Q:

Russell's Paradox

Let X be the set of all sets that do not contain themselves. Is X a member of X?

A:

In ZFC, either the axiom of foundation [as mentioned] or the axiom (scheme) of comprehension will prohibit this. The first, for obvious reasons; the second, since it basically says that for given z and first-order property P, you can construct { x ∈ z : P(x) }, but to generate the Russell set, you would need z = V (the class of all sets), which is not a set (i.e. cannot be generated from any of the given axioms).
In New Foundations (NF), "x ∉ x" is not a stratified formula, and so again we cannot define the Russell set. Somewhat amusingly, however, V is a set in NF.
In von Neumann--Bernays--Gödel set theory (NBG), the class R = { x : x is a set and x ∉ x } is definable. We then ask whether R ∈ R; if so, then also R ∉ R, giving a contradiction. Thus we must have R ∉ R. But there is no contradiction here, since for any given class A, A ∉ R implies either A ∈ A or A is a proper class. Since R ∉ R, we must simply have that R is a proper class.
Of course, the class R = { x : x ∉ x }, without the restriction, is simply not definable in NBG.
Also of note is that the above procedure is formally constructable as a proof in NBG, whereas in ZFC one has to resort to meta-reasoning.

