↪Philosopher19 ZFC is, I believe, set up specifically so that "a list can't list itself". That's how it avoids the various paradoxes. — Banno
ZFC is, I believe, set up specifically so that "a list can't list itself". That's how it avoids the various paradoxes. — Banno
Problems occur if you consider the elements of a set to not be themselves sets. Set theory only talks about sets. It does not, for example, talk about individuals.
The lists only list other lists... — Banno
If you're trying to argue that a "correct" set theory must allow for a universal set then I don't think you really understand mathematics. — Michael
I believe I understand Russell's paradox very well — Philosopher19
Given any set A and any set B, if for every set X, X is a member of A if and only if X is a member of B, then A is equal to B (the of axiom of extensionality) — Michael
It says that A and B are equal if every member of A is a member of B and every member of B is a member of A. — Michael
There exists a set B whose members are precisely those objects that satisfy the predicate — Michael
is predicate φ "A and B are equal if every member of A is a member of B and every member of B is a member of A"? If not, what is it? — Philosopher19
In Russell's paradox, φ is "sets that are not members of themselves". — Michael
are you essentially saying "there is a set that contains all sets that are not members of themselves"? If not, can you clarify?There exists a set B whose members are precisely those objects that satisfy the predicate — Michael
↪Philosopher19 Yes. Given the axiom schema of unrestricted comprehension there exists a set B whose members are sets that are not members of themselves. This leads to a contadiction. If B is not a member of itself then it should be a member of itself. — Michael
1 is contradictory if you say set B only contains all sets that are not members of themselves. — Philosopher19
Russell proved that these two axioms entail that there is a set that only contains all sets that are not members of themselves (the Russell set). — Michael
So when you say:
There exists a set B whose members are precisely those objects that satisfy the predicate
— Michael
are you essentially saying "there is a set that contains all sets that are not members of themselves"? If not, can you clarify? — Philosopher19
1) There exists a set whose members are sets that are not members of themselves
2) There exists a set that contains all sets that are not members of themselves — Philosopher19
Again, 1 is contradictory. Put it in clear language as to why the contradictoriness of 1 obliges us to reject 2 or to view the set of all sets as contradictory. — Philosopher19
When you say the axioms of naive set theory, are you referring to those notations that I asked you to put in clear language. — Philosopher19
If so, it seems to me you left half way through trying to clarity on it. — Philosopher19
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