Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
8  
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
2  
@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
add comment

4 Answers 4

up vote 11 down vote accepted

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.

share|improve this answer
3  
"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
1  
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
2  
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
show 3 more comments

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.

share|improve this answer
add comment

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.

share|improve this answer
add comment

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).

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.