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✓ 9 results · all verified · 0 also independently AI-judged

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Relations, Functions, and Quotients: Examples and Counterexamples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

ExampleConstruction: AI-adaptedVerification: AI-adaptedprecheck passverified 2026-08-06 (claude-opus-5)Open item page →

A four-pair relation on ${\emptyset, {\emptyset}, {\emptyset, {\emptyset}}}$ with its domain, range, field, inverse, restriction, image and self-composite computed

Example

Write $u : = \emptyset$ , $v : = {\emptyset}$ and $w : = {\emptyset, {\emptyset}}$ ; these are pairwise distinct, since $u$ has no element, $v$ has exactly the element $u$ , and $w$ has $v$ as an element while $v$ does not. Put $A : = {u, v} \cup {w}$ and

$R : = {(u, v), (v, w)} \cup {(w, u), (u, u)} .$

Then $R$ is a relation on $A$ with

$dom R = ran R = fld R = A, R^{- 1} = {(v, u), (w, v)} \cup {(u, w), (u, u)},$

$R ↾ {u} = {(u, v), (u, u)}, R [{u} \cup {v}] = A, R \circ R = {(u, w), (u, v)} \cup {(u, u), (v, u)} \cup {(w, v), (w, u)} .$

Facts & Assumptions

Given: $u : = \emptyset$ , $v : = {\emptyset}$ , $w : = {\emptyset, {\emptyset}}$ , $A : = {u, v} \cup {w}$ and $R$ as displayed.

[L1]

A relation is a set $R$ every element of which is an ordered pair (Relation, $dom R$ , $ran R$ , $fld R$ , and the specialisations "relation from $A$ to $B$ " and "relation on $A$ ").

[L2]

$dom R : = {a : \exists b (a, b) \in R}, ran R : = {b : \exists a (a, b) \in R}$ (Relation, $dom R$ , $ran R$ , $fld R$ , and the specialisations "relation from $A$ to $B$ " and "relation on $A$ ").

[L3]

$(b, a) \in R^{- 1}$ holds if and only if $(a, b) \in R$ ; $(a, c) \in S \circ R$ holds if and only if $(a, b) \in R$ and $(b, c) \in S$ for some $b$ ; and $(a, b) \in R ↾ A$ holds if and only if $(a, b) \in R$ and $a \in A$ (The inverse relation $R^{- 1}$ , the composite $S \circ R$ , and the restriction $R ↾ A$ ).

[L4]

$b \in R [A]$ 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).

[L5]

$(a, b) = (c, d)$ if and only if $a = c$ and $b = d$ ( $(a, b) = (c, d)$ if and only if $a = c$ and $b = d$ ).

[L6]

$(a, b) : = {{a}, {a, b}}$ (The Kuratowski ordered pair $(a, b) : = {{a}, {a, b}}$ ).

[L7]

${x, y}$ is the set whose elements are exactly $x$ and $y$ , and ${x} : = {x, x}$ (The unordered pair ${x, y}$ and the singleton ${x} = {x, x}$ ).

[L8]

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

[L9]

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

[L10]

$z \in A \times B$ holds if and only if $z = (a, b)$ for some $a \in A$ and some $b \in B$ (The Cartesian product $A \times B : = {z \in P (P (A \cup B)) : \exists a \in A \exists b \in B z = (a, b)}$ ).

[L11]

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

[L12]

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

Verification

technique · direct

1.1

$u$ , $v$ and $w$ are pairwise distinct and $A$ has exactly them as elements; every element of $R$ is an ordered pair with both coordinates in $A$ , so $R$ is a relation on $A$ , that is $R \subseteq A \times A$ .

L1L6L7L8L9L10L12

2.1

Reading off first coordinates gives $u$ , $v$ , $w$ , so $dom R = A$ ; reading off second coordinates gives $v$ , $w$ , $u$ , so $ran R = A$ ; and the field, being the union of the two, is $A$ as well.

L2L5L9L11step 1.1

2.2

Reversing each pair gives $R^{- 1} = {(v, u), (w, v)} \cup {(u, w), (u, u)}$ ; keeping the pairs whose first coordinate is $u$ gives $R ↾ {u} = {(u, v), (u, u)}$ ; and the elements related to $u$ or to $v$ are $v$ , $u$ and $w$ , so $R [{u} \cup {v}] = A$ .

L3L4L5L7L9L11step 1.1

2.3

For the self-composite, each pair $(a, b)$ of $R$ is matched with every pair of $R$ whose first coordinate is $b$ : from $(u, v)$ and $(v, w)$ comes $(u, w)$ ; from $(u, u)$ and the two pairs starting at $u$ come $(u, v)$ and $(u, u)$ ; from $(v, w)$ and $(w, u)$ comes $(v, u)$ ; and from $(w, u)$ and the two pairs starting at $u$ come $(w, v)$ and $(w, u)$ . These six pairs are exactly the displayed $R \circ R$ .

L3L5L9L11step 1.1

3.1

The domain, range, field, inverse, restriction, image and self-composite are as displayed.

step 2.1step 2.2step 2.3∎

ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passverified 2026-08-06 (claude-opus-5)Open item page →

$\emptyset$ is a relation on every set, is the unique equivalence relation on $\emptyset$ , is a function $\emptyset \to B$ for every $B$ , is a bijection $\emptyset \to \emptyset$ , and is not a surjection $\emptyset \to {\emptyset}$

Example

The empty set does the work of five different objects at once.

