Problem 1 - Applied Discrete Structures

Question

Let \(A=\\{0,2,3\\}, B=\\{2,3\\}, C=\\{1,4\\},\) and let the universal set be \(U=\\{0,1,2,3,4\\} .\) List the elements of (a) \(A \times B\) (e) \(A \times A^{c}\) (b) \(B \times A\) (f) \(B^{2}\) (c) \(A \times B \times C\) (g) \(B^{3}\) (d) \(U \times \emptyset\) (h) \(B \times \mathcal{P}(B)\)

Step-by-Step Solution

Answer

(a) \(A \times B = \{(0,2), (0,3), (2,2), (2,3), (3,2), (3,3)\}\)\n(b) \(B \times A = \{(2,0), (2,2), (2,3), (3,0), (3,2), (3,3)\}\)\n(c) \(A \times B \times C = \{(0,2,1), (0,2,4), (0,3,1), (0,3,4), (2,2,1), (2,2,4), (2,3,...\n(d) \(U \times \emptyset = \emptyset\)\n(e) \(A \times A^{c} = \{(0,1), (0,4), (2,1), (2,4), (3,1), (3,4)\}\)\n(f) \(B^2 = \{(2,2), (2,3), (3,2), (3,3)\}\)\n(g) \(B^3 = \{(2,2,2), (2,2,3), (2,3,2), (2,3,3), (3,2,2), (3,2,3), (3,3,2), (3,3,3)\}\)\n(h) \(B \times \mathcal{P}(B) = \{(2, \emptyset), (2, \{2\}), (2, \{3\}), (2, \{2,3\}), (3, \emptyset), (3, \{2\}), (3, \{3\}), (3, \{2,3\})\}\)

Step 1: Understand Set Notation

The problem presents several different operations with sets, namely Cartesian products and complements, and requires knowledge of the universal set. Remember that the Cartesian product of two sets \(X\) and \(Y\), denoted \(X \times Y\), consists of all ordered pairs \((x, y)\) such that \(x \in X\) and \(y \in Y\). The complement of a set \(X\) with respect to the universal set \(U\) is \(X^{c} = U - X\).

Step 2: Calculate Complement Sets

First calculate the complements needed for the problem using the universal set \(U = \{0,1,2,3,4\}\).- For \(A^{c}\): \(A = \{0,2,3\}\), so \(A^{c} = \{1,4\}\).- For \(B^{c}\): \(B = \{2,3\}\), so \(B^{c} = \{0,1,4\}\).

Step 3: Calculate A × B

Compute the Cartesian product \(A \times B\):\[A \times B = \{(a,b) \mid a \in A, b \in B\}\].Thus, \[A \times B = \{(0,2), (0,3), (2,2), (2,3), (3,2), (3,3)\}\].

Step 4: Calculate B × A

Compute the Cartesian product \(B \times A\):\[B \times A = \{(b,a) \mid b \in B, a \in A\}\].Thus, \[B \times A = \{(2,0), (2,2), (2,3), (3,0), (3,2), (3,3)\}\].

Step 5: Calculate A × B × C

Compute the Cartesian product \(A \times B \times C\). Since \(C = \{1,4\}\), list all possibilities:\[(a, b, c) \mid a \in A, b \in B, c \in C\].This results in 12 pairs:\[\{(0,2,1), (0,2,4), (0,3,1), (0,3,4),(2,2,1), (2,2,4), (2,3,1), (2,3,4),(3,2,1), (3,2,4), (3,3,1), (3,3,4)\}\].

Step 6: Calculate U × Empty Set

The Cartesian product \(U \times \emptyset\) is empty because there are no elements in \(\emptyset\) to form a pair with elements of \(U\). Thus, \(U \times \emptyset = \emptyset\).

Step 7: Calculate A × A^c

Compute the Cartesian product \(A \times A^{c}\) with \(A^{c} = \{1,4\}\):\[A \times A^{c} = \{(a, a^c) \mid a \in A, a^c \in A^{c}\}\].Thus, \[A \times A^{c} = \{(0,1), (0,4), (2,1), (2,4), (3,1), (3,4)\}\].

