Ordered Pairs and Cartesian Product of Sets

In earlier work with sets we treated the elements of a set as an unordered collection — $\{a, b\}$ and $\{b, a\}$ denote exactly the same set. Many situations, however, need us to combine two objects while keeping track of which one came first. A student's roll number paired with their marks, the $x$- and $y$-coordinates of a point in a plane, the day and the month of a date — in every one of these, swapping the two entries changes the meaning entirely. This is the idea an ordered pair captures.

Ordered pair. Given two elements $a$ and $b$ (not necessarily distinct, and not necessarily from the same set), the ordered pair $(a, b)$ is the pair taken in a specific order: $a$ first, $b$ second. Here $a$ is called the first component (or first coordinate) and $b$ the second component (or second coordinate).

Equality of ordered pairs. Two ordered pairs $(a, b)$ and $(c, d)$ are equal if and only if $a = c$ and $b = d$. In particular, $(a,b) = (b,a)$ only when $a = b$; in general $(a, b) \ne (b, a)$ whenever $a \ne b$. This equality rule is the working tool used to solve for unknowns hidden inside an ordered pair — as in Example 1, where equating $(x+1,\ y-2)$ to $(3,\ 1)$ gives one equation for each component.

Cartesian product of two sets. Let $A$ and $B$ be two non-empty sets. The Cartesian product of $A$ and $B$, written $A \times B$, is the set of all ordered pairs $(a, b)$ such that $a \in A$ and $b \in B$:

If either $A$ or $B$ is the empty set, $A \times B$ is defined to be the empty set.

Number of elements in $A \times B$. If $A$ and $B$ are finite sets with $n(A) = p$ and $n(B) = q$, then $A \times B$ has exactly $pq$ ordered pairs, so $n(A \times B) = n(A) \times n(B)$. This follows because each of the $p$ choices for the first component can be paired with each of the $q$ choices for the second component. The same counting rule gives $n(B \times A) = n(B) \times n(A) = pq$ as well, so $A \times B$ and $B \times A$ always have the same number of elements — though, as the next remark shows, the two sets themselves are not usually equal.

$A \times B$ versus $B \times A$. Because an ordered pair remembers order, $A \times B \ne B \times A$ in general (unless $A = B$, or one of $A, B$ is empty). For example, if $A = \{1, 2\}$ and $B = \{3\}$, then $A \times B = \{(1,3), (2,3)\}$ while $B \times A = \{(3,1), (3,2)\}$ — the same number of pairs, but different pairs.

Cartesian product of more than two sets. The idea extends naturally: $A \times B \times C = \{(a,b,c) : a \in A, b \in B, c \in C\}$ is a set of ordered triples, and $n(A \times B \times C) = n(A)\cdot n(B)\cdot n(C)$ when the sets are finite. This generalisation is exactly what is needed in the next section to build $\mathbb{R} \times \mathbb{R} \times \mathbb{R}$.