Cardinal Number Alephs: Countably Infinite?

[mathman]

Cardinal numbers (infinite) are $\aleph_0.\ \aleph_1$, etc. There are an infinite number of them. Is it known whether or not the number is countably infinite or not, or is it unknown?

 

[S.G. Janssens]

I believe that the selected answer here (together with the comments on it) explains why, at least in the usual sense, the number you are asking for is not defined. (The underlying object is not a set.)

 

[mathman]

I went through the link material, but I couldn't fully grasp why the object is not a set.

 

[SSequence]

Well, regarding the question in OP, here is a simple answer. One thing to note is that if one assumes that every set can be well-ordered, then the cardinalities can definitely be indexed with ordinals (and in a strictly increasing manner so to speak). So this is what I also assume below in the answer.
(I don't know anything about the situation where "every set can be well-ordered" is false ... this is why I mentioned this)

Cardinal numbers (infinite) are $\aleph_0.\ \aleph_1$, etc. There are an infinite number of them. Is it known whether or not the number is countably infinite or not, or is it unknown?

Yes, the "number" of cardinal numbers is not countably infinite. For example, the smallest cardinal number that serves as such an example is:
$\aleph_{\omega_1}$

Another way to look at it goes like this. The ordinal $\omega_1$ has the cardinality $\aleph_1$. And, by definition, $\omega_1$ is the smallest ordinal whose cardinality is $\aleph_1$ (and not $\aleph_0$). It seems that by this convention we would write $\omega_0=\omega$.

Similarly the smallest ordinal whose cardinality is $\aleph_2$ (but not $\aleph_1$) is written as $\omega_2$. So, essentially, $\omega_{\omega_1}$ is the smallest ordinal whose cardinality is $\aleph_{\omega_1}$. Note that $\omega_{\omega_1}$ can be defined as supremum of all $\omega_{\alpha}$'s, where we have $\alpha<\omega_1$.Edit:
This is essentially a digression, but one thing that I was thinking about was whether we can write $\aleph_{\omega_1}$ as $\aleph_{\aleph_1}$ or not. It seems to me that if we take the liberty of writing $\aleph_\alpha=\omega_\alpha$ more generally, then the answer should be yes for obvious reasons (but since I am not very well acquainted here, I don't really know whether this is considered acceptable usually).

But if we do that, we would have to distinguish between two different types of addition (for example) more carefully it seems (say something like $+_O$ and $+_C$). Since we obviously have $\omega_1+_O \omega_0 \ne \omega_1$ (but the analogous equation is defined as an equality for cardinal addition I think).

 

[scormus]

I think the answer to the question is linked to the truth, falsity or undecidability of the Generalized Continuum Hypothesis (GCH). If GCH is undecidable, then I think this is too.

 

[SSequence]

I am not clear as to why anything related to CH is required here.

Is $\aleph_{\omega_1}$ not well-defined in ZFC? I remember reading that stuff like this (and well-beyond) can be easily defined in ZFC. None of the advanced "large cardinal" stuff is remotely required to define this kind of cardinal number. And I would be quite surprised anyway if it couldn't be defined easily. After all $\omega_{\omega_1}$ is much smaller than even the first fixed point of the function $x \mapsto \omega_x$.

 

[scormus]

I am not clear as to why anything related to CH is required here.

Is $\aleph_{\omega_1}$ not well-defined in ZFC? I remember reading that stuff like this (and well-beyond) can be easily defined in ZFC. None of the advanced "large cardinal" stuff is remotely required to define this kind of cardinal number. And I would be quite surprised anyway if it couldn't be defined easily. After all $\omega_{\omega_1}$ is much smaller than even the first fixed point of the function $x \mapsto \omega_x$.

 

[scormus]

Thanks for your reply.
I'm not sure whether CH is strictly required either. But I am certain in my mind that clarifies the issue.
I'm also not sure about the argument about defining these cardinal numbers.

If GCH is true - OK we cannot know that - let's us suppose it is true, then we consider a sequence of sets starting with the natural numbers and repeatedly defining the power set. The first one corresponds to the reals, which are uncountable. To each power set there is an Aleph. If GCH is true then they are consecutive in the sense that there are no intermediate Alephs larger than the cardinality of any of the and its power set. Then the Alephs are countable.

But GCH is not provable (not even CH). If CH is not true we cannot determine how many Alephs between C(N) and C(P(N)), or even if the number is finite. I don't see how we could even determine whether they were countable between Pⁱ(N) and Pⁱ⁺¹(N).

FWIW, I cannot conceive of what kind of set would have cardinality greater than Aleph-0 but less than C(P(N)). I am uncomfortable with that possibility rather more than I am uncomfortable wit the unprovability of CH. I am happy to suppose CH and GCH are true, in which case the Alephs are countable.

BTW, I hope my nomenclature for natural numbers, cardinality, power (and repeated power) set.

 

[SSequence]

Thanks for your reply.
I'm not sure whether CH is strictly required either. But I am certain in my mind that clarifies the issue.
I'm also not sure about the argument about defining these cardinal numbers.

I am highly certain what I wrote specfically about $\aleph_{\omega_1}$ showing alephs are not countable is true. Still, I felt it is better to exercise some caution when it is an area whose technicalities/subtleties one isn't familiar with.

The thing is that if one assumes AC (and hence well-ordering of all sets) then there NO other cardinalities except of the form $\aleph_{\alpha}$, where $\alpha$ is an entirely arbitrary ordinal (but the entirety of all ordinals themselves isn't a set ... and I think perhaps this is what the linked thread in post#2 was about). So we can absolutely write something like:
$card(\omega_{\omega_1})=\aleph_{\omega_1}$
just like we can write:
$card(\omega)=\aleph_{0}$
$card(\omega_{1})=\aleph_{1}$
$card(\omega_{2})=\aleph_{2}$
etc.

If GCH is true - OK we cannot know that - let's us suppose it is true, then we consider a sequence of sets starting with the natural numbers and repeatedly defining the power set. The first one corresponds to the reals, which are uncountable. To each power set there is an Aleph. If GCH is true then they are consecutive in the sense that there are no intermediate Alephs larger than the cardinality of any of the and its power set. Then the Alephs are countable.

I am thinking that you are assuming AC ... the situation is much more complicated without it ... and honestly I have no idea about it in that case.
In any case, I am still skeptical of this. You are assuming that you can repeat powerset operation only countable number of times. But I don't know why this can be assumed. More specifically, I am not clear why any of this would have any bearing on existence of $\omega_{\omega_1}$ which I described as an example of set which shows that the alephs are not countable.

FWIW, I cannot conceive of what kind of set would have cardinality greater than Aleph-0 but less than C(P(N)). I am uncomfortable with that possibility rather more than I am uncomfortable wit the unprovability of CH. I am happy to suppose CH and GCH are true, in which case the Alephs are countable.

Once again, assume AC. suppose we have $card(\mathbb{R})=\aleph_2$. Here is a set with cardinality greater than Aleph-0 but less than Card(P(N)) ... $\omega_1$. More specifically $card(\omega_1)=\aleph_1$.
The only difference with CH being true is that there "exists" a bijection between $\mathbb{R}$ and $\omega_1$.P.S. Also assuming we are talking about traditional set theory here, not anything alternative.