Y

Y = {x||x|||P(x)|}

P(x) is the power set.



Describe Y.

>>8
I didn't mean that {A} isn't a set. For every set A in the class of sets you can form the set {A}. Therefore Y has a subclass in bijection (assuming this makes sense for classes in general) with the class of sets, hence Y is a proper class.

(Still sketchy arguments here.)