"We can formulate a consistent set theory based on this approach. However, it has some drawbacks. Most obviously, basic set operations become problematic. In particular, consider the following additional two axioms:
(S) For each set X, the singleton {X} exists.
(D) For all sets X, Y, the difference set X-Y={a: a in X, a not in Y} exists.
Both (S) and (D) are pretty unobjectionable, but they wind up contradicting (C):
Let p be the property "Does not contain itself." Let [p] be the corresponding set guaranteed by (C).
As usual, we must have [p] in [p].
But now consider the new set U=[p]-{[p]}. Clearly U contains every set which doesn't contain itself, and U is a proper subset of [p], which is a contradiction."
First all thanks for taking your time articulate your answerin the way you did.
So if I understand it well,
We could derive a contradiction as soon as we try to remove the set from itself via U= [p]-{[p]}. via axiom (S) and (D).
Well consider this;
Suppose we define a set [p]'
such that it contains almost all the sets that don't contain themselves expect for that set itself. (kinda like the set € but in the reverse direction)
clearly that set doesn't contain itself
also if [p]'would contain itself and it contains almost all the sets that don't contain themselves (from the definition we know that [p]' was the only set making that collection incomplete)
Then we could derive a contradiction,
namely [p]' does contain itself and [p]' doesn't contain itself
can the collection of almost all the sets that don't contain themselves in that set be completed?
Well , suppose that [p]' contains itself
Then we have a set of all the sets that don't contain themselves since it was the only set that made it as such that it was that collection was incomplete
how did adding [p]' to itself complete it ?
NOT by satisfying the property of "being not contained by itself" since that would be a contradiction.
well that is exactly why it is complete now exactly because
[p]' doesn't satisfy that property no more
but it DOES satisfy the property of "being contained by itself"
As we have already seen above with the set € that is not inconsistent.
Since it is not a set of all and only the sets that don't contain themselves. but it is a set of all the sets that contain themselves + [p]'
similary [p]' can be removed from itself again
so [p]' now contains almost* all the sets that don't contain themselves. (except for [p]')
Therefore it can be consistently contain itself and be removed from itself , to have an analogy we can have a set
off all cats + that set containing itself.
but we can also remove that set from itself and therefore
there would be only a set off all and only cats left over.