symplectic_manifold said:
I hope most of you are fairly well acquainted with Russell's paradox, which states that the concept of the set of all sets is contradictory within the framework of naive set theory.
Could anyone explain in the layman's language how this paradox is resolved in modern set theory?
It is resolved by showing that the Russell class {x:~(x e x)} does not exist.
~Ey(Ax(x e y <-> ~(x e x))) is a theorem. That is to say,
~Ey(y={x:~(x e x)}) is a theorem.
More formally ...
Stanford Encyclopedia of Philosophy:
"Russell’s Paradox
Russell’s paradox is the most famous of the logical or set-theoretical
paradoxes. The paradox arises within naive set theory by considering the set
of all sets that are not members of themselves. Such a set appears to be a
member of itself if and only if it is not a member of itself, hence the
paradox.
Some sets, such as the set of all teacups, are not members of themselves.
Other sets, such as the set of all non-teacups, are members of themselves.
Call the set of all sets that are not members of themselves S. If S is a
member of itself, then by definition it must not be a member of itself.
Similarly, if S is not a member of itself, then by definition it must be a
member of itself. Discovered by Bertrand Russell in 1901, the paradox
prompted much work in logic, set theory and the philosophy and foundations
of mathematics during the early part of the twentieth century."
If S={x:~(x e x)}, then (S e S) <-> ~(S e S) is the paradox.
"If S is not a member of itself, then by definition it must be a member of
itself", is not true.
That is, Fy -> (y e {x:Fx}), is invalid.
~(S e S) -> (S e S), is false.
~({x:~(x e x)} e {x:~(x e x)}) -> ({x:~(x e x)} e {x:~(x e x)}), is false.
Proof:
If we grant first order predicate logic, and add Russell's contextual
definition of Classes,
determined by some predicate, then the antinomy does not occur.
D1. G{x:Fx} = EyAx((x e y) <-> Fx .& Gy) Df.
1. (z e {x:Fx}) <-> EyAx((x e y) <-> Fx .& (z e y))
<-> EyAx((x e y) <-> Fx .& Fz)
<-> Ey(Ax((x e y) <-> Fx) & Fz)
<->.EyAx((x e y) <-> Fx) & Fz valid
2. (z e {x:Fx}) -> Fz is valid
3. Fz -> (z e {x:Fx}) is invalid
4. EyAx((x e y) <-> Fx) is invalid
5. ~Ax((x e y) <-> ~(x e x))
Proof:
Ax((x e y) <-> ~(x e x)) ->. (y e y) <-> ~(y e y)
Ax((x e y) <-> ~(x e x)) -> contradiction
~Ax((x e y) <-> ~(x e x))
6. ({x:~(x e x)} e {x:~(x e x)}) <-> EyAx((x e y) <-> ~(x e x) .& (y e y))
by D1.
<->Ey(Ax((x e y) <-> ~(x e x) &
(y e y))
<-> contradiction
by 5.
Therefore 6. ~({x:~(x e x)} e {x:~(x e x)}), is a theorem.
The answer to Russell's question 'Is the class of those classes that are not
members of themselves,
a member of itself or not?' is NO it is not.
Frege's axiom V, {x:Fx}={x:Gx}<->Ax(Fx<->Gx), is not required for this
proof.
I did not us the term "unrestricted comprehension" and the rest of this sentence I do not recognise as anything I said.
