The math procedes in a way that assumes an ultimate set will exist. — Mark Nyquist
I'm saying ultimately the Russell set does not ultimately exist. — Mark Nyquist
And you can't have a paradox if the defined mathematical object does not exist. — Mark Nyquist
And you can't have a paradox if the defined mathematical object does not exist. — Mark Nyquist
You seem to be arguing for the paradox after the paradox has been dismissed. — Mark Nyquist
You [Michael] seem to be arguing for the paradox after the paradox has been dismissed. — Mark Nyquist
In B, A is not a member of both A and B. — Philosopher19
So the Russell set as a rejected defined mathematical object does not exist and cannot be a paradox. — Mark Nyquist
If the Russell set doesn't exist there is no contradiction. — Mark Nyquist
It's my first look at anything in set theory so I don't have background. — Mark Nyquist
Finally, yes — Mark Nyquist
You've argued that there is a set of all sets, U.
If A is the set {A} then A is a member of both A and U. — Michael
If A is the set {A} then A is a member of both A and U. — Michael
p) In A, A is a member of itself/A.
q) In U, A is not a member of itself/A.
p and q are true by definition. — Philosopher19
I never said A can't be both a member of A and B. I said, in A, A is a member of A/itself, and in B, A is a member of B/other-than-itself. — Philosopher19
Your argument seems to be that A is not a member of B in A because B is not defined in A — Lionino
here is a demonstration of how a set can only be viewed as a member of itself in its own respective set:
v = any set
The v of all vs = the set of all sets
z = any set that is not 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
The z of all zs is a member of itself in the the z of all zs, 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 v of all vs. We can't treat two different things/references/contexts/standards/items 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
No, it's nonsense. That's not how set theory works.
1 is a member of N and R.
A is a member of A and B.
That's it. — Michael
So it's settled in your mind that you won't read even a single book or article about this subject, but that others should read your website? — TonesInDeepFreeze
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