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.

If $a \subseteq b$ then $⋃ a \subseteq ⋃ b$ ; if in addition $a \neq \emptyset$ then $⋂ b \subseteq ⋂ a$ ; and $c \subseteq ⋃ a$ for every $c \in a$ , while $⋂ a \subseteq c$ for every $c \in a$

Statement

Let $a$ and $b$ be sets.

(i) If $a \subseteq b$ then $⋃ a \subseteq ⋃ b$ .
(ii) If $a \subseteq b$ and $a \neq \emptyset$ , then $b \neq \emptyset$ , both intersections are defined, and $⋂ b \subseteq ⋂ a$ .
(iii) $c \subseteq ⋃ a$ for every $c \in a$ .
(iv) $⋂ a \subseteq c$ for every $c \in a$ .

Facts & Assumptions

Given: sets $a$ and $b$ .

[L1]

$⋃ x$ is the set whose elements are exactly the elements of the elements of $x$ (The union $⋃ x$ of a set, and the binary union $a \cup b : = ⋃ {a, b}$ ).

[L2]

For $x \neq \emptyset$ , $⋂ x$ is the set whose elements are exactly the sets belonging to every element of $x$ (The intersection $⋂ x$ of a nonempty set, the binary intersection $a \cap b : = ⋂ {a, b}$ , and disjointness).

[L3]

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

[L4]

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

Proof

technique · direct

1.1

Claim (i): assume $a \subseteq b$ and let $z \in ⋃ a$ ; then $z \in s$ for some $s \in a$ , and $s \in b$ because $a \subseteq b$ , so $z \in ⋃ b$ .

L1L3

1.2

Claim (ii): assume $a \subseteq b$ and $a \neq \emptyset$ . Then $a$ has a member, which is also a member of $b$ , so $b \neq \emptyset$ and both intersections are defined. If $z \in ⋂ b$ then $z$ lies in every member of $b$ , hence in every member of $a$ , so $z \in ⋂ a$ .

L2L3L4

1.3

Claim (iii): let $c \in a$ and $z \in c$ ; then $z$ lies in a member of $a$ , so $z \in ⋃ a$ , and therefore $c \subseteq ⋃ a$ .

L1L3

1.4

Claim (iv): let $c \in a$ ; then $a \neq \emptyset$ , so $⋂ a$ is defined, and every $z \in ⋂ a$ lies in every member of $a$ , in particular in $c$ ; therefore $⋂ a \subseteq c$ .

L2L3L4

2.1

Claims (i) to (iv) are established, which is the statement.

step 1.1step 1.2step 1.3step 1.4∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

10 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)
Union (set theory) (Wikipedia) (standard reference, not scraped)
Intersection (set theory) (Wikipedia) (standard reference, not scraped)

If a⊆ba⊆b then ⋃a⊆⋃b⋃a⊆⋃b; if in addition a≠∅a=∅ then ⋂b⊆⋂a⋂b⊆⋂a; and c⊆⋃ac⊆⋃a for every c∈ac∈a, while ⋂a⊆c⋂a⊆c for every c∈ac∈a