Find a subset of $(\mathbb{Q}, <)$ isomorphic to $Exp(\omega, \omega)$ as well ordered set

Question

I have tried to find an isomorphism $\phi$ in this way: Let $f \in \operatorname{Exp}(\omega, \omega)$ and let the $k$ pairs $(n_i,f(n_i))$ such that for every $0<i<k+1$, $f(n_i)\ne 0$. I'd like to close $\phi(f)$ between $n_k$ and $n_k +1$. For example, if exist an only pairs with $f(n)\not=0$, $\phi(f)=n+1- \frac{1}{f(n)}$. I think it's possible to build $ \phi $ for countable recursion but I can not figure out how.

Answer 1

The following is the best I could find.

The idea is to construct the images $S_n:=\text{Im}\left(\phi|_{[0,\omega^n]}\right)$ recursively, so that:

• $\phi(0)=0$ and $\phi(1)=1$;
• for any $n \ge 0$, if $\phi(\omega^n)=a_n$ (so that $S_n \setminus \{ a_n \}=\text{Im}(\phi) \cap [0,a_n)$ is isomorphic to $\omega^n$), the set $S_{n+1}$ is obtained by concatenating $\omega$ "rescaled copies" of $S_n \setminus \{ a_n \}$, each of which is contained in an interval of the form $\left[2a_n(1-2^{-k}),2a_n(1-2^{-k-1})\right]$, and finally $\phi(\omega^{n+1})=2a_n=a_{n+1}$.

This actually satisfies $\phi(f)\in [2^{n_k},2^{n_k+1}]$ (with the same notation you used), but it should have some other nice properties, like the following (I didn't prove all of them, but I strongly believe all of them to be true):

• $\phi(\omega^n)=2^n$;
• $\phi(\omega \cdot \alpha)=2 \phi(\alpha)$;
• $\phi$ is limit-preserving;
• if $\alpha=\omega^{n_1}+\omega^{n_2}+...+\omega^{n_r}$ with $n_1 \ge n_2 \ge ... \ge n_r$, then $\phi(\alpha)=\sum\limits_{i=1}^r 2^{n_i+1-i}$.