I was wondering if it was possible to formally prove that the set of all sets does not exist in ZFC by using a simple argument based on Cantor's Theorem:
- Assume for contradiction that the set of all sets exists (call it $S$).
- This means that its power set $\wp(S)$ exists as well.
- By Cantor's Theorem, we know $|S| < |\wp(S)|$.
- Since $|S| < |\wp(S)|$, we know that $\wp(S) – S \ne \emptyset$
- Since every element of $\wp(S)$ is a set, we know that $\wp(S) – S = \emptyset$.
- However, we know that $\wp(S) – S \ne \emptyset$.
- Therefore, we have a contradiction, so $S$ does not exist.
I have not seen this line of reasoning used before to show that the set of all sets does not exist. Is this proof correct?
Best Answer
It is correct. I posted about it a while ago in a comment somewhere here (on a question on whether there was a relation between Cantor's theorem and Russell's paradox). You can see a 140-characters-or-less summary here: https://twitter.com/andrescaicedo/status/263151880295813120
The usual proofs of both the fact that there is no set of all sets ("Russell's paradox"), and Cantor's theorem use a diagonal argument. There is actually a proof of Cantor's theorem that avoids diagonalization, due to Zermelo, see this answer on MO. This is why I find this argument interesting, as it gives us a diagonalization-free proof of Russell's paradox.
(For the answer to the question on MO mentioned at the end of the link above, see here.)