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Now let set $A$ be arbitrary. So set $A$ exists. Using axiom of pairing $\{A\}$ exists. And then using axiom of regularity, we can argue that $A \notin A$. So, there is some $x$ such that $x \notin A$. Is this a valid proof ?

Thanks

user9026
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  • "Prove", from what axioms? – Asaf Karagila Aug 29 '20 at 09:50
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    Oh no. This looks like the classic "Adding the Axiom of Regularity is how Russell's paradox is avoided". – Jonathan Schilhan Aug 29 '20 at 14:52
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    I think it is bad manners to invoke regularity to establish basic properties of sets. Here is one of the standard approaches avoiding it: First, argue that $A$ has fewer elements than $\mathcal P(A)$ (Cantor's theorem). Conclude. – Andrés E. Caicedo Aug 29 '20 at 15:25
  • @Andrés I feel like you, and also I, answered that in the past. – Asaf Karagila Aug 29 '20 at 19:44
  • I feel like there are a few other good duplicate options out there. E.g., https://math.stackexchange.com/questions/279832/a-proof-that-the-set-of-all-sets-does-not-exist-in-zfc-using-cantors-theorem and https://math.stackexchange.com/questions/2805893/fol-proof-that-there-is-no-set-of-all-sets and https://math.stackexchange.com/questions/2714068/there-is-not-a-set-of-all-sets and probably more. – Asaf Karagila Aug 29 '20 at 20:30

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This a perfectly valid proof. For completeness you could add some comment on why regularity implies $A \notin \{A\}$; namely, since $\{A\}$ is non-empty, by regularity there must be some $y \in \{A\}$ such that $y\cap \{A\} = \varnothing$, and since the only element in $\{A\}$ is $A$, we have that $A \cap \{A\} = \varnothing$, so $A \notin A$.

Rick
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