In the case of B = {A, 0}, is A a member of A/itself, or is A a member of B/non-itself? — Philosopher19

Both

Both — Michael

But in B, A is a member of B. In B, A is not a member of both A and B. So once again, in B, is A a member of itself or not a member of itself?

So once again, in B, is A a member of itself or not a member of itself? — Philosopher19

Both a member of itself and a member of B.

In B, A is not a member of both A and B. — Philosopher19

Yes it is.

Both a member of itself and a member of B. — Michael

Look at what you're saying:

In B, A is both a member of A and B.

"In B" does not equal to "in both A and B".
Do you see your contradiction? You have treated "in B" as the same as "in both A and B".

There's no such thing as "in A" and "in B".

It is just the case that the symbol "A" is defined recursively as "{A}" and that the symbol "B" is defined as "{A, 0}", which is the same as "{{A}, 0}" given the recursive definition of "A".

$A=\{A\}\\B=\{A,0\}=\{\{A\},0\}$

It is just the case that the symbol "A" is defined recursively as "{A}" and that the symbol "B" is defined as "{A, 0}", which is the same as "{{A}, 0}" given the recursive definition of "A". — Michael

And now you have strayed from clear language. What I have given you is clear should you choose to pay attention to it:

I asked you:

In the case of B = {A, 0}, is A a member of A/itself, or is A a member of B/non-itself? — Philosopher19

to which you said:

Both a member of itself and a member of B. — Michael

to which I said:

Look at what you're saying:

"In B" does not equal to "in both A and B".
Do you see your contradiction? You have treated "in B" as the same as "in both A and B". — Philosopher19

Just take a math lesson or two.

↪Philosopher19 A is a member of both A and B. This is basic set theory. Take a math lesson. — Michael

And in B (as opposed to in both A and B), A is not a member of itself because it is a member of B (and not A). This is basic.

And in B, A is not a member of itself. — Philosopher19

N is the set of natural numbers.
R is the set of real numbers.

Every natural number is a member of both N and R (every natural number is both a natural and a real number).

We don't say "in R, 1 is not a member of N".

N is the set of natural numbers.
R is the set of real numbers.

Every natural number is a member of both N and R. We don't say "in R, the natural numbers are not members of N". — Michael

I addressed a similar point in my reply to your non-math example. Every natural number can be both a member of N and R. I am not denying this. But what a natural number cannot be (and what N and R cannot be) is members of themselves. When you talked about A, you talked about something that was by definition a member of itself precisely because A = {A}. But in the case of B = {A, 0}, A is not a member of itself because now, by definition, A is a member of B.

So why is it that A can be both a member of B and C but not a member of both A and B?

So why is it that A can be both a member of B and C but not a member of both A and B? — Michael

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.

Yet you seem to think that in B, A is both a member of A and B (which is contradictory)

Explain to me the difference here, and why you disagree with scenario 2:

Scenario 1
B = {0, A}, where A = {1}

Scenario 2
B = {0, A}, where A = {A}

Scenario 1
— Michael

Scenario 2
B = {0, A}, where A = {A} — Michael

I have not disagreed with scenario 2. I have said that in B, A is not a member of itself precisely because it is a member B (as opposed to itself), and in A, A is a member of itself.

I believe I have already said more than enough. Check my replies to you because I believe I am now repeating myself where I shouldn't have to.

I have not disagreed with scenario 2. I have said that in B, A is not a member of itself precisely because it is a member B (as opposed to itself), and in A, A is a member of itself. — Philosopher19

And you’re confused. It’s not the case that “in A” it’s a member of one thing and “in B” it’s a member only of something else.

It’s the case that in scenario 2, A is a member of A and B.

Maybe the following will help. I don't know.

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

And you’re confused. It’s not the case that “in A” it’s a member of one thing and “in B” it’s a member only of something else. — Michael

Am I the one that's confused? So it's not the case that in A it's a member of one thing and in B it's a member of another thing? So it's not the case that in A it's a member of itself and in B it's not a member of itself?

These are blatantly obvious things. As far as I can see, only dogma and bias and insincerity to Truth would cause one to fail to see/recognise them.

