Cartesian Products of Sets: Definition, Properties

Cartesian Products of Sets: We are quite familiar with the term ‘product’. Mathematically, it means multiplying two or more quantities. For example: \(3\) multiplied by \(8\) gives \(24, 4\) multiplied by \(7\) gives \(28\). So, this is the case with numbers. Can we multiply two or more sets?

We know of operations like the union and intersection of two or more sets. We also know that the result of such operations is always a set. Here, let us understand what the Cartesian product of sets means and how the operation is performed with two or three sets.

What Is a Cartesian Product of Sets?

The Cartesian product is the product of two non-empty sets in an ordered fashion. 
Let \(A\) and \(B\) be two non-empty sets.
The set of all ordered pairs \((a, b)\) such that \(a \in A\) and \(b \in B\) is called the Cartesian product of the sets \(A\) and \(B\).
It is denoted as \(A \times B\).
Thus,
\(A \times B=\{(a, b): a \in A\) and \(b \in B\}\)
This is also called the cross product of sets.

Example (i):
Let \(A=\{2,4,6\}\) and \(B=\langle 1,2\}\)
Then \(A \times B=\{2,4,6\} \times\{1,2\}\)
\(=\{(2,1),(2,2),(4,1),(4,2),(6,1),(6,2)\}\)
And, \(B \times A=\{1,2\} \times\{2,4,6\}\)
\(=\{(1,2),(1,4),(1,6),(2,2),(2,4),(2,6)\}\)
Thus, from example (i), it is evident that for \(A \times B\), we take an element from set \(A\) and form ordered pairs with this as the first, and the element from set \(B\) as the second elements.
Next, we choose the second element from set \(A\), and with elements from set \(B\), we form ordered pairs. This process is continued till all the elements of set \(A\) are paired with all the elements in set \(B\).

Note:
If \(A\) or \(B\) is a null set, then we define \(A \times B=\phi\), where \(\phi\) denotes an empty set.

Example (ii):
Let \(A=\{a, b\}\) and \(B=\{1,2,3\}\), then find \(A \times B, B \times A, A \times A, B \times B\), and \((A \times B) \cap(B \times A)\).
Solution: Given: \(A=\{a, b\}\) and \(B=\{1,2,3\}\)
Then,
\(A \times B=\{(a, 1),(a, 2),(a, 3),(b, 1),(b, 2),(b, 3)\}\)
\(B \times A=\{(1, a),(1, b),(2, a),(2, b),(3, a),(3, b)\}\)
\(A \times A=\{(a, a),(a, b),(b, a),(b, b)\}\)
\(B \times B=\{(1,1),(1,2),(1,3),(2,1),(2,2),(2,3),(3,1),(3,2),(3,3)\}\)
\((A \times B) \cap(B \times A)=\phi\)

Cartesian Products of Three Sets

Let \(A, B\), and \(C\) be three non-empty sets. Then, \(A \times B \times C\) is the set of all ordered triplets having first element from set \(A\), second element from set \(B\) and the third element from set \(C\).
Cartesian product of three sets is defined as,
\(A \times B \times C=\{(a, b, c): a \in A, b \in B, c \in C\}\)

Example:
Let \(A=\{1,2\}, B=\{3,4\}\) and \(C=\{4,5,6\}\), then find \(A \times B \times C\).
Solution: Given: \(A=\{1,2\}, B=\{3,4\}, C=\{4,5,6\}\)
So, \(A \times B = \{ 1,2\}  \times \{ 3,4\}  = \{ (1,3),(1,4),(2,3),(2,4)\} \)
Then, \(A \times B \times C=(A \times B) \times C\)
\(=\{(1,3),(1,4),(2,3),(2,4)\} \times\{4,5,6\}\)
\(\therefore A \times B \times C=((1,3,4),(1,3,5),(1,3,6),(1,4,4),(1,4,5),(2,4,4),(2,4,5),(2,4,6)\}\)

Note:

