One doesn't have to provide much argument that the following claim onto itself is not self-contradictory:

(1) There exists a set such that every set is a member of it.

However, it does contradict the claim that:

(2) For any property and for any set, there is the subset of that set with the members of the subset being those with said property.

You can have (1) or you can have (2), but you can't have both. That is the basic upshot of Russell's paradox applied to sets.

One doesn't have to provide much argument that the following claim onto itself is not self-contradictory:

(1) There exists a set such that every set is a member of it. — GrandMinnow

Perhaps you could humor me and provide "much of an argument." Such a set would violate regularity/foundation. Are you perhaps making reference to Quine's New Foundations or some other alternate axiomitization of set theory? Reading the Wiki article in its entirety, or at least to the point where they said the category of NF sets is not Cartesian closed (*), convinced me that "much of an argument" must indeed be given to put your remark into its proper context. I could be wrong, curious what's in your mind with this post.

(*) This means that the sets of NF lack products and exponentials. There's not always a Cartesian product of two given sets; and/or there is not always the set of functions from one set to another. In computer science terms, you can't always curry functions. These are not sets as generally understood except perhaps by specialists in NF, as I understand it.

Of course it contradicts the axiom of regularity.

But I said ONTO ITSELF.

The formula

ExAy yex

is not a contradiction. It is consistent. Trivially, it has a model.

However, indeed it is in contradiction with certain axioms of set theory, as I mentioned in particular it contradicts the axiom schema of separation.

as I mentioned in particular it contradicts the axiom schema of separation. — GrandMinnow

You surely said no such thing, in particular or in general. Here is a quote of your entire post.

One doesn't have to provide much argument that the following claim onto itself is not self-contradictory:

(1) There exists a set such that every set is a member of it.

However, it does contradict the claim that:

(2) For any property and for any set, there is the subset of that set with the members of the subset being those with said property.

You can have (1) or you can have (2), but you can't have both. That is the basic upshot of Russell's paradox applied to sets. — GrandMinnow

I wonder if you can explain what you have in mind. You can't just say there's a set of all sets and that "One doesn't have to provide much argument" to justify it, and provide no argument, and then claim you said something you didn't say. Really wondering what your post was about. I did give the example of Quine's NF but evidently you're not talking about that. So what is the context of your remark? Not giving you a hard time for the sake of it but trying to get you to explain your cryptic remark, which is false without additional qualification. If you deny specification you haven't got a theory of sets, unless (as in NF) you stratify your sets. But why do you think you made a point about specification when you clearly didn't? Am I being unfair in challenging you here?

Separation and specification are subtly different according to Wiki. It's surely not the case that "one doesn't need to provide much argument" here. Quite the contrary IMO.

But I said ONTO ITSELF. — GrandMinnow

Where? Those words are clearly not in your post. Am I missing other posts of yours perhaps?

As the Wiki article on NF points out, your claim falls to Russell's paradox unless you stratify your sets, and that DOES require some explanation.

There is nothing cryptic in what I wrote and it is not false without further explanation.

And I did not say that there is a universal set. I said that "there is a universal set" is not onto itself a contradiction.

And I didn't claim that I said something I didn't say:

You QUOTED me saying that a universal set contradicts:

(2) For any property and for any set, there is the subset of that set with the members of the subset being those with said property.

And (2) IS the axiom schema of separation.

So surely I did mention that it contradicts the axiom schema of separation.

/

If someone says to me that his concept of set demands that there is a set of all sets, but he has not made fully clear what else is in his concept of sets, then I say "Fine, it is not onto itself contradictory that there is a set of all sets, but you will incur contradiction once you add certain other principles to your concept, such as the subset principle (which is given more explicitly as separation)."

That it doesn't take much argument to see that "There exists a U such that everything is a member of U" is not BY ITSELF contradictory, we just need to show a model M.

Let M be the model for the language with the 2-place relation 'e' as follows:

The universe of M is {0}, and e is interpreted as {<0 0>}.

The reason I didn't pedantically spell out that argument is that it takes but a nanosecond of reflection to see that yes, of course, "ExAy yex" has models.

Granted, that doesn't capture an ordinary concept of sets, but I am at first allowing, for sake of argument, that one my have whatever concept of set one may wish to have. And toward that end, I am mentioning that "there is a universal set" is not ONTO ITSELF a contradiction.

The point in the context of this thread is that we find someone who claims that there is a universal set, and I wish to emphasize that explaining what is wrong with that claim may be best understood as a two step process: First, being as generous as possible to the person that he may have any concept of set he wishes to have, we recognize that, ONTO ITSELF, "there is a universal set" is not self-contradictory, then Second, mention however that "there is a universal set" does contradict other plain and well agreed upon aspects of sets, such as separation.

