Relations

"Is a parent of." "Is greater than." "Maps to." Every one of these everyday links between two collections of things has a precise mathematical shape: a relation is nothing but a chosen set of ordered pairs, one entry taken from each collection. Before you can define a function — a rule assigning exactly one output to each input — you need the more general language of relations, and that language starts with the ordered pair and the Cartesian product. This article builds that language from scratch: what an ordered pair is and when two of them are equal, how the Cartesian product $A \times B$ collects every possible pairing of two sets, how a relation is simply a subset of that product, and how domain, co-domain, and range describe which pairs actually occur. Two worked examples, one in roster form and one from a rule, fix the ideas.

The core insight

A relation from set $A$ to set $B$ is any subset of the Cartesian product $A \times B$ , so it is completely described by which ordered pairs $(a,b)$ it contains.

The core idea

Ordered pairs. An ordered pair $(a,b)$ has a fixed first component $a$ and second component $b$ , and the order is part of its identity: two ordered pairs are equal exactly when both components match.

(a,b) = (c,d) \iff a = c \text{ and } b = d

where: $a, c$ are the first components and $b, d$ are the second components. In particular $(a,b) \neq (b,a)$ unless $a = b$ — this is what separates an ordered pair from a two-element set, where $\{a,b\} = \{b,a\}$ always.

Cartesian product. For sets $A$ and $B$ , the Cartesian product collects every possible ordered pairing of an element of $A$ with an element of $B$ :

A \times B = \{(a,b) : a \in A,\ b \in B\}

where: $a$ ranges over all of $A$ and $b$ ranges over all of $B$ , independently, so every combination appears exactly once. When $A$ and $B$ are finite, the size of the product follows a simple counting rule, $n(A \times B) = n(A) \cdot n(B)$ , proved below. The same idea extends to three sets: $A \times B \times C$ is the set of ordered triples $(a,b,c)$ , and $\mathbb{R} \times \mathbb{R} \times \mathbb{R} = \mathbb{R}^3$ is the set of all ordered triples of real numbers — the coordinate space used for points in three dimensions.

Relations. A relation $R$ from $A$ to $B$ is simply a subset of the Cartesian product:

R \subseteq A \times B

where: every element of $R$ is itself an ordered pair $(a,b)$ with $a \in A$ , $b \in B$ . If $(a,b) \in R$ , we say " $a$ is related to $b$ " and write $a\,R\,b$ . A relation can be presented in roster form (a list of pairs), by a set-builder rule (a condition the pairs must satisfy), as an arrow diagram between two sets, or as a graph of points in the plane — all four are the same underlying set of pairs, just displayed differently. Three sets travel with every relation:

The domain is the set of first components that actually occur in $R$ — the inputs that are used.
The co-domain is the set $B$ itself, fixed in advance when $R$ is defined, whether or not every element of $B$ is used.
The range is the set of second components that actually occur in $R$ — always a subset of the co-domain, but not necessarily all of it.

Figure

An arrow diagram from set A to set B showing three arrows landing on two of three elements of B, illustrating that the domain and range are the parts actually used while the co-domain is the whole target set

Derivation AThe counting rule for a Cartesian product

The goal here is to show why $n(A \times B) = n(A) \cdot n(B)$ always holds for finite sets, not just to state it, so that the rule can be trusted for any pair of finite sets rather than memorised as a formula that happens to work in small examples.

Step 1. Fix the first component. Suppose $n(A) = m$ and $n(B) = k$ . Pick any single element $a \in A$ . Pairing this one fixed $a$ with each element of $B$ in turn produces exactly $k$ distinct ordered pairs — one pair $(a,b)$ for every $b \in B$ , and no two of these $k$ pairs can coincide because their second components differ.

Step 2. Repeat for every choice of first element. There are $m$ possible choices of $a$ (all of $A$ ), and Step 1 shows each choice contributes its own block of exactly $k$ pairs. These blocks never overlap: a pair from the block built on $a_1$ and a pair from the block built on $a_2 \neq a_1$ differ in their first component, so they cannot be the same pair.

Step 3. Add the blocks. The full set $A \times B$ is the union of these $m$ non-overlapping blocks, each of size $k$ , so

n(A \times B) = \underbrace{k + k + \cdots + k}_{m \text{ times}} = m \cdot k = n(A) \cdot n(B)

Step 4. Sanity-check on a small case. With $A = \{1,2,3\}$ ( $m=3$ ) and $B=\{2,4\}$ ( $k=2$ ), the rule predicts $n(A \times B) = 3 \times 2 = 6$ , matching the 6 pairs listed by direct enumeration in the worked example below.