• For any three sets \(A, B\), and \(C\), we can write
  \(A \times B \times C=(A \times B) \times C=A \times(B \times C)\)
• If \(A_{1}, A_{2}, A_{3}, \ldots, A_{n}\) are \(n\) sets, then the Cartesian product of \(A_{1} \times A_{2} \times \ldots \times A_{n}\) of these \(n\) sets is the set of \(n\)-tuples of the form \(\left(a_{1}, a_{2}, a_{3}, \ldots, a_{n}\right)\), where \(a_{1} \in A, a_{2} \in A_{2}, \ldots, a_{n} \in A_{n}\), so
  \(A_{1} \times A_{2} \times \ldots \times A_{n}=\left\{\left(a_{1}, a_{2}, a_{3}, \ldots, a_{n}\right): a_{1} \in A_{1}, a_{2} \in A_{2}, a_{3} \in A_{3}, \ldots, a_{n} \in A_{n}\right\}\)

Tuples

The concept of ordered pair can be extended to more than two elements. An ordered \(n\)-tuple is a set of \(n\) objects with an associated order. Tuples are generally denoted by \(\left(a_{1}, a_{2}, \ldots, a_{n}\right)\). The element \(a_{j}(j=1,2, \ldots, n)\) is called \(j^{t h}\) entry or component, and \(n\) is the length of the tuple.

Cardinality of Cartesian Product

The cardinality of a finite set is the number of distinct elements present in that set.
If \(A\) and \(B\) are finite sets, then \(n(A \times B)=n(A) \times n(B)\)
Proof: Let \(A=\left\{a_{1}, a_{2}, a_{3}, \ldots, a_{m}\right\}\) and \(B=\left\{b_{1}, b_{2}, b_{3}, \ldots, b_{n}\right\}\) are two finite sets are having \(m\) and \(n\) elements respectively, then,

Thus, in the tabular representation of \(A \times B\) there are \(m\) rows, and each row has \(n\) distinct ordered pairs.
The Cartesian product \(A \times B\) has \(m \times n\) elements.

Therefore, 
\(n(A \times B)=m \times n=n(A) \times n(B)\)
Where,
\(n(A) \rightarrow\) number of distinct elements in set \(A\)
\(n(B) \rightarrow\) number of distinct elements in set \(B\)

Note:

• If one of the sets \(A\) or \(B\) is an infinite set, then \(A \times B\) is also an infinite set.
• If \(A, B\), and \(C\) are three finite sets, then \(n(A \times B \times C)=n(A) \times n(B) \times n(C)\)

Example:
Let \(A=\{1,2\}, B=\{3,4\}\) and \(C=\{4,5,6\}\), then find \(n(A \times B \times C)\).
Solution: Given: \(A=\{1,2\}, B=\{3,4\}\) and \(C=\{4,5,6\}\)
\(\Rightarrow n(A)=2 ; n(B)=2 ; n(C)=3\)
\(n(A \times B \times C)=n(A) \times n(B) \times n(C)\)
\(\Rightarrow n(A \times B \times C)=2 \times 2 \times 3\)
Therefore, \(n(A \times B \times C)=12\).

Graphical Representation

Let \(A\) and \(B\) be two non-empty sets. To represent \(A \times B\) graphically follow these steps.
Step 1: Draw two perpendicular lines.
Step 2: On the horizontal line, we represent the elements of set \(A\), and on the vertical line, the elements of \(\operatorname{set} B\).
Step 3: For \(a \in A, b \in B\), draw a vertical line through \(a\) and horizontal line through \(b\). These two lines will meet in a point which will denote the ordered pair \((a, b)\).
Step 4: Repeat the steps to mark points corresponding to each of the ordered pairs in \(A \times B\).
Example: Let \(A=\{1,2,3\}\) and \(B=\{2,4\}\), then find \(A \times B\) and show It graphically.
Solution: Given that, we have: \(A=\{1,2,3\}\) and \(B=\{2,4\}\)
Now, \(A \times B=\{(1,2),(1,4),(2,2),(2,4),(3,2),(3,4)\}\)
The set \(A \times B\) is represented on the graph as shown below.

Diagrammatic Representation

In order to represent \(A \times B\) by an arrow diagram, we first draw Venn diagrams representing sets \(A\) and \(B\) as shown in the example below. Now we connect each element of set \(A\) to each element of set \(B\).

Example:
If \(A=\{1,3,5\}\) and \(B=\{a, b\}\), then the following figure show the arrow diagram of \(A \times B\).

