CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck 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.

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 $\overset{ˉ}{p}$ and any set $x$ , there is a set $y$ whose elements are exactly the elements $z$ of $x$ for which $φ (z, \overset{ˉ}{p})$ holds (The Axiom Schema of Separation: for each formula $φ$ , $\forall \overset{ˉ}{p} \forall x \exists y \forall z (z \in y \leftrightarrow (z \in x \land φ (z, \overset{ˉ}{p})))$ ).

[L2]

The separated set is written ${z \in x : φ (z, \overset{ˉ}{p})}$ (Subset $x \subseteq y$ , proper subset $x ⊊ y$ , and the separation notation ${z \in x : φ (z)}$ ).

[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

1.1assume-contra

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

2.1L1L2step 1.1

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

3.1step 1.1step 2.1

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 $x$ .

4.1L3step 3.1discharge-contradiction∎

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

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 $a ∖ b$ , the symmetric difference $a △ b$ , and the complement $X ∖ a$ relative to a set $X$ rather than absolutely.

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

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

There is no set UU with y∈Uy∈U for every set yy

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

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