I would like to construct a hyperreal version of the Cantor set. Let $X_0$ be the interval $[0,1]$ in the hyperreal line, and for any $n$, let and let $X_{n+1}$ be the set of hyperreal numbers obtained if the middle third is removed from each interval that makes up $X_n$. Then $\cap_nX_n$ is a set whose standard part is the Cantor set. But unlike the Cantor set, it does not have empty interior; it still consists of intervals we can take middle thirds out of.
So let’s use transfinite recusion to iterate the process further. As before, let $X_0$ be the interval $[0,1]$. For any ordinal $\alpha$, let $X_{\alpha+1}$ be the set of hyperreal numbers obtained if the middle third is removed from each interval that makes up $X_\alpha$. And for any limit ordinal $\beta$, let $X_\beta = \cap_{\alpha<\beta}X_\alpha$.
Then my question is, does there exist an ordinal $\alpha$ such that $X_\alpha$ has empty interior? If so, what is the smallest such $\alpha$?
This construction just doesn't work. The set $\bigcap_n X_n$ does not consist of intervals, at least not intervals that have well-defined middle thirds. For example, one of these "intervals" is the set of all nonnegative infinitesimal hyperreals. This "interval" has a lower endpoint $0$, but no upper endpoint (there is no largest infinitesimal, or smallest non-infinitesimal), and there is no meaningful way to define a "middle third" of it.
Note that the natural "hyperreal version" of the Cantor set would be defined by internal recursion on the hypernatural numbers, not by transfinite recursion on the ordinals. By transfer, this will behave just like the ordinary Cantor set, at least as far as the relevant first-order logic is concerned. It can be described, for instance, as the set of hyperreals which have a (hypernatural-indexed) ternary expansion consisting of $0$s and $2$s. If you're using an ultrapower to define the hyperreals, then this set would just be the ultrapower of the Cantor set.