/

Whatever Wiki may say, 'the axiom schema of separation' and "the axiom schema of specification" are two names for the same schema, as I mentioned that schema previously. And I don't know what difference it makes for my comments, as I specifically used only "axiom schema of separation" anyway.

All that said, please slow down and read exactly what I post at exact face value. That will avoid time wasting squabbles about who said what about what.

The universe of M is {0}, and e is interpreted as {<0 0>}.

What do the angle brackets mean? Perhaps you can explain your example. I don't understand it.
— GrandMinnow

The reason I didn't pedantically spell out that argument is that it takes but a nanosecond of reflection to see that yes, of course, "ExAy yex" has models. — GrandMinnow

But without further qualification, those models are in no way sets. And you refuse to provide such context. And if you mean something like NF, that's a pretty sophisticated concept that does involve stratification in order to avoid Russell's paradox. I have given your posts quite a bit more than a nanosecond of reflection and I don't believe you are making your case in the least.

Granted, that doesn't capture an ordinary concept of sets, — GrandMinnow

We're in agreement. You just refuted your own point. I'll leave it at that. We're not going to reach mutual understanding because you're stretching a point for its own sake, failing to provide context, and you have already refuted the point you claimed to be making. You said "There exists a set such that every set is a member of it," but in the end you agree that your claim is not about sets as commonly understood; and you have failed to provide any context in which your claim could be taken about sets.

Ironically I gave you such a context, NF, but you don't want to go down that road. There's no other road you can take.

All that said, please slow down and read exactly what I post at exact face value. — GrandMinnow

I most definitely have. I gave the NF article a pretty good read. You should do the same to get a clue about what you think you're talking about. You've already agreed you're not talking about sets, which refutes your own point.

Would you please slow down. You're swinging your arms around wildly.

I didn't refute my own point. I never claimed that a demonstration of the consistency of ExAy yex would be faithful to ordinary concepts of set. Go back and read exactly what I posted: I said that it doesn't take much to show that "There exists a set such that every set is a member of it" is not self-contradictory; I did not claim that showing that fact would adhere to ordinary concepts of set.

And you say that I fail to provide context, when I just spent my time tying out for you the context per this thread.

You said "There exists a set such that every set is a member of it" — fishfry

No I did NOT. I said the sentence is not self-contradictory. I did not say or even suggest that it is true or even compatible with ordinary concepts of set. Really, it is becoming egregious of you that you're putting words in my mouth even after I just asked you not to do that.

I'm not picking up on your remarks about NF, because I don't need recourse to NF to support my own remarks.

/

Angle brackets, indicate ordered pair, as is ordinary notation. Also, incorrectly, you put the question about the meaning of angle brackets within my own quote.

/

To recap:

(1) ExAy yex

is consistent. And it takes little argument to see that it is consistent. And I did not claim such an argument would use only our ordinary concept of set. And I did not claim ExAy yex. And indeed I pointed out that ExAy yex contradicts separation (which, goes without saying, is part of our ordinary concept of set).

(2) The point of mentioning (1) in the context of this thread was described in a previous post.

But I said ONTO ITSELF.
— GrandMinnow

Where? Those words are clearly not in your post. Am I missing other posts of yours perhaps? — fishfry

Clearly there is a problem with the manner in which you are reading. My very first sentence in this thread:

"One doesn't have to provide much argument that the following claim onto itself is not self-contradictory."

And you even quoted me saying that.

/

And I don't know why you say that Wiki claims a subtle difference between separation and specification. The Wikipedia (which I don't rely on as authoritative in mathematics anyway; for example see the wildly incorrect article on the rule of existential instantiation) article says they are the same. Nor, as I mentioned, do I know why you even mention the matter.

One doesn't have to provide much argument that the following claim onto itself is not self-contradictory: — GrandMinnow

Pace, pace.

I have been doing that. I can't add anything to what I've said other than that you should carefully examine the proof of Russell's paradox. And you should carefully examine your own argument, to see that you repeatedly claim that x is a set but you never present an argument to that effect. — fishfry

I am being sincere when I say I have carefully examined his argument. I also don't think I can do more by way of furthering our discussion on this matter except perhaps to highlight the following:

All existing things exist. They cannot exist in non-existence/nothingness. Call that thing in which all things exist in, Existence. Call the set of all existing things, Existence. Existence is the set of all existing things (including itself because it Itself exists).

I find this outrageously paradoxical/absurd to deny. Do you not? Russell's paradox (which again, I would say is a misunderstanding of semantics and poor usage of labels) denies this very obvious truth.

I do not think it in any way reasonable to believe in the following absurdities:

Things can exist in non-existence
Not everything exists in Existence
There is no set of all existing things

It is paradoxical/unreasonable/absurd/irrational of us to believe in the 3 aforementioned absurdities. It is paradoxical to believe or embrace or accept any absurdity. From triangles having four sides, to married-bachelors existing.

