The true solution to Russell's paradox

Hi all

I won't explain the paradox. If you know the paradox then continue.

Is the set of all sets that are not members of themselves, a member of itself?

We're trying to non-paradoxically define the set of all sets that are not members of themselves.

There is the set of all sets. Call this x.

x includes all sets that are not members of themselves, as well as all sets that are members of themselves because x is the set of all sets.

There is the set of all sets that are members of themselves. Call this z. The set of all lists, is one set (a member of x). The set of all things that can be members of themselves (lists, folders, sets), is one set and this set is a member of x. x is a set. It contains all sets. Thus, only x can contain x. Therefore, clearly z = x as only x can be the set of all sets that are members of themselves.

There is the set of all sets that are not members of themselves. Call this y. If x does not exist and y is not a member of itself, then we have no set of all sets that are not members of themselves precisely because y (a set that is not a member of itself) is outside of the set of all sets that are not members of themselves. To say x does not exist is to be paradoxical. Thus, clearly, x exists. x encompasses all sets. Suppose y is the most encompassing set after x. y is not a member of itself because y is a set and is thus a member of the set of all sets (x). Thus, x is the set of all sets that are not members of themselves. Clearly, the original y in this paragraph, is x. y = x.

x, the set of all sets that are not members of themselves, is thus a member of itself.

All we need is a non-paradoxical set of all sets that are not members of themselves. We have this. We have x and x is clearly not paradoxical.

Thank you for reading,

Nyma