Maybe I'm missing something, but that seems to rely on the premise that either S = 0 or S = {S}, which is a premise we are not allowed.

I was being lazy because this discussion has gone on long enough. Should have been:

S = {S, …}

and

S = {x1, x2, …} where no xn = S

S might not be countable.

But, yes, we do have that either S in S or S not in S.

Either S in S or S not in S.

Suppose S in S. Then S not in S. So S not in S.

Suppose S not in S. Then S in S. So S in S.

So both S in S and S not in S.

Your p and q make no sense in set theory — Michael

I believe it makes perfect sense to say set x is only a member of itself in its own set, and I tired to prove it to you, and we even narrowed it down to p and q, but you have ceased to engage on the grounds that "p and q make no sense in set theory". This truth (the truth of a set only being a member of itself in its own set) has logical implications that would have shown (contrary to those who reject the universal set because they believe a set can be a member of itself outside of its own set) that the universal set is not contradictory in any way.

Russell’s paradox:

Assumption: S is the set of all sets that are not members of themselves.

Option 1:

S = {}

S is not a member of itself. But, as per the assumption above, it ought be a member of itself.

Option 2:

S = {S}

S is a member of itself. But, as per the assumption above, it ought not be a member itself.

Neither option 1 nor option 2 work. Therefore, the assumption is a contradiction. — Michael

I believe if I respond to this, we'll just go around in circles, and then perhaps we'll come back to the p and q point again. But instead of doing that, I will just say that my question remains unanswered:

Which is correct: p or q or both or neither? — Philosopher19

I will just say that my question remains unanswered — Philosopher19

I answered it. Neither p nor q make sense. @TonesInDeepFreeze has explained to you in depth that the sentence "A is a member of B in C" is meaningless in set theory.

the universal set is not contradictory in any way. — Philosopher19

This isn't about the universal set. This is about the Russell set. The Russell set is contradictory. It can neither include nor exclude itself without defying its own definition.

There are a number of set theories with a universal set, such as New Foundations and positive set theory.

No one knows what you mean by such locutions as "x is a member of itself only in its own set". You have not defined what it might mean.

Someone might as well say "x is not a member of itself not only outside itself" or similar nonsense and then require you to understand it though it is impossible to understand.

./

The assertion that there exists a set of which every set is a member allows Russell's paradox.

we'll just go around in circles, and then perhaps we'll come back to the p and q point again. — Philosopher19

Yes, going back to your p and q would be going back full circle yet again.

To break the circle requires that you give serious consideration to the fact that no one understands your phraseology but you. That would lead to you learning more about the subject so that you could communicate your ideas about it with other people.

I believe it makes perfect sense to say set x is only a member of itself in its own set — Philosopher19

No problem. If it makes sense to you, that's OK. Mathematicians don't have to agree with you. It's not like the fate of the world hinges upon this. It's OK to feel good about your own creation. Why argue with others?

Why argue with others? — jgill

It could be that arguing with someone is what it takes for them to be sincere to a truly perfect existence/being (or God/Goodness/Truth). It could be that that is what it takes for them to favour good/truth over evil/falsehood in a particular instance/moment. It could also be the opposite. I'd like to think I got into this discussion as a part of my efforts to favour/prioritise Goodness/Truth (or to strive in the cause of a truly perfect existence/being).