LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck 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.

For a set $x \neq \emptyset$ the collection ${z : \forall s (s \in x \to z \in s)}$ is a set, and it does not depend on the member of $x$ used to separate it

Statement

Let $x$ be a set with $x \neq \emptyset$ . Then there is a set whose elements are exactly the sets belonging to every member of $x$ ; that is, the class ${z : \forall s (s \in x \to z \in s)}$ is a set. Moreover, for every $B \in x$ the separated set ${z \in B : \forall s (s \in x \to z \in s)}$ is that same set, so the construction does not depend on which member of $x$ is used.

Facts & Assumptions

Given: a set $x$ with $x \neq \emptyset$ .

[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]

If every $z$ satisfies $z \in x$ if and only if $z \in y$ , then $x = y$ (The Axiom of Extensionality: $\forall x \forall y (\forall z (z \in x \leftrightarrow z \in y) \to x = y)$ ).

[L3]

There is exactly one set with no elements, written $\emptyset$ (There is exactly one set with no elements, written $\emptyset$ ).

[L4]

$x \subseteq y$ means that every element of $x$ is an element of $y$ (Subset $x \subseteq y$ , proper subset $x ⊊ y$ , and the separation notation ${z \in x : φ (z)}$ ).

Proof

technique · direct

1.1

If $x$ had no members it would be a set with no elements and hence equal to $\emptyset$ , contrary to hypothesis; so $x$ has a member, and we fix one, $B \in x$ .

L3givenchoose

2.1

Apply Separation to $B$ with the formula $φ (z, x) : = \forall s (s \in x \to z \in s)$ and the parameter $x$ : the collection $c_{B} : = {z \in B : \forall s (s \in x \to z \in s)}$ is a set, and $c_{B} \subseteq B$ .

L1L4step 1.1

3.1

For every $z$ , $z \in c_{B}$ holds exactly when $z \in B$ and $z$ belongs to every member of $x$ ; since $B$ is itself a member of $x$ , the second condition already forces $z \in B$ , so $z \in c_{B}$ holds exactly when $z$ belongs to every member of $x$ .

step 2.1step 1.1

4.1

The condition characterising the elements of $c_{B}$ in step 3.1 does not mention $B$ , so for any other member $B^{'}$ of $x$ the set $c_{B^{'}}$ has exactly the same elements as $c_{B}$ and equals it; the class is therefore a set and is independent of the member used to separate it.

L2step 3.1∎

Depends on

Used by

Dependency tree · two levels

5 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

B. Kaya, MATH 320 Set Theory (METU), §1.2 (standard reference, not scraped)
Intersection (set theory) (Wikipedia) (standard reference, not scraped)
C. Wilson, A Brief Introduction to ZFC (Chicago REU 2016), §2.3 (standard reference, not scraped)

For a set x≠∅x=∅ the collection { z:∀s (s∈x→z∈s) }{z:∀s(s∈x→z∈s)} is a set, and it does not depend on the member of xx used to separate it