Mathematical objects exist as brain state.

Brain; (mathematical objects)

These come in different forms with specific properties.

Like fixed,

Brain; (fixed mathematical objects)

Things like pi, i, e, trig functions, √2, √3.

Or defined,

Brain; (defined mathematical objects)

Like defining sets, setting parameters, variables in functions.

Not sure if this will help with Russell's paradox but in general, applying universal form and understanding mathematical objects will work in understanding paradoxes or contradictions

I might give it a try later.

I cover the basics of mathematical objects in my post on universal form. There is an example of the contradictions in time perseption given.

Am I the one that's confused? So it's not the case that in A it's a member of one thing and in B it's a member of another thing? So it's not the case that in A it's a member of itself and in B it's not a member of itself? — Philosopher19

Yes, you’re confused. A is a member of A and B. 1 is a member of N and R. That’s all there is to it.

But none of this addresses the fundamental problem with this discussion, and that is that this is Russell’s paradox:

1. x is a member of R if and only if x is not a member of x
2. Let x = R
3. R is a member of R if and only if R is not a member of R

You haven’t proved that (3) is not a contradiction. And you can’t because it is.

The solution to the paradox is known: construct a set theory with axioms that do not entail (1).

ZFC does this by not allowing a set to be a member of itself. New Foundations does this by restricting which sorts of sets can be members of themselves.

Yes, you’re confused. A is a member of A and B. 1 is a member of N and R. That’s all there is to it. — Michael

I believe I addressed this point both in my reply to your non-math example and in your N and R example and in my other replies. No point doing in doing it again. Plus there is my z and v example.

As you seem unwilling to accept facts about maths, let's use your own reasoning against you.

1. x is a member of A if and only if x is a member of x
2. Let x = B
3. B is a member of A if and only if B is a member of B

But according to your reasoning, (3) is a contradiction. Therefore (1) is a contradiction.

Your axioms lead to an inverse Russell paradox.

We can resolve this either by allowing that B is a member of both A and B, or by not allowing a set to be a member of itself.

1. x is a member of A if and only if x is a member of x — Michael

I have been saying to you that x is not a member of A if x is a member of x. I have not been saying 1. 1 is blatantly contradictory. So it seems to me that you have misrepresented/misunderstood what I have been saying to you.

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.

ZFC does this by not allowing a set to be a member of itself. — Michael

That's a common misconception.

Yes, ZFC has the axiom of regularity that implies that no set is a member of itself. But that doesn't avoid Russell's paradox.

Rather, Russell's paradox is avoided by not having unrestricted comprehension.

Even with the axiom of regularity, and even if there are no sets that are members of themselves, with unrestricted comprehension we would have the set of all sets that are not members of themselves, thus a contradiction.

Explicity:

1. ExAy(yex <-> ~yey) ... instance of unrestricted comprehension
2. Ex(xex <-> ~xex) ... from 1

Having the axiom of regularity, or any other axiom, does not block getting 2 from 1.

The way to not have 2 is not to have 1, irrespective of the axiom of regularity.

I looked at Russell's paradox again and am thinking the problem is in the use of defined mathematical objects.

I would classify the Russell set as a defined mathematical object. That means it is subject to a determination of if it exists or does not exist. The fact that paradoxes develop means it is a defined mathematical object that does not exist.

This is basic theory of predication. Fixed mathematical objects are determined by predication. Defined mathematical objects are arbitrary and defined by rules and ultimately may not exist.

In short....the Russell set never exists as the defining process procedes and once the problems are discovered the conclusion should be the Russell set is non-existent.

The fact that paradoxes develop means it is a defined mathematical object that does not exist. — Mark Nyquist

In a theory with unrestricted comprehension we define the set whose members are all and only those sets that are not members of themselves. And having that set implies a contradiction.

In modern set theory, we do not have unrestricted comprehension and there is no way to define a set whose members are all and only those sets that are not members of themselves.

Never had the set in the first place.

Never had the set in the first place. — Mark Nyquist

Who or what never had the set? And what is "the first place"? You, personally? People who don't work with set theory?