Nobody disagrees on that.whether something is a member of itself or not is determined by whether it is in its own set or not — Philosopher19
Again, your argument is nonsensical. It does not mean anything in mathematics.My argument is that A is not a member of itself in B because A is a member of B in B. — Philosopher19
This is just repeating the same semantic nonsense.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 — Philosopher19
↪TonesInDeepFreeze
You suggest the Russell set exists only based on the process of defining it. — Mark Nyquist
You accept that than the Russel set exists and is legitimate. I don't think it has a sound basis. It's based on definition and that's not proof of existence. You have a burden of proof. — Mark Nyquist
But we are trying to dispell the contradiction, not prove it.
If the Russell set doesn't exist there is no contradiction. — Mark Nyquist
What book or article in the subject have you read/researched? — TonesInDeepFreeze
Just enough to understand the problem. — Philosopher19
I see no point in continuing this discussion. — Philosopher19
So sorry you got mixed up about my view of the existence of the Russell set. — Mark Nyquist
I was developing an alternative method using the concept of mathematical objects as proposed, existent or non-existent. — Mark Nyquist
Would it be fair to say your view develops the Russell set as a proposed mathematical object and concludes that it is ultimately a non-existent mathematical object? — Mark Nyquist
I am relying on your intermediate conclusion that the Russell set does not exist to go straight to the final conclusion that if the Russell set does not exist — Mark Nyquist
a paradox does not exist — Mark Nyquist
Summary,
The Russell set does not exist.
Based on the proposed defined mathematical object failing by contradiction. — Mark Nyquist
In defining the Russell set, two or more (known to exist) defined mathematical objects are used to define the Russell set — Mark Nyquist
n — Mark Nyquist
In mathematics, it makes no sense to ask, "In which set does x have the property P?" Rather, we ask, "Does x have property P?" If x has property P, then that is not qualified by "x has property P in some sets but not others". So one can't give sensical answers to nonsensical questions such as 1) and 2) in the previous post. — TonesInDeepFreeze
The property P is instantiated based on what set the item x is in. — Philosopher19
1) In which list does L list itself? — Philosopher19
I said, in A, A is a member of A/itself, and in B, A is a member of B/other-than-itself — Philosopher19
It does not mean anything in mathematics.
In B, A is not a member of anything, A simply exists. Because it exists in B, it is a member of B. But that has no bearing on Russell's paradox. It is a semantic point. — Lionino
I don't know what you mean by that. I don't know what you mean by a property being instantiated in this context — TonesInDeepFreeze
In B, A is not a member of anything, A simply exists. — Lionino
Because it exists in B, it is a member of B — Lionino
It is a semantic point — Lionino
In terms of function and logic, there is no difference between "lists itself" and "is a member of itself". — Philosopher19
Let x in x. Let y not in x. So x not equal {x y}. x in {x y}.
So x in x, and x in a set other than x. — TonesInDeepFreeze
Does L list itself in L?
Does L list itself in LL? — Philosopher19
No, I explained the difference.
I'll say it again, a list is a sequence. A sequence is a function whose domain is an ordinal. So the members of a list are ordered pairs. The members of the range of a list are the items listed by the list. — TonesInDeepFreeze
I answered those questions exactly. — TonesInDeepFreeze
Do you see how you have contradicted yourself? — Philosopher19
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