# Cardinality of all cardinalities

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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It's not well-defined (like "cardinality of the set of all sets" etc). – Grigory M Aug 3 '10 at 18:27
As a side question, does anyone have any useful links (read: not wikipedia and easy to read links) on what cardinality is? I see it come up everywhere and would like to read up on it. Thanks – Tyler Hilton Aug 4 '10 at 13:10
@Affan: You could just ask a question. – kennytm Aug 4 '10 at 13:52
It would probably just get voted to close because its a "broad" topic, something wikipedia could do for me. – Tyler Hilton Aug 4 '10 at 14:02
@Affan well, wikipedia really does explain what cardinality is; and if after reading it you have some more specific question(s), you can ask them on this site – Grigory M Aug 4 '10 at 14:15

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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"every cardinality is equivalent to the set of all smaller cardinalities" Are you not confusing cardinalities with ordinals? If by "equivalent" you mean "has the same cardinality", then aleph_1 is not equivalent to the set {0,1,2,...,aleph_0}. – Samuel Aug 3 '10 at 20:45
@Samuel, you're right, I've mixed up between the two. Still, C is not a set, because a set cannot contain itself (en.wikipedia.org/wiki/Axiom_of_regularity), and in such a case I'm not sure what |C| even means. – Tomer Vromen Aug 3 '10 at 21:33
In fact, one can prove that C is not a set even if one drops the axiom of regularity. The idea is that one can prove directly that for any ordinal a, a is not an element of a. Then one considers X = the union of all sets in C. One can show that X is an ordinal which contains itself, giving the contradiction. – Jason DeVito Aug 3 '10 at 21:40
@Tomer: $C$ is not an ordinal (it does not contain $\aleph_0+1$), but it only contains ordinals (if we let $|X|$ be the least ordinal in bijection with $X$), so it doesn't contain itself. – Samuel Aug 3 '10 at 21:49
Based on all of the above, I think @KennyTM should accept another answer. I've thought of deleting the answer, but I wouldn't want the discussion in the comments to disappear. – Tomer Vromen Aug 4 '10 at 19:39

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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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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It should be noted that the C as defined above is not necessary a set of all cardinalities. If (CH)[en.wikipedia.org/wiki/Continuum_hypothesis] is false (i.e. not accepted as an axiom - it is neither provable nor disprovable in set theory) the C shown is not the all cardinalities. – Maciej Piechotka Aug 4 '10 at 21:42
I don't understand what this has to do with CH. If the Axiom of Choice holds, then every cardinal is an aleph. But even without this, the OP probably just means a class formed by taking one set of every cardinality, which can be seen not to be a set. Anyway, how do you get $\mathbb{Z}$ as an answer? – Pete L. Clark Aug 5 '10 at 2:19
I take it Maciej interprets $\{0,1,\ldots,\aleph_0,\aleph_1,\ldots\}$ to mean it contains $n$ and $\aleph_n$ for each natural $n$, whereas the original poster means it contains $\aleph_\alpha$ for each ordinal $\alpha$, that is, $C=\{|X|:X\text{ is a set}\}$, which is similar to a construction of a set of all sets in that it ranges over all sets. – Samuel Aug 5 '10 at 12:09
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 $C = {n: n \in \mathbb{N} \lor n = \aleph_n' \land n' \in \mathbb{N} }$. – Maciej Piechotka Aug 5 '10 at 14:46

### A simple analogy:

Let's imagine a world where all cardinalities are finite. The cardinalities are:

$$C = \{0, 1, 2, …\} = ℕ$$

(Note that while there are infinitely many cardinalities, each is finite.)

If $C$ has a cardinality, it is infinite:   $|C|=∞$

Leading to the contradiction that   $C = \{0, 1, 2, …; ∞\}$.

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How is this an explanation? Did you mean an analogy? – Andrés Caicedo Jul 7 '15 at 22:40
@Andres: My bad, I changed it. – Zaz Jul 8 '15 at 4:59
Please bear in mind that a lot of people viewing a simple question like this are here because they don't have much knowledge of Math. Math SE is for people studying math at any level. After an hour or two of researching set theory, I still cannot understand the accepted answer. Maybe I'll get it after putting in another couple of hours tomorrow. Sometimes people here forget that dumb people like me exist, who take 3-4 hours to grasp 1 idea of basic set-theory and need things explained to them in baby language, or as I like to call it, “layman's terms” – Zaz Jul 8 '15 at 5:18
The problem isn't that this answer is too simple, it's that it is wrong. All you've shown is that assuming every set is finite there is no set of all cardinalities, but that's very different from the actual question. Making things overly complicated is bad, but sometimes questions actually do have complicated answers. As evidence that this question doesn't have a very simple answer, note that in some set theories (such as New Foundations) we do have a set-of-all-sets, and (I believe) a set-of-all-cardinalities (appropriately defined). So this really is a technical question. – Noah Schweber Jul 8 '15 at 5:29
Of course, "technical" doesn't mean "people shouldn't understand it" - I'm all for making technical subjects accessible! But the analogy you give really isn't right (and I don't just mean that it isn't as rigorous as one might want). Not every question on this site has a short, simple answer. – Noah Schweber Jul 8 '15 at 5:33