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I'm trying to work through Patrick Suppes' Axiomatic Set Theory, which follows Zermelo-Fraenkel. He introduces the definition $\emptyset = \{x \mid x \neq x \}$, which I have no problem with, but then claims that $\{x \mid x = x \}$ is also the empty set, apparently to avoid Russel's paradox.

However, since in the text abstraction is performed by the axiom of separation, surely $\{x \mid x = x \}$ is merely a lazy version of $\{x \in A \mid x = x\}$ for some $A$, and should trivially resolve to $A$?

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    Wouldn't you avoid any incidences like $\{x|\dots\}$ in ZF? – 2017-02-24
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    It' snot the empty set, but rather it is no set at all. – 2017-02-24
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    That sounds very strange, and I suspect you may be misinterpreting something. Could you quote the point of the book that claims $\{x\mid x=x\} $ is the empty set, with some more context, please? – 2017-02-24
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    @user160738: It makes some sense in the context of ZF to use the notation $\{x\mid x\ne x\}$ for the set that the Axiom of Empty Set promises you will exist. Generally $\{x\mid \cdots \}$ do not define sets, but most of the axioms tell you that _some_ expressions of this form are actually satisfied by something, $\{x\mid \mathit{false}\}$ being one of them. – 2017-02-24
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    Thanks @HenningMakholm, I carefully reread the chapter and apparently I missed the preceding section where set comprehension is defined in such a way that effectively, any impossible set is defined to be $\emptyset$: $\{x: \phi (x)\} = y \iff [(\forall x)(x \in y \iff \phi (x)) \wedge \text{y is a set}] \vee [y = \emptyset \wedge (\lnot \exists B)(\forall x)(x \in B \iff \phi(x))]$ – 2017-02-24
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    The correct reference is : Patrick Suppes, [Axiomatic set theory](https://books.google.it/books?id=skTCAgAAQBAJ&pg=PA35), Dover (1960), page 35. – 2017-02-25
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    It is a "nitpicking" point is Suppes' textbook... The coontext is the Russell's paradox and the usul argument showing that the "unrestricted" comprehension axiom schema in incorrect: thus, the need of Separation Axiom (schema). This implies that the set-forming operator $\{x \mid \varphi(x) \}$ is not always defined : if $\varphi$ is a formula like $x=x$, we know that there is **no** *set* of all and only those $x$ such that $x=x$, and thus the set-forming operator cannot be used. 1/2 – 2017-02-25
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    But Suppes knows that in "usual" first-order logic *function* symbols are total, i.e. always defined; thus, he formally define it in such a way that we can always apply it to **any** formula (recall that the set-forming operator is a symbol that receives as input a formula a get as output a *term*, i.e. a "name" of a set). Thus, in the cases where there is **no** set $\{ x \mid \varphi(x) \}$, the formal definition adopt the convention that the symbol is a name for $\emptyset$. Is like convenionally defining that $\dfrac n 0$ is defined and equal to $0$ (see Suppes, page 19). 2/2 – 2017-02-25
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    @MauroALLEGRANZA Thanks, that makes sense. I'll change the reference, not sure how I got that wrong! – 2017-02-26

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There is no such thing as a set $A=\{x|x=x\}$ because we would have $A=A$ and so $A\in A$, violating the axiom of regularity.

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    We don't need the axiom of regularity to prove there is no such set; we can define a Russell's set $\{x:x\notin x\}$ from $A$ and the axiom of separation. – 2017-02-25