The set of all sets is really the set of all members, which is also how we use the word "all" in the first place. — Fire Ologist
Isn't the set of all sets equivalent to the set of all members? — Fire Ologist
There aren't actually any sets within the set of all sets. There are only members. — Fire Ologist
I am not skipping. You say:
So whether a set x is a member of itself or not a member of itself, if there is a set of all sets then x is a member of that set of all sets.
— TonesInDeepFreeze — Philosopher19
So whether a set x is a member of itself or not a member of itself, if there is a set of all sets then x is a member of that set of all sets.
— TonesInDeepFreeze
This point implies that a set v can have more than one set that is a member of itself, as a member of itself — Philosopher19
you perhaps fairly said "the z of all zs has no apparent meaning to me" — Philosopher19
L = the list of all lists
LL = the list of all lists that list themselves
Is L a member of itself in L? — Philosopher19
The v of all vs [...] The z of all zs — Philosopher19
the issue lies in correctly determining what it is for something to be a member of itself and what it is for something to be not a member of itself. — Philosopher19
What you skipped is my refutation (posted twice) of your claim that a set can't be both a member of itself and a member of another set. — TonesInDeepFreeze
"is a member of itself, as a member of itself" has no apparent meaning to me. — TonesInDeepFreeze
You said that it is perhaps fair to say that such locutions have no apparent meaning, but then you proceed to post them again — TonesInDeepFreeze
The above logically implies you are rejecting it. — Philosopher19
If it's a member of itself, it's not a member of other than itself. — Philosopher19
You need to show a meaningful difference between 1 and 2 — Philosopher19
z and v were clearly defined. — Philosopher19
I get where you're coming from. I believe the issue lies in correctly determining what it is for something to be a member of itself... — Philosopher19
All we need is a non-paradoxical set of all sets that are not members of themselves. We have this. — Philosopher19
So revised: Let x not equal b. Let b not be a member of x. Let x be a member of x. But x is a member of {x b}. So x is a member of x and x is a member of a set different from x, viz {x b}.
That refutes your claim. — TonesInDeepFreeze
Wrong. You cannot produce a valid demonstration that — TonesInDeepFreeze
I'm not sure I get what I'm saying. — Fire Ologist
Help me out. Besides the set of all sets, what is an example of a set that is a member of itself? — Fire Ologist
Is there any way you can provide an example of a set that IS a member of itself, other than the set of all sets, in plain language? I can't think of one. — Fire Ologist
Is there any way you can provide an example of a set that IS a member of itself, other than the set of all sets, in plain language? — Fire Ologist
How does this show there is a meaningful/semantical difference between 1 and 2? — Philosopher19
Call the set of all sets the v of all vs. — Philosopher19
I think you can understand this if, for a few moments, you clear your mind of the voice in it that keeps saying "I am right. I know I am right. I must be right. All the logicians and mathematicians are wrong and I am right", then very carefully, very slowly, consider: — TonesInDeepFreeze
it is unclear to me as to what it's doing — Philosopher19
I believe it deliberately strays from what is clear simple language to try and force something that cannot be forced (perhaps due to dogma). — Philosopher19
B) When a set is a member of itself, it is not a member of another set. — Philosopher19
because it is a member of itself as opposed to another set). — "Philosopher19
THAT is dogma. You have no proof of it, and I gave an exact disproof of it. — TonesInDeepFreeze
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
These are two different claims:
1. A is not a member of itself
2. A is a member of some other set
Given this:
A={A}B={A,0}
(1) is false and (2) is true. — Michael
Regarding Russell's paradox, it is simply this:
1. x is a member of R if and only if x is not a member of x.
Is R a member of R?
Either answer entails a contradiction, and so (1) is a contradiction. Given that naive set theory entails (1), naive set theory is shown to be inconsistent. — Michael
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
A is a member of both A and B.
I'll explain it to you in non-math terms:
I am a member of the football team and a member of the tennis team.
These are two different claims:
1. I am not a member of the football team
2. I am a member of a non-football team
(1) is false and (2) is true. — Michael
I repeat:
A) When a set is not a member of itself, it is a member of another set
B) When a set is a member of itself, it is not a member of another set — Philosopher19
But a set can either be a member of itself or a member of other than itself. — Philosopher19
A) When a set is not a member of itself, it is a member of another set
B) When a set is a member of itself, it is not a member of another set — Philosopher19
Evidently, in A, A is a member of itself.
Evidently, in B, A is not a member of itself because A is a member of B. — Philosopher19
You are not a set. You can never be a member of yourself. But a set can either be a member of itself or a member of other than itself. And it is logically the case that when a set is a member of itself, it is a member of itself. And when it is a member of other than itself (as was the case with A in B), it is not a member of itself. — Philosopher19
When a set is a member of itself, it is not a member of another set — Philosopher19
If A = {A} and if B = {A, 0} then A is a member of A and a member of B. — Michael
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