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
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，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
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} 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
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? 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
Q:既然朴素集合论存在悖论，你是如何做作业的？ 但菇看不到… 4口，1①，43，t夏，30Q0 Jni jmmjtedncn Set Theory:Axioms and Operations 2021 11 25 7/40
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Q : 既然朴素集合论存在悖论，你是如何做作业的? Jun Ma (majun@nju.edu.cn) Set Theory: Axioms and Operations 2021 年 11 月 25 日 7 / 40