Consider the actual reality (not just a hypothetical possibility) that the mathematicians thoroughly studied the subject matter down to its finest details and understand its rigorous axiomatization, including that set theoretic proofs are machine checkable, while on the other hand, it appears you have not read the first page in a textbook on mathematical logic or set theory. — GrandMinnow

I sort of thought I should tame myself and not say that I'm right and many famous mathematicians and philosophers are wrong. If I am wrong, then I am an idiot and I apologise for wasting people's time (including yours). I would like to highlight the following:

All existing things exist. They cannot exist in nothingness/non-existence. They all exist in something. Call this thing Existence. Call the set of all existing things, Existence. Existence is the set of all existing things (including Itself because it Itself exists).

I find this outrageously paradoxical/absurd to deny. Do you not? Russell's paradox (which again, I would say is a misunderstanding of semantics and poor usage of labels) denies this very obvious truth.

I do not think it in any way reasonable to believe in the following absurdities:

Things can exist in non-existence
Not everything exists in Existence
There is no set of all existing things

It is paradoxical/unreasonable/absurd/irrational of us to believe in the 3 aforementioned absurdities. It is paradoxical to believe or embrace or accept any absurdity. From triangles having four sides, to married-bachelors existing.

My solution to the paradox, put differently;

There exists sets that are not members of themselves. All these sets, are sets. They are therefore a member of the set of all sets. The set of all sets need not be not a member of itself to contain all sets that are not members of themselves. Why should it? The set of all sets contains itself because it is a set. Semantically/logcailly there is only one set that is a member of itself, that set is the set of all sets: Existence. There is only one Existence. We are not Existence, we are members of Existence. We are not members of ourselves. We are members of Existence. We are ourselves and Existence is Itself. I will explain:

We exist because Existence exists. Existence exists because Existence exists.
The set of all sets contains all sets that are not members of themselves (which includes us), as well as itself. There is no set of all sets that are members of themselves because there is only one Existence. You cannot have a set of Existences because it is absurd for non-existence to separate two Existences from each other. You have a set of existing things existing in Existence, which itself exists. If a set is not a member of itself and not a member of the set of all sets, then that set is absurd. That set is not a set. The empty set that Frege detailed, contains all absurdities such as round-squares. They are not members of Existence. They are not members of the set of all sets.

Kind regards,
Nyma

All existing things exist. They cannot exist in non-existence. They all exist in something. Call this thing Existence. — Philosopher19

I'm a plain language person. So given the level of technical detail from fishfry et al, I'm jumping in here with more than a bit of trepidation. But fools rush in - I'll try to analyze what I think you're saying in my own clumsy way:

All existing things exist. — Philosopher19

Seems like a tautology to me, but just for completeness we need to extend the property of existence to energy fields & spacetime as well. Spacetime exists.

They cannot exist in non-existence. — Philosopher19

Not quite sure what you're getting at here - it seems like you're saying "Things do not have the property of non-existence"? But this falls out of the definitions of the words. So at best you're simply re-stating your first sentence in different words. Nothing wrong with that. :smile:

They all exist in something. Call this thing Existence. — Philosopher19

And here is where we go astray. I'm seeing two inter-related problems. The first is calling this "thing" Existence.. Using the word Existence leads to confusion - let's use the word universe. So now we have:

They all exist in something. Call this thing the universe.

Next using the word "something" looks wrong to me. The universe is not a thing. The universe does not contain itself. So to make this work, let's rephrase these two sentences like this:

They all exist in the universe.

This works. In fact we can now combine the two revised sentences:

All existing things exist in the universe.

That works. Then finally we have this:

Call the set of all existing things, Existence. Existence is the set of all existing things (including Itself because it Itself exists). — Philosopher19

And here is the third point where we disagree. I am far from an expert in Set Theory - but the basic concepts are clear and comprehensible to the average person. Sets are not real. Sets do not have the property of existence - they are mathematical constructs with mathematical properties according to various mathematical rules. The universe is not the set of all things that have the property of existence.

The universe IS all things.

Or put differently by this very smart philosopher guy I once read :razz:

The universe is all that is the case.

If I am wrong, then I am an idiot — Philosopher19

I agree with the hypothesis of that sentence, I disagree with the conclusion.

All existing things exist.
— Philosopher19
Seems like a tautology to me, but just for completeness we need to extend the property of existence to energy fields & spacetime as well. Spacetime exists. — EricH

I see. All triangular things are triangular because they have the property of being triangular.
All existing things exist because they have the property of existing. If x exists, then x has the property of existing. Do we agree on this?

Not quite sure what you're getting at here - it seems like you're saying "Things do not have the property of non-existence"? But this falls out of the definitions of the words. So at best you're simply re-stating your first sentence in different words. — EricH

Things that do not exist, do not have the property of existing. They are therefore not members of Existence. They are non-existent.

