whether something is a member of itself or not is determined by whether it is in its own set or not — Philosopher19

Nobody disagrees on that.

My argument is that A is not a member of itself in B because A is a member of B in B. — Philosopher19

Again, your argument is nonsensical. 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.

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

This is just repeating the same semantic nonsense.

What book or article in the subject have you read/researched?

↪TonesInDeepFreeze
You suggest the Russell set exists only based on the process of defining it. — Mark Nyquist

You are so very mixed up that you are getting this all completely reversed.

No, I do not at all suggest that such a set exists. Rather, we are giving you proofs that such a set does NOT exist.

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

Same as above. You're mixed up and have it reversed.

We very clearly do NOT think there is such a set. Rather, we PROVE that there is no such set.

But we are trying to dispell the contradiction, not prove it.

If the Russell set doesn't exist there is no contradiction. — Mark Nyquist

Again, you're mixed up, and likely unfamiliar with proof by contradiction.

We prove that the assumption "there is a set whose members are all and only those sets that are not members of themselves" implies a contradiction thus that that assumption is false, which is to say that there does NOT exist such a set.

If you continue to insist that we're saying that such a set does exist, after three posters have already explained, in detail and in different ways, your misunderstanding, then I'm guessing you're trolling.

EDIT: I see now that after the quoted post, you replied further that you recognize that you don't understand. So, fair enough.

What book or article in the subject have you read/researched? — TonesInDeepFreeze

I watched a YouTube video amongst other things. Didn't read a book on it. Had a look at wiki and Stanford but maybe or maybe not with massive amounts of focus. Just enough to understand the problem. Also had an exchange with a math lecturer who specialises in set theory (but we disagreed with each other fundamentally, but perhaps the exchange helped me better understand the other's position)

In any case, given the responses I have seen, I see no point in continuing this discussion.

Peace TonesInDeepFreeze

Just enough to understand the problem. — Philosopher19

Just enough to think you understand it. But more than enough for you to completely mangle it.

I see no point in continuing this discussion. — Philosopher19

You've said that about twenty times already, yet you continue. But you are right. If you are unwilling to carefully read toward understanding, not just skim for an opportunity to misrepresent, an article such as the one at the Stanford Encyclopedia then there is no point for you to be posting about it.

It might be jolly for you to say what you think is the very first incorrect sentence in the Stanford article.

Funny thing is that, though I might be mistaken, I suspect that there is an error in the article (though it is not material).

The article says that intuitionistically:

~ReR -> ReR implies ReR.

But I don't see how that is so.

Yes, ~ReR -> ReR implies ~~ReR. But we can't infer ReR from ~~ReR.

So, the only way I could think of doing an intutionistic proof is not by way of:

ReR & ~ReR

but rather by way of:

~ReR & ~~ReR

So sorry you got mixed up about my view of the existence of the Russell set.

We both are saying the Russell set ultimately does not exist. If you missed it I was developing an alternative method using the concept of mathematical objects as proposed, existent or non-existent.

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? If so we should have nothing to disagree on.

Also, as relates to the philosophy of mathematics, my view reaches back fully to physical brain state. So for me it was an exercise in exploring that and as a simple result I got the same answer you did.

I am relying on your intermediate conclusion that the Russell set does not exist to go straight to the final conclusion that the Russell set does not exist....a paradox does not exist

I'm not at all saying the Russell set shouldn't be explored. How else would you know?.

Summary,
The Russell set does not exist.
Based on the proposed defined mathematical object failing by contradiction.

Additionally,
In defining the Russell set, two or more (known to exist) defined mathematical objects are used to define the Russell set and produce a failed defined mathematical object (non-existent).....Which I think is an interesting result

Reference,
I covered mathematical objects as brain state 4 days ago in my post on Universal Form.

So sorry you got mixed up about my view of the existence of the Russell set. — Mark Nyquist

I didn't get mixed up. You were mixed up.

I was developing an alternative method using the concept of mathematical objects as proposed, existent or non-existent. — Mark Nyquist

I had no comment on that. Rather, your replies regarding the mechanics of the ordinary proof were confused.

