Does the set of all sets that do not contain themselves,

>>8786807
Are you using ZFC for these sets?

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Suppose such a set A exists. Since A is a set, its complement C is a set. If C contains itself, then it can not be a set*, and then A is not a set, so C must be in A. Similarly, every set in C must be in A, but C is a complement of A, so C must be empty. Now A consists of all sets, and thus contains itself contradicting its own definition. Such a set, therefore, does not exist.

*follows from
>Foundation (Regularity): Each nonempty class is disjoint from one of its elements.

>>8786873
good bait

>>8786807
If a barber cuts the hair of all the people in the town except those who cut their own hair, does he cut his own hair?

>>8786873
Any literature that captures that statement not using 400 pages?

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>>8787842
No idea, sorry. It was just a train of thought before going to sleep. Maybe the intro chapter of Dugundji's book on topology. That's where I learned my axioms from.

>>8786807
There exists such a class in NBG, but not a set.

>>8788945
>just a train of thought before going to sleep

So in other words, what you basically did was just pull a proof out of your ass? Well I can do that too.

If C contains itself it doesn't. If C doesn't contain itself it does. Hence this whole problem is fucked and indicates that you can't just make up sets in set theory and expect them to always make sense.

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>>8789480
I did. But it follows logically from the idea that a set can't have itself as its element. Don't you ever prove stuff on your own?

>>8789672
>don't you ever prove stuff on your own
Hahahahahahahahahaha, no.

>>8789701
>>8789672
just kidding. I'm simply saying that it's obvious such a set doesn't exist

>>8786807
I thought every set is a subset of itself? So how can there be a set of sets that do not contain themselves

>>8789721
This is linguistics

>>8786807
There is no set that can match those requirements, set theory is incomplete.