Calculus

Set Theory

Set Theory Logo

Cartesian Product of Sets

Solved Problems

Example 1.

Given \(A = \{1,2,5\}\) and \(B = \{1,2\}.\) Find the following sets:

  1. \({A \times B}\)
  2. \({B \times A}\)
  3. \({A^2}\)
  4. \({B^2}\)

Solution.

  1. By definition, the Cartesian product \({A \times B}\) contains all possible ordered pairs \(\left({a,b}\right)\) such that \(a \in A\) and \(b \in B.\) Therefore, we can write
    \[A \times B = \left\{ {1,2,5} \right\} \times \left\{ {1,2} \right\} = \left\{ {\left( {1,1} \right),\left( {1,2} \right),\left( {2,1} \right),\left( {2,2} \right),\left( {5,1} \right),\left( {5,2} \right)} \right\}.\]
  2. Similarly we find the Cartesian product \({B \times A}:\)
    \[B \times A = \left\{ {1,2} \right\} \times \left\{ {1,2,5} \right\} = \left\{ {\left( {1,1} \right),\left( {1,2} \right),\left( {1,5} \right),\left( {2,1} \right),\left( {2,2} \right),\left( {2,5} \right)} \right\}.\]
  3. The Cartesian square \(A^2\) is defined as \({A \times A}.\) So, we have
    \[{A^2} = A \times A = \left\{ {1,2,5} \right\} \times \left\{ {1,2,5} \right\} = \left\{ {\left( {1,1} \right),\left( {1,2} \right),\left( {1,5} \right),\left( {2,1} \right),\left( {2,2} \right),\left( {2,5} \right),\left( {5,1} \right),\left( {5,2} \right),\left( {5,5} \right)} \right\}.\]
  4. The Cartesian square \(B^2\) is given by
    \[{B^2} = B \times B = \left\{ {1,2} \right\} \times \left\{ {1,2} \right\} = \left\{ {\left( {1,1} \right),\left( {1,2} \right),\left( {2,1} \right),\left( {2,2} \right)} \right\}.\]

Example 2.

Given \(A = \{a,b,c\}\) and \(B = \{b,c\}.\) Find the following sets:

  1. \(\left( {A \times B} \right) \cap \left( {B \times A} \right)\)
  2. \(\left( {A \times B} \right) \cup \left( {B \times A} \right)\)

Solution.

  1. We calculate the Cartesian products \({A \times B}\) and \({B \times A}\) and then determine their intersection:
    \[A \times B = \left\{ {a,b,c} \right\} \times \left\{ {b,c} \right\} = \left\{ {\left( {a,b} \right),\left( {a,c} \right),\left( {b,b} \right),\left( {b,c} \right),\left( {c,b} \right),\left( {c,c} \right)} \right\},\]
    \[B \times A = \left\{ {b,c} \right\} \times \left\{ {a,b,c} \right\} = \left\{ {\left( {b,a} \right),\left( {b,b} \right),\left( {b,c} \right),\left( {c,a} \right),\left( {c,b} \right),\left( {c,c} \right)} \right\},\]
    \[\left( {A \times B} \right) \cap \left( {B \times A} \right) = \left\{ {\left( {b,b} \right),\left( {b,c} \right),\left( {c,b} \right),\left( {c,c} \right)} \right\}.\]
  2. The union of the Cartesian products \({A \times B}\) and \({B \times A}\) is given by:
    \[\left( {A \times B} \right) \cup \left( {B \times A} \right) = \left\{ {\left( {a,b} \right),\left( {b,a} \right),\left( {a,c} \right),\left( {c,a} \right),\left( {b,b} \right),\left( {b,c} \right),\left( {c,b} \right),\left( {c,c} \right)} \right\}.\]

Example 3.

Suppose \(A = \{x,y\},\) \(B = \{1,2\}\) and \(C = \{2,3\}.\) Determine the sets:

  1. \(A \times \left( {B \cup C} \right)\)
  2. \(\left( {A \times B} \right) \cup \left( {A \times C} \right)\)

Solution.

