Theorem(概括原则) For any predicate v(x),there is a set X: X={x|(x)}. Definition (Russell's Paradox) (x)≌“x夫x” R={x|x生x} Q:R∈R? Hengfeng Wei (fweinju.edu.cn Set Theory:Axioms and Operations 2019年11月26日6/38Theorem (概括原则) For any predicate ψ(x), there is a set X: X = {x | ψ(x)}. Definition (Russell’s Paradox) ψ(x) ≜ “x /∈ x” R = {x | x /∈ x} Q : R ∈ R ? Hengfeng Wei (hfwei@nju.edu.cn) Set Theory: Axioms and Operations 2019 年 11 月 26 日 6 / 38