I'd say it's not just incomplete. It's contradictory in the sense that it logically implies "a list can't list itself". That's like saying a shape can't be a triangle (which is contradictory because a shape can be a triangle).

ZFC is, I believe, set up specifically so that "a list can't list itself". That's how it avoids the various paradoxes.

Unfortunately, this thread is the definition of insanity.

↪Philosopher19 ZFC is, I believe, set up specifically so that "a list can't list itself". That's how it avoids the various paradoxes. — Banno

But you don't solve a paradox or contradiction by seeking refuge in another.

It is more damning/problematic to reject the set of all sets than to accept ZFC, but any statement, belief, or theory that has a contradiction in it, is wrong by definition.

In any case, I believe I presented a true solution to Russell's paradox. The full writing is here if you are interested:

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

Ok. I think is right. I'll leave you to it.

Russell's paradox is a mathematical proof that the axiom schema of unrestricted comprehension leads to a contradiction. As such, early naive set theories had to be abandoned.

Zermelo–Fraenkel set theory is the most used replacement, and doesn't allow for a universal set.

New Foundations is an alternative replacement that does allow for a universal set.

This isn't really anything to do with philosophy. It's just about the internal consistency of some set of mathematical axioms. Some lead to a contradiction, as Russell's paradox shows, and so their axioms must change.

If you're trying to argue that a "correct" set theory must allow for a universal set then I don't think you really understand mathematics.

ZFC is, I believe, set up specifically so that "a list can't list itself". That's how it avoids the various paradoxes. — Banno

Yes, the axiom of regularity and the axiom of pairing entail that no set is an element of itself.

Problems occur if you consider the elements of a set to not be themselves sets. Set theory only talks about sets. It does not, for example, talk about individuals.

The lists only list other lists... — Banno

The question of what "a set" is, the definition of "set", becomes an issue when you consider the possibility of an empty set. If a "set" is taken to be the type, category, or definition which indicates the criteria of membership, then an empty set is possible. If "set" is supposed to refer to the group or collection of elements itself, then an empty set is impossible because that would be a non-existent collection of elements. The "non-existent collection" could not be understood by that definition, because that would mean "set" is understood by the category, not the elements in membership, So it is at this point, when we consider the possibility of an empty set, that we need to make a judgement about the relationship between individuals and sets.

If you're trying to argue that a "correct" set theory must allow for a universal set then I don't think you really understand mathematics. — Michael

I believe I understand Russell's paradox very well, but I am not a mathematician.

I believe I understand Russell's paradox very well — Philosopher19

So you understand the below?

Axiom of extensionality:
${\displaystyle \forall x\,\forall y\,(\forall z\,(z\in x\iff z\in y)\implies x=y)}$

Axiom schema of unrestricted comprehension:
${\displaystyle \exists y\forall x(x\in y\iff \varphi (x))}$

Substitute ${\displaystyle x\notin x}$ (the Russell set) for ${\displaystyle \varphi (x)}$:
${\displaystyle \exists y\forall x(x\in y\iff x\notin x)}$

Therefore:
${\displaystyle y\in y\iff y\notin y}$

The conclusion is a contradiction. Therefore the premises are inconsistent. In this case, the problematic premise is the second premise, which is the axiom schema of unrestricted comprehension.

I don't understand the notation you have used. If you were to put it into words, I could reply in kind.

The axiom of extensionality
Given any set A and any set B, if for every set X, X is a member of A if and only if X is a member of B, then A is equal to B:
${\displaystyle \forall x\,\forall y\,(\forall z\,(z\in x\iff z\in y)\implies x=y)}$

The axiom schema of unrestricted comprehension
There exists a set B whose members are precisely those objects that satisfy the predicate $φ$:
${\displaystyle \exists y\forall x(x\in y\iff \varphi (x))}$

Russell's paradox
Let $φ(x)$ be ${\displaystyle x\notin x}$ ($x$ is not a member of $x$):
${\displaystyle \exists y\forall x(x\in y\iff x\notin x)}$

Therefore:
${\displaystyle y\in y\iff y\notin y}$

This is a contradiction. Therefore the axiom of extensionality and the axiom schema of unrestricted comprehension are inconsistent.

