0

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 $|A|^{<\omega}$?

Alexdanut
  • 2,807
  • It's the set of finite sequences of ordinals all less than the cardinal $|A|$. – BrianO Jan 03 '21 at 11:27
  • @BrianO Oh, I see, it actually makes sense in the context of that question. So, just to make sure that I understand this correctly, it is the set of all finite sequences of elements from $A$, right? I am not sure if this is what you mean by "finite sequences of ordinals". – Alexdanut Jan 03 '21 at 11:30
  • In the context of the answer where it was found, $|A}^{\lt\omega}$ seems to be a cardinal number rather than a set of finite sequences. – bof Jan 03 '21 at 12:39
  • @bof I think that it is the cardinality of that set of finite sequences. – Alexdanut Jan 03 '21 at 13:45
  • 1
    In the context of that question, both make sense, but $A^{<\omega}$ would make a bit more.

    The actual usage there shows that the author means $|A|^{<\omega}$ to be a cardinal: he writes that proving $|A|^{<\omega} = |A|$ for $A$ infinite requires AC.

     – BrianO Jan 03 '21 at 19:11

1 Answers1

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.

Mark Kamsma
  • 12,935
  • Thank you very much! Logic is definitely not my area, but I needed the result I linked in an expository paper I am writing for university, that's why I wanted to make sure that I properly understand everything. – Alexdanut Jan 03 '21 at 13:59