PropositionStatement: 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 $f : A \to B$ and $S, T \subseteq A$ : $f [S \cup T] = f [S] \cup f [T]$ , $f [S \cap T] \subseteq f [S] \cap f [T]$ and $f [S] ∖ f [T] \subseteq f [S ∖ T]$ ; both inclusions are equalities for all $S$ and $T$ if and only if $f$ is injective

Statement

Let $f : A \to B$ and let $S, T \subseteq A$ . Then

(i) $f [S \cup T] = f [S] \cup f [T]$ ;
(ii) $f [S \cap T] \subseteq f [S] \cap f [T]$ ;
(iii) $f [S] ∖ f [T] \subseteq f [S ∖ T]$ .

Moreover the inclusions in (ii) and (iii) are equalities for all $S, T \subseteq A$ if and only if $f$ is injective.

Facts & Assumptions

Given: a function $f : A \to B$ and subsets $S, T$ of $A$ .

[L1]

$b \in R [A]$ holds if and only if $(a, b) \in R$ for some $a \in A$ (The image $R [A]$ and the preimage $R^{- 1} [B]$ of a set under a relation).

[L2]

We write $f : A \to B$ , and say $f$ is a function from $A$ to $B$ , when $f$ is a function with $dom f = A$ and $ran f \subseteq B$ (A function is a relation $f$ with $(a, b) \in f$ and $(a, c) \in f$ implying $b = c$ ; $f : A \to B$ , the value $f (a)$ , domain and codomain).

[L3]

$f$ is injective (one-to-one) if $f (x) = f (y)$ implies $x = y$ , for all $x, y \in A$ (Injection, surjection, bijection).

[L4]

$z \in a \cup b$ holds if and only if $z \in a$ or $z \in b$ ( $⋃ \emptyset = \emptyset$ , $⋃ {a} = a$ , $⋃ {a, b} = a \cup b$ , $⋂ {a} = a$ , and $⋂ {a, b} = a \cap b$ ).

[L5]

$z \in a \cap b$ holds if and only if $z \in a$ and $z \in b$ ( $⋃ \emptyset = \emptyset$ , $⋃ {a} = a$ , $⋃ {a, b} = a \cup b$ , $⋂ {a} = a$ , and $⋂ {a, b} = a \cap b$ ).

[L6]

$z \in a ∖ b$ holds exactly when $z \in a$ and $z \notin b$ (The difference $a ∖ b$ , the symmetric difference $a △ b$ , and the complement $X ∖ a$ relative to a set $X$ ).

[L7]

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

[L8]

${x} : = {x, x}$ , the singleton of $x$ , is the set whose only element is $x$ (The unordered pair ${x, y}$ and the singleton ${x} = {x, x}$ ).

[L9]

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

[L10]

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

Proof

technique · direct

1.1L1L2L9

Membership criterion: for $S \subseteq A$ , $y \in f [S]$ holds exactly when $y = f (s)$ for some $s \in S$ , since $(s, y) \in f$ with $s \in S \subseteq A$ says $y = f (s)$ .

2.1L4L7step 1.1

Claim (i): $y \in f [S \cup T]$ exactly when $y = f (x)$ for some $x$ lying in $S$ or in $T$ , which is exactly $y \in f [S]$ or $y \in f [T]$ .

2.2L5L9step 1.1

Claim (ii): if $y \in f [S \cap T]$ then $y = f (x)$ for some $x$ in both $S$ and $T$ , so $y \in f [S]$ and $y \in f [T]$ .

2.3L6L9step 1.1

Claim (iii): if $y \in f [S] ∖ f [T]$ then $y = f (s)$ for some $s \in S$ , and no element of $T$ is sent to $y$ ; in particular $s \notin T$ , so $s \in S ∖ T$ and $y \in f [S ∖ T]$ .

2.4L3L5L6L7step 1.1

If $f$ is injective, both inclusions are equalities. For (ii): $y \in f [S] \cap f [T]$ gives $y = f (s) = f (t)$ with $s \in S$ and $t \in T$ , and injectivity gives $s = t \in S \cap T$ , so $y \in f [S \cap T]$ . For (iii): $y \in f [S ∖ T]$ gives $y = f (s)$ with $s \in S$ and $s \notin T$ ; if $y$ were $f (t)$ for some $t \in T$ then injectivity would give $s = t \in T$ , so $y \notin f [T]$ and $y \in f [S] ∖ f [T]$ .

2.5L3L5L8L10step 1.1

Conversely, suppose $f$ is not injective, so $f (a) = f (a^{'})$ for some $a, a^{'} \in A$ with $a \neq a^{'}$ . Take $S : = {a}$ and $T : = {a^{'}}$ . Then $S \cap T = \emptyset$ , so $f [S \cap T] = \emptyset$ , while $f (a)$ lies in $f [S] \cap f [T]$ ; the inclusion in (ii) is therefore proper for this pair.

3.1step 2.1step 2.2step 2.3step 2.4step 2.5∎

Claims (i), (ii) and (iii) hold in general, and steps 2.4 and 2.5 show that the two inclusions are equalities for all $S$ and $T$ exactly when $f$ is injective.

Depends on

Used by

Dependency tree · two levels

19 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

Image (mathematics) (Wikipedia) (standard reference, not scraped)
Bijection, injection and surjection (Wikipedia) (standard reference, not scraped)
B. Kaya, MATH 320 Set Theory (METU), §2.2 (standard reference, not scraped)

For f:A→Bf:A→B and S,T⊆AS,T⊆A: f[S∪T]=f[S]∪f[T]f[S∪T]=f[S]∪f[T], f[S∩T]⊆f[S]∩f[T]f[S∩T]⊆f[S]∩f[T] and f[S]∖f[T]⊆f[S∖T]f[S]∖f[T]⊆f[S∖T]; both inclusions are equalities for all SS and TT if and only if ff is injective