The math procedes in a way that assumes an ultimate set will exist.

I'm saying ultimately the Russell set does not ultimately exist.

Edit: rather never exists and ultimately does not exist...

And you can't have a paradox if the defined mathematical object does not exist.

The math procedes in a way that assumes an ultimate set will exist. — Mark Nyquist

"The math" refers to what mathematics?

After Russell discovered the paradox, mathematicians replaced the systems that allowed the paradox with systems that don't allow the paradox. With set theory since the early 20th century, we have systems that don't have the paradox, as indeed the systems prove that there does not exist a set whose members are all and only the sets that are not members of themselves.

I'm saying ultimately the Russell set does not ultimately exist. — Mark Nyquist

And you can't have a paradox if the defined mathematical object does not exist. — Mark Nyquist

Then you agree with set theory.

And you can't have a paradox if the defined mathematical object does not exist. — Mark Nyquist

You misunderstand the paradox.

Naive set theory allows the Russell set. The Russell set is a contradiction. Therefore, naive set theory is inconsistent.

That's all it is.

We don't need to bring up mathematical realism.

You seem to be arguing for the paradox after the paradox has been dismissed.

You seem to be arguing for the paradox after the paradox has been dismissed. — Mark Nyquist

I'm arguing that Philosopher19 doesn't understand set theory, and that his attempted "solution" to the Russell paradox makes no sense.

The Russell paradox has already been "solved": see ZFC, for example.

Okay.

You [Michael] seem to be arguing for the paradox after the paradox has been dismissed. — Mark Nyquist

I don't know what you mean by "dismissed" but, mathematics came up with a system in which the paradox does not occur.

Okay. — Mark Nyquist

Great. That's progress.

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

That is incoherent and has no apparent meaning.

"A is a not a member of of both A and B" does not take a qualifier "In B".

In general: Let P be any formula of set theory and B be any set. We don't say "In B, P". Putting "In B" before "P" has no meaning and indeed is not even grammatical in this context.

Philosopher19 has some word salad dogma tossing around in his head. His first step should be to at least learn how to coherently express whatever it is he's trying to say.

Defined mathematical objects can be subdivided:

Proposed defined mathematical object
Confirmed defined mathematical object...exists
Rejected defined mathematical object ... non-existent

So the Russell set as a rejected defined mathematical object does not exist and cannot be a paradox.

So the Russell set as a rejected defined mathematical object does not exist and cannot be a paradox. — Mark Nyquist

A paradox occurs when there is a contradiction or highly counter-intuitive statement that
follows from premises or principles that we regard as themselves true, reasonable or intuitive.

I suggest that the supposed Russell set is not itself a paradox.

But rather, the paradox is that from the premise "for any property of sets, there exists the set whose members are all and only those sets that have that property" we get "There exists the set whose members are all and only those sets that have the property of not being members of themselves (i.e., the supposed Russell set)", which implies the contradiction "The Russell set is a member of itself and the Russell set is not a member of itself".

There, the premise that up to 1901 we thought was true, reasonable and intuitive is "for any property of sets, there exists the set whose members are all and only those sets that have that property" but the contradiction that follows from it is "The Russell set is a member of itself and the Russell set is not a member of itself".

So we see that what we thought back around 1901 to be true, reasonable and intuitive, and that we moreover had as an axiom around that time, is actually worse than false as it implies an outright contradiction. Therefore, since Russell's paradox was announced, mathematicians have eschewed using the aforementioned premise, have eschewed using it as an axiom.

Note that this can be seen to be a matter even more basic than set theory. We can see formulate it even without mentioning 'set' or 'member':

For any 2-place relation R, there is no x such that for all y, y bears R to x if and only if y does not bear R to y.

For example, famously:

There is no Mr. X such that for everyone, they bear the relation with Mr. X of being shaved by Mr. X if and only if they bear the relation with themselves of not being shaved by themself.

More simply: There is no one who shaves all and only those who do not shave themself.

Or:

There is no one who loves all and only those who do not love themself.

Doesn't matter what the binary (2-place) relation is: 'member of' or 'shaves' or 'loves', etc.

You suggest the Russell set exists only based on the process of defining it.

I'm saying it does not exist.

Which is it.

In my development of the issues these are mathematical objects that don't exist outside of brain state.
One method of establishing existence is predication such as for fixed mathematical objects.

Defined mathematical objects are not based on predication....because we set the definition process. I'm saying for an object to be a legitimate mathematical objects it needs to pass an existence test. Creating paradoxes is a failure of an existence test.

Didn't we go over this here?

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.