$\emptyset$ is a relation, and a relation on $A$ for every set $A$ ; its domain, range and field are all $\emptyset$ .
$\emptyset$ is the only relation on $\emptyset$ , and it is an equivalence relation on $\emptyset$ ; so $\emptyset$ carries exactly one equivalence relation.
$\emptyset$ is a function $\emptyset \to B$ for every set $B$ , and it is the only one.
$\emptyset$ is a bijection $\emptyset \to \emptyset$ .
$\emptyset$ is not a surjection $\emptyset \to {\emptyset}$ , even though it is an injective function $\emptyset \to {\emptyset}$ .

The last two together are the reason a codomain belongs to the declaration $f : A \to B$ rather than to the set $f$ : one and the same set is a bijection under one declaration and a non-surjection under another. The empty function is also the unique element of the empty product.

Facts & Assumptions

Given: the set $\emptyset$ and arbitrary sets $A$ and $B$ .

[L1]

A relation is a set $R$ every element of which is an ordered pair (Relation, $dom R$ , $ran R$ , $fld R$ , and the specialisations "relation from $A$ to $B$ " and "relation on $A$ ").

[L2]

$dom R : = {a : \exists b (a, b) \in R}, ran R : = {b : \exists a (a, b) \in R}$ (Relation, $dom R$ , $ran R$ , $fld R$ , and the specialisations "relation from $A$ to $B$ " and "relation on $A$ ").

[L3]

A function is a relation $f$ such that $(a, b) \in f$ and $(a, c) \in f$ imply $b = c$ (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).

[L4]

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).

[L5]

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

[L6]

$f$ is surjective (onto) if for every $b \in B$ there is some $x \in A$ with $f (x) = b$ (Injection, surjection, bijection).

[L7]

A binary relation $\sim$ on $A$ is an equivalence relation when it is reflexive, symmetric and transitive (Equivalence relation, equivalence class, and the quotient set $A / \sim$ ).

[L8]

$R$ is reflexive on $A$ when $(a, a) \in R$ for every $a \in A$ (Reflexive, irreflexive, symmetric, asymmetric, antisymmetric, transitive, and connex relations on a set).

[L9]

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

[L10]

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

[L11]

$A \times B = \emptyset$ if and only if $A = \emptyset$ or $B = \emptyset$ ( $A \times (B \cup C) = (A \times B) \cup (A \times C)$ , $A \times (B \cap C) = (A \times B) \cap (A \times C)$ , $A \times (B ∖ C) = (A \times B) ∖ (A \times C)$ , $(A \cap B) \times (C \cap D) = (A \times C) \cap (B \times D)$ ; $A \times B = \emptyset$ if and only if $A = \emptyset$ or $B = \emptyset$ ; and for nonempty $A$ and $B$ , $A \times B \subseteq C \times D$ if and only if $A \subseteq C$ and $B \subseteq D$ ).

[L12]

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

[L13]

[L14]

if $I = \emptyset$ then $\prod_{i \in I} A_{i} = {\emptyset}$ ( $\prod_{i \in \emptyset} A_{i} = {\emptyset}$ ; if $A_{j} = \emptyset$ for some $j \in I$ then $\prod_{i \in I} A_{i} = \emptyset$ ; and for $I = {j}$ the evaluation $f \mapsto f (j)$ is a bijection $\prod_{i \in I} A_{i} \to A_{j}$ ).

[L15]

an element of $\prod_{i \in I} A_{i}$ is a function with domain $I$ that takes its value at each index inside the member carried by that index (The product $\prod_{i \in I} A_{i} : = {f : I \to ⋃_{i \in I} A_{i} ∣ f (i) \in A_{i} for every i \in I}$ ).

Verification

technique · direct

1.1

$\emptyset$ has no elements, so "every element is an ordered pair" holds vacuously and $\emptyset$ is a relation; $\emptyset \subseteq A \times A$ for every $A$ , so it is a relation on every set; and no set satisfies the defining conditions for its domain or its range, so both are $\emptyset$ , hence so is its field.

L1L2L9L12

1.2

A relation on $\emptyset$ is a subset of $\emptyset \times \emptyset$ , which is $\emptyset$ , so $\emptyset$ is the only one. It is reflexive on $\emptyset$ , symmetric and transitive, since each condition quantifies over elements of $\emptyset$ ; hence it is the unique equivalence relation on $\emptyset$ .

L7L8L9L11L12L13

2.1

$\emptyset$ is single valued vacuously, has domain $\emptyset$ and range $\emptyset \subseteq B$ , so $\emptyset : \emptyset \to B$ for every $B$ ; and any function with domain $\emptyset$ has no elements, so it is $\emptyset$ .

L3L4L9L12step 1.1

3.1

As a function $\emptyset \to \emptyset$ it is injective, since the injectivity condition quantifies over elements of the domain, and surjective, since the surjectivity condition quantifies over elements of the codomain and $\emptyset$ has none; so it is a bijection.

L5L6L9step 2.1

3.2

As a function $\emptyset \to {\emptyset}$ it is still injective, for the same reason, but not surjective: $\emptyset$ is an element of ${\emptyset}$ and no element of the domain is sent to it.

L5L6L9L10step 2.1

3.3

The empty function is the unique element of the empty product: $\prod_{i \in \emptyset} A_{i} = {\emptyset}$ , and its one element is a function with domain $\emptyset$ .

