0

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

Question author Kyle-cronin | Source

# Answer

1

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.

Answer author Domenic

#### Share this

Tickanswer.com is providing the only single recommended solution of the question Russell's Paradox [closed] under the categories i.e math , paradox , set-theory , . Our team of experts filter the best solution for you.

## You may also add your answer