ZFC replaces the axiom schema of unrestricted comprehension with the axiom of regularity and the axiom of pairing and as such is consistent. This doesn't allow for a universal set.

New Foundations restricts the axiom schema of comprehension by allowing only stratifiable formula for $φ(x)$. This allows for a universal set.

Given any set A and any set B, if for every set X, X is a member of A if and only if X is a member of B, then A is equal to B (the of axiom of extensionality) — Michael

Are you saying that A and B only contain X as members of themselves and they contain nothing other than X as members of themselves? If you are, I don't see it included in the above. If you are not, then either A or B can contain members other than X such that A is not equal to B despite both A and B have all Xs as members of themselves

It says that A and B are equal if every member of A is a member of B and every member of B is a member of A.

It says that A and B are equal if every member of A is a member of B and every member of B is a member of A. — Michael

Ok

There exists a set B whose members are precisely those objects that satisfy the predicate — Michael

is predicate φ "A and B are equal if every member of A is a member of B and every member of B is a member of A"? If not, what is it?

is predicate φ "A and B are equal if every member of A is a member of B and every member of B is a member of A"? If not, what is it? — Philosopher19

In Russell's paradox, $φ$ is "sets that are not members of themselves".

In Russell's paradox, φ is "sets that are not members of themselves". — Michael

Thanks.

So when you say:

There exists a set B whose members are precisely those objects that satisfy the predicate — Michael

are you essentially saying "there is a set that contains all sets that are not members of themselves"? If not, can you clarify?

Yes. Given the axiom schema of unrestricted comprehension there exists a set B whose members are sets that are not members of themselves.

This leads to a contradiction. If B is not a member of itself then it is a member of itself.

Therefore, we must reject the axiom schema of unrestricted comprehension (or the axiom of extensionality, but that would be far too problematic).

↪Philosopher19 Yes. Given the axiom schema of unrestricted comprehension there exists a set B whose members are sets that are not members of themselves. This leads to a contadiction. If B is not a member of itself then it should be a member of itself. — Michael

There is a difference between:

1) There exists a set whose members are sets that are not members of themselves
2) There exists a set that contains all sets that are not members of themselves

2 is not contradictory at all whereas 1 could be contradictory.

2 is not contradictory because by definition, the set of all sets contains all sets that are not members of themselves and it is a member of itself. Where is the contradiction here?

1 is contradictory if you say set B only contains all sets that are not members of themselves. Again, the only set that by definition can contain all sets that are not members of themselves, is the set of all sets and it does not just contain all sets that are not members of themselves, it contains itself too.

All sets that are not members of themselves have to be a member of something don't they? It's like saying all existing non-self-contingent things have to be contingent on something don't they?

1 is contradictory if you say set B only contains all sets that are not members of themselves. — Philosopher19

Yes, that's the premise. Let R be the set of all sets that are not members of themselves. This is the Russell set.

If R is not a member of itself then it is a member of itself. This is a contradiction. Therefore we must abandon the axiom schema of unrestricted comprehension.

ZFC replaces the axiom schema of unrestricted comprehension with the axiom of regularity and the axiom of pairing. These entail that no set is an element of itself, and so doesn't allow for a universal set.

Whereas NF restricts the axiom schema of comprehension by allowing only stratifiable formulas. This doesn't allow for a Russell set but does allow for a universal set.

There's really nothing to argue here. Russell's paradox is an undeniable proof that naive set theory is inconsistent. The "resolution" is to replace the problematic axiom with others, which is what has been done. ZFC does it one way, NF another way, and others in other ways.

I am a lay person trying to learn formal logic, so this post is probably a bit of a lower level than the rest of the posts. Studying Russel's paradox make me think I understand the difference between a subset and a member better. Take these examples:

( A is a subset of A ) = ( {1,2,3} = {1,2,3} )

(A is a member of A ) = ( {1,2, A} = {1,2, {1,2, {A} }, } = {1,2,{1,2,{1,2,{1,2{...}}}}}

From this example, you can see that defining a set as a member of itself immediately leads to infinite recursion.

