Your p and q make no sense in set theory — Michael
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
Which is correct: p or q or both or neither? — Philosopher19
I will just say that my question remains unanswered — Philosopher19
the universal set is not contradictory in any way. — Philosopher19
we'll just go around in circles, and then perhaps we'll come back to the p and q point again. — Philosopher19
I believe it makes perfect sense to say set x is only a member of itself in its own set — Philosopher19
Why argue with others? — jgill
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