We're not talking about models here, though. We're talking about domains of discourse. — fishfry

model theory is not relevant to this conversation as I understand it. — fishfry

The question was:

wouldn’t the statement or conclusion «there is no set of all sets» be all-inclusive in one way or another if it really is true? — Sunner

"all inclusive in one way or another" is not definite. My choice was to give a mathematically definite framework for the the question. I surely do not presume to guarantee that I know what the poster in particular had in mind, but I responded correctly vis-a-vis the best way I know to make the question mathematically definite.

in general, there is an operation of unrestricted complement. — fishfry

Not in set theory (as I see you agree). And (just to be clear) in class theory, only with sets, not proper classes.

The complement of the set {1,2,3}, within set theory, is the collection of all sets that are not {1,2,3}. That complement is a well-defined collection, but it's not a set. — fishfry

It is a proper class (as you mention also) in class theory. In set theory, it doesn't exist. Clearly, the context of my posts was set theory.

I looked up Domain of Discourse on Wikipedia, and they did indeed say a domain of discourse is a set. I assumed they were mistaken, and were simply using "set" in its everyday, casual meaning, without regard for the issues of set-hood versus proper classes.

So I agree that if I looked it up, I'd find at least one source, namely Wiki, that claims a domain is a set. I just think they're wrong, and gave many examples to show why. — fishfry

Wikipedia is unreliable for mathematics (and other subjects). But it happens to be correct on this matter. And you don't need to wonder whether it's just a fluke of Wikipedia. Look in any textbook in mathematical logic or model theory. Or any PDF book or class notes on the Internet. The universe (aka 'the domain of discourse') for a model is a set.

any model of set theory has as a set, not a proper class, as its domain of discourse. For any model, the universal quantifier ranges over the members of the domain of discourse, which is a set.
— TonesInDeepFreeze

You are saying that when I make a statement such as, "Every set has a powerset," I am really saying:

1. I assume ZF is consistent.

2. By Gödel's completeness theorem, if ZF is consistent it has a model, which is a set.

3. The powerset axiom is implicitly quantifying over that set. — fishfry

No, I am definitely not saying that.

Phrases such as "quantifier ranges over" are not definite. To pin them down to a definite mathematical formulation, we turn to the mathematical logic. The method of models gives us definite formulations. A quantifier ranges over a universe. But what is that universe? Well, a universe is the carrier set for a model. So, per any given model, the quantifier ranges over the universe of that model.

In what I said, there is no need for an assumption that set theory has a model or is consistent. There is a crucial between (1) a model for the language of a theory and (2) a model of the theory:

(1) M is a model for a language L iff [fill in the definition here, in outline: M is a non-empty set together with a mapping of the non-logical symbols to (elements of the universe, n-ary functions on the universe, and n-rary relations on the universe)]

(2) M is a model of a theory T iff (M is a model for the language of T & every theorem of T is true in M)

Of course, if a theory is inconsistent, then there are no models of that theory. But whether or not a theory is consistent, there are models for the language of the theory.

To mention that a quantifier ranges over a universe per a model requires only (1) and no consideration whether any given model is or is not a model of the theory.

set theory can not prove its own consistency. — fishfry

Yes, if set theory is consistent, then set theory does not prove that set theory is consistent. Nothing I've said contradicts that.

The claim that every set has a powerset is true whether or not set theory has a model. All that is required is the axiom of powersets. — fishfry

No, a sentence is true or not depending on what model we're looking at. Truth is defined per models. The power set axiom is true in some models but false in other models (as you agree). It is true in any model of set theory (since set theory includes the power set axiom) but it is false in other models (ones that are not models of set theory).

Indeed, "Every set has a powerset" is NOT a semantic claim; it's a syntactic one. It follows from the axiom of powersets. There are models lacking the axiom of powersets where the claim is false. — fishfry

A sentence is an uninterpreted syntactical object. Aside from what the word 'claim' means, a sentence is interpreted per models.

And the phrase "models lacking the axiom of powersets" doesn't make sense. Models don't have axioms. Rather, axiomatizations of theories have axioms. What you might mean is "models in which the power set axiom is false".

