A model-theoretic question re: Nelson and exponentiation

Solution 1:

Here is a proof of the weak claim. Let $X$ be any nonstandard model of PA with an initial segment $I$ containing all standard numbers which is closed under addition and multiplication but not exponentiation. Let $C$ be the set of $x\in X$ such that $x\leq n^i$ for some standard $n$ and some $i\in I$. Note that $C$ is closed under addition and multiplication: if $x\leq n^i$ and $y\leq m^j$, then $xy$ and $x+y$ are both at most $(m+n)^{i+j}$. So, we can take $X$ as a model of PA' with this $C$, and your set $M$ will be $C$.

Note also that $C$ is closed under exponentiation to elements of $I$, since if $x\leq n^i$ then $x^j\leq (n^i)^j=n^{ij}$. It follows that your set $E$ contains $I$. On the other hand, if $e\in E$, then in particular $2^e\in C$ so $2^e\leq n^i$ for some standard $n$ and some $i\in I$. But we have $(2^m)\geq n$ for some standard $m$, and so $n^i\leq 2^{mi}$ and so $e\leq mi$. Thus $e\in I$.

Thus your set $E$ for this model is just $I$. Since $I$ was chosen to not be closed under exponentiation, this proves the weak claim.