I cannot write x = {x, y, z}. — Philosopher19
You cannot have a set of ALL sets that are not members of themselves because it will result in at least one set not being included in the set. In other words, x will have to be in x, but it can't. — Philosopher19
If x, y, and z are sets that are members of themselves, and I form a set of these three sets, to represent this, I can write something like: p = {x, y, z}. I cannot write x = {x, y, z}. — Philosopher19
You cannot have a set of ALL sets that are not members of themselves — Philosopher19
You cannot have a set of ALL sets that are members of themselves — Philosopher19
That is incorrect. With the axiom of regularity, that set is the empty set. And without the axiom of regularity, it would still be consistent for there to be a non-empty set of all sets that are members of themselves. For example, allow that there is just one set S that is a member of itself. Then the set of all sets that are members of themselves is {S}. — TonesInDeepFreeze
I do not believe there could be a set of all sets that contain themselves — fishfry
Such a set would be subject to Russell's paradox. — fishfry
With regularity, It's the empty set.
And we can't derive a contradiction by dropping an axiom, so such a set is consistent also without regularity. But it would be inconsistent with set theory without regularity if every set were a member of itself. — TonesInDeepFreeze
You would like to form the set R={x:x∉x}R={x:x∉x} but you haven't got an existing set to start wit — fishfry
Set theory does not allow taking a universal complement like that. — TonesInDeepFreeze
Your math notation in your previous post does not format form me. — TonesInDeepFreeze
Anyway, {x | ~xex} is not at question. There is no such set. — TonesInDeepFreeze
I don't think you mean {x | ~xex}. We're talking about {x | xex}. — TonesInDeepFreeze
Is it still bad? — fishfry
Then we're in agreement and you have conceded my point, since that is exactly the set you claim exists. — fishfry
Write out your claim formally and you'll get exactly what you just wrote. — fishfry
that also is not a legal set specification. — fishfry
In order to form the set of all sets that are members of themselves, you have to start with some existing set and then apply specification to the predicate "x element of x". — fishfry
I thought Russell's paradox was meant to undermine set theory. — TheMadFool
Looks okay now. — TonesInDeepFreeze
No, I don't. — TonesInDeepFreeze
It doesn't depend on the abstraction operator. I could just as well write the whole conversation without the abstraction operator. — TonesInDeepFreeze
To prove the existence of sets having a certain property, we can only use the axioms. But the axioms don't say that other sets don't exist, except as we can prove from the axioms that there do not exist sets of a certain property. — TonesInDeepFreeze
Again, the axioms don't prove that there does not a exist a set whose members are all and only those sets that are members of themselves. Indeed, with regularity, the axioms prove that here does exist such a set. — TonesInDeepFreeze
I thought Russell's paradox was meant to undermine set theory. — TheMadFool
What follows as of necessity? — TheMadFool
I believe the liar sentence too is treated in a similar way - banished from the world of propositions.
Russell's paradox shows the contradiction in set theory with unrestricted comprehension. After Russell's note, we moved to a set theory that does not have unrestricted comprehension. — TonesInDeepFreeze
Then do so. Let me see it. — fishfry
But there is NOT a SET of all sets that are members of themselves, not even the empty set. — fishfry
So you are saying that some set exists that's not given by the axioms? — fishfry
You can't invoke unrestricted comprehension. — fishfry
All you've done is remind me that when I tell people that "the empty set is the set of all purple, flying elephants," I'm violating the axiom schema of specification. — fishfry
To prove the existence of a set, we don't always have to do it directly from separation. We have union, power set, etc. But in this case I did prove it from just separation, extensionality, and regularity, as separation was used in the previous result that ExAy ~yex. — TonesInDeepFreeze
If y and z are members of x, then you actually can write it (if a set can be a member of itself). (I'm refering to the part where you say x is a member of itself). — Amalac
No, you know it.
(1) is set theory proving there is no set whose members are all and only those sets that are not members of themselves.
(2) is Tarski's theorem. — TonesInDeepFreeze
If x, y, and z are sets that are members of themselves, and I form a set of these three sets, to represent this, I can write something like: p = {x, y, z}. I cannot write x = {x, y, z}. — Philosopher19
But the key part of my post is that you cannot have a set of all sets that are members of themselves because it will result in at least one set being a member of itself twice. This is the overlooked part of Russell's paradox. — Philosopher19
you cannot have a set of all sets that are members of themselves — Philosopher19
it will result in at least one set being a member of itself twice — Philosopher19
Here you say they are members of themselves. If they are members of themselves, then x can be contained in x, right? — Amalac
'member of itself twice' has no apparent mathematical meaning. — TonesInDeepFreeze
That's the part I believe everyone has overlooked. I will try and show this clearly:
x, y and z, are sets that are not members of themselves. I am trying to form a set of these three sets that are not members of themselves. — Philosopher19
If x, y, and z are sets that are members of themselves, and I form a set of these three sets, to represent this, I can write something like: p = {x, y, z}. I cannot write x = {x, y, z}. — Philosopher19
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