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This question already has an answer here:

Does there exist set that contains all the cardinal numbers?

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marked as duplicate by Asaf Karagila, Ross Millikan, Henning Makholm, Jason DeVito, Cameron Buie Feb 16 '13 at 16:54

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  • $\begingroup$ This has been asked several times before. Once recently. $\endgroup$ – Asaf Karagila Feb 16 '13 at 16:39
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Assume $C$ was the set of all cardinals. Then $\bigcup C$ would be a cardinal exceeding all cardinals in $C$ which is a contradiction.

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    $\begingroup$ While true this argument presupposes knowledge in cardinal and ordinal arithmetics. I have a hard time seeing how someone familiar with ordinals will not see the answer to this question immediately. $\endgroup$ – Asaf Karagila Feb 16 '13 at 16:42
  • $\begingroup$ @AsafKaragila When one is learning something for the first time "obvious" things aren't usually so obvious. $\endgroup$ – Rudy the Reindeer Feb 16 '13 at 16:42
  • $\begingroup$ This would also imply that there is no function that maps n to $\aleph_n$? $\endgroup$ – PyRulez Feb 16 '13 at 16:43
  • $\begingroup$ @PyRulez The cardinals are not limited to the $\aleph_n$'s, they go way way beyond that. $\endgroup$ – Asaf Karagila Feb 16 '13 at 16:43
  • $\begingroup$ Wait, how is that from n to $ \aleph_n$? $f(3) \neq \aleph_3$! $\endgroup$ – PyRulez Feb 16 '13 at 16:46
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Set of cardinals is well ordered by $\in$. Now, as a corollary we get Burali-Forti theorem, which says that there is no the set of all ordinal numbers. As a corollary from a corollary we can prove that, there is no set that contains all the ordinals. Proof: Let $A$ be a set that contains all the ordinals. You can prove that $\{x\in A : x \ \text{is ordinal}\}$ is a set, which contradicts Burali-Forti theorem.

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