I'll copy from the Wikipedia article:

The term "naive set theory" is used in various ways. In one usage, naive set theory is a formal theory, that is formulated in a first-order language with a binary non-logical predicate $\in$, and that includes the axiom of extensionality:

$\forall x \, \forall y \, ( \forall z \, (z \in x \iff z \in y) \implies x = y)$

and the axiom schema of unrestricted comprehension:

$\exists y \forall x (x \in y \iff \varphi(x))$

for any formula $\varphi$ with the variable $x$ as a free variable inside $\varphi$. Substitute $x \notin x$ for $\varphi(x)$ to get:

$\exists y \forall x (x \in y \iff x \notin x)$

Then by existential instantiation (reusing the symbol $y$) and universal instantiation we have:

$y \in y \iff y \notin y ,$

a contradiction. Therefore, this naive set theory is inconsistent.

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

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

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

This is what you're failing to understand.

According to naive set theory, the Russell set "exists".

The Russell set doesn't "exist", because the Russell set is a contradiction.

Therefore, naive set theory is inconsistent.

The solution is to fix the inconsistencies with naive set theory, which was done in ZFC (and others).

Finally, yes, that is what I was getting at.

It's my first look at anything in set theory so I don't have background.

It's my first look at anything in set theory so I don't have background. — Mark Nyquist

Then you really shouldn't comment, because by your own admission you don't understand the problem.

Finally, yes — Mark Nyquist

This was explained in the very first comment of this discussion by fishfry.

Okay. I'll go over and sit in the corner and read some set theory textbooks.

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. — Michael

It seems evident to me that you've not been paying attention to what I've been saying. I've already given more than one relevant reply to this. I'm not going to repeat myself.

Fine, I will more or less repeat myself but this is probably the last time.

If A is the set {A} then A is a member of both A and U. — Michael

p) In A, A is a member of itself/A.
q) In U, A is not a member of itself/A.

p and q are true by definition.

If definitional truths are not enough (and they really should be), here is a demonstration of how a set can only be viewed as a member of itself in its own respective set:

v = any set
The v of all vs = the set of all sets
z = any set that is not 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

In the z of all zs, the z of all zs 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 v of all vs. We can't treat two different things/references/contexts/standards 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?)

p) In A, A is a member of itself/A.
q) In U, A is not a member of itself/A.

p and q are true by definition. — Philosopher19

No, it's nonsense. That's not how set theory works.

1 is a member of N and R.
A is a member of A and B.

That's it.

Just take a few actual lessons in set theory.

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. — Philosopher19

That is tautological. As pointed out before there is no such as "in A". Yes, within the set A, A is a member of A. That is a tautalogy, cut it out to "Yes, A is a member of A" and you get the same meaning. So it is wrong to say the quote above because it is the same as saying that A can't be both a member of A and B.

Your argument seems to be that A is not a member of B in A because B is not defined in A. That doesn't make sense because sets do not include definitions of other sets. Even if they did somehow, by your argument: even in B, A would not be a member of B because B is not defined within B either.

I'm saying whether something is a member of itself or not is determined by whether it is in its own set or not. This matter of pure reason/definition seems to have been disregarded and faulty notation/language has then been used to falsely conclude there is no universal set. It seems to me foundationally corrupt theories have been built on a misunderstanding and so much has been invested in them that there is either unwillingness to let them go, or there is so much dogma that the obviously true is not paid attention to.

Your argument seems to be that A is not a member of B in A because B is not defined in A — Lionino

My argument is that A is not a member of itself in B because A is a member of B in B. To preserve this truth and take into account what set theorists seem to be focusing on, I wrote the following:

here is a demonstration of how a set can only be viewed as a member of itself in its own respective set:

v = any set
The v of all vs = the set of all sets
z = any set that is not 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

The z of all zs is a member of itself in the the z of all zs, 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 v of all vs. We can't treat two different things/references/contexts/standards/items 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

For the whole of what I'm saying, read:

http://godisallthatmatters.com/2021/05/22/the-solution-to-russells-paradox-and-the-absurdity-of-more-than-one-infinity/

No, it's nonsense. That's not how set theory works.

1 is a member of N and R.
A is a member of A and B.

That's it. — Michael

Already responded to it at least twice. Won't repeat myself.

So it's settled in your mind that you won't read even a single book or article about this subject, but that others should read your website?

Won't repeat myself. — Philosopher19

Oh, yes you will. I guarantee it.

So it's settled in your mind that you won't read even a single book or article about this subject, but that others should read your website? — TonesInDeepFreeze

I believe I understand the matter well enough. I don't feel it sincere to Truth/Goodness to read/research any more on it than I already have. The link to what I've written on Russell's paradox and Infinity is there should anyone choose to read it. Whether they should read it or not, is not for me to say. It may be sincere to Truth/Goodness for them to read it, it may be sincere to Truth/Goodness for them not to read it. It may even be that Truth/Goodness is not their priority. I don't know, I don't have their self-awareness.