CorollaryStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passverified 2026-08-06 (claude-opus-5)

How statement and proof provenance work

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.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

There is no set $U$ with $y \in U$ for every set $y$

Statement

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

Facts & Assumptions

Given: nothing beyond the axioms cited below.

[L1]

For any parameters $\bar p$ and any set $x$ , there is a set $y$ whose elements are exactly the elements $z$ of $x$ for which $\varphi(z,\bar p)$ holds (The Axiom Schema of Separation: for each formula $\varphi$ , $\forall \bar p\,\forall x\,\exists y\,\forall z\,(z \in y \leftrightarrow (z \in x \wedge \varphi(z,\bar p)))$ ).

[L2]

The separated set is written $\{\, z \in x : \varphi(z,\bar p) \,\}$ (Subset $x \subseteq y$ , proper subset $x \subsetneq y$ , and the separation notation $\{\, z \in x : \varphi(z) \,\}$ ).

[L3]

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 $x$ ).

Proof

technique · contradiction

1.1

Suppose $U$ is a set with $y \in U$ for every set $y$ .

assume-contra

2.1

Separation applied to $U$ with the formula $\varphi(x) := x \notin x$ gives the set $R := \{\, x \in U : x \notin x \,\}$ , whose elements are exactly the $x \in U$ with $x \notin x$ .

L1L2step 1.1

3.1

Every set $x$ satisfies $x \in U$ , so for every set $x$ the condition " $x \in U$ and $x \notin x$ " reduces to $x \notin x$ ; hence $x \in R$ holds if and only if $x \notin x$ , for every set $x$ .

step 1.1step 2.1

4.1

Step 3.1 produces exactly the set that cannot exist, so the supposition is untenable and no such $U$ exists.

L3step 3.1discharge-contradiction∎

Remarks

The class of all sets is not a set. The formula $x = x$ has a class abbreviation, and this corollary says that abbreviation is not a set; that is why a complement is always taken relative to a set $X$ in The difference $a \setminus b$ , the symmetric difference $a \triangle b$ , and the complement $X \setminus a$ relative to a set $X$ rather than absolutely.

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 5 results over 3 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

B. Kaya, MATH 320 Set Theory (METU), Theorem 3 (standard reference, not scraped)
Russell's paradox (Wikipedia) (standard reference, not scraped)
Zermelo-Fraenkel set theory (Wikipedia) (standard reference, not scraped)

There is no set UUU with y∈Uy \in Uy∈U for every set yyy