This is a very famous and a very old paradox.I was graced upon it just recently.

Starting like Paul R. Halmos does with set-theoretic Axiom Of Specification

‘To every set A and to every condition S(x) there corresponds a set B whose elements are exactly those elements x of A for which S(x) holds.’

Now for the proof part of it.

{i’ll show not(ϵ) as ϵ` for simplicity of discourse}

let S(x) be not(x ϵ x) or restated as xϵ`x.

It follows that whatever the set A may be , if B= {xϵA: xϵ`x} , then for all y ,

yϵB iff (yϵA and yϵ`y) …………….(#)

Can it be that BϵA? Halmos proceeds to prove that the answer is no.

If BϵA, then either BϵB also or else Bϵ`B.If BϵB, then by (#) the assumption BϵA yields Bϵ`B a contradiction again.If Bϵ`B then by (#) again the assumption BϵA yields BϵB another contradiction.