The intersection of sets A and B, written \(A \cap B \) is the set of elements common to both \(A\) and \(B\), represented as
\[ A \cap B = \{x|x\in A \,and\,x\in B\} \]
Example: Find the intersection of the following sets, \(A = \{1, 2, 3, 5, 8 \}, \,B=\{1, 2, 3, 4 \} \)
Solution: \(A \cap B =\{1, 2, 3 \}\) Elements that are in both sets A and B are in the intersection.
The union of sets A and B, written \(A \cup B\), is the set of elements belonging to either of the sets, represented as:
\[ A \cup B = \{x|x\in A \,or\,x\in B\} \]
Example: Find the union of the following sets, \(A = \{1, 2, 3, 5, 8 \}, \,B=\{1, 2, 3, 4 \} \)
Solution: \(A \cup B =\{1, 2, 3, 4, 5, 8 \}\) All elements that in sets A and B are in the union.
The difference of sets A and B, represented \(A - B \), is the set of elements belonging to set A and not to set B,
\[ A - B = \{x|x\in A \,and\,x\notin B\} \]
Example: Find the difference of the sets, \(A = \{1, 2, 3, 5, 8 \}, \,B=\{1, 2, 3, 4 \} \), \(A - B \), and \(B - A\)
Solution: \(A - B =\{5, 8 \}\) All elements that are in set A but not in set B.
\(B - A =\{4\}\) All elements that are in set B but not in set A.
In the ordered pair \(\left(a,b\right)\), a is called the first component and b is called the second component. Switching components does not give you the same statement. \(\left(a,b\right)\ \neq \left(b,a\right)\).
The Cartesian product of sets A and B, written \(A \times B\), is: \[A \times B = \{(a, b)|a\in A \,and\, b\in B\} \]
Example: Let \(A = \{1, 2 \}, \, and \,B=\{a, b, c \} \)
Then \(A \times B = \{(1,a), (1,b), (1,c), (2,a), (2,b), (2,c) \}\)
The cardinal number of a Cartesian product is the product of the cardinal numbers of each set.
If \(n\left(A\right)=a\) and \(n\left(B\right)=b\), then
\begin{align} n\left(A \times B\right) &= n\left(B \times A\right) \\ &= n\left(A \right) \cdot n\left(B \right) = n\left(B \right) \cdot n\left(A \right) \\ &= ab = ba \end{align}
Example: Let \(A = \{1, 2 \}, \, and \,B=\{a, b, c \} \), then \(n(A) = 2\) and \(n(B) = 3\). \(n(A \times B) = 6\)