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

No. The way you say it is kind of along the lines of what I say, but I don't bring along notions of "proposed object" and "ultimately non-existent".

Rather, as I explained, it is a proof by contradiction, keeping it precise and simple, without the extraneous "proposed object" and "ultimately non-existent":

Toward a contradiction, suppose there is an x such that for all y, y is a member of x if and only if y is not a member of y.

Then x is a member of x, and x is not a member of x.

Therefore there is no x such that for all y, y is a member of x if and only if y is not a member of y.

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

What? What you call my "intermediate conclusion" is not intermediate, and it is the same as what you call the "final conclusion".

The theorem ("the final conclusion") proved is:

There is no x such that for all y, y is a member of x if and only if y is not a member of y.

Put more informally, "There is no set whose members are all and only those sets that are not members of themselves".

Put another way, the name "The Russell set", which is supposed to name a set whose members are all and only those sets that are not members of themselves, does not properly refer to anything.

a paradox does not exist — Mark Nyquist

The paradox was that, from the ordinary view (once given as an axiom) that for every property there is the set of those things having that property, we derive a contradiction.

When we eschew that view (and the axiom that captures it) that for every property there is the set of things having that property, we are not burdened with the contradiction that that view (and axiom) entails.

Summary,
The Russell set does not exist.
Based on the proposed defined mathematical object failing by contradiction. — Mark Nyquist

That's not how I would say it, but it's close enough. I won't split hairs to quibble with it.

In defining the Russell set, two or more (known to exist) defined mathematical objects are used to define the Russell set — Mark Nyquist

I don't know what you're referring to.

The definition is:

R = {y | y is not a member of y}

i.e.

R is the set of all and only those y such that y is not a member of y.

But it's an improper definition, because there is no set whose members are all and only those y such that y is not a member of y.

I used intermediate because there are two questions to this problem based on two phases.

Phase one, discovery,
Does the Russell set exist?
Requires exploration.
No.

Phase two, end point,
Is the Russell set a paradox?
Given: the Russell set does not exist.
Since the Russell set does not exist we now know it cannot be a paradox.

You can say my method is extraneous because you resolve it using your own method.

But you are using the ideas subconsciously.
And you default to 'not a paradox' when you reach your intermediate conclusion (is a contradiction therefore non-existent).

Also, there is word confusion in contradiction and paradox so be careful of that.

A little more.....You have,

A discovery phase where the question is does a proposed mathematical object exist or not exist.

And

An end point phase where you are given the state.... a(n) existent mathematical object or non-existent mathematical object. Is an existent mathematical object a paradox? No. Is a non-existent mathematical object a paradox? No.

So I don't see how a paradox can exist. The contradiction in your intermediate result is only a basis for determining non-existence.

Also the problem is misnamed because no true paradox ever exists. Because the discovery phase is hypothetical.

n — Mark Nyquist

I thought you meant an intermediate step in the proof. I have no comment on your characterization of phases.

The problem is not misnamed. I explained why it is a paradox from certain assumptions or axioms. Also, to be clear, 'paradox' is not a technical mathematical term. It's a general term for the situations I mentioned: From supposedly acceptable assumptions, principles or axioms, we derive a contradiction or even just a highly counter-intuitive implication.

Russell gets too much credit for suggesting a set that cannot be.

True, because he didn't discover it, Ernst Zermelo did.

thank you. I shall Google.

If I'm not mistaken, the importance of Russell's role is that he noted that the paradox applies to Frege's system.

I have this in my notes:

Russell discovered the paradox while studying Cantor's argument that there is no function from a set onto its power set. Zermelo had discovered the paradox earlier in 1900 or 1901 (cf. "Zermelo's discovery of the "Russell Paradox"" by Rang & Thomas in Hist Math 8 pp. 15-22).