They all exist in something. Call this thing the universe. — EricH

From what I gather, the universe had a beginning. It could not have had a beginning in non-existence/nothingness. Thus it had a beginning in some existing thing/entity/being. Do you agree with this? I would also assert that this existing thing/entity/being, is necessarily actual infinity or truly infinite. I say necessarily because if this thing/entity/being was finite, we'd run into the paradox of something coming from nothing. For example if this thing was temporally finite, then that implies that it came from nothing. If this thing was spatially finite, then that implies it is surrounded by non-existence (which implies that non-existence exists). Since non-existence does not exist, the notion of being surrounded by non-existence is absurd. Hence why this thing/being/entity must necessarily be truly infinite.

The universe IS all things. — EricH

Since the universe itself is an existing thing or whatever we choose to call it, it is an existing thing or whatever we choose to call it. It is existing. It is not nothing. It is not a non-existent thing like a married bachelor. The universe does not denote the whole of Existence because it had a beginning. True infinity necessarily denotes the whole of Existence because it logically ensures there is no non-existence. By this I mean, true/actual infinity cures us of the paradox of something coming from nothing or non-existence existing. Since actual/true infinity is infinite through and through, it becomes meaningful and non-paradoxical to say that infinity contains itself. By this I mean the infinitesimal is infinite and it is contained within the infinite. This is essentially saying infinity contains infinity.

I agree with the hypothesis of that sentence, I disagree with the conclusion — EricH

I've come here and claimed that I've truly solved Russell's paradox. If I said I think I may have solved Russell's paradox and I was wrong, then maybe I'm not an idiot. But if I said I've truly solved Russell's paradox and claim that all famous philosophers and mathematicians were misguided whilst I am not (which is what I did), and I am actually wrong, then I am arrogant and therefore an idiot. But if I am right, then I'm neither arrogant nor an idiot. I'm truthful and accurate with regards to the description of what I am, and what other philosophers and mathematicians were regarding this matter.

This post is not addressed to any specific person.

[...] "ExAy yex" has models.
— GrandMinnow

But without further qualification, those models are in no way sets. — fishfry

I am GrandMinnow. I hadn't gotten around to answering the above.

Here is a model of "ExAy yex":

<{0} {<'e' {<0 0>}>}>

And its domain is a set, and the model itself is a set.

If one is familiar with mathematical logic, then one should recognize that is a model of "ExAy yex". It is quite trivial. But, to me, the very fact that it is so trivial is interesting.

But if one is not familiar with mathematical logic, then one would not understand the above. I am going to explain it exactly as I can and with as much detail as is feasible (even if painstakingly pedantic) in the context of posting.

(1) My original point was that "ExAy yex" is consistent but that it is not consistent with the axiom schema of separation. Note that "ExAy yex" is not consistent with this instance of the axiom schema of separation: AzEyAx(xey <-> (xez & ~xex)).

I mentioned that to point out that ruling out a universal set is not just a matter of looking alone at the notion of a universal set but rather that the notion of a universal set is not consistent with the notion of comprehension.

Then I said that "ExAy yex" has models. Note that if a sentence has a model, then the sentence is consistent.

(2)

(a) Df: a set of formulas T is consistent iff T does not prove a contradiction.

(b) Df: a formula P is consistent iff {P} is consistent.

If I recall correctly, (b) is fairly standard, but perhaps some people prefer to provide only (a).

I mention that only to ward against quibbles that originally I said:

"ExAy yex" is consistent

instead of

{"ExAy yex"} is consistent.

(3) Let 'N' stand for the set of natural numbers.

Df: a first order language L is determined by a signature <F B t> such that the intersection of F and B is empty, and F is the set of function symbols for L, and B is the set of relation symbols for L, and t is a function from FuB into N. For any function or relation symbol s, t(s) is called "the arity of s". We will leave 'first order' tacit in the rest of the post.

Df: L is a language for (or 'of') a set of formulas T iff L is a language that has at least all the function symbols and relation symbols that occur in T, and with the same arity they have in the formulas. Of course, this makes sense only if the formulas in the set don't have a symbol s that occurs in one formula as a function symbol of arity n but in another formula as a function symbol of arity m not equal n, or as a relation symbol; and mutatis mutandis for relation symbols. .

By "the language of set theory" we mean the language determined by the signature <0 {'e' '="} {<'e' 2> <'=' 2>}>. (Note: the appearance of '0' there does not imply that '0' is a symbol in the language, but rather that the set of function symbols is empty). This language has no function symbols and only two relation symbols: 'e' and '='. Any other function symbols or relation symbols used for doing set theory are not in the language for set theory but rather they are in a language of set theory extended by definitions. With the method of definitions, any formula that has defined symbols can be reverted mechanically to a certain formula that does not have the defined symbols.