Here, \(A \times B=\{(1, a),(1, b),(3, a),(3, b),(5, a),(5, b)\}\)

Properties of Cartesian Products

Property 1: The Cartesian product is non-commutative. This means that the Cartesian product of two sets is not the same when the sets are interchanged.
\(A \times B \neq B \times A\)
Property 2: \(A \times B=B \times A\), if only \(A=B\)
Property 3: \(A \times B=\emptyset\), if either \(A=\emptyset\) or \(B=\emptyset\)
Property 4: The Cartesian product is non-associative. This means that when three non-empty sets are multiplied, the product changes when the sets are grouped differently.
\((A \times B) \times C \neq A \times(B \times C)\)
Property 5: The Cartesian product is distributive over intersection.
\(A \times(B \cap C)=(A \times B) \cap(A \times C)\)
Property 6: The Cartesian product is distributive over union.
\(A \times(B \cup C)=(A \times B) \cup(A \times C)\)
Property 7: The Cartesian product is distributive over difference.
\(A \times(B-C)=(A \times B)-(A \times C)\)
Property 8: If \(A \subseteq B\), then \(A \times C \subseteq B \times C\) for any set \(C\).
Property 9: The cardinality of the Cartesian Product is defined as the number of elements in \(A \times B\) and is equal to the product of cardinality of both sets \(A\) and \(B\) i.e., \(n(A \times B)=n(A) \times n(B)\).

Solved Examples

Q.1. Let \(A=\{1,3,5,6\}\) and \(B=\{2,4\}\), then find \(A \times B\) and \(B \times A\).
Ans: Given: \(A=\{1,3,5,6\}\)
\(B=\{2,4\}\)
Then, \(A \times B=\{1,3,5,6\} \times\{2,4\}\)
\(\therefore A \times B=\{(1,2),(1,4),(3,2),(3,4),(5,2),(5,4),(6,2),(6,4)\}\)
Similarly, \(B \times A=\{2,4\} \times\{1,3,5,6\}\)
\(\therefore B \times A=\{(2,1),(2,3),(2,5),(2,6),(4,1),(4,3),(4,5),(4,6)\}\)

Q.2. Let \(A=\{1,2,3\}, B=\{3,4\}\) and \(C=\{1,3,5\}\), then find \(A \times(B \cap C)\).
Ans: Given that, we have: \(A=\{1,2,3\}, B=\{3,4\}\) and \(C=\{1,3,5\}\)
Now, \(B \cap C=\{3,4\} \cap\{1,3,5\}\)
\(=\{3\}\)
So, \(A \times(B \cap C)=\{1,2,3\} \times\{3\}\)
\(=\{(1,3),(2,3),(3,3)\}\)

Q.3. Let \(A=\{1,2,3\}\) and \(B=\{x: x \in N, x\) is prime less than \(5\}\). Find \(A \times B\) and \(B \times A\).
Ans: Given: \(A=\{1,2,3\}\)
\(B=\{x: x \in N, x\) is prime less than \({\text{5\} =}}\left\{ {{\text{2,3}}} \right\}\)
Then, \(A \times B=\{1,2,3\} \times\{2,3\}\)
\(\therefore A \times B=\{(1,2),(1,3),(2,2),(2,3),(3,2),(3,3)\}\)
Similarly, \(B \times A=\{2,3\} \times\{1,2,3\}\)
\(\therefore B \times A=\{(2,1),(2,2),(2,3),(3,1),(3,2),(3,3)\}\)

Q.4. Let \(A \times B=\{(a, 1),(a, 4),(a, 2),(b, 2),(b, 4),(b, 1)\}\), then find \(B \times A\).
Ans: We have, \(A \times B=\{(a, 1),(a, 4),(a, 2),(b, 2),(b, 4),(b, 1)\}\)
So, \(B \times A\) can be obtained from \(A \times B\) by interchanging the entries or components of ordered pairs in \(A \times B\)
\(\therefore B \times A=\{(1, a),(4, a),(2, a),(2, b),(4, b),(1, b)\}\)

Q.5. Let \(A\) and \(B\) are two sets such that \(n(A)=5\) and \(n(B)=2\). If \(a, b, c, d, e\) are distinct elements and \((a, 2),(b, 3),(c, 2),(d, 3),(e, 2)\) are in \(A \times B\), then find \(A\) and \(B\).
Ans: Since \((a, 2),(b, 3),(c, 2),(d, 3),(e, 2)\) are the elements of \(A \times B\), we know that, \(A=\{a, b, c, d, e\}\) and \(B=\{2,3\}\).
It is also given that, \(n(A)=5\) and \(n(B)=2\)
Therefore, \(a, b, c, d, e \in A\) and \(n(A)=5 \Rightarrow A=\{a, b, c, d, e\}\)
Similarly, \(2,3 \in B\) and \(n(B)=2 \Rightarrow B=\{2,3\}\)
Hence, \(A=\{a, b, c, d, e\}\) and \(B=\{2,3\}\)