And "every set has a powerset" is just an English way of saying the power set axiom.

Perhaps we're arguing about syntactic versus semantic domains. — fishfry

I have never read of a "syntactic domain". I don't know what you mean by it.

On the other hand, we can always define unary predicate symbols. For example, in set theory, we can have the predicate symbol 'G' defined by

Gx <-> [fill in the requirements for x being a group]

Then we have universally quantified conditionals:

Ax(Gx -> P)

Note the quantifier ranges over the universe, but it happens that the formula it applies to is a conditional in which x being a group is the antecedent.

So that is a relativization of P to groups.(That's a simplification. Relativizations are recursive so that P and its subformulas are themselves relativized.]

For example, we define a unary predicate symbol 'L' and 'V':

Lx <-> [fill in the requirements for x being constructible]

Vx <-> x=x

So the relativizations such as:

Ax(Lx -> P)
read as "P holds for constructible sets".

and

V = L
for
Ax Lx

And per a given model, 'L' will map to a subset of the universe, and 'V' will map to a subset of the universe (and if the model is a model of "Vx <-> x=x" then V maps to the subset of the universe that is the universe itself).

that impressively buzzword-compliant paragraph — fishfry

I didn't use the terminology to impress anyone with buzzwords. And they are not buzzwords. They are terminology of mathematics.

there is no mention of the domain of discourse. So again, none of this is relevant. — fishfry

It's about models. A model is a domain of discourse along with a function on the nonlogical symbols. So where I mentioned a model, there is a domain of discourse associated with that model. And the paragraph was not meant to address the original question, but rather to give an idea of how notions of proper classes as models are (as I hope I recall correctly) reducible to the syntactical approach of relativizations.

You're perfectly correct that ZFC is consistent if ZF is, but what has that got to do with the conversation? — fishfry

It was added merely as an illustration of an important theorem that comes from relativizations.

For example, in set theory, we have a defined predicate 'is an natural number' (I'll use 'B'). So theorems such as:

Ax(Bx -> (x is even or x is odd))

And in the intended interpretation, 'B' maps to the set of natural numbers that is a subset of the universe for the model. But for another model, it might be a different universe, and 'B' might map to a different set from the set of natural numbers.

Or consider first order PA.

In the intended interpretation, the universe is the set of natural numbers, so the quantifier ranges over the natural numbers. But there are non-standard models (not just for the language but even of PA), so the quantifier ranges over a set very different from the set of natural numbers.

It is only by per a model that the domain of discourse is definite, and so it is only per a model that it is definite what the quantifier ranges over.

@TonesInDeepFreeze

Note the quantifier ranges over the universe, but it happens that the formula is a conditional in which being x being a group is the antecedent.

So that is a relativization of P to groups. — TonesInDeepFreeze

Well yes, we agree on that. But there is no set of all groups! The class of groups is a proper class. So you seem to be conceding my point.

In any event, much of the rest of your post is pretty technical and I'm not sure how it bears on the question. I did find this MathSE thread:

How do I quantify over the proper class of all the cardinal numbers? where there's a lot of learned back and forth about quantifying over proper classes.

The consensus seems to be that when we quantify over all sets or all cardinals or whatever, we are really restricting to the particular predicate that defines the object in question. Which doesn't help us, since the extension of a predicate is a class and not necessarily a set.

And then there is a comment from Asaf Karagila, a professional set theorist and prolific SE contributor. He says:

No, quantifiers quantify over everything. All sets. Bounded quantifiers are shorthands to make it clearer that we are only interested in a particular set. But you can bound them to a class just as easily.

As far as I'm concerned, if Asaf says we can quantify over classes, we can quantify over classes. That is good enough for me.

As far as whether a domain of discourse must necessarily be a set, this seems like a matter of which definition we choose. When we make general statements about sets or groups or cardinals, we are quantifying over proper classes. Whether you call that a domain of discourse or not seems like a question of semantics.

