A function on ordinals.

Question

Denote by $L$ the set of all non-zero countable limit ordinals. Is there an injection $f\colon \omega_1\times L\to L$ such that $f(\alpha,\beta)>\beta$ for all $(\alpha,\beta)$ in the domain of $f$?

Answer 1

Partition $L$ as $\{L_\xi:\xi<\omega_1\}$, where $|L_\xi|=\omega_1$ for each $\xi<\omega_1$. For $\xi<\omega_1$ let $\eta_\xi=\min L_\xi$; we may assume that $\langle\eta_\xi:\xi<\omega_1\rangle$ is strictly increasing. If $\{\alpha_\xi:\xi<\omega_1\}$ is a strictly increasing enumeration of $L$, then $\alpha_\xi\le\eta_\xi$ for each $\xi<\omega_1$. Necessarily $\eta_0=\alpha_0=\omega$.

Let $X=\{\xi<\omega_1:\eta_\xi=\alpha_\xi\}$, let $L_0'=L_0\cup X$, and for $\xi<\omega_1$ let

$$L_\xi'=\begin{cases} L_\xi\setminus\{\eta_\xi\},&\text{if }\xi\in X\\ L_\xi,&\text{otherwise} \end{cases}$$

and $\eta_\xi'=\min L_\xi'$; clearly $\eta_\xi'>\alpha_\xi$ for $\xi>0$. For each $\xi<\omega_1$ enumerate $L_\xi'=\{\beta_\gamma^\xi:\gamma<\omega_1\}$, and define

$$f:\omega_1\times L\to L:\langle\gamma,\alpha_\xi\rangle\mapsto\beta_\gamma^\xi\;;$$

then $f$ is a bijection, and $f(\gamma,\alpha_\xi)>\alpha_\xi$ unless $\gamma=\xi=0$, in which case $$f(0,\alpha_0)=f(0,\omega)=\omega=\alpha_0\;;$$ a single exception at $\omega$ is unavoidable if $f$ is to be a bijection.