Let A be the set of all integers.
Let B be the set of all positive integers.

Every member of B is also a member of A, but some members of A are not members of B (i.e. the negative integers).

B is a subset of A.
A is a superset of B.

Yes, I understood that just fine before reading this post. I am at the level where I can understand much of the formal symbols I have seen in this post, but would have trouble writing them myself. The biggest problem is that since I am learning this stuff randomly rather than as a part of a formal curriculum, there are many holes on my knowledge, and I do not know where to look to fill those holes. I am reading an introductory text on formal logic now, and it is tedious not only because of the tedious nature of the subject, but also because I have to go through a lot that I already know before encountering something that I do not know yet.

But you haven't addressed my point.

We are in agreement that p) you cannot have a set that only contains all sets that are not members of themselves. I am saying this has nothing to do with the fact that you can have q) a set that contains all sets that are not members of themselves. The set of all sets contains all sets that are not members of themselves (plus itself). Again, where is the contradiction in this?

It is clear that p is contradictory. You have not shown how this logically obliges us to view the set of all sets as contradictory (you have only shown how p is contradictory. See your post). And you have also not shown any contradiction in q.

This is where you show that you don't understand the problem.

Naive set theory accepts both the axiom of extensionality and the axiom schema of unrestricted comprehension.

The axiom schema of unrestricted comprehension entails that there is a set that only contains all sets that are not members of themselves (the Russell set). In conjunction with the axiom of extensionality this is a contradiction, and so naive set theory is shown to be inconsistent.

This is all Russell's paradox does.

In response to this, naive set theory had to be changed. Specifically, it had to reject the axiom schema of unrestricted comprehension.

ZFC replaces this axiom with two others; the axiom of regularity and the axiom of pairing. These axioms entail that no set is a member of itself, and so that there is no universal set.

NF restricts the axiom schema of comprehension to permit only stratifiable formulas. This does allow for a universal set.

It's not entirely clear what you're trying to argue. Is it that Russell's paradox doesn't prove naive set theory is inconsistent? Is it that ZFC allows for a universal set? You'd be wrong on both counts. Is it that NF is "better" than ZFC? I'm not entirely sure such a claim would make sense.

Or are you trying to argue that, independently of any set theory, a set that contains all sets that are not members of themselves is possible, therefore only set theories that allow for this are "correct"? That's putting the cart before the horse.

Russell proved that these two axioms entail that there is a set that only contains all sets that are not members of themselves (the Russell set). — Michael

I followed your original notation and tried to get clarity on it. We came to the following:

So when you say:
There exists a set B whose members are precisely those objects that satisfy the predicate
— Michael
are you essentially saying "there is a set that contains all sets that are not members of themselves"? If not, can you clarify? — Philosopher19

To which you answered "yes". To which I highlighted to you a clear difference:

1) There exists a set whose members are sets that are not members of themselves
2) There exists a set that contains all sets that are not members of themselves — Philosopher19

Again, 1 is contradictory. Put it in clear language as to why the contradictoriness of 1 obliges us to reject 2 or to view the set of all sets as contradictory.

Again, 1 is contradictory. Put it in clear language as to why the contradictoriness of 1 obliges us to reject 2 or to view the set of all sets as contradictory. — Philosopher19

The axioms of naive set theory entail (1). Therefore, the axioms of naive set theory are inconsistent.

This is all Russell's paradox shows. Again, you're showing that you don't even understand the problem.

When you say the axioms of naive set theory, are you referring to those notations that I asked you to put in clear language. If so, it seems to me you left half way through trying to clarity on it.

When you say the axioms of naive set theory, are you referring to those notations that I asked you to put in clear language. — Philosopher19

Yes.

If so, it seems to me you left half way through trying to clarity on it. — Philosopher19

I'm not interested in teaching you mathematics. I am simply explaining to you that Russell proved that the axioms of naive set theory are inconsistent. That's it. It's not a debatable issue.

Given your responses in the other discussion (the infinity one). I see no point in continuing this discussion.

Peace