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I would like to know whether the number of infinity sets of different cardinalities is uncountable or countable. Is the proof simple? Thanks.

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The argument in shows that the collection of cardinalities cannot form a set. If it did, you could take the powerset of their union. – Qiaochu Yuan Aug 22 '10 at 23:26

4 Answers 4

Technically, neither one is true: the class of cardinalities is too large to form a set. (However, having to choose between "countable" and "uncountable", certainly "uncountable" seems better: there is at least an uncountable set of cardinalities of sets.)

I think this is equivalent to a previous question on this site, but anyway: please see Fact 20 of

and let me know if you have any further questions.

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If by the number of cardinalities you mean the set of all cardinalities - then there is none.

The class of all cardinal numbers is a proper class, as it contains a copy of the class of ordinal numbers - which is not a set.

So to answer you question, it's not countable. It's so big it's not even a set.

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Contrary to the responses thus far (and with no disrespect intended to their authors), I think we can say quite unequivocally that there are strictly more than countably-many sets with distinct infinite cardinalities, and that (ipso facto) there are uncountably many such sets.

The answers before mine fixate on the idea of cardinality itself as the concept of quantity, and correctly point out that that there isn't a "set of infinite cardinalities". (There is a class of infinite cardinalities, if you like that sort of thing, but perhaps not everyone likes talking about proper classes.) Because there cannot be a set of infinite cardinalities, the reasoning goes, there cannot be a cardinality (a well-defined sense of "the number of") of infinite cardinalities. However, the original question isn't about "cardinality" per se: it's about a less formal notion of "the number of something", which we conventionally define by cardinality in the case of sets. As we are not talking about "the set of X", however, perhaps we shouldn't be using this definition of number. What then?

Well, the original question isn't asking us precisely how many infinite cardinalities there are: just whether this number is countable or uncountable --- a dichotomy of ordering; whether it is less-than-or-equal-to the "number" of natural numbers (i.e. whether it can be placed in bijective correspondance with a subset of the natural numbers), or not. This is, as far as I'm aware, the definition of the terms "countable" and "uncountable".

As the number of infinite cardinalities cannot be put into one-to-one correspondence with any subset of the natural numbers, it follows that the number of infinities is uncountable. The fact that they cannot be put into a set is irrelevant, in my opinion: at worst, one only needs to identify a single uncountable set U of infinities, note that there is no surjection onto it from the natural numbers, and note incidentally that there are more infinities outside of the set U anyway. Any cardinal ℵC (identified in the usual way with an ordinal), for C itself an uncountable ordinal, will suffice for U.

Thus, there are uncountably many different infinities, QED.

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I honestly don't see anything in this answer that was not already addressed in mine. In particular, if you follow the link, you'll find that the word "cardinality" is not used at all. – Pete L. Clark Aug 23 '10 at 12:57
Also, in all the many treatments I have seen, the terms "countable", "uncountable" and "bijective correspondence" inherently apply to sets. Do you have a reference for your alternative definition? – Pete L. Clark Aug 23 '10 at 13:03
@Pete: you did indicate that "uncountable" would be the better term to apply, but I feel that you put too much emphasis on it being somehow not technically correct. --- The reference to a particular uncountable cardinal in my last paragraph was in anticipation of the complaint about "bijective correspondance" being between sets (because the bijection itself is usually taken to be a set as well); this basically recapitulates your answer, but without the hedging. And "countability" necessarily applies to sets. But for uncountability, I don't see why we should refrain from the obvious extension. – Niel de Beaudrap Aug 23 '10 at 18:41

For any set $X,$ the set $P(X)$ of all subsets of $X$ has a bigger cardinality than $X$ itself (for $X$ is finite this is easy, for $X$ infinite you need a clever argument from Cantor, obtaineble in any set theory text). You can do this again, obtaining:

card(X) < card(P(X)) < card(P(P(X))) < ...

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This, however, doesn't answer the question - what happens after $\aleph_0$ steps. – Asaf Karagila Aug 23 '10 at 12:52

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