# Cardinality of a set that consists of all existing cardinalities

1. What is the easiest way to prove (if possible, without using ordinals etc. as my current math understanding of set theory counts only cardinals, and countable & uncountable sets) that the number of cardinalities that exists is not countable (that is, can't be put into bijection with $\mathbb{N}$)?

2. What exactly does it mean that the set of all cardinals is so big that it's not even a set, but a class? Where does contradiction that does not allow it to be a set arise? I have read Pete Clark's notes, but am not quite sure how #20 leads up to that conclusion.

I have taken a look at the following topics:

But still can't quite find/understand the answer.

• Some similar discussion on this thread as well. math.stackexchange.com/questions/10085/… – user17762 Nov 20 '10 at 2:35
• I have removed the tag "large cardinals", because "large cardinals" is a technical concept, it does not mean "big sizes". – Andrés E. Caicedo Nov 20 '10 at 2:41

There is no "number of cardinalities". As you say, there are so many that they cannot form a set.

Suppose that ${\mathcal A}$ is a set whose elements are sets, with the property that if $A,B\in{\mathcal A}$ and $A\ne B$, then $|A|\ne|B|$, i.e., $A,B$ have different cardinalities. Let $C=\bigcup{\mathcal A}$, i.e., $C=\bigcup_{A\in{\mathcal A}}A$. Clearly, $|C|\ge|A|$ for each $A\in{\mathcal A}$. Let $D={\mathcal P}(C)$ be the power set of $C$. Then $|D|>|C|$ so $|D|>|A|$ for any $A\in{\mathcal A}$. This proves that there cannot be a set of all cardinalities, because given any such set, we just found a new cardinality different from all the ones in the set.

Of course, if ${\mathcal A}$ is countable, this shows that the ''number'' of cardinalities is not countable.

There is a small remark that may be worth making. The argument works just as well if we do not require that all sets in ${\mathcal A}$ have different cardinalities, but simply that for any prescribed cardinality we want to consider, there is at least one set in ${\mathcal A}$ of that size (but there may more than one). This is slightly more general, but there is also a technical advantage, namely, in this form, the argument does not depend in any version of the axiom of choice.

(Finally: I just checked Pete's nice note that is linked to in the body of the question. His fact 20 there is essentially the argument I've shown here.)

• Here is an interesting technical question in the absence of the axiom of choice: Given any set $X$, is there a set $Y$ with $|Y|=|X|$, whose elements are sets, and such that if $a,b\in Y$ and $a\ne b$, then $|a|\ne|b|$? The answer has to be no, at least consistently. – Andrés E. Caicedo Nov 20 '10 at 2:46
• Thank you, this seems like a very trivial yet powerful proof! – InterestedGuest Nov 20 '10 at 2:50