L14L15step 2.1

4.1

All five descriptions hold of the single set $\emptyset$ , and the last two differ only in the declared codomain.

step 1.1step 1.2step 2.1step 3.1step 3.2step 3.3∎

CounterexampleConstruction: AI-adaptedVerification: AI-adaptedprecheck passverified 2026-08-06 (claude-opus-5)Open item page →

Relations $R$ and $S$ on a two-element set with $S \circ R \neq R \circ S$

Statement refuted

Refuted claim: $S \circ R = R \circ S$ for all relations $R$ and $S$ . With $u : = \emptyset$ , $v : = {\emptyset}$ , $A : = {u, v}$ and

$R : = {(u, v)}, S : = {(v, u)},$

the two composites are $S \circ R = {(u, u)}$ and $R \circ S = {(v, v)}$ , which are different. Composition of relations is associative ( $T \circ (S \circ R) = (T \circ S) \circ R$ , $(S \circ R)^{- 1} = R^{- 1} \circ S^{- 1}$ , $(R^{- 1})^{- 1} = R$ , $dom (R^{- 1}) = ran R$ , and $Δ_{B} \circ R = R = R \circ Δ_{A}$ for a relation $R$ from $A$ to $B$ ) but not commutative.

Facts & Assumptions

Given: $u : = \emptyset$ , $v : = {\emptyset}$ , $A : = {u, v}$ , $R : = {(u, v)}$ and $S : = {(v, u)}$ .

[L1]

$(a, c) \in S \circ R$ holds if and only if $(a, b) \in R$ and $(b, c) \in S$ for some $b$ (The inverse relation $R^{- 1}$ , the composite $S \circ R$ , and the restriction $R ↾ A$ ).

[L2]

A relation is a set $R$ every element of which is an ordered pair (Relation, $dom R$ , $ran R$ , $fld R$ , and the specialisations "relation from $A$ to $B$ " and "relation on $A$ ").

[L3]

$(a, b) = (c, d)$ if and only if $a = c$ and $b = d$ ( $(a, b) = (c, d)$ if and only if $a = c$ and $b = d$ ).

[L4]

$(a, b) : = {{a}, {a, b}}$ (The Kuratowski ordered pair $(a, b) : = {{a}, {a, b}}$ ).

[L5]

${x, y}$ is the set whose elements are exactly $x$ and $y$ , and ${x} : = {x, x}$ (The unordered pair ${x, y}$ and the singleton ${x} = {x, x}$ ).

[L6]

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

[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]

$T \circ (S \circ R) = (T \circ S) \circ R$ ( $T \circ (S \circ R) = (T \circ S) \circ R$ , $(S \circ R)^{- 1} = R^{- 1} \circ S^{- 1}$ , $(R^{- 1})^{- 1} = R$ , $dom (R^{- 1}) = ran R$ , and $Δ_{B} \circ R = R = R \circ Δ_{A}$ for a relation $R$ from $A$ to $B$ ).

[L9]

Counterexample

technique · direct

1.1

$u \neq v$ , because $v$ has an element and $u$ has none; and $R$ and $S$ are relations on $A$ , each with one pair, both of whose coordinates lie in $A$ .

L2L5L6L9

2.1

$S \circ R$ : the only pair of $R$ is $(u, v)$ , and the only pair of $S$ whose first coordinate is $v$ is $(v, u)$ , so the only element of $S \circ R$ is $(u, u)$ .

L1L3L5L7step 1.1

2.2

$R \circ S$ : the only pair of $S$ is $(v, u)$ , and the only pair of $R$ whose first coordinate is $u$ is $(u, v)$ , so the only element of $R \circ S$ is $(v, v)$ .

L1L3L5L7step 1.1

3.1

$(u, u)$ and $(v, v)$ are different, since equality of ordered pairs would force $u = v$ ; so the two composites have different elements and $S \circ R \neq R \circ S$ , while the associativity of composition is unaffected.

L3L4L7L8step 1.1step 2.1step 2.2∎

ExampleConstruction: AI-adaptedVerification: AI-adaptedprecheck passverified 2026-08-06 (claude-opus-5)Open item page →

A two-cell partition of a three-element set, the equivalence relation it induces listed pair by pair, and the quotient set recovered from it

Example

Write $u : = \emptyset$ , $v : = {\emptyset}$ and $w : = {\emptyset, {\emptyset}}$ , which are pairwise distinct, and put $A : = {u, v} \cup {w}$ . The collection ${{u, v}, {w}}$ has nonempty cells, covers $A$ , and its two cells are disjoint. The equivalence relation it induces on $A$ , namely " $a$ and $b$ lie in the same cell", is

$\sim = {(u, u), (u, v)} \cup {(v, u), (v, v)} \cup {(w, w)},$

with classes $[u] = [v] = {u, v}$ and $[w] = {w}$ , so that $A / \sim = {{u, v}, {w}}$ is the original collection again.

Facts & Assumptions

Given: $u : = \emptyset$ , $v : = {\emptyset}$ , $w : = {\emptyset, {\emptyset}}$ , $A : = {u, v} \cup {w}$ , and $\sim$ as displayed.

[L1]

reflexive: $a \sim a$ for every $a \in A$ (Equivalence relation, equivalence class, and the quotient set $A / \sim$ ).

[L2]

symmetric: $a \sim b$ implies $b \sim a$ , for all $a, b \in A$ (Equivalence relation, equivalence class, and the quotient set $A / \sim$ ).

[L3]

transitive: $a \sim b$ and $b \sim c$ imply $a \sim c$ , for all $a, b, c \in A$ (Equivalence relation, equivalence class, and the quotient set $A / \sim$ ).