Q.6. If \(A=\{-1,3,4\}\) and \(B=\{2,3\}\), then represent \(A \times B\) graphically.
Ans: Given that, we have: \(A=\{-1,3,4\}\) and \(B=\{2,3\}\)
Now, \(A \times B=\{-1,3,4\} \times\{2,3\}\)
\(=\{(-1,2),(-1,3),(3,2),(3,3),(4,2),(4,3)\}\)

Q.7. Let \(A=\{1,3,5\}, B=\{x, y\}\), then represent \(A \times B\) and \(B \times A\) by using arrow diagram.
Ans: We have, \(A=\{1,3,5\}, B=\{x, y\}\)
Now, \(A \times B=\{1,3,5\} \times\{x, y\}\)
\(=\{(1, x),(1, y),(3, x),(3, y),(5, x),(5, y)\}\)
The following arrow diagram represents \(A \times B\)

Similarly, \(B \times A = \left\{ {x,y} \right\} \times \left\{ {1,3,5} \right\}\)
\(=\{(x, 1),(x, 3),(x, 5),(y, 1),(y, 3),(y, 5)\}\)
The following arrow diagram represents \(B \times A\).

Summary

Cartesian product of two sets \(A\) and \(B\) results in a set of ordered pairs whose first element is from set \(A\) and the second element is from set \(B\). Similarly, Cartesian product of three sets can also be obtained as a set of ordered triplets. The number of elements in a Cartesian product is equal to the product of the number of elements in set \(A\) and \(B\). The article also describes the various properties of the Cartesian product of two or three sets, along with graphical and diagrammatic representation. Although Cartesian products are not commutative, and not associative, the products are distributive over other operations such as union and intersection.

Know More About Relations and Their Representations Here

FAQs

Q1. What is the Cartesian product of three sets? 
Ans: If \(A, B\) and \(C\) are three sets. Then, \(A×B×C\) is the set of all ordered triplets having the first element from the set \(A\), second element from the set \(B\) and third element from the set \(C\). This is represented as, \(A \times B \times C=\{(p, q, r): p \in A, q \in B, r \in C\}\)

Q2. How do you find the Cartesian product of two sets?
Ans: In Mathematics, the Cartesian product of two sets \(A\) and \(B\) is defined as the set of all ordered pairs \((x, y)\) such that \(x\) belongs to set \(A\) and \(y\) belongs to set \(B\). This is represented as, \(A \times B=\{(x, y): x \in A\) and \(y \in B\}\)

Q3. What is the Cartesian product of a set with the empty set?
Ans: The Cartesian product of sets \(A\) and \(B\) is defined as the set of all ordered pairs \((x, y)\) such that \(x\) belongs to set \(A\) and \(y\) belongs to set \(B\). So, the Cartesian product of any set with an empty set will always be an empty set, as an empty set does not contain any elements.

Q4. Is Cartesian product distributive?
Ans: For any three sets \(A, B\) and \(C\), the Cartesian product is distributive over union is \(A \times(B \cup C)=(A \times B) \cup(A \times C)\)

Q5. What is the Cartesian product of \(n\) sets?
Ans: Let \(A_{1}, A_{2}, A_{3}, \ldots, A_{n}\) are \(n\) sets, then the Cartesian product of \(A_{1} \times A_{2} \times \ldots \times A_{n}\) of these \(n\) sets is the set of \(n\)-tuples of the form \(\left(a_{1}, a_{2}, a_{3}, \ldots, a_{n}\right)\), where \(a_{1} \in A, a_{2} \in A_{2}, \ldots, a_{n} \in A_{n}\), so,\(A_{1} \times A_{2} \times \ldots \times A_{n}=\left\{\left(a_{1}, a_{2}, a_{3}, \ldots, a_{n}\right): a_{1} \in A_{1}, a_{2} \in A_{2}, a_{3} \in A_{3}, \ldots, a_{n} \in A_{n}\right\}\)