(4)

Df: a model M for a language L with signature <F B t> is a tuple <U V> such that U is a nonempty set and V is a function on FuB that assigns to each f in F a t(f)-place function on U, and to each R in B a t(R)-place relation on U.

So for first order logic with identity: Except for the relation symbol '=', V may map any function symbol to any function on U as long as that function is the arity for the function symbol, and V may map any relation symbol to any relation on U as long as that relation is the arity for the relation symbol. And '=' is always assigned the identity relation on U.

So a model M for the language of set theory is a tuple <U V> such that U is a nonempty set and V('e') is a 2-place relation on U.

Nota bene: A model for the language of set theory is not required to map 'e' to the membership relation on U. If a model M doesn't map 'e' to the membership relation, then M does not adhere to our intuitions of what set theory is about, but M is still a model for the language of set theory.

(5) Df: a sentence P is true in a model M for a language L iff [fill in the recursive definition here that is too detailed for this post].

Df: a set of sentences T has a model iff there is a model M for the language of T such that every member of T is true in M.

Df. a sentence P has a model iff there is a model M for a language for {P} such that P is true in M.

So "ExAy yex" has a model iff there is a model M for a language with 'e' such that "ExAy yex" is true in M.

If a sentence has a model then there is no upper bound to the number of models the sentence has, so, a fortiori, if a sentence has a model then it has at least two models. So "the sentence has a model" implies "the sentence has models". Moreover, I will trivially show two models for "ExAy yex" anyway. I mention this only to ward against a quibble that originally I said 'models' plural.

(5) If a sentence has a model, then the sentence is consistent. So to prove that a sentence is consistent, it suffices to prove that the sentence has a model.

(6) There is a model M such that "ExAy yex" is true in M.

'=' does not occur in "ExAy yex". So a model of "AxEy yex" is:

<{0} {<'e' {<0 0>}>}>

I did not fill in the definition of 'true in the model' previously in this post, because it is too detailed for this post. But here is an intuitive account regarding the above:

The universe is {0}.

The symbol 'e' maps to the relation {<0 0>}.

Nota bene: This is not an interpretation of 'e' that we have in mind for our intuitive meaning of 'e'. But a model does not have to conform to our intuitive meanings of the symbols. For the purpose of modeling the sentence, we can interpret 'e' as standing for any 2-place relation on the domain.

Uninterpreted, "ExAy yex" says that there is an x in whatever is the domain, such that every y in whatever is the domain bears whatever is the relation symbolized by 'e' to x.

With the interpretation above, the domain is {0} and the relation symbolized by 'e' is {<0 0>}.

And every y in the domain (the only y in the domain is 0) bears the relation {<0 0>} to 0. So there is an x (viz. 0) in the domain, such that every y in the domain bears the relation {<0 0>} to x.

Or, including '=' as a symbol, here is a model for the language of set theory that is a model of "AxEy yex":

<{0} {<'e' {<0 0>}> <'=' {<0 0>}>}>

Nota bene: Trivially there are theorems of set theory that are false in this model. So this model is not a model of set theory. My claim has never been that there is a model of "ExAy yex" that is a model of set theory. The model I show is a model for the language of set theory, but it is not a model of set theory. It doesn't need to be a model of set theory.

And to make it 'models' plural, trivially here's another:

<{1} {<'e' {<1 1>}> <'=' {<1 1>}>}>

And another that is not isomorphic with those:

<{0 1} {<'e' {<0 1> <1 1>}> <'=' {<0 0>}>}>

those models are in no way sets. — fishfry

'those models' there refers to models I claimed to exist, I but I had not specified them.

The domains of the models are sets. And the models themselves are sets:

For example, <{0} {<'e' {<0 0>}>}> is an ordered pair, and ordered pairs are sets.

Apologies for what may be a naive question, but 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?

What is your definition of 'all-inclusive' in this context?

The theorem is:

~ExAy yex

"It is not the case that there exists an x such that every y is a member of x."

or, in context, "There is no set of which every set is a member."

I made a mistake in this thread. I didn't know it then, but I know now, that I used the word 'onto' in a way that is not standard English, which got tangled into a larger disagreement.

I said that the statement "There is a universal set" is not, onto itself. a contradiction.

I should have said that the statement "There is a universal set" is not, in and of itself, a contradiction.

The correct English there is not 'onto itself' but 'in and of itself'.

It seems that the other poster might have thought I meant 'onto' in the sense of a surjection, which I did not mean and would not even make sense in that context. But I don't blame the other poster on that particular point, since it was my mistake in English.

However, my substantive point, as I explained it clearly in other passages not with the word 'onto', stands:

ExAy yex

(read in context as, "There is a set of which every set is a member").

is not a contradiction, but it is inconsistent with the axiom schema of separation.

