clears it up. $\endgroup$ – cmn1 Dec 22 '14 at 17:26
  • $\begingroup$ I still don't understand. If $E_N$ is the the specific end extension given by Theorem 1, then of course it must be proper. However proving that every model of PA has a (not necessarily proper) end extension which is a model of PA is trivial, because every model of PA is an elementary end extension of itself. (That's why the theorem specifies that $E_N$ is proper.) I still don't know what $|E_N|\leq\omega$ means. $\endgroup$ – Henry Dec 22 '14 at 18:02

I think I understand the question well enough to answer now.

  1. Every model is a (proper, by definition) elementary end-extension of itself, so it's only interesting to prove that there are also elementary end-extensions which aren't proper.
  2. It follows easily from the ordinary proof of Macdowell-Specker that when $M$ is a countable model of PA, $M$ has a countable proper elementary end-extensions.
  3. It follows that if $M$ is countable, there must also be uncountable elementary end-extensions: let $M=M_0$, given $M_\alpha$, let $M_{\alpha+1}$ be a proper elementary end-extension of $M_\alpha$, and when $\lambda$ is a limit ordinal, le $M_\lambda=\bigcup_{\alpha<\lambda}M_\alpha$. Then $M_{\omega_1}$ is an uncountable elementary end-extension of $M$.

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