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 Hengenng Wei hkweionjn.ed.cn Set Theory:Axioms and Operations 2019 11 26 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
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
Theorem(概括原则) For any predicate v(x),there is a set X: X={x|(x)} Definition(Russell's Paradox) (x)≌“x生x” R={x|x华x} 4口，1①，43，t夏，30Q0 Hengfeng Wei (fweinju.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
Theorem(概括原则) For any predicate v(x),there is a set X: X={x|(x)} Definition (Russell's Paradox) (x)≌“x生x” R={x|x生x} Q:R∈R? 4口，1①，43，t夏，30Q0 Hengfeng Wei (fweinju.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
Q:既然朴素集合论存在悖论，你是如何做作业的？ 但装看不到… 4口，￥0,43，t夏，里Q0 Hengong Wei Chiweinjned.cn Set Theory:Axioms and Operations 2019 11 26 7/38
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Q : 既然朴素集合论存在悖论，你是如何做作业的? Hengfeng Wei (hfwei@nju.edu.cn) Set Theory: Axioms and Operations 2019 年 11 月 26 日 7 / 38