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
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
in general, there is an operation of unrestricted complement. — fishfry
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
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
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
set theory can not prove its own consistency. — fishfry
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
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
Perhaps we're arguing about syntactic versus semantic domains. — fishfry
that impressively buzzword-compliant paragraph — fishfry
there is no mention of the domain of discourse. So again, none of this is relevant. — fishfry
You're perfectly correct that ZFC is consistent if ZF is, but what has that got to do with the conversation? — fishfry
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
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.
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
But there is no set of all groups! — fishfry
The class of groups is a proper class. So you seem to be conceding my point. — fishfry
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
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
Whether you call that a domain of discourse or not seems like a question of semantics. — fishfry
I can't comment on that quote without a link to it. — TonesInDeepFreeze
I guess you mean "question of semantics" in the sense of how we use words. — TonesInDeepFreeze
"Then what are your rigorous, mathematical definitions of 'domain of discourse' — TonesInDeepFreeze
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. — TonesInDeepFreeze
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
We frequently quantify over proper classes — fishfry
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
Not sure what contradiction "mathematics indeed does not tolerate." The referent of this paragraph is unclear. — fishfry
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
So, I really don't know what your trip is. — TonesInDeepFreeze
he understood the opposite of the correct answer! — fishfry
I've been feeling the same way about you. — fishfry
No sentence is everywhere true — Sunner
every sentence is both true and false — Sunner
Is the relevant Russell set incoherent? — Agent Smith
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
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. — Michael
Where do we have a paradox in what I have proposed? — Philosopher19
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), — Fire Ologist
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
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
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
Yes. But a set, by definition, cannot contain itself — Fire Ologist
Agree, but wouldn't it also, in a naive sense, have to also be finite, because it is now an "encompassing" container? — Fire Ologist
Or I guess I'm now saying a set of infinite, ever-increasing members, never gets to be a set — Fire Ologist
In Zermelo–Fraenkel set theory the axiom of regularity rules out a set being a member of itself. — Banno
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