Step 8: Calculate B²

Compute \(B \times B\): \[B^2 = B \times B = \{(b_1, b_2) \mid b_1 \in B, b_2 \in B\}\].Thus, \[B^2 = \{(2,2), (2,3), (3,2), (3,3)\}\].

Step 9: Calculate B³

Compute \(B \times B \times B\):\[B^3 = B \times B \times B = \{(b_1, b_2, b_3) \mid b_1, b_2, b_3 \in B\}\].Thus, \[B^3 = \{(2,2,2), (2,2,3), (2,3,2), (2,3,3),(3,2,2), (3,2,3), (3,3,2), (3,3,3)\}\].

Step 10: Calculate B × Power Set of B

The power set \(\mathcal{P}(B)\) of \(B = \{2,3\}\) consists of \(\{\emptyset, \{2\}, \{3\}, \{2,3\}\}\).Compute \(B \times \mathcal{P}(B)\):\[(b, S) \mid b \in B, S \in \mathcal{P}(B)\].Thus, \[B \times \mathcal{P}(B) = \{(2, \emptyset), (2, \{2\}), (2, \{3\}), (2, \{2,3\}),(3, \emptyset), (3, \{2\}), (3, \{3\}), (3, \{2,3\})\}\].

Key Concepts

Set Complement

A set complement, often expressed as \( A^c \), refers to the elements that are part of the universal set \( U \) but not in the set \( A \). It is calculated by subtracting the elements of \( A \) from the universal set. For example, if \( A = \{0,2,3\} \) and \( U = \{0,1,2,3,4\} \), the complement \( A^c \) includes elements that are present in \( U \) but not in \( A \). Thus, \( A^c = \{1,4\} \). It is essential to first define the universal set to accurately determine complements because it serves as the reference set from which elements are "excluded" to find a complementary set. Remember, the complement can vary if \( U \) changes, so always confirm your universal set before calculating.

Universal Set

The universal set \( U \) is a foundational concept in set theory. It is the set that contains all possible elements of the discussion, encompassing every member of all sets we are working with in a problem. In the given exercise, \( U = \{0,1,2,3,4\} \), which means all calculations, including set complements, Cartesian products, and elsewhere, are based upon this set.

• When we find complements, such as \( A^c \) for any set \( A \), we always refer back to \( U \) as it represents the whole universe of possibilities within that context.
• The universal set is an essential piece of context that helps set bounds for set operations and ensures complete understanding of the problem.

Neglecting the correct definition of \( U \) can lead to incorrect results, so always ensure it is clearly defined at the start of any problem.

Power Set

A power set of a given set \( B \), denoted by \( \mathcal{P}(B) \), is the set of all possible subsets of \( B \), including both the empty set and \( B \) itself. For instance, for \( B = \{2,3\} \), the power set \( \mathcal{P}(B) \) consists of these subsets:

• \( \emptyset \)
• \( \{2\} \)
• \( \{3\} \)
• \( \{2, 3\} \)

The power set has a total of \( 2^n \) elements, where \( n \) is the cardinality (number of elements) of \( B \). In this scenario, \( B \) has 2 elements, so the power set includes 4 subsets. The concept of power sets allows us to explore all possible combinations of a set’s elements, forming the basis for more advanced topics like set algebra and logic.

Ordered Pairs

Ordered pairs form the basis of the Cartesian product operation. Each ordered pair is a collection of two elements where the order of these elements is significant. Notationally, \( (x, y) \) describes pairs where \( x \) is the first element from one set and \( y \) is the second element from another. The importance of order is crucial as \( (x, y) \) is distinct from \( (y, x) \) unless \( x = y \).

In the exercise problem, several solutions involve Cartesian products generating ordered pairs from sets. For example, \( A \times B = \{(0,2), (0,3), (2,2), (2,3), (3,2), (3,3)\} \) showcases all possible ordered pair combinations when selecting an element \( a \) from \( A \) and \( b \) from \( B \).

• Ordered pairs are fundamental in defining more complex structures like relations and functions.
• Understanding how these pairs work helps navigate through various mathematical fields, including graph theory and geometry.

In a mathematical context, recognizing the significance and operations of ordered pairs facilitates a clear understanding of relations across different sets.