[L4]

$[a] : = {b \in A : a \sim b} \subseteq A$ (Equivalence relation, equivalence class, and the quotient set $A / \sim$ ).

[L5]

$A / \sim : = {[a] : a \in A}$ (Equivalence relation, equivalence class, and the quotient set $A / \sim$ ).

[L6]

The equivalence classes of an equivalence relation are nonempty, cover $A$ , and are pairwise equal or disjoint (The equivalence classes of an equivalence relation are nonempty, cover $A$ , and are pairwise equal or disjoint; conversely every such cover arises from exactly one equivalence relation).

[L7]

${x, y}$ is the set whose elements are exactly $x$ and $y$ , and ${x} : = {x, x}$ (The unordered pair ${x, y}$ and the singleton ${x} = {x, x}$ ).

[L8]

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

[L9]

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

[L10]

[L11]

$(a, b) = (c, d)$ if and only if $a = c$ and $b = d$ ( $(a, b) = (c, d)$ if and only if $a = c$ and $b = d$ ).

[L12]

$(a, b) : = {{a}, {a, b}}$ (The Kuratowski ordered pair $(a, b) : = {{a}, {a, b}}$ ).

[L13]

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

[L14]

A relation is a set $R$ every element of which is an ordered pair (Relation, $dom R$ , $ran R$ , $fld R$ , and the specialisations "relation from $A$ to $B$ " and "relation on $A$ ").

[L15]

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

Verification

technique · direct

1.1

$u$ , $v$ and $w$ are pairwise distinct: $u$ has no element, $v$ has exactly $u$ , and $w$ has $v$ as an element while $v$ does not. So $A$ has exactly those three elements, and $\sim$ is a relation on $A$ , each of its five listed pairs having both coordinates in $A$ .

L7L8L9L10L12L14L15

2.1

$\sim$ is an equivalence relation on $A$ : it contains $(u, u)$ , $(v, v)$ and $(w, w)$ , so it is reflexive on $A$ ; its pairs with distinct coordinates are $(u, v)$ and $(v, u)$ , and each of those has its reverse present, so it is symmetric; and every composable pair of its members has its composite present, since chains through $u$ and $v$ stay inside ${u, v}$ and $w$ relates only to itself.

L1L2L3L11L13step 1.1

3.1

The classes are as stated: $[u] = {b \in A : u \sim b} = {u, v}$ , $[v] = {u, v}$ by the pairs $(v, u)$ and $(v, v)$ , and $[w] = {w}$ . Hence $A / \sim$ has exactly the two elements ${u, v}$ and ${w}$ , which is the collection we started from.

L4L5L7L11L13step 1.1step 2.1

4.1

The collection is therefore a partition of $A$ in the sense of the classes being nonempty, covering $A$ and pairwise equal or disjoint, and it is recovered as the quotient set of the equivalence relation it induces.

L6step 2.1step 3.1∎

ExampleConstruction: AI-adaptedVerification: AI-adaptedprecheck passverified 2026-08-06 (claude-opus-5)Open item page →

A two-to-one function on a four-element set, its kernel, the quotient, and the canonical factorisation computed explicitly

Example

Write $u : = \emptyset$ , $v : = {\emptyset}$ , $w : = {\emptyset, {\emptyset}}$ and $x : = {{\emptyset}}$ ; these four are pairwise distinct. Put $A : = {u, v} \cup {w, x}$ and $B : = {u, v}$ , and let

$f : = {(u, u), (v, u)} \cup {(w, v), (x, v)},$

so $f : A \to B$ takes the value $u$ twice and the value $v$ twice. Then

$\ker f = {(u, u), (u, v)} \cup {(v, u), (v, v)} \cup {(w, w), (w, x)} \cup {(x, w), (x, x)},$

$A / \ker f = {{u, v}, {w, x}}, f [A] = B,$

and the canonical factorisation $f = ι \circ \overset{ˉ}{f} \circ π$ has $π (u) = π (v) = {u, v}$ , $π (w) = π (x) = {w, x}$ , $\overset{ˉ}{f} ({u, v}) = u$ , $\overset{ˉ}{f} ({w, x}) = v$ , and $ι = Δ_{B}$ .

Facts & Assumptions

Given: $u, v, w, x$ , $A$ , $B$ and $f$ as displayed.

[L1]

$\ker f$ is a set, is a relation on $A$ , and is an equivalence relation on $A$ (For a function $f : A \to B$ the relation $\ker f : = {(a, a^{'}) \in A \times A : f (a) = f (a^{'})}$ is an equivalence relation on $A$ ).

[L2]

there is a function $\overset{ˉ}{f} : A / \sim \to f [A]$ with $\overset{ˉ}{f} ([a]) = f (a)$ for every $a \in A$ , and it is a bijection (Every function $f : A \to B$ factors as $f = ι \circ \overset{ˉ}{f} \circ π$ with $π : A \to A / \ker f$ the surjective quotient map, $\overset{ˉ}{f} : A / \ker f \to f [A]$ the bijection $[a] \mapsto f (a)$ , and $ι : f [A] \to B$ the inclusion injection; $\overset{ˉ}{f}$ is the only function making the factorisation hold).

[L3]

[L4]

$[a] : = {b \in A : a \sim b} \subseteq A$ (Equivalence relation, equivalence class, and the quotient set $A / \sim$ ).

[L5]

$A / \sim : = {[a] : a \in A}$ (Equivalence relation, equivalence class, and the quotient set $A / \sim$ ).