Worked example

Problem. $A = \{1, 2, 3\}$ , $B = \{2, 4\}$ . Find $A \times B$ and $n(A \times B)$ , then let $R = \{(a,b) \in A \times B : a < b\}$ and find $R$ , its domain, co-domain, and range.

Step 1. List every ordered pair with first entry from $A$ and second entry from $B$ :

A \times B = \{(1,2),(1,4),(2,2),(2,4),(3,2),(3,4)\}

By the counting rule, $n(A \times B) = n(A) \cdot n(B) = 3 \times 2 = 6$ , which matches the 6 pairs listed.

Step 2. Test each pair of $A \times B$ against the condition $a < b$ : $(1,2)$ satisfies it since $1<2$ ; $(1,4)$ satisfies it since $1<4$ ; $(2,2)$ fails since $2 \not< 2$ ; $(2,4)$ satisfies it since $2<4$ ; $(3,2)$ fails since $3 \not< 2$ ; $(3,4)$ satisfies it since $3<4$ .

Step 3. Collect the pairs that pass:

R = \{(1,2),(1,4),(2,4),(3,4)\}

Step 4. Read off domain, co-domain, and range. The domain is the set of distinct first entries in $R$ : $\{1,2,3\}$ . The co-domain is the target set $B$ itself: $\{2,4\}$ . The range is the set of distinct second entries in $R$ : $\{2,4\}$ . Here the range happens to equal the co-domain, though that is not required in general.

Competitive stretch

A second relation on $\mathbb{R}$ is defined by $R' = \{(x,y) : y = x^2,\ x \in \{-2,-1,0,1,2\}\}$ . Evaluating $y = x^2$ at each given $x$ gives $x=-2 \Rightarrow y=4$ , $x=-1 \Rightarrow y=1$ , $x=0 \Rightarrow y=0$ , $x=1 \Rightarrow y=1$ , $x=2 \Rightarrow y=4$ , so $R' = \{(-2,4),(-1,1),(0,0),(1,1),(2,4)\}$ . The domain is $\{-2,-1,0,1,2\}$ , but the range collapses repeated values into $\{0,1,4\}$ — even though $4$ and $1$ each arise from two different $x$ -values, each appears only once in the range because the range is a set.

Common traps

Confusing an ordered pair with a two-element set: $(a,b) \neq \{a,b\}$ , since order matters for pairs — $(1,2)\neq(2,1)$ — but not for sets, since $\{1,2\}=\{2,1\}$ .
Listing repeated values in the range: the range is a set, so a second component that occurs twice, as in the competitive stretch above, is written once, not twice.
Assuming range equals co-domain automatically: the co-domain is chosen in advance as the whole target set $B$ , while the range depends only on which pairs actually belong to $R$ ; range is always a subset of co-domain but need not be all of it.
Miscounting $n(A \times B)$ by adding instead of multiplying: the counting rule is $n(A) \cdot n(B)$ , never $n(A) + n(B)$ .

Ask the tutor: Want to practise turning a word description straight into a relation and pinning down its domain and range?

Why does order matter in an ordered pair but not in a set?

An ordered pair is built to record which element came from which set — the first slot is reserved for the element of $A$ , the second for the element of $B$ , so swapping them changes the meaning entirely, like swapping latitude and longitude. A set has no slots at all, just membership, so $\{a,b\}$ and $\{b,a\}$ describe the exact same collection. Keeping this distinction straight is exactly why $(a,b) \neq \{a,b\}$ in general.

Is every relation also a function?

No. A relation is any subset of $A \times B$ at all — it can leave some elements of $A$ unused, and it can pair one element of $A$ with several different elements of $B$ . A function is a much more restrictive relation: every element of the domain set must appear as a first component exactly once, giving one and only one output. So every function is a relation, but most relations are not functions — the formal test for this comes later in the "Functions" sub-topic.

How is the co-domain different from the range if they can look the same in examples?

The co-domain is a choice you make when defining the relation — you fix a target set $B$ before deciding which pairs go into $R$ . The range is a consequence — it is whatever second components happen to show up once $R$ is actually written down. They can coincide, as in the first worked example, purely by chance of which pairs satisfied the condition; nothing forces them to. Always check the range by listing the pairs, never by assuming it matches the stated co-domain.

Relations

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads