Question

Let \(A=\\{0,2,4,6,8\\}\) and \(B=\\{1,3,5,7\\}\) have universal set \(U=\\{0,1,2, \ldots, 8\\} .\) Find: (a) \(\bar{A}\) (b) \(\bar{B}\) (c) \(A \cap \bar{A}\) (d) \(A \cup \bar{A}\) (e) \(A-\bar{A}\) (f) \(\overline{A \cup B}\) (g) \(\bar{A} \cap \bar{B}\) (h) \(\overline{A \cap B}\) (i) \(\bar{A} \times B\)

Short answer

(a) \(\bar{A}\) = \{1,3,5,7\}, (b) \(\bar{B}\) = \{0,2,4,6,8\}, (c) \(A \cap \bar{A}\) = \(\emptyset\), (d) \(A \cup \bar{A}\) = \{0,1,2,3,4,5,6,7,8\}, (e) \(A-\bar{A}\) = \{0,2,4,6,8\}, (f) \(\overline{A \cup B}\) = \(\emptyset\), (g) \(\bar{A} \cap \bar{B}\) = \(\emptyset\), (h) \(\overline{A \cap B}\) = \{0,1,2,3,4,5,6,7,8\}, (i) \(\bar{A} \times B\) = \{(1,1), (1,3), (1,5), (1,7), (3,1), (3,3), (3,5), (3,7), (5,1), (5,3), (5,5), (5,7), (7,1), (7,3), (7,5), (7,7)\}

Step-by-step solution

Finding the complement of sets A and B

The complement of A \(\bar{A}\) are all elements in U not in A. U = \{0,1,2,3,4,5,6,7,8\}, A = \{0,2,4,6,8\}. So, \(\bar{A}\) = \{1,3,5,7\}. On the other hand, \(\bar{B}\) = U-B = \{0,2,4,6,8\}

Finding intersection and union of A and its complement

\(A \cap \bar{A}\) represents elements common to both A and \(\bar{A}\). Since A and \(\bar{A}\) are complementary sets, they do not have any common elements. So, \(A \cap \bar{A}\) = \(\emptyset\) (the empty set). The union of A and \(\bar{A}\) represents all elements that are in A or \(\bar{A}\) or in both. As the universal set U forms the union of A and \(\bar{A}\), we get \(A \cup \bar{A}\) = U = \{0,1,2,3,4,5,6,7,8\}

Finding the difference between A and its complement

The set difference \(A - \bar{A}\) represents the elements that are in A but not in \(\bar{A}\). As \(A - \bar{A}\) = A, because all the elements of A are not present in \(\bar{A}\), we get \(A - \bar{A}\) = \{0,2,4,6,8\}

Finding the complement of the union of A and B

The union of A and B, \(A \cup B\) is \{0,1,2,3,4,5,6,7,8\}, which is similar to universal set U. Thus, the complement of \(A \cup B\), denoted as \(\overline{A \cup B}\), is \(\emptyset\)

Finding the intersection of the complements of A and B

\(\bar{A} \cap \bar{B}\) represents elements common to \(\bar{A}\) and \(\bar{B}\). As \(\bar{A}\) and \(\bar{B}\) do not have any common elements, \(\bar{A} \cap \bar{B} = \emptyset\).

Finding the complement of the intersection of A and B

The intersection of A and B, \(A \cap B\) is \(\emptyset\) as A and B do not have common elements. Therefore, the complement of \(A \cap B\), \(\overline{A \cap B}\), is the set of all elements in U which is equal to \{0,1,2,3,4,5,6,7,8\}

Finding Cartesian product of the complement of A and B

The Cartesian product of the complement of A (\(\bar{A}\)) and B, \(\bar{A} \times B\) represents all ordered pairs \((\bar{a},b)\), where \(\bar{a}\) is in \(\bar{A}\) and b is in B. So \(\bar{A} \times B\) = \{(1,1), (1,3), (1,5), (1,7), (3,1), (3,3), (3,5), (3,7), (5,1), (5,3), (5,5), (5,7), (7,1), (7,3), (7,5), (7,7)\}

Key concepts

Complement of a Set

In set theory, the complement of a set consists of elements not in the original set but within a larger set, known as the universal set. For instance, if we have a universal set \( U \), and a subset \( A \), the complement of \( A \) (represented as \( \bar{A} \) or \( A' \) ) includes all the elements in \( U \) that are not in \( A \). For clear understanding, imagine you've a box of assorted chocolates. If \( A \) represents all the milk chocolates in the box, then \( \bar{A} \) would represent all the other types of chocolates, but not the milk chocolates. From the given exercise, \( \bar{A} \) = {1, 3, 5, 7}, which are precisely the numbers in the universal set that are not in \( A \).

Intersection of Sets

The intersection of sets contains all the elements that are common to all the sets being considered. Mathematically, if you have sets \( A \) and \( B \) then \( A \cap B \) would include items that appear in both \( A \) and \( B \). For a real-world example, if one circle represents people who like pineapple on pizza, and another represents those who like ham on pizza, their intersection would be people who enjoy a 'Hawaiian' style, with both toppings. In our exercise, \( A \cap \bar{A} \) yields no common elements, thus resulting in an empty set, \( \emptyset \).

Union of Sets

The union of sets combines all distinct elements found in each of the sets. Symbolically, it's denoted by \( A \cup B \) for sets \( A \) and \( B \) and includes every element that is in \( A \), \( B \), or both. Consider a pizza menu. If one section lists vegetarian options and another lists meat options, the union of these sets would be a complete list of all pizzas available, without repeats. For our example, \( A \cup \bar{A} \) is the same as the universal set \( U \), which means it includes all numbers from 0 to 8.

Set Difference

The set difference denotes elements that belong to one set but not another. It's expressed as \( A - B \) for two sets \( A \) and \( B \) and consists of only those elements that are in \( A \) but not in \( B \). A simple analogy might be if you have two grocery lists, \( A \) represents your list, and \( B \) is what's already in the pantry. \( A - B \) is essentially what you still need to buy that's not in the pantry. From our exercise \( A - \bar{A} \) resulted in the set \( A \) itself since none of its elements are in \( \bar{A} \).

Cartesian Product

The Cartesian product of two sets forms ordered pairs where the first element comes from the first set, and the second element comes from the second. Denoted by \(( A \times B \), it is the set of all possible combinations you can make from the two sets. To visualize it: imagine you have two dials, one with colors and another with shapes. Turning each dial creates a pair of color and shape; the complete set of these combinations is the Cartesian product. Following the logic in the solved exercise, \( \bar{A} \times B \) gives us all possible pairings between numbers in set \( \bar{A} \) and set \( B \).

Universal Set

The universal set is the parent set that typically contains all objects under consideration for a particular discussion. It encompasses all other sets and their elements. In simple terms, if we are in a classroom, the universal set might be all the people there, including students and faculty. Every subset—like students, faculty, or staff—is contained within this larger set. In our exercise, \( U \) is defined as {0, 1, 2, ..., 8}, which includes all single digit numbers that are relevant to sets \( A \) and \( B \) and their complements. This concept is crucial as it forms the reference where we find the complement of any set.