Why the following initial segment is NOT a model of PA

Question

Let $M$ be a non-standard model of PA, fix an $a \in M$ \ $\mathbb{N}$

Consider the collection:

$$I:=\{b\in M\ |\ b<a^n,\ \text{for some }n \in \mathbb{N}\}$$

It is easy to see that this is infact an initial segment (smallest one containing $a$ in fact)

This is an example given by Richard Kaye in his book on PA (Chapter 6)

I'm have trouble understanding his reason on why this is NOT a model for PA.

Any help or simple explanation is deeply appreciated.

Cheers

Answer 1

As mentioned in the comments, this is because exponentiation is definable in PA. Specifically, what we will use here is that there is a definable binary operation $(x,y)\mapsto x^y$ which PA proves has each of the following properties:

• For each standard $n\in\mathbb{N}$, $x^n$ is the product of $n$ copies of $x$ with itself (this is a separate theorem for each $n$).
• If $x>1$ and $y<z$, then $x^y<x^z$.

Now suppose $I$ were a model of PA and consider the element $a^a$ of $I$. Since $n<a$ for all standard $n$, $a^n<a^a$ for all standard $n$, and $a^n$ has its usual meaning. But this is a contradiction, since this would mean $a^a\not\in I$.