Theorem(概括原则) For any predicate v(x),there is a set X: X={x|(x)} Definition(Russell's Paradox) b(x)≌“xtx” 4口￥0,43，t夏里Q0 Hengfeng Wei (hfweinju.edu.cn)Set Theory:Axioms and Operations 2019年11月26日6/38. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Theorem (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