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
as I mentioned in particular it contradicts the axiom schema of separation. — GrandMinnow
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
But I said ONTO ITSELF. — GrandMinnow
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
Granted, that doesn't capture an ordinary concept of sets, — GrandMinnow
All that said, please slow down and read exactly what I post at exact face value. — GrandMinnow
You said "There exists a set such that every set is a member of it" — fishfry
But I said ONTO ITSELF.
— GrandMinnow
Where? Those words are clearly not in your post. Am I missing other posts of yours perhaps? — fishfry
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
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
All existing things exist. They cannot exist in non-existence. They all exist in something. Call this thing Existence. — 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.All existing things exist. — 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 cannot exist in non-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 Existence. — Philosopher19
Call the set of all existing things, Existence. Existence is the set of all existing things (including Itself because it Itself exists). — Philosopher19
If I am wrong, then I am an idiot — Philosopher19
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
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
They all exist in something. Call this thing the universe. — EricH
The universe IS all things. — EricH
I agree with the hypothesis of that sentence, I disagree with the conclusion — EricH
[...] "ExAy yex" has models.
— GrandMinnow
But without further qualification, those models are in no way sets. — fishfry
those models are in no way sets. — fishfry
As a layman, it is interesting to hear that no domain of discourse is truly universal. — Sunner
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
A domain of discourse is a set. — TonesInDeepFreeze
And there is no set of all sets that are not in that domain of discourse. — TonesInDeepFreeze
A domain of discourse is a set.
— TonesInDeepFreeze
I'm afraid I can't agree. — fishfry
The unrestricted complement of a set always exists. It just may not be a set. — fishfry
On the contrary, domains of discourse are often truly universal. They're just not always sets. — fishfry
Well, we are quantifying over the collection of all sets. — fishfry
There are set theories in which classes are formalized — fishfry
The official, formal definition of a 'model' is that the domain of discourse is a set. — TonesInDeepFreeze
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
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
In mathematical logic, a domain of discourse is a set. You may look it up anywhere. — TonesInDeepFreeze
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
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
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
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
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