If the infinite cardinals aleph-null, aleph-two, etc. continue indefinitely, is there any meaning in the idea of aleph-aleph-null?

Question

Apologies if this isn't a sensible question, I really don't know too much about these infinite cardinals aside from the basics. I did, however, think that the idea of the "aleph-null"th aleph number was interesting enough to base my username on and my own attempts did not prove fruitful, so I was wondering if anyone here could shed some light. Thanks!

For clarity: I'm asking about $\aleph_{\aleph_0}$ . Thanks!

P.S. I was somewhat unsure about the tags for this, sorry if I accidentally placed it in the wrong category.

@GitGud yes, I'll put it in that formatting in the body of the question for clarity — aleph_aleph_null, Mar 17 '14 at 10:42
FYI: make sure that your actual question is in the body of your post. It's confusing to have it only in the title. — RBarryYoung, Mar 18 '14 at 00:47

score 24 · Accepted Answer · answered Mar 17 '14 at 11:37

I'd like to elaborate on some of the fine points that Arthur raised.

The $\aleph$ numbers (also the $\beth$ numbers) are used to denote cardinals. However one of the key features of cardinals is that we can say "the next cardinal", and we can say which cardinal came first and which came second. These are ordinal properties.

Note that the least cardinal greater than $\aleph_{\aleph_0}$ also has countably many [infinite] cardinals smaller than itself. But since $\aleph_0+1=\aleph_0$, what sense would that make?

So we are using the ordinals. It's a fine point, because the finite notions coincide, the finite cardinals are the finite ordinals, and it's not until we reach the infinite ordinals that we run into the difference between $\omega$ and $\aleph_0$.

Therefore, instead of $\aleph_{\aleph_0}$ we have $\aleph_\omega$, then we have $\aleph_{\omega+1}$ and so on and so forth. After we have gone through uncountably many of these we finally have $\aleph_{\omega_1}$, where $\omega_1$ is the least uncountable ordinal -- which corresponds to $\aleph_1$.

And so on and so forth. For every ordinal $\alpha$ we have $\aleph_\alpha$ is the unique cardinal that the infinite cardinals below it have the same order type as $\alpha$.

score 23 · Answer 2 · answered Mar 17 '14 at 10:39

23

Yes, mostly. It's the least cardinal for which there are infinitely many infinite cardinals below. And it is usually denoted $\aleph_\omega$. (Where $\omega$ denotes the least infinite ordinal; of course, $\omega = \aleph_0$, but we use $\omega$ to indicate that we are interested in ordinal properties of this object. Another description for $\aleph_\omega$ is that it is the $\omega$'th infinite cardinal, or the unique cardinal such that the order-type of the family of all infinite cardinals below it is $\omega$.)

answered Mar 17 '14 at 10:39

user642796

52,188

And is the smallest non numerable cardinal with numerable cofinality. – Martín-Blas Pérez Pinilla Mar 17 '14 at 10:46
@Martin: Nonsense. Every limit ordinal has countable cofinality. – Asaf Karagila Mar 17 '14 at 11:36
@Martín: Are you trying to say that $\aleph_\omega$ is the smallest uncountable cardinal with countable cofinality? If so, while true it doesn't really help explain what the subscript has to do with anything. – user642796 Mar 17 '14 at 11:40
@Asaf, you are the expert here, but for all the cardinals $\aleph_n$, $0<n<\omega$: cf$(\aleph_n)=\aleph_n$. – Martín-Blas Pérez Pinilla Mar 17 '14 at 11:41
@ArthurFischer, right (numerable=countable, non numerable=uncountable, true?). – Martín-Blas Pérez Pinilla Mar 17 '14 at 11:44
2

@Martin: I'm just messing with you. :-) It's consistent without the axiom of choice that for every $\alpha$, $\operatorname{cf}(\aleph_\alpha)=\omega$. :-) I should have put a smiley face at the end of my previous comment... – Asaf Karagila Mar 17 '14 at 12:03
2

@Asaf, I always forgot that you are the non-AC guy here. – Martín-Blas Pérez Pinilla Mar 17 '14 at 12:08
@Martin: Yeah, I sometime that these jokes a bit too far. :-) – Asaf Karagila Mar 17 '14 at 12:44

score 2 · Answer 3 · answered Jan 14 '17 at 23:07

I realize this question has been answered but I would like to provide a way of visualizing $\aleph_\omega$ and $\aleph_{\omega_\omega}$:

Start with $\aleph_0$ and consider the sequence $\aleph_0, \aleph_1,... \aleph_n,...$ for $n \in \omega.$

This is a strictly monotone increasing sequence that has a limit, namely $\aleph_\omega$, which would make $\aleph_\omega$ the smallest uncountable limit cardinal. This number does have applications. Example: if GHC holds and $\aleph_\lambda = \beth_\lambda$ for all $\lambda$, then $|V_{\omega +\omega}|=\beth_\omega=\aleph_{\omega}$. In other words $\aleph_\omega$ would be the cardinality of the von Neumann universe $V_{\omega +\omega}$ which is a universe sufficient for most ordinary mathematics, and is a model of Zermelo set theory.

As to $\aleph_{\omega_\omega}$, consider the sequence $\aleph_\omega, \aleph_{\omega+1},...,\aleph_{\omega^2},...,\aleph_{\omega^\omega},...,\aleph_{\epsilon_0},...$

Notice that this also is a strictly monotone increasing sequence which is countable, the supremum of which is a $\aleph_{\omega_1}$. Continuing on like this yields $\aleph_{\omega_1}, \aleph_{\omega_2},...,\aleph_{\omega_n},...$ the limit of which is $\aleph_{\omega_\omega}$

score 1 · Answer 4 · answered Mar 17 '14 at 10:47

1

Yes, note that $\aleph_1$ can be interpreted as there is one cardinal ($\aleph_0$) smaller than it. In the same way, there are 100 cardinal numbers smaller than $\aleph_{100}$ ($\aleph_0, \aleph_1, \dots, \aleph_{99}$).

The smallest infinite list of cardinals is hence $\aleph_{\aleph_0}$ also denoted $\aleph_{\omega}$.

We may even continue with $\aleph_{\aleph_{\aleph_0}}$ (the smallest infinite list of the smallest infinite lists of cardinals) and so on.

answered Mar 17 '14 at 10:47

naslundx

9,720

I don't understand $\aleph_{\aleph_{\aleph_0}}$. – Asaf Karagila Mar 17 '14 at 12:31
@AsafKaragila Consider the infinte list starting with $\aleph_{\aleph_0}$ with the even larger cardinals $\aleph_{\aleph_1}, \aleph_{\aleph_2}, \aleph_{\aleph_3}, \dots$ all the way up to $\aleph_{\aleph_{\aleph_0}}$. – naslundx Mar 17 '14 at 13:32
5

(1) The indexing is done by ordinals. Not by cardinals, as Arthur pointed out, and I elaborated. (2) I understand what $\aleph_{\omega_\omega}$ would be, but I'm unclear as to the remark in the parenthesis. – Asaf Karagila Mar 17 '14 at 13:37

If the infinite cardinals aleph-null, aleph-two, etc. continue indefinitely, is there any meaning in the idea of aleph-aleph-null?

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