[L6]

$b \in R [A]$ 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).

[L7]

${x, y}$ is the set whose elements are exactly $x$ and $y$ , and ${x} : = {x, x}$ (The unordered pair ${x, y}$ and the singleton ${x} = {x, x}$ ).

[L8]

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

[L9]

$f$ is surjective (onto) if for every $b \in B$ there is some $x \in A$ with $f (x) = b$ (Injection, surjection, bijection).

[L10]

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

[L11]

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

[L12]

$(a, b) = (c, d)$ if and only if $a = c$ and $b = d$ ( $(a, b) = (c, d)$ if and only if $a = c$ and $b = d$ ).

[L13]

$dom R : = {a : \exists b (a, b) \in R}, ran R : = {b : \exists a (a, b) \in R}$ (Relation, $dom R$ , $ran R$ , $fld R$ , and the specialisations "relation from $A$ to $B$ " and "relation on $A$ ").

Verification

technique · direct

1.1

The four sets are pairwise distinct: $u$ has no element; $v$ and $x$ each have exactly one, namely $u$ and $v$ respectively, which differ; and $w$ has both $u$ and $v$ , so it differs from all three. Hence $A$ has exactly four elements and $B$ exactly two.

L7L8L10L11

2.1

$f$ is a function $A \to B$ : its four pairs have distinct first coordinates, so it is single valued; its domain is $A$ and its range is $B$ . It is surjective, since $u = f (u)$ and $v = f (w)$ , so $f [A] = B$ .

L3L6L9L12L13step 1.1

3.1

$\ker f$ is as displayed: a pair $(a, a^{'})$ of elements of $A$ lies in it exactly when $f (a) = f (a^{'})$ , and $f$ takes the value $u$ exactly at $u$ and $v$ , and the value $v$ exactly at $w$ and $x$ ; so the pairs are those with both coordinates in ${u, v}$ or both in ${w, x}$ . By the general lemma it is an equivalence relation on $A$ .

L1L12step 1.1step 2.1

4.1

The classes are $[u] = [v] = {u, v}$ and $[w] = [x] = {w, x}$ , so $A / \ker f$ has exactly the two elements ${u, v}$ and ${w, x}$ , and the quotient map sends $u$ and $v$ to the first and $w$ and $x$ to the second.

L4L5L7step 3.1

5.1

The canonical factorisation is as displayed: $\overset{ˉ}{f}$ sends ${u, v} = [u]$ to $f (u) = u$ and ${w, x} = [w]$ to $f (w) = v$ , and it is a bijection onto $f [A] = B$ ; the inclusion $ι$ is the identity on $B$ because $f [A] = B$ ; and composing gives back $f$ .

L2L6step 2.1step 3.1step 4.1∎

CounterexampleConstruction: AI-adaptedVerification: AI-adaptedprecheck passverified 2026-08-06 (claude-opus-5)Open item page →

A symmetric and transitive relation on a two-element set that is not reflexive on it

Statement refuted

Refuted claim: a symmetric and transitive relation on a set $A$ is reflexive on $A$ . Write $u : = \emptyset$ and $v : = {\emptyset}$ , put $A : = {u, v}$ and

$R : = {(u, u)} .$

$R$ is a relation on $A$ that is symmetric and transitive, and it is not reflexive on $A$ , because $(v, v) \notin R$ .

The failure is located exactly at the point of $A$ that $R$ does not touch: $R$ is reflexive on its own field ${u}$ , and symmetry and transitivity constrain $R$ only there.

Facts & Assumptions

Given: $u : = \emptyset$ , $v : = {\emptyset}$ , $A : = {u, v}$ and $R : = {(u, u)}$ .

[L1]

$R$ is reflexive on $A$ when $(a, a) \in R$ for every $a \in A$ (Reflexive, irreflexive, symmetric, asymmetric, antisymmetric, transitive, and connex relations on a set).

[L2]

$R$ is symmetric when $(a, b) \in R$ implies $(b, a) \in R$ , for all $a, b \in A$ (Reflexive, irreflexive, symmetric, asymmetric, antisymmetric, transitive, and connex relations on a set).

[L3]

$R$ is transitive when $(a, b) \in R$ and $(b, c) \in R$ imply $(a, c) \in R$ , for all $a, b, c \in A$ (Reflexive, irreflexive, symmetric, asymmetric, antisymmetric, transitive, and connex relations on a set).

[L4]

A relation is a set $R$ every element of which is an ordered pair (Relation, $dom R$ , $ran R$ , $fld R$ , and the specialisations "relation from $A$ to $B$ " and "relation on $A$ ").

[L5]

$dom R : = {a : \exists b (a, b) \in R}, ran R : = {b : \exists a (a, b) \in R}$ (Relation, $dom R$ , $ran R$ , $fld R$ , and the specialisations "relation from $A$ to $B$ " and "relation on $A$ ").

[L6]

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

[L7]

${x, y}$ is the set whose elements are exactly $x$ and $y$ , and ${x} : = {x, x}$ (The unordered pair ${x, y}$ and the singleton ${x} = {x, x}$ ).

[L8]

[L9]

$(a, b) = (c, d)$ if and only if $a = c$ and $b = d$ ( $(a, b) = (c, d)$ if and only if $a = c$ and $b = d$ ).

[L10]

$(a, b) : = {{a}, {a, b}}$ (The Kuratowski ordered pair $(a, b) : = {{a}, {a, b}}$ ).

