# Ordinal exponentiation being closed in infinite initial ordinals

I previously asked this question about ordinal addition being closed in any infinite initial ordinal $\omega_\gamma$, and now I am stuck on showing the same thing for ordinal exponentiation.

Per one of several helpful comments on that question, my solution in the case of addition was to first show that $|\alpha + \beta| = |\alpha| + |\beta|$ for any ordinals $\alpha$ and $\beta$. This was fairly easy to show and the desired result follows almost immediately from this. I was also able to show the analogous result for ordinal multiplication.

However, in the case of exponentiation it is evidently not true in general that $|\alpha^\beta| = |\alpha|^{|\beta|}$. As a counterexample $|2^\omega| = |\omega| = \aleph_0$ whereas $|2|^{|\omega|} = 2^{\aleph_0}$, which we know is uncountable. So I am not sure how to show my result for exponentiation.

One of the comments on the addition questions also suggested showing that an ordinal is closed under addition if an only if it is an ordinal power of $\omega$ and then showing that every infinite cardinal is such an ordinal power of $\omega$. I have been thinking that a similar approach might work for exponentiation but I'm not certain what the pattern is here. Clearly both $2$ and $\omega$ are closed under exponentiation (and nothing between them is) but I'm thinking that the next ordinal with this property is actually $\varepsilon_0$. I am also thinking that the pattern might involve ordinal tetration, for example all ordinals of the form $^\alpha 2$ for nonzero $\alpha$, where $^\alpha \beta$ denotes ordinal tetration defined recursively by $$^\alpha \beta = \begin{cases} 1 & \alpha = 0 \\ \beta^{^\gamma \beta} & \alpha = \gamma + 1 \\ \sup_{\gamma < \alpha} {^\gamma \beta} & \alpha \text{ is a nonzero limit ordinal} \end{cases}$$ for any ordinal $\beta$. With this definition we have that $^1 \omega = \omega$, $^2 \omega = \omega^\omega$, $^3 \omega = \omega^{\omega^\omega}$, etc., and that $\varepsilon_0 = {^\omega \omega}$.

Any ideas on how I might begin to approach this?

• With ordinal exponentiation, one can actually verify that $2^\omega=\omega$. Note, simply that $2^n$ is finite for $n<\omega$, and therefore $2^\omega\leq\omega$ (the other inequality is easy). This is why it is often a good idea to reserve $\omega$ for ordinals and $\aleph_0$ for cardinals, especially when basic arithmetic is involved. Sep 9, 2017 at 22:03

It's not true that $|\alpha^\beta|=|\alpha|^{|\beta|}$, but it is true that if one of $\alpha$ and $\beta$ is infinite, then $|\alpha^\beta|\leq\max(|\alpha|,|\beta|)$ (with equality except in the trivial cases $\alpha\leq 1$ or $\beta=0$). You can prove this by induction in much the same way as for addition and multiplication.

• The result clearly follows if I can show this as a lemma. However, I am having trouble showing it by induction (on $\beta$) for the case when $\beta$ is a limit ordinal. That is, for limit ordinal $\beta$, show that $|\alpha^\beta| \leq \max(|\alpha|, |\beta|)$ given that $|\alpha^\gamma| \leq \max(|\alpha|, |\gamma|)$ for all $\gamma < \beta$. Based on the definition of ordinal exponentiation clearly $\alpha^\beta = \sup_{\gamma < \beta} \alpha^\gamma$. (1 of 2)
– kyp4
Sep 14, 2017 at 23:38
• However, in general, it is not true that $|\beta| \leq \kappa$ for cardinal number $\kappa$ even if $\beta = \sup_{\gamma < \beta} \gamma$ and $|\gamma| \leq \kappa$ for all $\gamma < \beta$. As an example where this fails consider $\beta = \omega_1$ and $\kappa = \aleph_0$. (2 of 2)
– kyp4
Sep 14, 2017 at 23:38
• Right, but here you know that the $\beta$ indexing the limit is also at most $\kappa$. In other words, if $\kappa\geq|\alpha|$ and $\kappa\geq|\beta|$, then $\alpha^\beta$ is the limit of a sequence of at most $\kappa$ ordinals, each of which is at most $\kappa$. Sep 15, 2017 at 2:02
• I went on a little odyssey that involved educating myself on the Axiom of Choice, and it seems that to prove this requires the Axiom of Choice, at least if we are to use the definition of supremum of a set of ordinals as the union of the set. See this question. Am I mistaken here or is there some other way we can prove this without using choice?
– kyp4
Oct 4, 2017 at 1:24
• You can prove it without Choice. One way to do so is to give an explicit description of $\alpha^{\beta}$ as a well-ordered set. Namely, it is the order-type of the set of functions $\beta\to\alpha$ which are $0$ at all but finitely many points, with the reverse lexicographic order (you can prove this description by induction). You can then compute the cardinality of this set by the usual methods of cardinal arithmetic. Oct 4, 2017 at 1:40