It is only by models that the domain of discourse is definite, and so it is only by models that it is definite what the quantifier ranges over. — TonesInDeepFreeze

This post crossed paths with mine. As far as I'm concerned if I have Asaf on my side I'm happy. And when we say "every set has a powerset," we are quantifying over the proper class of sets. And again, as far as what a domain of discourse is, that seems like a matter of semantics. If you say it has to be a set, then so be it. When we say every set has a powerset, we are quantifying over a proper class; but if you don't want to call that a domain of discourse, that's ok by me.

But there is no set of all groups! — fishfry

Why are you exclaiming that to me? Of course I agree.

The class of groups is a proper class. So you seem to be conceding my point. — fishfry

There is no class of all groups in set theory. Set theory has only classes that are sets. In class theory, there is the class of all groups.

To say I "concede" that is like saying I concede that 0+0=0.

In any event, much of the rest of your post is pretty technical and I'm not sure how it bears on the question. — fishfry

My post is nothing that isn't in, or discernible from, introductory mathematical logic.

And it bears on the questions in exactly the way I explained.

/

I can't comment on that quote without a link to it.

Except, as you've presented it alone, "quantifiers quantify over everything", I say:

Please rigorously, mathematically define "quantifier over" and "everything".

I mentioned a rigorous, mathematical usage. My posts are correct in that context. And that context is definitely relevant to addressing the original question in a rigorous, mathematical way.

Anyway, if we have a model of class theory (or even of set theory for that matter), then, yes, the sentences "Ax x is a class" or "Ax(x is a set or x is a proper class)" are true in the model. But, ironic though it may be, a model of class theory has a set as its domain of discourse. Class theory is a first order theory; and a model of a first order theory has a set as its universe.

As far as whether a domain of discourse must necessarily be a set, I don't see how that can be possible. When we make general statements about sets or groups or cardinals, we are quantifying over proper classes. — fishfry

I explained that exactly in my post.

Moreover, in an earlier post, I proved that allowing a universe for a model to be a proper class implies a contradiction. You can go back to look at that.

Whether you call that a domain of discourse or not seems like a question of semantics. — fishfry

I guess you mean "question of semantics" in the sense of how we use words.

I don't presume to say what people may mean by 'domain of discourse'. I just gave an answer to the original question by using 'domain of discourse' in the sense of a universe for a model. The poster himself didn't use 'domain of discourse'. Rather, I first used it, and in the sense of rigorous mathematics as developed in mathematics. If someone else means something else by 'domain' of discourse' or 'quantify over' then of course I can't ensure that my use agrees, while meanwhile I would say, "Then what are your rigorous, mathematical definitions of 'domain of discourse' and 'quantifies over'?"

I can't comment on that quote without a link to it. — TonesInDeepFreeze

I hotlinked it. Here's the link.

https://math.stackexchange.com/questions/2724236/how-do-i-quantify-over-the-proper-class-of-all-the-cardinal-numbers

I guess you mean "question of semantics" in the sense of how we use words. — TonesInDeepFreeze

Yes.

"Then what are your rigorous, mathematical definitions of 'domain of discourse' — TonesInDeepFreeze

As I stated up front earlier, my usage is informal.

my usage is informal. — fishfry

And my rigorous, mathematical and standard use and explanations are not refuted (or whatever your disagreement is supposed to be) by your own informal usage.

The original question was informal. The original question was in invitation to explain a seeming contradiction. That merits a response that is rigorous and definitive, in order to appreciate that mathematics indeed does not tolerate such a contradiction, not just by informal hand waving, so that when we look at the matter with exactness, we do show that the seeming contradiction actually is avoided.

And my rigorous, mathematical and standard use and explanations are not refuted (or whatever your disagreement is supposed to be) by your own informal usage. — TonesInDeepFreeze

As I'm sure I've agreed several times. If you don't want to call quantification over a proper class a domain of discourse, I'm fine with that. We frequently quantify over proper classes, however you call it.