[L11]

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

[L12]

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

Counterexample

technique · direct

1.1

$u \neq v$ , because $v$ has the element $u$ and $u$ has none; so $A$ has exactly the two elements $u$ and $v$ .

L6L7

2.1

$R$ is a relation on $A$ : its only element is the ordered pair $(u, u)$ , and both coordinates lie in $A$ , so $R \subseteq A \times A$ .

L4L8L10step 1.1

3.1

The three characteristic sets of $R$ are $dom R = ran R = fld R = {u}$ , since $(u, u)$ is its only pair.

L5L7L9L11L12step 2.1

3.2

$R$ is symmetric: the only pair in $R$ is $(u, u)$ , whose reversal is itself. It is transitive: the only composable pair of members is $(u, u)$ with $(u, u)$ , and the required conclusion $(u, u) \in R$ holds.

L2L3L9step 2.1

4.1

$R$ is reflexive on $fld R$ : the only element of ${u}$ is $u$ , and $(u, u) \in R$ .

L1L7step 3.1

5.1

$R$ is not reflexive on $A$ : $v \in A$ , and $(v, v) \neq (u, u)$ because $v \neq u$ , so $(v, v) \notin R$ . The failure is therefore confined to the single element of $A$ lying outside $fld R$ ; symmetry and transitivity say nothing about such a point, which is exactly why they do not imply reflexivity on $A$ .

L1L9step 1.1step 2.1step 3.1step 3.2step 4.1∎

False statementConstruction: Literature-sourcedVerification: AI-adaptedprecheck passverified 2026-08-06 (claude-opus-5)Open item page →

FALSE: a symmetric and transitive relation on a set $A$ is reflexive on $A$ , so reflexivity is redundant in the definition of an equivalence relation

Statement

False statement. If a relation $R$ on a set $A$ is symmetric and transitive, then it is reflexive on $A$ ; consequently the reflexivity clause in the definition of an equivalence relation follows from the other two and could be dropped.

Facts & Assumptions

Given: the claim above.

[L1]

$R$ is reflexive on $A$ when $(a, a) \in R$ for every $a \in A$ (Reflexive, irreflexive, symmetric, asymmetric, antisymmetric, transitive, and connex relations on a set).

[L2]

$R$ is symmetric when $(a, b) \in R$ implies $(b, a) \in R$ , for all $a, b \in A$ (Reflexive, irreflexive, symmetric, asymmetric, antisymmetric, transitive, and connex relations on a set).

[L3]

[L4]

A binary relation $\sim$ on $A$ is an equivalence relation when it is reflexive, symmetric and transitive (Equivalence relation, equivalence class, and the quotient set $A / \sim$ ).

[L5]

$R$ is a relation on $A$ when $R \subseteq A \times A$ (Relation, $dom R$ , $ran R$ , $fld R$ , and the specialisations "relation from $A$ to $B$ " and "relation on $A$ ").

[L6]

[L7]

$(a, b) : = {{a}, {a, b}}$ (The Kuratowski ordered pair $(a, b) : = {{a}, {a, b}}$ ).

[L8]

${x, y}$ is the set whose elements are exactly $x$ and $y$ , and ${x} : = {x, x}$ (The unordered pair ${x, y}$ and the singleton ${x} = {x, x}$ ).

[L9]

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

Refutation

technique · direct

1.1

The argument that makes the claim look right: given $a \in A$ , take any $b$ with $(a, b) \in R$ ; symmetry gives $(b, a) \in R$ , and transitivity applied to $(a, b)$ and $(b, a)$ gives $(a, a) \in R$ .

L1L2L3

1.2

The witness: put $u : = \emptyset$ , $v : = {\emptyset}$ , $A : = {u, v}$ and $R : = {(u, u)}$ . Here $u \neq v$ , because $v$ has an element and $u$ has none.

L8L9

2.1

The gap in step 1.1 is the phrase "take any $b$ with $(a, b) \in R$ ": no hypothesis supplies such a $b$ . Symmetry and transitivity are conditional on pairs that are already in $R$ , so they constrain $R$ only at points that $R$ relates to something, and say nothing whatever about a point of $A$ that $R$ leaves untouched.

L2L3step 1.1

2.2

$R$ is a relation on $A$ : its only element is the ordered pair $(u, u)$ , whose coordinates both lie in $A$ .

L5L6L7step 1.2

2.3

$R$ is symmetric, since its only pair is its own reversal, and transitive, since the only composable pair of its members is $(u, u)$ with $(u, u)$ , whose conclusion $(u, u) \in R$ holds.

L2L3step 1.2

3.1

$R$ is not reflexive on $A$ : $v$ is an element of $A$ and $(v, v) \notin R$ , since the only element of $R$ is $(u, u)$ and $v \neq u$ .

L1L7L8step 1.2step 2.2

4.1

The claim is therefore false, and with it the conclusion drawn from it: the reflexivity clause in the definition of an equivalence relation is not redundant, since $R$ satisfies the other two clauses on $A$ and is not an equivalence relation on $A$ .

