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Let $C = \{0, 1, 2, \ldots, \aleph_0, \aleph_1, \aleph_2, \ldots\}$. What is $\left|C\right|$? Or is it even well-defined?

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    @GrigoryM I think this is a duplicate of https://math.stackexchange.com/questions/1712964/attempt-at-proving-the-class-of-all-cardinals-is-a-proper-class. Also, if Affan posted a question asking what a cardinal number is, it would be a duplicate because it has already been asked and answered at https://math.stackexchange.com/questions/53770/defining-cardinality-in-the-absence-of-choice.2018-02-20

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C is not a set - it is infact a proper class. If C were a set, then |C| would be defined. It then follows that |C| would be the largest cardinality, since there is a total order between all the cardinalities, and |C| > \kappa for every cardinality $\kappa$ (every cardinality is equivalent to the set of all smaller cardinalities). But 2^{|C|} > |C| and so there cannot be a largest cardinal.

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    It's not true that every cardinal number is the cardinality of the set of all smaller cardinal numbers. The cardinality of the set of all cardinal numbers less than $\aleph_1$ is $\aleph_0$. It is true that every ordinal number is the order type of the set of all ordinal numbers smaller than it.2018-05-22
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This is "Fact 20" on page 10 of

http://math.uga.edu/~pete/settheorypart1.pdf

These are notes on infinite sets from the most "naive" perspective (e.g., one of the facts is that every infinite set has a countable subset, so experts will see that some weak form of the Axiom of Choice is being assumed without comment. But this is consistent with the way sets are used in mainstream mathematics). It is meant to be accessible to undergraduates. In particular, ordinals are not mentioned, although there are some further documents -- replace "1" in the link above with "2", "3" or "4" -- which describe such things a bit.

But I don't see why it is necessary or helpful to speak of ordinals (or universes!) to answer this question.

Added: to be clear, I wish to recast the question in the following way:

There is no set $C$ such that for every set $X$, there exists $Y \in C$ and a bijection from $X$ to $Y$.

This is easy to prove via Cantor's diagonalization and it sidesteps the "reification problem for cardinalities", i.e., we do not need to say what a cardinality of a set is, only to know when two sets have the same cardinality. I believe this is appropriate for a general mathematical audience.

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    Thanks for those notes. They are very useful. It is hard to find expositions which don't just identify cardinals with ordinals (Levy is an exception). But pedagogically, it seems that the concept of cardinal should be distinguished from the concept of ordinal, since a set's having an ordinal implies its wellorderability.2016-12-06
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The most common way to define the cardinal number $|X|$ of a set $X$ is as the least ordinal which is in bijection with $X$. Then $C$ is an unbounded class of ordinals, and any such is necessarily a proper class. Since $C$ is not a set, it does not have a cardinality.

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I may be wrong but it seems that C is not a set of all sets or anything similar. It has IMHO a trivial bijection into $\mathbb{Z}$ and therefore into $\mathbb{N}$.

I haven't met any paradox to forbid $|C| \in C$ ($\{1\}$ is such set).

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    The $C = {|X|: X is a set}$ is indeed not correctly defined. However that's not the usual meaning of the $\ldots$. I assumed that the meaning is $C = {n: n \in \mathbb{N} \lor n = \aleph_n' \land n' \in \mathbb{N} }$.2010-08-05