False statementConstruction: Literature-sourcedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passverified 2026-08-06 (claude-opus-5)

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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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FALSE: for every formula $\varphi$ of the language of set theory there is a set $\{\, x : \varphi(x) \,\}$

Statement

False statement. For every formula $\varphi(x)$ of the language of set theory (The first-order language of set theory: $\in$ , $=$ , formulas with parameters, and class abbreviations) there is a set whose elements are exactly the sets satisfying $\varphi$ ; that is, every instance of

$\exists y\,\forall x\,\bigl(x \in y \leftrightarrow \varphi(x)\bigr)$

holds. This is the unrestricted comprehension schema, and it is the principle The Axiom Schema of Separation: for each formula $\varphi$ , $\forall \bar p\,\forall x\,\exists y\,\forall z\,(z \in y \leftrightarrow (z \in x \wedge \varphi(z,\bar p)))$ deliberately weakens.

Facts & Assumptions

Given: the claim above, asserted for every formula of the language (The first-order language of set theory: $\in$ , $=$ , formulas with parameters, and class abbreviations).

[L1]

There is no set $R$ such that, for every set $x$ , $x \in R$ holds if and only if $x \notin x$ (There is no $R$ with $x \in R \leftrightarrow x \notin x$ for every $x$ ).

[L2]

There is no set $U$ such that $y \in U$ for every set $y$ (There is no set $U$ with $y \in U$ for every set $y$ ).

Refutation

technique · contradiction

1.1

Suppose the schema holds for every formula of the language.

assume-contra

2.1

Instantiate it at the formula $\varphi(x) := x \notin x$ : there is a set $R$ such that, for every set $x$ , $x \in R$ holds if and only if $x \notin x$ .

step 1.1

3.1

No such set exists, so the supposition fails and the schema is false. Instantiating instead at $\varphi(x) := x = x$ produces a set with every set as an element, which is impossible for the same underlying reason.

L1L2step 2.1discharge-contradiction∎

Remarks

What survives. Restricting the schema so that the separated elements are drawn from a set already in hand gives The Axiom Schema of Separation: for each formula $\varphi$ , $\forall \bar p\,\forall x\,\exists y\,\forall z\,(z \in y \leftrightarrow (z \in x \wedge \varphi(z,\bar p)))$ , which is consistent as far as anything on this page can tell and is what every construction here uses. Separation and Replacement build subsets of sets already in hand, which is exactly what blocks Russell's construction runs Russell's argument against the restricted schema and obtains a theorem instead of a contradiction.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 6 results over 4 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

Russell's paradox (Wikipedia) (standard reference, not scraped)
Zermelo-Fraenkel set theory (Wikipedia) (standard reference, not scraped)
B. Kaya, MATH 320 Set Theory (METU), §1.1 (standard reference, not scraped)

FALSE: for every formula φ\varphiφ of the language of set theory there is a set { x:φ(x) }\{\, x : \varphi(x) \,\}{x:φ(x)}