The original question was informal. The original question was in invitation to explain a seeming contradiction. That merits a response that is rigorous and definitive, in order to appreciate that mathematics indeed does not tolerate such a contradiction, not just by informal hand waving, so that when we look at the matter with exactness, we do show that the seeming contradiction actually is avoided. — TonesInDeepFreeze

I thought your responses to the recent OP @Sunner were too detailed and technical to be of use at the level the question was being asked. And, frankly -- not really wanting to get back into this -- wrong. The Russell class DOES define a perfectly good collection. That was the question. @Sunner had the impression (rightly or wrongly) that you said it didn't. I pointed out that it did. Perhaps OP misinterpreted what you said. In that case you were right, and I added clarity. So everyone can be happy, yes?

Not sure what contradiction "mathematics indeed does not tolerate." The referent of this paragraph is unclear.

https://math.stackexchange.com/questions/2724236/how-do-i-quantify-over-the-proper-class-of-all-the-cardinal-numbers — fishfry

(1) I don't know exactly which post(s) and passage(s) he is saying "no" to.

(2) I understood everything in that thread prior to his post. And, modulo details of phrasing, I agree with the replies that, in set theory, we simply use the predicate 'is a cardinal'. That is what I said myself in my posts above.

(3) "bounded quantifiers" there is just another way of saying what I said: The antecedent of the conditional stipulates that a set has a certain definable property. And I said (though I don't know whether that poster agrees) that the quantifier itself ranges over the universe, not just those sets having the property specified in the antecedent of the conditional.

So, if I understand that poster, he's mostly in agreement with me, not you.

You had the notion that the quantifier ranges over groups. But the quantifier ranges over the universe.

As I understand him, his remarks mostly coincide with mine:

"quantifiers quantify over everything."

If 'everything' means 'everything in the domain of discourse' then I agree.

"All sets."

I don't know what 'all sets' means. But if it means 'all sets in the universe', then I agree.

"Bounded quantifiers are shorthands to make it clearer that we are only interested in a particular set."

That seems to be what I'm saying about the antecedent of the conditional.

"But you can bound them to a class just as easily."

I don't know what he means by 'bound them to a class'. But, as I keep saying, we can relativize (use a conditional) to a proper class through a defined unary predicate. Also, another poster there suggested using a defining formula, which is tantamount to what I've said.

We frequently quantify over proper classes — fishfry

That's your informal understanding. I can't comment with real definiteness, because your informality doesn't provide a clear, definite meaning of 'quantify over'. Meanwhile, I have given the rigorous sense in which I use 'quantify over'.

your responses to the recent OP Sunner were too detailed and technical to be of use at the level the question was being asked. — fishfry

So we've moved from your substantive criticisms, which were incorrect, to a criticism of the pedagogy.

My answer to the poster's question is somewhat (though not terribly) technical to avoid confusions from oversimplification or vague hand waving. Meanwhile, you quoted me about that, why I was as technical as I was, but replied merely to say "too technical". Argument by mere assertion is what that is. Moreover, if a poster lacks familiarity with set theory and mathematical logic, then we can bet that there is no ideal answer for him; no answer that wouldn't be either too vague to be responsible to the mathematics or too technical that he wouldn't quickly understand it. (Though, this particular poster did basically grasp the explanation.) I choose to err on the side of being correct, and then to leave it to the poster to follow up himself by learning more about the subject.

Not sure what contradiction "mathematics indeed does not tolerate." The referent of this paragraph is unclear. — fishfry

It refers to the original question, as I said exactly that's what it refers to - in the very first sentence of the paragraph.

And your criticism is belied by the fact that the poster himself explicitly said that my answer was clear and helpful, and his followup questions do show that he basically understood my answer.

First you challenged me on substantive points in my posts, and you were incorrect. Most particularly, a universe for a model is a set.

Then you griped that my answer to the poster was too technical and not helpful, which contradicts what the poster himself said.

So, I really don't know what your trip is.

And your criticism is belied by the fact that the poster himself explicitly said that my answer was clear and helpful, and his followup questions do show that he basically understood my answer. — TonesInDeepFreeze

Yes but he understood the opposite of the correct answer!

So, I really don't know what your trip is. — TonesInDeepFreeze

I've been feeling the same way about you.

I'll quit while I'm behind here.

he understood the opposite of the correct answer! — fishfry

You haven't shown that my answer is incorrect. Nor has anyone said what other answer is "the" correct answer.

