Theorem(概括原则) For any predicate v(x),there is a set X: X={x|(x)} 4口，1①，43，t夏，30Q0 Jni jmmjtedncn Set Theory:Axioms and Operations 2021 11 256/40. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 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 ? Jun Ma (majun@nju.edu.cn) Set Theory: Axioms and Operations 2021 年 11 月 25 日 6 / 40