Question: Let $f: \Bbb{N}\rightarrow \text{Ord}$ (where "Ord" is the set of ordinals) be defined inductively by: $$f(0)=\omega\\ f(n^+)=\omega^{f(n)}$$ Let $\epsilon_0=\{\sup f(i) : i \in \Bbb{N}\}.$ If $\alpha,\beta<\epsilon_0$ then show that:
$\alpha + \beta < \epsilon_0$
$\alpha\beta <\epsilon_0$
$\alpha^\beta < \epsilon_0$
Attempt at solution: The result seems clear if both $\alpha$ and $\beta$ are finite. So, assume $\beta$ is a nonzero limit ordinal. Then by definition:
$\alpha+\beta=\sup\{\alpha+\xi : \xi<\beta\} < \alpha + \epsilon_0 = \sup\{\alpha+\xi : \xi <\epsilon_0\}=\epsilon_0$
$\alpha\beta = \sup\{\alpha\xi : \xi <\beta\} < \alpha \epsilon_0 < \epsilon_0 \bullet \epsilon_0 = \epsilon_0$
$\alpha^\beta = \sup\{\alpha^\xi : \xi < \beta\}<\alpha^{\epsilon_0} = \epsilon_0$ because $\sup(\alpha,\alpha^2,...)=\epsilon_0$
Am I on the right track?
Any help greatly appreciated :)