Is the collection of all cardinalities a set or a proper class? Does anybody ever think about the problem?
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Collection of all cardinalities is indeed a proper class. To see this, note that there is at least as many cardinal numbers as ordinal numbers, because map $\alpha\rightarrow\aleph_\alpha$ is an injection.
Wojowu
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It is a proper class. There are several ways to see this. One is: suppose $X$ were the set of all cardinals (=initial ordinals). The ordinals - in particular, the cardinals - are well-ordered, so we may add them together (indexed by this well-order) to form a single "super-cardinal;" but it's easy to check that this super-cardinal is larger than any element of $X$.
This is essentially the same reasoning as the Burali-Forti paradox http://en.wikipedia.org/wiki/Burali-Forti_paradox, which shows that the class of ordinals is a proper class.
Noah Schweber
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Your argument requires, in explicit way, well ordering principle, which is independent of $ZF$. You might want to point this out in the answer. – Wojowu May 17 '15 at 05:38
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No it doesn't - the initial ordinals are all ordinals, so they're well-ordered with no assumption. – Noah Schweber May 17 '15 at 05:40
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1But cardinals need not be initial ordinals. :-) – Asaf Karagila May 17 '15 at 05:40
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In the absence of choice, one needs to be clear what a "cardinality" is - I've shown that the cardinalities of well-orderable sets are never a set, so a fortiriori the cardinalities in general do not form a set, whatever we take a "cardinality" to be. – Noah Schweber May 17 '15 at 05:41
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The collection of all cardinal numbers is a proper class. If it were a set $X$, then $(\sup X)^+$ would be a cardinal number greater than every element of $X$, which is a contradiction.
(The supremum of a set of cardinals is given by its union, and the cardinal successor $\kappa^+$ of a cardinal $\kappa$ is defined as the Hartogs number of $\kappa$.)
Trevor Wilson
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