  1. First we find the union of the sets \(B\) and \(C:\)
    \[B \cup C = \left\{ {1,2} \right\} \cup \left\{ {2,3} \right\} = \left\{ {1,2,3} \right\}.\]
    Then the Cartesian product of \(A\) and \(B \cup C\) is given by
    \[A \times \left( {B \cup C} \right) = \left\{ {x,y} \right\} \times \left\{ {1,2,3} \right\} = \left\{ {\left( {x,1} \right),\left( {x,2} \right),\left( {x,3} \right),\left( {y,1} \right),\left( {y,2} \right),\left( {y,3} \right)} \right\}.\]
  2. Compute the Cartesian products of given sets:
    \[A \times B = \left\{ {x,y} \right\} \times \left\{ {1,2} \right\} = \left\{ {\left( {x,1} \right),\left( {x,2} \right),\left( {y,1} \right),\left( {y,2} \right)} \right\}.\]
    \[A \times C = \left\{ {x,y} \right\} \times \left\{ {2,3} \right\} = \left\{ {\left( {x,2} \right),\left( {x,3} \right),\left( {y,2} \right),\left( {y,3} \right)} \right\}.\]
    Now we can find the union of the sets \(A \times B\) and \(A \times C:\)
    \[\left( {A \times B} \right) \cup \left( {A \times C} \right) = \left\{ {\left( {x,1} \right),\left( {x,2} \right),\left( {x,3} \right),\left( {y,1} \right),\left( {y,2} \right),\left( {y,3} \right)} \right\}.\]
    We see that
    \[A \times \left( {B \cup C} \right) = \left( {A \times B} \right) \cup \left( {A \times C} \right).\]
    This identity confirms the distributive property of Cartesian product over set union.

Example 4.

Suppose \(A = \{a,b\},\) \(B = \{4,6\}\) and \(C = \{5,6\}.\) Determine the sets:

  1. \(A \times \left( {B \cap C} \right)\)
  2. \(\left( {A \times B} \right) \cap \left( {A \times C} \right)\)

Solution.

  1. Find the intersection of the sets \(B\) and \(C:\)
    \[B \cap C = \left\{ {4,6} \right\} \cap \left\{ {5,6} \right\} = \left\{ 6 \right\}.\]
    The Cartesian product of \(A\) and \(B \cap C\) is written as
    \[A \times \left( {B \cap C} \right) = \left\{ {a,b} \right\} \times \left\{ 6 \right\} = \left\{ {\left( {a,6} \right),\left( {b,6} \right)} \right\}.\]
  2. Compute the Cartesian products:
    \[A \times B = \left\{ {a,b} \right\} \times \left\{ {4,6} \right\} = \left\{ {\left( {a,4} \right),\left( {a,6} \right),\left( {b,4} \right),\left( {b,6} \right)} \right\}.\]
    \[A \times C = \left\{ {a,b} \right\} \times \left\{ {5,6} \right\} = \left\{ {\left( {a,5} \right),\left( {a,6} \right),\left( {b,5} \right),\left( {b,6} \right)} \right\}.\]
    The intersection of the two sets is given by
    \[\left( {A \times B} \right) \cap \left( {A \times C} \right) = \left\{ {\left( {a,6} \right),\left( {b,6} \right)} \right\}.\]
    So, we have validated the distributive property of Cartesian product over set intersection:
    \[A \times \left( {B \cap C} \right) = \left( {A \times B} \right) \cap \left( {A \times C} \right).\]

Example 5.

Find the Cartesian product \(A \times \mathcal{P}\left( A \right)\) if \(A = \left\{ {0,1} \right\}.\)

Solution.

