2.2 Cartesian Product of Sets

Suppose A is a set of 2 colours and B is a set of 3 objects, i.e.,
\(
\mathrm{A}=\{\text { red, blue }\} \text { and } \mathrm{B}=\{b, c, s\},
\)
where \(b, c\) and \(s\) represent a particular bag, coat and shirt, respectively. How many pairs of coloured objects can be made from these two sets?
Proceeding in a very orderly manner, we can see that there will be 6 distinct pairs as given below:
\(
\text { (red, } b \text { ), (red, } c \text { ), (red, } s \text { ), (blue, } b \text { ), (blue, } c \text { ), (blue, } s \text { ). }
\)
Thus, we get 6 distinct objects.
Let us recall from our earlier classes that an ordered pair of elements taken from any two sets \(\mathrm{P}\) and \(\mathrm{Q}\) is a pair of elements written in small brackets and grouped together in a particular order, i.e., \((p, q), p \in \mathrm{P}\) and \(q \in \mathrm{Q}\). This leads to the following definition:

Given two non-empty sets \(\mathrm{P}\) and \(\mathrm{Q}\). The cartesian product \(\mathrm{P} \times \mathrm{Q}\) is the set of all ordered pairs of elements from \(P\) and \(Q\), i.e.,
\(
\mathrm{P} \times \mathrm{Q}=\{(p, q): p \in \mathrm{P}, q \in \mathrm{Q}\}
\)
If either \(P\) or \(Q\) is the null set, then \(P \times Q\) will also be empty set, i.e., \(P \times Q=\phi\) From the illustration given above we note that
\(
\mathrm{A} \times \mathrm{B}=\{(\text { red }, b) \text {, (red, } c \text { ), (red, } s \text { ), (blue, } b), \text { (blue, } c \text { ), (blue, } s)\} \text {. }
\)

Notes: 

(i) Two ordered pairs are equal, if and only if the corresponding first elements are equal and the second elements are also equal.

(ii) If there are \(p\) elements in \(\mathrm{A}\) and \(q\) elements in \(\mathrm{B}\), then there will be \(p q\) elements in \(\mathrm{A} \times \mathrm{B}\), i.e., if \(n(\mathrm{~A})=p\) and \(n(\mathrm{~B})=q\), then \(n(\mathrm{~A} \times \mathrm{B})= n(\mathrm{~A}) \times n(\mathrm{~B}) = p q\).

Similarly, if we have three sets A, B & C, then their cartesian product is as follows:

\(n(\mathrm{~A} \times \mathrm{B} \times \mathrm{C})= n(\mathrm{~A}) \times n(\mathrm{~B}) \times n(\mathrm{~C})\)

(iii) If \(\mathrm{A}\) and \(\mathrm{B}\) are non-empty sets and either \(\mathrm{A}\) or \(\mathrm{B}\) is an infinite set, then so is \(\mathrm{A} \times \mathrm{B}\).

(iv) \(\mathrm{A} \times \mathrm{A} \times \mathrm{A}=\{(a, b, c): a, b, c \in \mathrm{A}\}\). Here \((a, b, c)\) is called an ordered triplet.

(v) \(A \times\{\text { infinite set }\}=\{\text { infinite set }\} \text { where } A \text { is } \text { non-empty set. }\)

(vi) \(A \times \phi=\phi\)

Example 1:

If \((x+1, y-2)=(3,1)\), find the values of \(x\) and \(y\).
Solution:

Since the ordered pairs are equal, the corresponding elements are equal. Therefore \(\quad x+1=3\) and \(y-2=1\). Solving we get \(x=2\) and \(y=3\).

Example 2: If \(\mathrm{P}=\{a, b, c\}\) and \(\mathrm{Q}=\{r\}\), form the sets \(\mathrm{P} \times \mathrm{Q}\) and \(\mathrm{Q} \times \mathrm{P}\). Are these two products equal?

Solution: By the definition of the cartesian product,
\(
\mathrm{P} \times \mathrm{Q}=\{(a, r),(b, r),(c, r)\} \text { and } \mathrm{Q} \times \mathrm{P}=\{(r, a),(r, b),(r, c)\}
\)
Since, by the definition of equality of ordered pairs, the pair \((a, r)\) is not equal to the pair \((r, a)\), we conclude that \(\mathrm{P} \times \mathrm{Q} \neq \mathrm{Q} \times \mathrm{P}\).
However, the number of elements in each set will be the same.