Frege proposed the fix by using the following axiom schema (here couched set theoretically) instead of unrestricted comprehension:

For any formula, P in which x does not occur free:

ExAy(yex <-> (y not=x & P))

But the above axiom schema, with identity theory, is inconsistent with Exy x not=y, as became known to Russell and to Frege, and Lesniewski provided a proof in 1938 (cf. Fixing Frege by Burgess - pg 32-34; "On Frege's Way Out" by Quine (in Mind 1955 and in Selected Logic Papers - pg 151))

The proof is pretty cool, and it's interesting that it's much much more complicated than the easy Russell proof.

Of course it was Zermelo who provided a robust fix with:

AzExAy(yex <-> (y in z & P))

accompanied with the rest of the existence axioms.

Hi Michael

In case you are interested in discussing this further:

L = The list of all lists
LL = The list of all lists that list themselves

1) In which list does L list itself?
2) In which list is L a member of itself?

Can you answer both questions consistently and non-contradictorily?

I mentioned part of this before, but it was skipped:

In ordinary mathematics: A list is a sequence. A sequence is a function whose domain is an ordinal. A function is a certain kind of set of ordered pairs. So, usually we're not looking to see whether a list is a member of a list or not. Rather, usually, we would look to see whether a list is a member of the range of a list.

So, suppose there is a set S whose members are all and only the lists. (No such set exists in ordinary mathematics that does not have unrestricted comprehension, but suppose we have unrestricted comprehension.)

Let L be a sequence whose range is S. (L is a "list of all lists")

Let K be a sequence whose range is {x | x in S & x in range(x)}. (K is a "list of all the lists that list themselves")

Is L in L? No.

Is L in range(L)? Yes.

Is L in K? No.

Is L in range(K)? Yes.

Is K in L? No.

Is K in range(L)? Yes.

Is K in K? No.

Is K in range(K)? Yes.

Next question:

So what?

/

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.

That is another way of saying what Michael said in the post to which the previous post is in response.

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. 1 and 2 were solid meaningful questions that highlight precisely this point. L is listed in both L and LL. In L it has the property of being a member of itself/L, in LL it has the property of being a member of other than itself/L. L is not LL. L is L.

The property P is instantiated based on what set the item x is in. — Philosopher19

I don't know what you mean by that. I don't know what you mean by a property being instantiated in this context. I referred to any property. As I said: If x has property P, then that is not qualified by "x has property P in some sets but not others".

In a theory that does not disallow self-membership, a set may be a member of itself and it is a member of other sets too. An example has been shown many times already:

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.

This has been gone over already, but now we come back around full circle.

1) In which list does L list itself? — Philosopher19

This is the same tautological nonsense as before:

I said, in A, A is a member of A/itself, and in B, A is a member of B/other-than-itself — Philosopher19

Same thing:

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

L is listed in both L and LL. In L, it has the property P of being a member of itself/L. As in the fact that L is in L is what instantiates the property P (P = self-membership) in L's case.

In terms of function and logic, there is no difference between "lists itself" and "is a member of itself". Listing oneself and being a member of oneself are both matters of self-reference. So:
Does L list itself in L?
Does L list itself in LL?

So on the one hand you say:

In B, A is not a member of anything, A simply exists. — Lionino

On the other hand you say:

Because it exists in B, it is a member of B — Lionino

Do you see how you have contradicted yourself?

And then you say

It is a semantic point — Lionino

Yes, proper attention has not been paid to the semantics of "member of self" and "not member of self". Whether a set is a member of itself or not, is determined by what item it is and in what set it is in.

L lists itself in L. L does not list itself in LL. L is a member of itself in L/itself. It is not a member of itself in LL/not-itself.

In terms of function and logic, there is no difference between "lists itself" and "is a member of itself". — 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.

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

I answered those questions exactly.

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 believe this has nothing to do with what I said.

I answered those questions exactly. — TonesInDeepFreeze

I believe the exact answer to "Does L list itself in L?" is yes.

I believe the conversation won't progress beyond this point.

Peace

It has everything to do with what you said.

And I said that the exact answer to "Does L list itself?" is yes.

Progress will begin upon you paying attention to the replies you've received.

Do you see how you have contradicted yourself? — Philosopher19

I have not because those two are different sentences.