The power set of \(A\) is written in the form

\[\mathcal{P}\left( A \right) = \mathcal{P}\left( {\left\{ {0,1} \right\}} \right) = \left\{ {\varnothing,\left\{ 0 \right\},\left\{ 1 \right\},\left\{ {0,1} \right\}} \right\}.\]

Hence, the Cartesian product \(A \times \mathcal{P}\left( A \right)\) is given by

\[A \times \mathcal{P}\left( A \right) = \left\{ {0,1} \right\} \times \left\{ {0,\left\{ 0 \right\},\left\{ 1 \right\},\left\{ {0,1} \right\}} \right\} = \left\{ {\left( {0,\varnothing} \right),\left( {0,\left\{ 0 \right\}} \right),\left( {0,\left\{ 1 \right\}} \right),\left( {0,\left\{ {0,1} \right\}} \right),\left( {1,\varnothing} \right),\left( {1,\left\{ 0 \right\}} \right),\left( {1,\left\{ 1 \right\}} \right),\left( {1,\left\{ {0,1} \right\}} \right)} \right\}.\]

Example 6.

Find the Cartesian product \(\left\{ {1,2,3} \right\} \times \mathcal{P}\left( {\left\{ a \right\}} \right).\)

Solution.

The power set \(\mathcal{P}\left( {\left\{ a \right\}} \right)\) consists of one element and contains two subsets:

\[\mathcal{P}\left( {\left\{ a \right\}} \right) = \left\{ {\varnothing,\left\{ a \right\}} \right\}.\]

The Cartesian product of the sets \(\left\{ {1,2,3} \right\}\) and \(\mathcal{P}\left( {\left\{ a \right\}} \right)\) is given by

\[\left\{ {1,2,3} \right\} \times \mathcal{P}\left( {\left\{ a \right\}} \right) = \left\{ {1,2,3} \right\} \times \left\{ {\varnothing,\left\{ a \right\}} \right\} = \left\{ {\left( {1,\varnothing} \right),\left( {1,\left\{ a \right\}} \right),\left( {2,\varnothing} \right),\left( {2,\left\{ a \right\}} \right),\left( {3,\varnothing} \right),\left( {3,\left\{ a \right\}} \right)} \right\}.\]

Example 7.

Let \(A = \left\{ {{a_1}, \ldots ,{a_n}} \right\}.\) Compute the cardinality of the set \(\mathcal{P}\left( {{A^m}} \right).\)

Solution.

If the set \(A\) has \(n\) elements, then the \(m\text{th}\) Cartesian power of \(A\) will contain \(n^m\) elements:

\[\left| {{A^m}} \right| = \left| {\underbrace {A \times \ldots \times A}_m} \right| = \underbrace {\left| A \right| \times \ldots \times \left| A \right|}_m = \underbrace {n \times \ldots \times n}_m = n^m.\]

Then the cardinality of the power set of \(A^m\) is

\[\left| {\mathcal{P}\left( {{A^m}} \right)} \right| = {2^{n^m}}.\]

Example 8.

Let \(X = \left\{ {x,y} \right\}.\) Compute the cardinality of the set \(\mathcal{P}\left( {\mathcal{P}\left( X \right)} \right) \times \mathcal{P}\left( X \right).\)

Solution.

First we find the power set of \(X:\)

\[\mathcal{P}\left( X \right) = \mathcal{P}\left( {\left\{ {x,y} \right\}} \right)= \left\{ {\varnothing,\left\{ x \right\},\left\{ y \right\},\left\{ {x,y} \right\}} \right\}.\]

We see that \(\mathcal{P}\left( X \right)\) contains \(4\) elements:

\[\left| {\mathcal{P}\left( X \right)} \right| = \left| {\mathcal{P}\left( {\left\{ {x,y} \right\}} \right)} \right| = {2^2} = 4.\]

It is clear that the power set of \(\mathcal{P}\left( X \right)\) will have \(16\) elements:

\[\left| {\mathcal{P}\left( {\mathcal{P}\left( X \right)} \right)} \right| = {2^4} = 16.\]

Consider now the Cartesian product:

\[\left| {\mathcal{P}\left( {\mathcal{P}\left( X \right)} \right) \times \mathcal{P}\left( X \right)} \right| = \left| {\mathcal{P}\left( {\mathcal{P}\left( X \right)} \right)} \right| \times \left| {\mathcal{P}\left( X \right)} \right| = 16 \times 4 = 64,\]

so the cardinality of the given set is equal to \(64.\)

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