In the case of B = {A, 0}, is A a member of A/itself, or is A a member of B/non-itself? — Philosopher19
Both — Michael
So once again, in B, is A a member of itself or not a member of itself? — Philosopher19
In B, A is not a member of both A and B. — Philosopher19
Both a member of itself and a member of B. — Michael
It is just the case that the symbol "A" is defined recursively as "{A}" and that the symbol "B" is defined as "{A, 0}", which is the same as "{{A}, 0}" given the recursive definition of "A". — Michael
In the case of B = {A, 0}, is A a member of A/itself, or is A a member of B/non-itself? — Philosopher19
Both a member of itself and a member of B. — Michael
Look at what you're saying:
"In B" does not equal to "in both A and B".
Do you see your contradiction? You have treated "in B" as the same as "in both A and B". — Philosopher19
↪Philosopher19 A is a member of both A and B. This is basic set theory. Take a math lesson. — Michael
And in B, A is not a member of itself. — Philosopher19
N is the set of natural numbers.
R is the set of real numbers.
Every natural number is a member of both N and R. We don't say "in R, the natural numbers are not members of N". — Michael
So why is it that A can be both a member of B and C but not a member of both A and B? — Michael
Scenario 1
— Michael
Scenario 2
B = {0, A}, where A = {A} — Michael
I have not disagreed with scenario 2. I have said that in B, A is not a member of itself precisely because it is a member B (as opposed to itself), and in A, A is a member of itself. — Philosopher19
z = any set that is not the set of all sets
v = any set
The v of all vs = the set of all sets
The z of all zs = The not-the-set-of-all-sets-set of all not-the-set-of-all-sets-sets — Philosopher19
Take z to be any set that is not the set of all sets, and take v to be any set. The z of all zs is a member of itself as a z (as in in the z of all zs it is a member of itself). But it is not a member of itself in the v of all vs, precisely because in the v of all vs it is a member of the v of all vs as opposed to a member of itself. If we view the z of all zs as a z, it is a member of itself. If we view the z of all zs as a v, it is a member of the set of all sets. You can't view it as both a member of the z of all zs and a member of the v of all vs at the same time. That will lead to contradictions. In other words, we can't treat two different standards/contrexts as one (as in are we focused on the context/items of vs or the context/items of zs to determine what is a member of itself or not a member of itself?) — Philosopher19
And you’re confused. It’s not the case that “in A” it’s a member of one thing and “in B” it’s a member only of something else. — Michael
Am I the one that's confused? So it's not the case that in A it's a member of one thing and in B it's a member of another thing? So it's not the case that in A it's a member of itself and in B it's not a member of itself? — Philosopher19
Yes, you’re confused. A is a member of A and B. 1 is a member of N and R. That’s all there is to it. — Michael
1. x is a member of A if and only if x is a member of x — Michael
ZFC does this by not allowing a set to be a member of itself. — Michael
The fact that paradoxes develop means it is a defined mathematical object that does not exist. — Mark Nyquist
Never had the set in the first place. — Mark Nyquist
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