L4step 2.1step 2.3step 3.1∎

CounterexampleConstruction: AI-adaptedVerification: AI-adaptedprecheck passverified 2026-08-06 (claude-opus-5)Open item page →

A function $f$ and sets $S, T$ with $f [S \cap T] ⊊ f [S] \cap f [T]$

Statement refuted

Refuted claim: $f [S \cap T] = f [S] \cap f [T]$ for every function $f : A \to B$ and all $S, T \subseteq A$ . Write $u : = \emptyset$ and $v : = {\emptyset}$ , and take

$A : = {u, v}, B : = {u}, f : = {(u, u), (v, u)}, S : = {u}, T : = {v} .$

Then $S \cap T = \emptyset$ , so $f [S \cap T] = \emptyset$ , while $f [S] = f [T] = {u}$ and therefore $f [S] \cap f [T] = {u}$ . The inclusion $f [S \cap T] \subseteq f [S] \cap f [T]$ of 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 is proper here, and it is proper exactly because $f$ is not injective.

Facts & Assumptions

Given: $u : = \emptyset$ , $v : = {\emptyset}$ , $A : = {u, v}$ , $B : = {u}$ , $f : = {(u, u), (v, u)}$ , $S : = {u}$ and $T : = {v}$ .

[L1]

$f [S \cap T] \subseteq f [S] \cap f [T]$ (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).

[L2]

$b \in R [A]$ 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).

[L3]

[L4]

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

[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]

${x, y}$ is the set whose elements are exactly $x$ and $y$ , and ${x} : = {x, x}$ (The unordered pair ${x, y}$ and the singleton ${x} = {x, x}$ ).

[L7]

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

[L8]

$(a, b) : = {{a}, {a, b}}$ (The Kuratowski ordered pair $(a, b) : = {{a}, {a, b}}$ ).

[L9]

$dom R : = {a : \exists b (a, b) \in R}, ran R : = {b : \exists a (a, b) \in R}$ (Relation, $dom R$ , $ran R$ , $fld R$ , and the specialisations "relation from $A$ to $B$ " and "relation on $A$ ").

[L10]

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

[L11]

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

Counterexample

technique · direct

1.1

$u \neq v$ , since $v$ has an element and $u$ has none; so $A$ has exactly the two elements $u$ and $v$ , and $S$ and $T$ are the two singletons inside it.

L6L7

2.1

$f$ is a function $A \to B$ : its two pairs have distinct first coordinates, so it is single valued; its domain is $A$ and its range is ${u} = B$ . It is not injective, since $f (u) = u = f (v)$ while $u \neq v$ .

L3L4L8L9step 1.1

2.2

$S \cap T$ has no element, since an element would have to equal both $u$ and $v$ ; so $S \cap T = \emptyset$ and $f [S \cap T] = \emptyset$ , no element of $\emptyset$ being related to anything.

L2L5L6L7step 1.1

3.1

$f [S] = {u}$ and $f [T] = {u}$ , since $f (u) = u$ and $f (v) = u$ ; hence $f [S] \cap f [T] = {u}$ .

L2L5L6step 2.1

4.1

The general inclusion holds and is proper here: $u$ lies in $f [S] \cap f [T]$ and not in $f [S \cap T]$ , which has no element at all.

L1L7L10L11step 2.2step 3.1∎

ExampleConstruction: AI-generatedVerification: AI-adaptedprecheck passverified 2026-08-06 (claude-opus-5)Open item page →

For any indexed family the product $\prod_{i \in I} P (X_{i})$ contains the constant function with value $\emptyset$ , and $\prod_{i \in I} {i}$ has exactly one element

Example

Two families whose products can be shown nonempty by writing an element down, with no choice principle involved.

Let $(X_{i})_{i \in I}$ be any indexed family. Then $(P (X_{i}))_{i \in I}$ is an indexed family, and the constant function $c : = {(i, \emptyset) : i \in I}$ is an element of $\prod_{i \in I} P (X_{i})$ . So that product is nonempty for every $I$ and every family, even when some $X_{i}$ is empty.
Let $I$ be any set and let $({i})_{i \in I}$ be the family carrying the singleton of the index at each index. Then $\prod_{i \in I} {i} = {Δ_{I}}$ : its only element is the identity relation on $I$ .

Neither construction selects anything: in the first the value is the same set at every index, and in the second the value at $i$ is forced to be $i$ .

Facts & Assumptions

Given: an indexed family $(X_{i})_{i \in I}$ and a set $I$ .

[L1]

[L2]

An indexed family with index set $I$ is a function $A$ with $dom A = I$ (An indexed family $(A_{i})_{i \in I}$ is a function with domain $I$ ; ${A_{i} : i \in I}$ is its range).

[L3]

$z \in P (x)$ holds if and only if $z \subseteq x$ (The power set $P (x) = {z : z \subseteq x}$ ).

[L4]

(i) $\emptyset \subseteq x$ ( $\emptyset \subseteq x$ , $x \subseteq x$ , inclusion is transitive, and $x = y$ if and only if $x \subseteq y$ and $y \subseteq x$ ).

[L5]

[L6]

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

[L7]

$Δ_{A}$ is a function with $dom Δ_{A} = A$ and $Δ_{A} (a) = a$ for every $a \in A$ (If $f$ and $g$ are functions then $g \circ f$ is a function with domain $f^{- 1} [dom g]$ and $(g \circ f) (x) = g (f (x))$ there; $Δ_{A}$ is a function with $Δ_{A} (a) = a$ ; and $f \circ Δ_{A} = f = Δ_{B} \circ f$ for $f : A \to B$ ).

[L8]

$(a, b) \in Δ_{A}$ holds if and only if $a = b$ and $a \in A$ (The identity relation $Δ_{A} = {(a, b) \in A \times A : a = b}$ and the membership relation $\in_{A} = {(a, b) \in A \times A : a \in b}$ ).