What I mean is just that the sentence «there is no set of which every set is a member» clearly says something about every set. But how does this not define some kind of collection or set?

The theorem is:

~ExAy yex

"It is not the case that there exists an x such that every y is a member of x."

or, in context, "There is no set of which every set is a member."

That is not a definition. That is a statement that there does not exist a set having a certain property (the property of having every set as a member).

But that theorem is also equivalently:

Ax~Ay yex

"For every x, it is not the case that for every y, y is a member of x."

or, in context, "For every set, it is not the case that every set is a member of it."

That also is not a definition, but it is a universal quantification. For any domain of discourse for a model of the languge, the quantifier ranges over all members of that domain of discourse.

And, yes, that domain of discourse is a set. So, naturally, and fair enough, one might ask, "But then isn't that domain of discourse the set of all sets?"

The answer is 'no'. Given any domain of discourse, it is a set, but we don't stipulate that every set is a member of it. Indeed, we can't do that, since there is no set that has every set as a member.

Moreover, using a given set theory (for example, ZFC), we cannot, using only that particular set theory, specify a particular model (nor domain of discourse for the model) of the theory (per the second incompleteness theorem). But, for example, using a theory such as ZFC+"exists an inaccessible cardinal", we can specify a model (with a domain of discourse) of ZFC (but not of ZFC+"exists an inaccessible cardinal"); and that model would be a set, but still not a set that has every set as a member.

In sum: The universal quantifier ranges over some domain of discourse that is a set; but that set itself does not have every set as a member. So when informally we couch the universal quantifier as "For all sets", to be more accurate, we actually must mean, "For all sets in the domain of discourse" (and, as mentioned, that domain of discourse does not have every set as a member).

Thank you for the clear and helpful answers (here and elsewhere on the site), in contrast to my own questions ;)
As a layman, it is interesting to hear that no domain of discourse is truly universal.
So when we say, explicitly or otherwise, that «not everything is within this domain of discourse»,
that which is inside that domain and that which is not, when taken together still do not form a universal domain of discourse but only a larger one.

Your question is insightful. You're thinking along the right lines. But there is no set of all things not in the domain of discourse.

We take only relative complements of sets:

df. T\S = {y | yeT & ~yeS}

That's fine. The set of all things in T but not in S.

But we don't have:

{y | ~yeS}

That is, we don't have the set of all things not in S.

In set theory, given any set S, there does not exist the set of all sets that are not in S. Because if there were, then the binary union of S with the set of all sets that are not in S would be the set of all sets:

thm: ~ExAy(yex <-> ~yeS)
proof: Suppose ExAy(yex <-> ~yeS). Then it would be {y | ~yeS}. Then S u {y | ~yeS) = the set of all sets.

A domain of discourse is a set. And there is no set of all sets that are not in that domain of discourse.

So we might be tempted to say there are more things outside the domain of discourse than in it. Except, since there is not a set of all things outside the domain of discourse, we can't even give the totality of things outside a domain of discourse a cardinality even to use expressions such as "more".

As a layman, it is interesting to hear that no domain of discourse is truly universal. — Sunner

On the contrary, domains of discourse are often truly universal. They're just not always sets.

Let's take as an example the simplest and most important thing we can say about sets.

Two sets are equal if and only if they have exactly the same elements.

That is, the sets $\{1,2, 3\}$ and $\{2, 3, 1\}$ are exactly the same set. Sets are characterized by their membership, without regard to order.

What domain are we quantifying over when we make this statement? We are saying, "For all sets X, and for all sets Y, if X and Y have the same elements, then X = Y."

Well, we are quantifying over the collection of all sets. Twice, for that matter. And as Russell showed us, the collection of all sets is not a set. It is a perfectly well-defined collection. It just isn't a set.

So in fact we can, and commonly do, quantify over domains that are not sets.

What are these collections that are "too big" to be sets? They're called proper classes.

First, a class is any collection defined by a property, or predicate. So, "Is a set" is a property that's true or false about any given individual. The collection of all things for which the property is true, is a class.

From the Wikipedia page on Classes:

In set theory and its applications throughout mathematics, a class is a collection of sets (or sometimes other mathematical objects) that can be unambiguously defined by a property that all its members share. — Wikipedia

Some classes are sets. Others are "too big" to be sets, such as the class of all sets. Those are the proper sets.

Yet, we can still use a proper set as a domain of discourse. We do that every time we state a property of sets. "For every two sets, their union exists." Quantifying over the proper class of all sets. "Every set has a power set." Quantifying over a the proper class of all sets.

It's a perfectly everyday occurence in math to use a proper class as the domain of discourse. It's so commonplace that we don't even notice ourselves doing it.

In ZF (standard set theory), there are no official classes, so the usage is informal. There are set theories in which classes are formalized, but those set theories are not usually encountered except by specialists.