The problem that deserved an answer (in my words):

Set theory says, "For all sets, it is not the case that it has every set as a member" (i.e. there is no set of all sets). But that refers to all sets. So isn't what is referred to by "all sets" a set that has every set as a member, which contradicts that there is no set of all sets?

My answer is basically: The meaning of "for all" is per models, such that for any given model there is a universe that is a set, but it is not the non-existent set of all sets.

That is perfectly correct and it shows that the seeming contradiction is not actually a contradiction.

To refute my answer would require, refuting at least one of these claims:

(1) "for all" is understood per models. Correct. Just look it up in any textbook in mathematical logic.

(2) The universe for a model is a set. Correct. Just look it up in any textbook in mathematical logic.

(3) The universe for a model is not the non-existent set of all sets. Correct. There is no such set, so no set, including a universe for a model, is the non-existent set of all sets.

I've been feeling the same way about you. — fishfry

I haven't, as you have, gone on and on challenging someone who is making correct mathematical statements. I have no trip, so nothing for you to wonder about.

«No sentence is everywhere true» and «every sentence is both true and false» would both seem to make themselves untrue, leaving one with something universal and some consistency. These very informal thoughts, and the seeming tension between them in set theory is why I found this interesting in the first place, so the in-depth answers from you both are much appreciated!

No sentence is everywhere true — Sunner

What does that mean?

Do you mean that there is no sentence that is true in all models?

But there are sentences that are true in all models.

every sentence is both true and false — Sunner

No, there are sentences that are true in all models, sentences that are false in all models, and sentences that are true in some models but false in other models. And, for any given model, there is not sentence that is both true and false in that model.

you have to convert it into ontological terms, because as it stands, the sets in Russell’s paradox are empty referents, and because of this, the conclusion of Russell’s Paradox has no significant value whatsoever, and definitely doesn’t rule out the possibility of an Absolute Being.

Is the relevant Russell set incoherent?

Yes I agree, I only tried to show my fascination over how one inevitable or primal notion (that of something universal) is tamed by another (that of consistency) in the context of set theory and domains of discourse.

Is the relevant Russell set incoherent? — Agent Smith

Your question is incoherent.

Consider the following two lists:

The list of all lists (Call this L)
The list of all lists that list themselves (Call this LL)

Both the above lists list themselves (put differently, both lists are members of themselves).

Importantly, they are only members of themselves in their own respective lists. As in the list of all lists only lists itself in the list of all lists (L is only a member of itself in L). It does not list itself in the list of all lists that list themselves (L is not a member of L/itself in LL precisely because it is a member of LL).

The point I'm trying to make:

You cannot have a set of all sets that are members of themselves with all its members actually being members of themselves whilst they are members of it.

This should resolve the subset issue and we should no longer contradictorily say "the set of all sets is contradictory".

The Russell set is the set of all sets that are not members of themselves.

To my understanding, the subset issue was because you could have a set of all sets that are members of themselves. Since you could have this you should also have been able to have a set of all sets that are not members of themselves so that there are no inconsistencies in the subset level. I have shown that you cannot have a set of all sets that are members of themselves.

Wanting to have a set of all sets that are not members of themselves that is itself not a member of itself is a contradictory thing to want.

Only the set of all sets can contain all sets that are not members of themselves precisely because all sets are members of it and not themselves (of course it is a member of itself)

My full writing on Russell's paradox can be found here if interested:

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

To my understanding, the subset issue was because you could have a set of all sets that are members of themselves. Since you could have this you should also have been able to have a set of all sets that are not members of themselves — Philosopher19

I don't see how that follows.

Wanting to have a set of all sets that are not members of themselves that is itself not a member of itself is a contradictory thing to want. — Philosopher19

That's exactly what Russell was proving.

That's exactly what Russell was proving. — Michael

So let's say he proved you could not have a set that contains all sets that are not members of themselves and nothing more.

Did he prove that the set of all sets is contradictory?
Did he prove that you could not have a set that contains all sets that are not members of themselves? Because by definition/logic, the set of all sets contains all sets that are not members of themselves (as well as itself).

