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.

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

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    $\begingroup$ 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. $\endgroup$ – Andrés E. Caicedo Nov 20 '10 at 2:46
  • $\begingroup$ Thank you, this seems like a very trivial yet powerful proof! $\endgroup$ – InterestedGuest Nov 20 '10 at 2:50

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