Of interest to this thread is the Russell class, $R = \{x : x \notin x \}$.

Russell showed that $R$ can not be a set. But it's a perfectly well defined proper class. It's the collection of all the things that satisfy the property $x \notin x$.

To sum up: Sometimes domains of discourse are sets, as when we say, "All positive integers other than 1 have a unique factorization into prime powers." Here, the domain of discourse is the positive integers, because we explicitly stated that. The statement becomes false if we change the domain to the real numbers, for example.

Other times, the domain of discourse is a proper class that is too big to be a set. For example, when we say, "Every set is completely characterized by its elements," we are quantifying over the universe of sets; which as Russell showed, is not a set. But it's still a proper class, and may be spoken of and used as a domain of discourse.

A domain of discourse is a set. — TonesInDeepFreeze

I'm afraid I can't agree. Many obvious counterexamples come readily to mind. We literally could not do math without quantifying over domains that are not sets.

* Every set is in bijective correspondence with itself, via the identity map. Quantifying over the proper class of all sets.

* The identity element of every group generates a 1-element subgroup. Quantifying over the proper class of all groups

* Every vector space has a basis. Quantifying over the proper class of all vector spaces.

* Every topological space contains at least two open sets. Quantifying over the proper class of all topological spaces.

Quantifying over proper classes is so common that we don't even notice it.

And there is no set of all sets that are not in that domain of discourse. — TonesInDeepFreeze

Correct. But there is a class of all such sets. That class is not a set. It's a proper class, characterized by the property of being "not in that domain of discourse" that was being discussed.


A final example involving complements of sets, relative and otherwise, that shows how you can annoy your teacher.

You are asked, "What is the complement of the set of even numbers?" You answer, "The odd numbers, of course." Being a good student, you implicitly assumed that the domain of discourse is the set of integers.

But the literally correct answer is: "Everything that is not an even number." That includes the odd numbers as well as Captain Ahab, the Andromeda galaxy, and the Mormon Tabernacle Choir. The complement of the even numbers, without any further domain restriction, is the proper class of everything in the universe, abstract entitied included, that are not even numbers.

The unrestricted complement of a set always exists. It just may not be a set. If the teacher wanted the complement of the even numbers relativized to the set of integers, they should have said so!

A domain of discourse is a set.
— TonesInDeepFreeze

I'm afraid I can't agree. — fishfry

The official, formal definition of a 'model' is that the domain of discourse is a set.

The unrestricted complement of a set always exists. It just may not be a set. — fishfry

The context of my remark was set theory. In that context, there is no operation of absolute complement but only relative complement.

On the contrary, domains of discourse are often truly universal. They're just not always sets. — fishfry

In mathematical logic, a domain of discourse is a set. You may look it up anywhere.

Well, we are quantifying over the collection of all sets. — fishfry

Not formally. Formally, 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.

There are set theories in which classes are formalized — fishfry

Right. But even with those theories, the domain of discourse for a model for the language of the theory is a set.

Even a class theory such as NBG has only models that have a domain of discourse that is a set.

Moreover, if we tried to allow a proper class to be a domain of discourse, we'd get a contradiction:

For example, suppose we are doing model theory in a class theory in which there are proper classes. Okay, so far. Now suppose U is a proper class and, for simplicity, we have a language with just one nonlogical symbol. And let R be the relation on U that, per the model, is assigned to the nonlogical symbol.

Then we have the structure <U R>. But then, unpacking the ordered tuple by the definition of tuples (such as Kuratowski), we get that U is a member of a class, which contradicts that U is a proper class.

If you look at textbooks in mathematical logic, model theory, and set theory, you will see that without exception the definition of a model stipulates that its domain of discourse is a set.

Note: I put strikethrough there to accommodate the following post:

There also is the notion of proper classes as models, or more specifically, inner models. However, I think (I am rusty on this) that when we state this formally, it actually reduces to the syntactical method of relativization, so that when we say L is an inner model of set theory, we mean something different from the plain notion of a model. If I recall correctly, roughly speaking, relative to a theory T, saying 'sentence P is true in "class model" M' reduces to: In the language for T, we define a unary predicate symbol 'M', and P relativized to M is provable in T. So, for example, when considering the consistency of the axiom of choice relative to ZF, we find that the axiom of choice is true in the constructible universe L ("L is a model of AC"), which, in one way of doing this, reduces to: In the language of set theory, define a unary predicate symbol 'L', then we show that AC relativized to L is a theorem of ZF. So if we have a model D of ZF, then the submodel that is D intersected with L (the intersection of a set with a proper class is a set) is
a model of ZFC. That entails the consistency of ZFC relative to the consistency of ZF. But I am rusty here, so I may be corrected.