If Russell proved anything, it's that you can't create a new set that contains absolutely all sets that are not members of themselves and nothing more.

Again, the set of all sets contains absolutely all sets that are not members of themselves as well as itself.

Where do we have a paradox in what I have proposed? — Philosopher19

Your own conclusion is the answer to this question.

The set of all sets which contains all these sets, is a member of itself because it truly is a set. — Philosopher19

If all sets are contained in the set of all sets (that are not members of themselves and nothing more), then still, no sets are on any other level than the other sets, all of them now being contained together inside one container (that is what contained in means)... except the set of all sets, itself also being a set. But if it is a set, it would have to be contained along with the other sets. But it's not. But it is. This is paradoxical.

I don't think there is a way completely out of the paradox. You don't undo this paradox. It manufactures itself as we speak about it from any direction.

To skip to the end explained a bit further below, the set of all sets becomes the definition of what a set is, hiding in an example of what a set is. The paradox arises from the fact that the set of all sets serves as an example of one of the sets, and the definition of any/all of the sets at the same time. The set of all sets, is itself a setting of objects into a mentally constructed container. But when we are seeking to contain all sets (mentally constructed containers) in a set, it could equally be said we are seeking to define what a set is. If we say what a set in-itself is (if we define a set), we say something of all sets. We have created a set whose members include something of all sets, or we have created a definition that applies to all sets, namely the definition of all sets.

This doesn't resolve the logical problem; it merely restates it. But I think shows how the assertion "set of all sets" points to an edge or limit where logic itself and the language used to communicate mental constructs and logic, are distinct from each other, and here, unable to overlap. Basically, we can't say what we mean here, but we somehow still know something and know what we mean.

Stepping back, a 12-year-old child who understands what a set is, can look at 5 different sets on a blackboard (numbers, letters, shapes, etc) and can easily point to all 5 of the members of the "set of all sets on the blackboard". Thus the concept of "the set of all sets" is simple, easy, useful, logical, functional, even for children. But then you ask the child "But what about the set of all sets itself? Isn't that now one of the sets on the blackboard? What happens when you point that one out too? You've just added another set to the blackboard, making your prior answer of "5" wrong. Or you failed to show how the set of all sets on the blackboard is itself one of the sets on the blackboard and given the answer '6' in the first place." Now a grown, seasoned, pioneering mathematician and logician is perplexed. Thus, the concept of the set of all sets is both really child's play, and seemingly impossible to penetrate for a wise old professor.

But I think if you look at it from other directions, (like a child perhaps), I think we start to see why even the child can make easy use of the "set" (one example being the set of all sets) despite the fact that these sets can be made to appear and disappear both within and without themselves when we say "the set of all sets."

What is a set? A set is both 1) a membership (usually of multiple members but not necessarily, but comprised of membership nonetheless), and 2) their gathering as one containing reference. Four penguins and four seals in a zoo - the set of penguins, which is a unity as one containing reference, is the multiplicity of four member individual penguins. This set is not a penguin itself, because it is a set, and this set must not be a penguin because it has to sit beyond the penguins in order to contain them all as a set. But this set evaporates if we remove all of the penguins, because it is a set of four penguins.

These are the moving parts here. Sets must be distinct from their members in order to be sets of any members. And when looking for a set of penguins, we see that the set is not only distinct from its members, the set is distinct in kind - it's not a penguin. Sets are not their own members. But sets must have members, or be comprised of membership. (Please ignore the empty set here, or pretend the empty set has one member, the object "nothingness".)

So what are we actually doing when we say "set of all sets"? Are we taking all sets, turning them into member objects like penguins, and then stepping outside these objects to make something that is different in kind to those objects, calling it a set, namely a set of all sets? Are we just misuing the word "set" somewhere when we say "set of all sets"? Or have we left the sets alone and created a new class of set so that the set of all sets is different in kind from all of the sets that are its members?

