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Is there any meaningful definition to afford for $\aleph_r$ (as in cardinality) where $r\in\mathbb{R}^+$? $r\in\mathbb{C}$? What about $\aleph_{\aleph_0}$? Can we iterate this? $\aleph_{\aleph_{\aleph_{\cdots}}}$

I may be throwing in bunch of rather naive/basic questions, for I haven't learnt much about infinite cardinalities. If I am referring to bunch of stuff abundantly dealt in established areas, please kindly point out.

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The $\aleph$ numbers are well-ordered. This means that every non-empty set has a minimal element. Furthermore, they are linearly ordered.

This means that any indexing imposed on them should at least have these two properties. The real numbers are not well-ordered (consider the subset $(0,1)$, or even $\mathbb R$ itself) and the complex numbers are not even ordered in any natural sense.

The idea behind having a well-ordering is to say what is the next cardinality. Given a set, we can easily tell what is the least $\aleph$ which is larger. In the natural numbers (and their generalization, the ordinals) we have a successor function which does that, so it is a good ground to use when indexing cardinalities. We don't have a nice successor function for the real numbers, or for any dense ordering for a matter of fact.

It is possible to have $\aleph_{\aleph_0}$, but there is a minor problem here. $\aleph_0$ is the notation discussing size, whereas $\aleph_\alpha$ is the cardinal that the cardinals below it have order type $\alpha$. So we write $\aleph_\omega$, where $\omega$ is the least infinite ordinal. This is a limit cardinal, which means that it is not a successor of any cardinal -- but there are smaller $\aleph$'s nonetheless.

This of course can be reiterated, but we need to use the ordinal form, rather the cardinal form. Namely, every $\aleph$ number is also an ordinal. $\aleph_\alpha$ is the actually the ordinal $\omega_\alpha$, where these ordinals are defined recursively as ordinals which are $(1)$ infinite; and $(2)$ do not have an injection into any smaller ordinal. The least is $\aleph_0$ and it is the cardinality of the natural numbers, which is the ordinal $\omega$.

So if we wish to iterate, $\aleph_0\to\aleph_{\aleph_0}\to\aleph_{\aleph_{\aleph_0}}\to\dots$ we actually need to do it as following: $$\aleph_0\to\aleph_{\omega}\to\aleph_{\omega_\omega}\to\ldots$$


That been said, without the axiom of choice it is consistent to have sets whose cardinality is not an $\aleph$ number, namely sets which cannot be well-ordered. It is consistent that there is a collection of sets which is ordered (by inclusion) like the real numbers, and no two sets have the same cardinality (there is no bijection between two distinct sets).


Some threads which may interest the reader:

  1. "Homomorphism" from set of sequences to cardinals?
  2. Non-aleph infinite cardinals
  3. What are Aleph numbers intuitively?
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Set theorists abuse the notation, though, don't they? For example, they might speak of the fixed point of the $x\mapsto \aleph_x$ function, when what they really mean is the fixed point of the function that takes an ordinal $x$ to the smallest ordinal with cardinality $\aleph_x$. Or is there some other way to say this? –  MJD Oct 18 '12 at 2:01
    
@MJD: Th class of cardinals is definable, so it is not really an abuse of notation. The fixed point issue is simply to say that there are $\alpha$ cardinals (as an order type) below $\alpha$. –  Asaf Karagila Oct 18 '12 at 2:08
    
@MJD: Did my edit answer your question better? –  Asaf Karagila Oct 18 '12 at 23:06
    
What does $\aleph_{\omega_\omega}$ mean? –  MJD Oct 18 '12 at 23:55
    
@MJD: The unique initial ordinal such that the order type of the initial ordinals below it is as the ordinal $\omega_\omega$. Alternatively we can say that this is the least $\aleph$ for which there are $\aleph_\omega$ strictly smaller cardinals. –  Asaf Karagila Oct 18 '12 at 23:57
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