I see no point in continuing this discussion. — Philosopher19
Theorem 1.29 (Russell’s Paradox). There is no set R = {x : x ∉ x}.
Proof. If R = {x : x ∉ x} exists, then R ∈ R iff R ∉ R, which is a contradiction. — Open Logic: Complete build
Proof. If R = {x : x ∉ x} exists, then R ∈ R iff R ∉ R, which is a contradiction. — Open Logic: Complete build
What has not been shown to me is how this logically obliges us to view the set of all sets as contradictory. — Philosopher19
How do we set up a set theory which avoids falling into Russell’s Paradox, i.e., which avoids making the inconsistent claim that R = {x : x ∈/ x} exists? Well, we would need to lay down axioms which give us very precise conditions for stating when sets exist (and when they don’t). — On the next page...
And this shows that you have not understood R = {x : x ∉ x}: — Banno
Because the choice is between the whole of the remainder of that project being founded on an error unnoticed by more than a century of study by logicians world wide; and your being mistaken. — Banno
What has not been shown to me is how this logically obliges us to view the set of all sets as contradictory. — Philosopher19
Posts are missing from this thread, including some of my own. What happened? — TonesInDeepFreeze
What evidence do you have that fishfry left because of this thread? — TonesInDeepFreeze
The philosophical claims themselves are errant for not even being correctly about what they are supposed to be about. — TonesInDeepFreeze
So true. The OPs lay out belief systems in one form or another, and sometimes they don't budge. Which I find acceptable in Metaphysician Undercover's pronouncements, for he dwells with the ancients as they ponder space, time, and points and curves - although he balks at 1+4=5 and has little patience with Weierstrass and his limit ideas: admittedly useful, but fundamentally flawed. But I see where he is coming from there. Others, like this thread, are more or less unmovable in their opinions, which clash with standard mathematics. How you deal with the frustration of offering knowledge to those unwilling to accept it is admirable. — jgill
I believe no one until today has brought Weierstrass to my attention. — Metaphysician Undercover
The set of all sets encompasses all sets that are not members of themselves (precisely because they are members of it and not themselves) as well as itself (precisely because it is a set). — Philosopher19
A subset of "all sets that are members of themselves" and "all sets that are not members of themselves" is contradictory — Philosopher19
The z of all zs i — Philosopher19
We need a meaningful distinction between "member of self" and "not member of self" — Philosopher19
We need a set of all sets (math/logic would be incomplete without it, if not contradictory) — Philosopher19
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
Or does it contain 2 more members, total of 38, being the prior sets called "numerical" and "alphabetical" plus their members? — Fire Ologist
If we call it the set of two sets — Fire Ologist
Aren't we just overcounting this new "set of all sets" if we count the sets within it, and not just the members of those sets — Fire Ologist
A set of all sets has as members all the sets — TonesInDeepFreeze
I've refuted that claim. You skip the refutation. — TonesInDeepFreeze
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
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
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