Hyperfinite intervals are uncountable but nonstandard models of Peano arithmetic can be countable? My understanding is, by Lowenheim-Skolem I can find a countable nonstandard model of Peano Arithmetic. On the other hand, I have just encountered the following argument:
For $\alpha \in {}^*\mathbb{N}\setminus\mathbb{N}$ the map $\psi:(0,1)\to [1,\alpha]$ sending $r$ to $\min\{\beta | r < \beta / \alpha\}$ is well-defined (notice the set we take the minimum from is internal) and injective, so the interval $[1,\alpha]$ has cardinality continuum.
I'm confused as to how these statements are consistent.
 A: The argument for ${}^*\mathbb{N}$ uses more about ${}^*\mathbb{N}$ than just that it is a nonstandard model of Peano arithmetic.  Specifically, it uses the fact that ${}^*\mathbb{N}$ embeds in a set ${}^*\mathbb{R}$ which also contains $\mathbb{R}$ and has enough nice properties to be able to define $\psi$ and prove it is injective.  (I believe it suffices that the relation $\leq$ and operations $+$ and $\cdot$ extend to ${}^*\mathbb{R}$ to make $({}^*\mathbb{R},{}^*\mathbb{N},+,\cdot,\leq)$ an elementary extension of $(\mathbb{R},\mathbb{N},+,\cdot,\leq)$, but I have not checked the details.)  There is no reason to think that this should be true with ${}^*\mathbb{N}$ replaced by an arbitrary nonstandard model of Peano arithmetic (and in fact, your argument proves that it is not).
More concretely, note that if $\beta$ and $\alpha$ are in an arbitrary nonstandard model of PA and $r\in(0,1)$, then we can make sense of the statement $\beta/\alpha<r$ using rational approximations to $r$.  However, if $r$ is irrational, this involves infinitely many different rational numbers and so cannot be stated using a single formula in the language of PA.  As a result, $\{\beta:r<\beta/\alpha\}$ may not be definable in the language of PA and so you cannot use the PA induction schema to conclude it has a least element.  So, in an arbitrary nonstandard model of PA, you cannot prove that $\psi$ is well-defined.  The difference for ${}^*\mathbb{N}$ is that it satisfies induction over a larger language, one that includes the operations of ${}^*\mathbb{R}$ and not just ${}^*\mathbb{N}$.