Crack it open again. What is a set? A set is a form of "all". You have four penguins and four seals in a zoo, and someone asks "how many penguins are in the zoo?" The answer can be to count the members of the set of penguins, or it can be to count all of the penguins. You don't need to clarify the "set of all penguins" to come up with same answer. You can count "the set of penguins" or "all penguins" and conduct the same operation. All of the penguins is the same thing as the set of penguins. Therefore, a "set" is a semantically distinct but nonetheless an equivalent form of "all". "All" seals means the same thing as "the set of" seals here. Part of the essence of "set" is the notion of "all" or part of the notion of "all" is to create a "set".

Now apply this to the proposition "the set of all sets." It becomes the "all of all alls". This just sounds like poetry in need of analytics to clarify. The set of all sets is the be-all end-all of alls, cried the poet!

But I think there is some analytic clarity here. Think now of encircling members as an action we will call "setting" things; instead of fixing a set as a stagnant "x", think of it as an action of "containing". Setting as an action can be made distinct from a stagnant "all" which the setting action constructs. (I could do this by all-ing a stagnant set, but did I just actually say "all-ing"? Hope I don't have to do that to this conversation! But the fact of this temptation shows how we are at an edge or limit between what is logical and clear, but what can't be communicated in language.).

Now, the definition of "setting" is "the act of identifying all members as a set." When we say "the set of" in reference to anything, we are in the act of drawing a container, we are containing members by distinguishing those members from non-members, but we are acting, we are "setting" the membership. We have to sift through the 8 animals in the zoo, identify each individual uniquely, and then by drawing the container, by setting the membership, we claim "the set of penguins has four members."

This becomes as metaphysical as it is logical/mathematical. Now we are talking about "the all" and the "the individual identity of a single member" and "sameness" of membership and "distinctness" from non-membership and the action versus the thing acted upon versus the thing thereby constructed, namely the "set" which is the same as the "all".

Step back one more time. What is a set? It's a construction. It's a mental construction. It takes even physical objects (penguins) as members, but, of them, (as in "set of"), makes a mental construct. So a set of, or the act of setting, becomes the equivalent of making an idea of, or defining a limit or container. Now, we can analogically see that the "set of penguins" is equivalent to the definition of one of those penguins, equivalent to those defining characteristics that both identify each individual penguin as they do place all penguins as member of the set of penguins.

Setting becomes defining, or a set becomes a sort of quantifiable, demonstrable way of making a definition.

Applying this to setting itself, as when setting "sets", the container for all containers, therefore, is also the definition of all containers.

So the set of all sets means the same thing, or serves the same purpose as the definition of any set. A set, is like a definition; a definition is a statement about all of example members; so the set of all sets, is a mathematical way of denoting the definition of all sets. This is why the child blows right through this. If you understand what a set is, you can easily populate the set of all sets.

This doesn't resolve the paradox. It maybe explains how we, like the 12-year-old above, already live with it. Setting is defining, so when setting all sets, in a practical sense, we are defining what all sets are. In a logical sense, we are still creating a set that can't be a member of itself, but at the same time is a member of itself.

If all sets are contained in the set of all sets (that are not members of themselves and nothing more), — Fire Ologist

But I am not saying this. I am saying the set of all sets semantically/logically/rationally contains all sets and it is a member of itself (because it is itself a set). So the set of all sets consists of one set that is a member of itself (itself) and many sets that are not members of themselves (all sets other than the set of all sets). Where is the paradox in this?

Consider semantics. We did not create or make up the semantic that something can be a member of itself or a member of other than itself (or that triangles are triangular or that the set of all sets encompasses all sets). We are aware that such is the nature of Existence (that triangles are triangular and that the set of all sets encompasses all sets) and have expressed awareness of it. The contradiction lies in wanting a set that contains all sets that are not members of themselves that is itself not a member of itself (which appears to be the set that Russell was talking about when he asked is the set of all sets that are not members of themselves a member of itself or not). Such a thing is by definition, contradictory.

we are defining what all sets are. In a logical sense, we are still creating a set that can't be a member of itself, but at the same time is a member of itself. — Fire Ologist

Crack it open again. What is a set? A set is a form of "all". — Fire Ologist

the "all of all alls" — Fire Ologist

I don't see anything wrong with the "all of all alls". You have alls of various sizes with one all encompassing absolutely all alls. By definition, this all that contains absolutely all alls has to be infinite.