The official, formal definition of a 'model' is that the domain of discourse is a set. — TonesInDeepFreeze

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

I'll admit that for me, domain of discourse is an informal phrase meaning, "the collection over which we are quantifying," rather than a formal or technical definition. So there may be subtleties I'm missing.

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

My main point in posting was to address this concern of @Sunner:

What I mean is just that the sentence «there is no set of which every set is a member» clearly says something about every set. But how does this not define some kind of collection or set? — Sunner

I wanted to assure @Sunner that indeed, there is a collection defined by the phrase, "the collection of all sets." The collection is just not a set.

The unrestricted complement of a set always exists. It just may not be a set.
— fishfry

The context of my remark was set theory. In that context, there is no operation of absolute complement but only relative complement. — TonesInDeepFreeze

Fair enough. But in general, there is an operation of unrestricted complement. I pointed that out and gave an example.

Even within set theory, there are unrestricted complements. 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. Of course the relativized complement of {1,2,3} in the powerset of the integers is a set. But the unelativized complement is NOT a set, even in the context of set theory. It's a proper class.

In mathematical logic, a domain of discourse is a set. You may look it up anywhere. — TonesInDeepFreeze

I semi-agree. In prepping my post 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.

Well, we are quantifying over the collection of all sets.
— fishfry

Not formally. Formally, 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

Oh my goodness. I do see your point, but I can't agree with it.

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.

I simply can not believe that this is the implicit chain of logic behind every universal statement about sets. Indeed, the assumption of consistency is NOT part of set theory. Set theory can not prove its own consistency. 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.

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.

Perhaps we're arguing about syntactic versus semantic domains. "Every set has a powerset" is a purely formal statement in the language of set theory. It does not talk about models at all. And it does quantify over the universe of sets, which we know is not a set.

Likewise it can not possibly be the case that when we say, "The binary operation of every group is associative," we are implicitly quantifying over a "set" of all groups. There is no such set, and I can not believe there's a group theorist living who would agree with your point of view. Of course I have not asked them. But nobody carries around this implicit belief that universal statements about groups quantify over a mythical set of all groups, which provably does not exist. There is no set of all groups and we are not quantifying over it when we make general statements of groups.

Rather, we are quantifying over a proper class. From where I sit, you are being a bit unreasonable in your claim that there's a set of all groups that we're implicitly quantifying over. That's just not true.

There are set theories in which classes are formalized
— fishfry

Right. But even with those theories, the domain of discourse for a model for the language of the theory is a set.

Even a class theory such as NBG has only models that have a domain of discourse that is a set. — TonesInDeepFreeze

Nice to know. Not relevant to our discussion here IMO.

Moreover, if we tried to allow a proper class to be a domain of discourse, we'd get a contradiction:

For example, suppose we are doing model theory in a class theory in which there are proper classes. Okay, so far. Now suppose U is a proper class and, for simplicity, we have a language with just one nonlogical symbol. And let R be the relation on U that, per the model, is assigned to the nonlogical symbol.

Then we have the structure <U R>. But then, unpacking the ordered tuple by the definition of tuples (such as Kuratowski), we get that U is a member of a class, which contradicts that U is a proper class. — TonesInDeepFreeze

I haven't sufficient technical knowledge, but for sake of discussion I'll concede your point that if we work in NBG or Morse-Kelley set theory, domains are sets. But that's beyond the scope of the discussion. In everyday math, domains of discourse frequently are proper classes. "Every set has a powerset" quantifies over the proper class of sets; NOT, as you seem to claim, over a set model of sets whose existence depends on assuming the consistency of ZF. That's just not right.

There also is the notion of proper classes as models, or more specifically, inner models. However, I think (I am rusty on this) that when we state this formally, it actually reduces to the syntactical method of relativization, so that when we say L is an inner model of set theory, we mean something different from the plain notion of a model. If I recall correctly, roughly speaking, relative to a theory T, saying 'sentence P is true in "class model" M' reduces to: In the language for T, we define a unary predicate symbol 'M', and P relativized to M is provable in T. So, for example, when considering the consistency of the axiom of choice relative to ZF, we find that the axiom of choice is true in the constructible universe L ("L is a model of AC"), which, in one way of doing this, reduces to: In the language of set theory, define a unary predicate symbol 'L', then we show that AC relativized to L is a theorem of ZF. So if we have a model D of ZF, then the submodel that is D intersected with L (the intersection of a set with a proper class is a set) is
a model of ZFC. That entails the consistency of ZFC relative to the consistency of ZF. But I am rusty here, so I may be corrected. — TonesInDeepFreeze

I was never well-oiled enough to even aspire to being rusty in these subjects. I'm sure we're far beyond any considerations relevant to @Sunner. And in that impressively buzzword-compliant paragraph, there is no mention of the domain of discourse. So again, none of this is relevant. You're perfectly correct that ZFC is consistent if ZF is, but what has that got to do with the conversation?