Every Seat in a Cinema Hall Is an Ordered Pair of Row and Number

Seat $(D,12)$ in a hall is row $D$, seat number $12$. Swap the parts and you get nonsense or a different seat. An ordered pair $(a,b)$ has a first and a second component, and

$$(a,b)=(c,d)\text{exactly when}a=c\text{ and }b=d$$

Ordered pairs let us pair up elements of two sets. This part covers Cartesian products, equal pairs, relations and their domains and ranges.

**$A\times B$ is the set of all ordered pairs $(a,b)$ with $a\in A$ and $b\in B$, and it has $n(A)\times n(B)$ elements.

Worked example.** $A=\{1,2\}$ and $B=\{x,y,z\}$.

$$A\times B=\{(1,x),(1,y),(1,z),(2,x),(2,y),(2,z)\}n(A\times B)=2\times 3=6$$

$B\times A$ contains pairs such as $(x,1)$, so **$A\times B\ne B\times A$ here.

Extending the idea.

- $\mathbb{R}\times \mathbb{R}$** is the set of all points $(x,y)$ in the plane
- **$\mathbb{R}\times \mathbb{R}\times \mathbb{R}$** is the set of all points $(x,y,z)$ in space
- **For $A=\{0,1\}$**, $A\times A\times A$ has $2^3=8$ ordered triples

An everyday example. **A restaurant offering $3$ main courses and $4$ desserts** has $3\times 4=12$ possible meal combinations — the Cartesian product of the two menus.

The boundary case. If either set is empty, the Cartesian product is empty, since there is nothing to pair with.

Two ordered pairs are equal only when their first components are equal and their second components are equal, so equate them separately; to recover sets from a product, collect all first components and all second components.

Worked example 1. $(x+1,y-2)=(3,1)$.

$$x+1=3\Rightarrow x=2y-2=1\Rightarrow y=3$$

Worked example 2. $A\times B=\{(a,1),(a,2),(b,1),(b,2)\}$.

$$A=\{a,b\}B=\{1,2\}$$

Worked example 3. $A\times A$ has $9$ elements, two of which are $(-1,0)$ and $(0,1)$. Then $n(A)=3$, and the components seen are $-1$, $0$ and $1$, so

$$A=\{-1,0,1\}$$

An everyday example. **A delivery rider finding house $(5,3)$ on a city grid must match both the street number and the house number — getting one right is not enough.

The substance. $(2,3)\ne (3,2)$**, even though $\{2,3\}=\{3,2\}$ as sets.

**A relation from $A$ to $B$ is any subset of $A\times B$, and it can be listed as pairs, drawn as arrows from elements of $A$ to elements of $B$, or described by a rule.

Worked example 1.** $A=\{1,2,3,4\}$, $B=\{1,4,9,16,25\}$ and $R=\{(x,y):y=x^2\}$.

Roster form: $R=\{(1,1),(2,4),(3,9),(4,16)\}$.

Arrow diagram: draw two ovals listing $A$ and $B$, with arrows $1\to 1$, $2\to 4$, $3\to 9$ and $4\to 16$.

Worked example 2. On $A=\{1,2,3,4,5,6\}$, the relation $R=\{(a,b):b=a+1\}$ is

$$R=\{(1,2),(2,3),(3,4),(4,5),(5,6)\}$$

Set-builder form is the rule itself: $\{(a,b):a,b\in A,b=a+1\}$.

An everyday example. Pairing each student with the school house they belong to is a relation from the set of students to the set of houses.

The substance. Even the empty set is a relation, since $\varnothing$ is a subset of every Cartesian product.

**The domain is the set of first components, the co-domain is the whole set $B$, the range is the set of second components actually used, and there are $2^{mn}$ relations from a set of $m$ elements to a set of $n$ elements.

Worked example 1.** For $R=\{(1,1),(2,4),(3,9),(4,16)\}$ from $A$ to $B=\{1,4,9,16,25\}$:

$$\text{domain}=\{1,2,3,4\}\text{co-domain}=\{1,4,9,16,25\}\text{range}=\{1,4,9,16\}$$

Worked example 2. $R=\{(x,y):x+2y=8,x,y\in \mathbb{N}\}$.

$$y=1:x=6y=2:x=4y=3:x=2y=4:x=0\notin \mathbb{N}$$
$$R=\{(6,1),(4,2),(2,3)\}\text{domain}=\{2,4,6\}\text{range}=\{1,2,3\}$$

Counting relations. If $n(A)=2$ and $n(B)=3$, then $n(A\times B)=6$, and each of the $6$ pairs is in or out:

$$2^6=64\text{ relations}$$

An everyday example. Matching students to blood groups: the co-domain is all four groups, but the range contains only the groups that some student actually has.

The substance. The range is always a subset of the co-domain, and may be smaller.