In Zermelo–Fraenkel set theory the axiom of regularity rules out a set being a member 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...

The contradiction lies in wanting a set that contains all sets that are not members of themselves that is itself not a member of itself (which appears to be the set that Russell was talking about when he asked is the set of all sets that are not members of themselves a member of itself or not). Such a thing is by definition, contradictory. — Philosopher19

Yes. But a set, by definition, cannot contain itself. The set is the act of containing. The set doesn't come to be until something else (members) are contained. Triangles and triangularity aren't equal beings. Triangularity can be predicated of something that is not a triangle. So we aren't trying to force the set of all sets to be a member of itself, just as much as we are not trying to force a sub-set as a member set to simultaneously be the set that includes itself with other member sub-sets. We just keep recognizing and restating that the set of all sets already has to not be itself a container because it is one of the members, while it has to not be one of the members, because it is a container.

I say this is because, as we keep drawing sets, and keep getting bigger, by the time we get to the most inclusive set, the set of all sets, if we want to use logic, we need to stop thinking of the set of all sets as a set. At that point, we've reached a new kind of thing, at the end of members, just like we do when we create any set (We go, penguin, penguin, penguin, and then new thing, set of penguins). When those members are sets themselves, we reach the definition of all sets, or the concept that all sets share. The set of all sets is an empty way of exemplifying the definition of "set".

I don't see anything wrong with the "all of all alls". You have alls of various sizes with one all encompassing absolutely all alls. By definition, this all that contains absolutely all alls has to be infinite. — Philosopher19

Agree, but wouldn't it also, in a naive sense, have to also be finite, because it is now an "encompassing" container? A container that ever-grows because its members ever-multiply is not a container at all. And we arrive where we started. Again. Or I guess I'm now saying a set of infinite, ever-increasing members, never gets to be a set.

I admit that this subject clearly needs careful focus that I've never done. Russell himself didn't resolve this - I doubt I can.

But I don't think it can be resolved because resolutions are logical, and with the assertion "set of all sets", we stand at the edge of all things logical facing, raw assertion - if we retreat, we remain logical and ignore the issue; but if we press on beyond the edge, trying to explain the shape of this edge, we find that logic alone, so reliable on the way to the edge, no longer works and the things we say are difficult to make sense of.

Yes. But a set, by definition, cannot contain itself — Fire Ologist

I think I understand where you're coming from. I agree that something cannot be the container of itself in the way that you mean "container of itself". But I see a meaningful difference between something being a member of itself, and something containing itself in the way that you mean. To highlight this difference to you, consider the following:

I make a list of all things in my room. The list is in my room so I list the list in the list. This is clearly meaningful is it not? The list meaningfully lists itself does it not? This is what I mean by contain itself (as in in the sense of being a member of itself). Not in the sense of a box that contains a box which is itself.

Agree, but wouldn't it also, in a naive sense, have to also be finite, because it is now an "encompassing" container? — Fire Ologist

I don't think it follows that because x is encompassing, x is finite. If everything is in the Infinite, this does not mean that there is an end to the Infinite, but it does mean the Infinite encompasses all things.

Or I guess I'm now saying a set of infinite, ever-increasing members, never gets to be a set — Fire Ologist

If something (like a number sequence) goes no forever, it does not mean that it reaches infinity or that it is an infinite set/sequence. Does it reach infinity for it to be classed as an infinite set? Can we count to infinity to say it reaches infinity to be classed as an infinite set? So we cannot say 1, 2, 3 ad infinitum is an infinite set, precisely because infinity will not be reached for it to be classed as an infinite set.

There is no beyond infinity. Infinity is complete. It is precisely because Infinity is Infinite that something can go on forever or increase increasingly. But again, this does not make it Infinite/Complete. It just means it's a part of the Infinite/Complete.

In Zermelo–Fraenkel set theory the axiom of regularity rules out a set being a member of itself. — Banno

That's like saying a list can't list itself. Is this in itself not enough to conclude that the Z-F set theory is inadequate/incomplete?

Of course it is incomplete. Gödel.