Ordinal and cardinal arithmetic

Question

For ordinal exponentiations can we write ?

$$ω<ω^ω<ω^{ω^ω}<\cdots$$

I know, if $\omega$ is the first infinite cardinal,we have power set, which is

For example, $\aleph_0<{\aleph_0}^{\aleph_0}<{\aleph_0}^{{\aleph_0}^{\aleph_0}}\cdots$

So, if $\omega$ is the first infinite ordinal is this also correct?

$$ω<ω^ω<ω^{ω^ω}<\cdots$$

Answer 1

It may help to understand the following cardinal arithmetic fact to understand why the first displayed-math is not often seen. Observe $$2^\omega\le \omega^\omega\le (2^\omega)^\omega=2^{\omega\times\omega}=2^\omega.$$ Thus instead of writing $\omega^\omega$ (cardinal arithmetic) one usually writes $2^\omega$. Therefore what is usually seen instead of what you have written is $$\omega<P(\omega)<P(P(\omega))<\cdots$$ where $P$ is the powerset. This is exactly the Beth Numbers: https://en.wikipedia.org/wiki/Beth_number

Answer 2

The operations and relations on cardinals and ordinals are defined in different ways, we write $\alpha<\beta$ to imply that there is an injection from $\alpha$ into $\beta$ but not otherwise. When we write $\alpha<\beta$ for ordinals it normally means that there is an order preserving map from $\alpha$ to an initial segment of $\beta$, or equivalentely $\alpha\in\beta$.

The exponentiantion of cardinals $\alpha^\beta$ is the cardinality of functions from $\beta$ to $\alpha$, when the exponentiantion of the ordinals $\alpha^\beta$ is defined by transfinite recursion, so you have to be clear of what you mean in the question...

If you assume that $\omega^\omega$ is the set of all functions from $\omega$ to itself, then its cardinality is greater than $\omega$. Clearly, in this case we also have that the ordinal correspondent to $\omega^\omega$ has as initial segment $\omega$, but not otherwise, then $\omega<\omega^\omega$ in both notations.

Considering the usual notation of $\omega^\omega$ beign equal to $\sup\{\omega^n\}$ it would be countable, thus the cardinality of $\omega$ and $\omega^\omega$ would be the same. But considering the usual order notation for ordinals we would have $\omega<\omega^\omega$ for $\omega\in\omega^\omega$.

It is similar for the whole chain of $\omega,\omega^\omega,\omega^{\omega^\omega}$...