Example 3: Let \(A=\{1,2,3\}, B=\{3,4\}\) and \(C=\{4,5,6\}\). Find
(i) \(\mathrm{A} \times(\mathrm{B} \cap \mathrm{C})\)
(ii) \((\mathrm{A} \times \mathrm{B}) \cap(\mathrm{A} \times \mathrm{C})\)
(iii) \(\mathrm{A} \times(\mathrm{B} \cup \mathrm{C})\)
(iv) \((\mathrm{A} \times \mathrm{B}) \cup(\mathrm{A} \times \mathrm{C})\)

Solution: (i) By the definition of the intersection of two sets, \((B \cap C)=\{4\}\).
Therefore, \(\mathrm{A} \times(\mathrm{B} \cap \mathrm{C})=\{(1,4),(2,4),(3,4)\}\).

(ii) Now \((\mathrm{A} \times \mathrm{B})=\{(1,3),(1,4),(2,3),(2,4),(3,3),(3,4)\}\)
and \((\mathrm{A} \times \mathrm{C})=\{(1,4),(1,5),(1,6),(2,4),(2,5),(2,6),(3,4),(3,5),(3,6)\}\)
Therefore, \((\mathrm{A} \times \mathrm{B}) \cap(\mathrm{A} \times \mathrm{C})=\{(1,4),(2,4),(3,4)\}\).

(iii) Since, \(\quad(\mathrm{B} \cup \mathrm{C})=\{3,4,5,6\}\), we have
\(
\mathrm{A} \times(\mathrm{B} \cup \mathrm{C})=\{(1,3),(1,4),(1,5),(1,6),(2,3),(2,4),(2,5),(2,6),(3,3),
\) \((3,4),(3,5),(3,6)\}\).

(iv) Using the sets \(\mathrm{A} \times \mathrm{B}\) and \(\mathrm{A} \times \mathrm{C}\) from part (ii) above, we obtain
\(
\begin{aligned}
&(\mathrm{A} \times \mathrm{B}) \cup(\mathrm{A} \times \mathrm{C})=\{(1,3),(1,4),(1,5),(1,6),(2,3),(2,4),(2,5),(2,6), \\
&(3,3),(3,4),(3,5),(3,6)\}
\end{aligned}
\)

Example 4: If \(P=\{1,2\}\), form the set \(P \times P \times P\).

Solution: We have, \(\mathrm{P} \times \mathrm{P} \times \mathrm{P}=\{(1,1,1),(1,1,2),(1,2,1),(1,2,2),(2,1,1),(2,1,2),(2,2,1)\), \((2,2,2)\}\).

Example 5: If \(\mathbf{R}\) is the set of all real numbers, what do the cartesian products \(\mathbf{R} \times \mathbf{R}\) and \(\mathbf{R} \times \mathbf{R} \times \mathbf{R}\) represent?

Solution: The Cartesian product \(\mathbf{R} \times \mathbf{R}\) represents the set \(\mathbf{R} \times \mathbf{R}=\{(x, y): x, y \in \mathbf{R}\}\) which represents the coordinates of all the points in two dimensional space and the cartesian product \(\mathbf{R} \times \mathbf{R} \times \mathbf{R}\) represents the set \(\mathbf{R} \times \mathbf{R} \times \mathbf{R}=\{(x, y, z): x, y, z \in \mathbf{R}\}\) which represents the coordinates of all the points in three-dimensional space.

Example 6: If \(\mathrm{A} \times \mathrm{B}=\{(p, q),(p, r),(m, q),(m, r)\}\), find \(\mathrm{A}\) and \(\mathrm{B}\).

Solution:
\(
\begin{aligned}
&\mathrm{A}=\text { set of first elements }=\{p, m\} \\
&\mathrm{B}=\text { set of second elements }=\{q, r\}
\end{aligned}
\)

Example 7: If set A = {5,6}, form the set A x A x A (ordered triplets)

Solution: 

\(
\begin{aligned}
A \times A \times A =&\{5,6\} \times\{5,6\} \times\{5,6\}\\
=&\{(5,5,5),(5,5,6),(5,6,5),(5,6,6)\\
&(6,5,5),(6,5,6),(6,6,5),(6,6,6)\}
\end{aligned}
\)

Example 8: The Cartesian product \(A \times A\) has 9 elements among which are found \((-1,0)\) and \((0,1)\). Find the set \(A\) and the remaining elements of \(A \times A\).

Solution: 
\(n(A \times A)=9\).
\(n(A) \times n(A)=9\).
\(n(A)=3\)
\(A =\{-1,0,1\} \)
\(A \times A=\{-1,0,1\} \times\{-1,0,1\}\)
\(A \times A={(-1,-1), (-1,0), (-1,-1), (0,-1), (0,0), (0,1), (1,-1), (1,0), (1,1)}\)