Does there exist a set of all cardinals?

Question

Does there exist set that contains all the cardinal numbers?

This has been asked several times before. Once recently. – Asaf Karagila Feb 16 '13 at 16:39 — Asaf Karagila, Feb 16 '13 at 16:39

score 7 · Accepted Answer · answered Feb 16 '13 at 16:40

7

Assume $C$ was the set of all cardinals. Then $\bigcup C$ would be a cardinal exceeding all cardinals in $C$ which is a contradiction.

answered Feb 16 '13 at 16:40

Rudy the Reindeer

46,359

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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. – Asaf Karagila Feb 16 '13 at 16:42
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@AsafKaragila When one is learning something for the first time "obvious" things aren't usually so obvious. – Rudy the Reindeer Feb 16 '13 at 16:42
This would also imply that there is no function that maps n to $\aleph_n$? – Christopher King Feb 16 '13 at 16:43
@PyRulez The cardinals are not limited to the $\aleph_n$'s, they go way way beyond that. – Asaf Karagila Feb 16 '13 at 16:43
Wait, how is that from n to $ \aleph_n$? $f(3) \neq \aleph_3$! – Christopher King Feb 16 '13 at 16:46
How is that a map from $\omega$ to the $\aleph$'s? – Asaf Karagila Feb 16 '13 at 16:47
@PyRulez: The set ${\aleph_n\mid n\in\omega}$ is not the set of all cardinals. The map $n\mapsto\aleph_n$ is injective and well-defined. But $\aleph_\omega$ is a cardinality above all the $\aleph_n$'s. – Asaf Karagila Feb 16 '13 at 16:47
What is above that? – Christopher King Feb 16 '13 at 16:48
@PyRulez $\aleph_{\omega + 1}$ for example... – Rudy the Reindeer Feb 16 '13 at 16:49
@PyRulez Sorry, I misread what you wrote above: I don't see how this implies that the map $n \mapsto \aleph_n$ is not injective (assuming that's what you meant, but maybe I still don't understand you). – Rudy the Reindeer Feb 16 '13 at 16:52
@Matt: I think the question was whether or not the map is even defined (the image of a set is a set, if $\omega$ is a set but ${\aleph_n\mid n\in\omega}$ is not then the function is not well-defined. – Asaf Karagila Feb 16 '13 at 17:02

score 0 · Answer 2 · answered Feb 16 '13 at 16:53

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.

Does there exist a set of all cardinals?

2 Answers2