# Exponentially-closed ordinals

I have solved two exercises which asked what kind of ordinals satisfied the following conditions:

1. $\beta + \alpha =\alpha$ for all $\beta< \alpha$

2. $\beta\cdot \alpha=\alpha$ for all $\beta < \alpha$

Turns out the first works for $\alpha= \omega^{\xi}$ and the second fo $\alpha=\omega^{\omega^{\xi}}$.

I was wondering: what about exponentiation? I don't think it would be possible, but then how could I prove it?

• Well, you couldn't. For starters, ordinal arithmetic on infinite ordinals preserves cardinality. So every initial ordinal (read: cardinal) is closed under exponentiation, as a set of ordinals. And by the usual Löwenheim–Skolem arguments, it has a club of points which are also closed under exponentiation. Jun 7 '18 at 21:24
• @bof: If you're editing, why not edit the $*$ into $\cdot$? :) Jun 7 '18 at 21:31
• There is no such caracterization for exponentiation because what you ask is that for such an ordinal $\alpha$, if $\alpha>\omega$, then $\omega^{\alpha}=\alpha$. Those ordinals are called $\varepsilon$-numbers and are indeed stable under exponentiation. The case $\alpha\leq \omega$ should be easier to figure out. Mind that the result depends on the precise condition, e.g. here given your first condition, $0$ should also be a solution, and given your second condition, only $0$ is solution. Jun 8 '18 at 4:05

(1). If $0\ne \alpha=\cup \alpha$ then $\beta^{\alpha}=\cup \{ \beta^{\gamma}: \gamma <\alpha \}.$
(2). Let $a_0=\omega$ and $a_{n+1}=\omega^{(a_n)}$ for $n\in \omega.$ Let $\alpha =\cup \{a_n:n\in \omega\}$.
Confirm that $\beta^{\gamma}<\alpha$ if $\beta, \gamma <\alpha.$ By (1) we obtain $1<\beta <\alpha\implies \beta^{\alpha}=\alpha.$
This $\alpha$ is the least epsilon-number. That is, $\omega^{\alpha}=\alpha.$
BTW in your Q, condition 2, that $\beta<\alpha \implies \beta \cdot \alpha=\alpha,$ cannot hold unless $\alpha =0,$ because of the case $\beta=0.$ What it should say is $0<\beta<\alpha \implies ...$ (etc).