Proving Nothing Contains Everything: Set Theory and Russell's Paradox

[Hurkyl]

Your post would be easier to follow if you used the quote feature...

(In my writing, I use "class" in standard set-theoretic fashion, rather than the way you are using it)

Whether 'U exists' or not, {x:x∉x} is not a member of any set/class...{x:x∉x} does not exist.

It is true that "the class {x:x∉x} is not a set" is a theorem of ZFC.

It is also (vacuously) true that "If the class of all sets is itself a set, then the class {x:x∉x} is a set" is also a theorem of ZFC.

There is no entity that is equal to {x:x∉x} is provable within FOPL=
~EyAx(xRy <-> ~(xRx)) is a theorem. That is, ~EyAx(x e y <-> ~(x e x)) is true.

"~S is a theorem" does not imply "S is not a theorem". You are implicitly assuming consistency -- an assumption that fails for the hypotheses "ZFC + there is a set of all sets".

Not to be flippant, but you do know what a "proof by contradiction" is, right?

I guess you have not heard about "purported" entities.
Any purported entity described by a contradictory predication does not exist.
Non-referring descriptions are not unheard of.

I'm more familiar with the phrase "well-formed" -- and a set of symbols that violates the grammar of the formal language of interest.

In this case, however, {x:x∉x} is well-formed, denoting a class of sets. On the hypothesis that there is a set of sets, direct application of the axiom of subsets implies that {x:x∉x} actually defines a set.

Now, {x:x∉x} isn't what was originally written. What was originally written was

{x∈U | x ∉ x }​

which is plain ordinary set-builder notation. Given any set S and any formula P, the notation

{x∈S | P(x) }​

denotes a set, specifically the particular set whose existence is guaranteed by the axiom of subsets.



For example ... The present king of France, describes a purported person who does not exist.
That natural number between 2 and 3, describes a purported natural number which does not exist.
The x such that Fx & ~Fx, describes a purported object which does not exist.

The 'purported' set {x:x∉x} does not exist.

Classes are sets, ..collections of entities. If {x:x∉x} does not exist, then surely there is no set or class that it is equal to.

If you mean "set", then why the heck are you using the word "class", particularly since you are not using the word to mean what (standard) set-theory means by the word?

{x:x=x} is the class/set of all classes/sets

Class. Not a set. And {x:x=x} = {x:x∉x}.

What?! You have used this expression in your reply,

The main thing I was complaining about is that the types don't make sense -- you are applying the operator "&" when the left argument is a class and the right argument is a predicate.

Of course, I would also complain about the odd logic you are using -- one that allows strings of symbols that would ordinarily be ill-formed, instead using some sort of modal operator to capture the notion. I know such things are possible, but such a thing is definitely not first-order predicate logic.

 

[xxxx0xxxx]

how do we prove in set theory that ,nothing contains everything

The proof is by contradiction, see Suppes, 1960. It is the result of Russell's paradox, because the existence of the set of everything would make the axiom schema of separation unnecessary, and in fact reduces the schema of separation to the schema of abstraction; but the schema of abstraction leads to Russell's paradox which is contradictory.

BTW, this is a result found in ZF and NBG; In NBG the class of all things exists, but it is not a set.