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My guess is that it's the cardinality of $\omega_{\omega_{\omega}}$. Since $\aleph_0$ is the cardinality of $\omega$ and $\omega$ is a limit cardinal, so it must be that $\aleph_0=\omega$. (Is this true in general?) Then I can just substitute the subscript $\aleph_0$ with $\omega$, and I do the same thing again. Is this correct?

gen-ℤ ready to perish
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Sid Caroline
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  • I think you meant to write $\omega$ is a limit ordinal. Are you working in a context in which the cardinals are not represented as ordinals? Please be more specific about your problem with the notation. – Rob Arthan Aug 26 '17 at 23:29
  • See also https://math.stackexchange.com/questions/715321/if-the-infinite-cardinals-aleph-null-aleph-two-etc-continue-indefinitely-is/ – Asaf Karagila Aug 27 '17 at 01:17

2 Answers2

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Without any further context, the cardinality of $\omega_{\omega_\omega}$ seems to be the most reasonable guess at the meaning, but the notation is not standard. The subscript in $\aleph_{n}$ is supposed to be an ordinal number, not a cardinal number, so the standard notation for this would be $\aleph_{\omega_\omega}$.

If this interpretation makes sense in context, I would assume it is just a misnaming of $\aleph_{\omega_\omega}$, unless you're positively sure the $\aleph_{\aleph_{\aleph_0}}$ notation comes from someone who definitely knows what they're talking about.

(If you're asking due to the comment on your earlier question, I'm pretty sure Duncan just meant $\omega_{\omega_\omega}$ -- but not because there's anything special about that ordinal; any other large-looking ordinal could have been used to make the point he was making).

hmakholm left over Monica
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    Assuming Wikipedia and AC, the cardinality of a set X is the least ordinal number α such that there is a bijection between X and α, so one could interpret $\aleph_\alpha$ to be an ordinal. – Simply Beautiful Art Aug 26 '17 at 22:53
  • @SimplyBeautifulArt: Yes, but in this particular case there's the extra wrinkle that there are two different notations $\omega_\alpha$ and $\aleph_\alpha$ for the same set, with the strong convention that you write $\omega_\alpha$ whenever you're thinking of it as an ordinal, and (usually) write $\aleph_\alpha$ when you're thinking of it as a cardinal number. The existence of that convention is what makes me declare this particular case to be likely to be a typo. – hmakholm left over Monica Aug 26 '17 at 22:55
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    @SimplyBeautifulArt: The issue here is context. The successor cardinal of $\aleph_\alpha$ is $\aleph_{\alpha+1}$, but $\aleph_0+1=\aleph_0$, because the $\aleph$ notation implies cardinal arithmetic. Therefore, the successor of $\aleph_{\aleph_0}$ should not be denoted by $\aleph_{\aleph_0+1}$. So if the successor of $\aleph_{\aleph_0}$ is $\aleph_{\omega_\omega+1}$, might as well call the bastard by its name and use $\aleph_{\omega_\omega}$ from the get go. – Asaf Karagila Aug 27 '17 at 01:12
  • (Oops, I had too many $\omega$'s in the subscript of my previous comment. $\aleph_{\aleph_0}$ corresponds, of course, to $\aleph_\omega$ and not $\aleph_{\omega_\omega}$...) – Asaf Karagila Aug 27 '17 at 13:58
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It would be more appropriate to write this as "$\aleph_{\omega_\omega}$" - the index of an expression is supposed to be an ordinal, not a cardinal. And this is indeed the cardinality of the ordinal $\omega_{\omega_\omega}$.

However, cardinals are usually defined as initial ordinals, so e.g. $\aleph_0=\omega$; it's just that we often use two different names for these entities, to distinguish between their roles in a mathematical argument. So technically the expression is fine as written, if a bit odd.

Noah Schweber
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