A question about what does $|A|^{<\omega}$ mean for a set

Question

I was reading this answer https://math.stackexchange.com/a/346249/629594 and I am not really sure what $|A|^{<\omega}$ means (I am not so familiar with things that concern ordinals and cardinals). I think that it means the cardianlity of the set $A^k$, where $k$ is a finite cardinal. Am I right? Furthermore, could we write $|A|^{<|\mathbb{N}|}$ instead of $|A|^{<\omega}$?

Answer 1

By $A^{< \omega}$ we mean the set of all finite sequences in $A$. By $|A|$ we mean the cardinality of $A$. Assuming the axiom of choice, which people usually do, this cardinality is represented by an ordinal: the least ordinal that is in bijective correspondence with $A$. So technically $|A|^{< \omega}$ would be the set of all finite sequences in that ordinal, as is mentioned in the comments.

Via the bijection between $A$ and $|A|$ you might as well think of $|A|^{<\omega}$ as the set of all finite sequences in $A$. The point of writing $|A|^{<\omega}$ instead of $A^{<\omega}$ is that we want to stipulate that we ultimately only interested in cardinalities.

This is also true in the context of the question you linked: ultimately they just want $|A|^{<\omega} = |A|$, so they are really only interested in the cardinality of $|A|^{<\omega}$. Perhaps it would be more accurate to write $||A|^{<\omega}|$, but that becomes messy so convention is to not do this. We could also write $|A^{<\omega}|$, but for some reason this does not seem to be standard.

Edit: I left some remaining points in your original question untouched. So it is not quite $A^k$ for some finite $k$, but rather $A^{<\omega} = \bigcup_{k < \omega} A^k$. The second point: yes $|A|^{<\omega} = |A|^{<|\mathbb{N}|}$, this would just be different notation for the same thing (since $|\mathbb{N}| = \omega$), but this notation is again a bit more messy and nonstandard.