[L9]

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

[L10]

[L11]

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 $φ$ , $\forall \overset{ˉ}{p} \forall x \exists y \forall z (z \in y \leftrightarrow (z \in x \land φ (z, \overset{ˉ}{p})))$ ).

[L12]

[L13]

$dom R : = {a : \exists b (a, b) \in R}, ran R : = {b : \exists a (a, b) \in R}$ (Relation, $dom R$ , $ran R$ , $fld R$ , and the specialisations "relation from $A$ to $B$ " and "relation on $A$ ").

[L14]

$z \in ⋃_{i \in I} A_{i}$ holds if and only if $z \in A_{i}$ for some $i \in I$ ( $⋃_{i \in I} A_{i} : = ⋃ {A_{i} : i \in I}$ , and $⋂_{i \in I} A_{i} : = ⋂ {A_{i} : i \in I}$ for $I \neq \emptyset$ ).

Verification

technique · direct

1.1

$(P (X_{i}))_{i \in I}$ is an indexed family: separating inside $I \times P (P (⋃_{i \in I} X_{i}))$ with the formula $\exists i \exists w (z = (i, w) \land i \in I \land w = P (X_{i}))$ gives a set, which is single valued and has domain $I$ ; the ambient set contains each $P (X_{i})$ , since every subset of $X_{i}$ is a subset of $⋃_{i \in I} X_{i}$ .

L2L3L5L11L12L13L14

1.2

Similarly $({i})_{i \in I}$ is an indexed family, obtained by separating inside $I \times P (I)$ , since ${i} \subseteq I$ for $i \in I$ .

L2L3L5L9L11L12L13

2.1

The constant function $c : = {(i, \emptyset) : i \in I}$ is a set, by separating inside $I \times {\emptyset}$ ; it is single valued, has domain $I$ , and $c (i) = \emptyset$ for every $i \in I$ . Since $\emptyset \subseteq X_{i}$ , we have $\emptyset \in P (X_{i})$ for every $i$ , so $c$ lies in $\prod_{i \in I} P (X_{i})$ and that product is nonempty.

L1L3L4L5L6L9L11L12L13step 1.1

2.2

$Δ_{I}$ is a function with domain $I$ and $Δ_{I} (i) = i$ , and $i$ is the only element of ${i}$ , so $Δ_{I} \in \prod_{i \in I} {i}$ . Conversely any $f$ in that product has domain $I$ and $f (i) \in {i}$ , hence $f (i) = i$ for every $i \in I$ , so $f$ and $Δ_{I}$ are functions with the same domain agreeing everywhere and are equal.

L1L5L7L8L9step 1.2

3.1

Both products are therefore nonempty, and the second has exactly one element; when $I = \emptyset$ both statements agree with the general computation of the empty product, whose single element is the empty function.

L10step 2.1step 2.2∎

Relations, Functions, and Quotients: Examples and Counterexamples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

5 · Examples, counterexamples and false statements

A four-pair relation on {∅,{∅},{∅,{∅}}}{∅,{∅},{∅,{∅}}} with its domain, range, field, inverse, restriction, image and self-composite computed

Example

Facts & Assumptions

Verification

∅∅ is a relation on every set, is the unique equivalence relation on ∅∅, is a function ∅→B∅→B for every BB, is a bijection ∅→∅∅→∅, and is not a surjection ∅→{∅}∅→{∅}

Example

Facts & Assumptions

Verification

Relations RR and SS on a two-element set with S∘R≠R∘SS∘R=R∘S

Statement refuted

Facts & Assumptions

Counterexample

A two-cell partition of a three-element set, the equivalence relation it induces listed pair by pair, and the quotient set recovered from it

Example

Facts & Assumptions

Verification

A two-to-one function on a four-element set, its kernel, the quotient, and the canonical factorisation computed explicitly

Example

Facts & Assumptions

Verification

A symmetric and transitive relation on a two-element set that is not reflexive on it

Statement refuted

Facts & Assumptions

Counterexample

FALSE: a symmetric and transitive relation on a set AA is reflexive on AA, so reflexivity is redundant in the definition of an equivalence relation

Statement

Facts & Assumptions

Refutation

A function ff and sets S,TS,T with f[S∩T]⊊f[S]∩f[T]f[S∩T]⊊f[S]∩f[T]

Statement refuted

Facts & Assumptions

Counterexample

For any indexed family the product ∏i∈IP(Xi)∏i∈I​P(Xi​) contains the constant function with value ∅∅, and ∏i∈I{i}∏i∈I​{i} has exactly one element

Example

Facts & Assumptions

Verification

Sources

A four-pair relation on ${\emptyset, {\emptyset}, {\emptyset, {\emptyset}}}$ with its domain, range, field, inverse, restriction, image and self-composite computed

$\emptyset$ is a relation on every set, is the unique equivalence relation on $\emptyset$ , is a function $\emptyset \to B$ for every $B$ , is a bijection $\emptyset \to \emptyset$ , and is not a surjection $\emptyset \to {\emptyset}$

Relations $R$ and $S$ on a two-element set with $S \circ R \neq R \circ S$

FALSE: a symmetric and transitive relation on a set $A$ is reflexive on $A$ , so reflexivity is redundant in the definition of an equivalence relation

A function $f$ and sets $S, T$ with $f [S \cap T] ⊊ f [S] \cap f [T]$

For any indexed family the product $\prod_{i \in I} P (X_{i})$ contains the constant function with value $\emptyset$ , and $\prod_{i \in I} {i}$ has exactly one element