The problem defines a universal set $U$ and several subsets $A, B, C, D, E, F$. The task is to compute various set operations based on these sets. Specifically, we need to find: (a) $A \cup B$, $A \cap B$, $U^c$, $A \setminus B$, $B \setminus A$, $D \setminus E$, $F \setminus D$ (b) $A \Delta B$, $C \Delta D$, $E \Delta F$ (c) $A \cap (B \cup E)$, $(A \setminus B)^c$, $(A \cap D) \setminus B$, $(B \cap F) \cup (C \cap E)$, $B \times F$

Discrete MathematicsSet TheorySet OperationsUnionIntersectionComplementDifferenceSymmetric DifferenceCartesian Product
2025/4/24

1. Problem Description

The problem defines a universal set UU and several subsets A,B,C,D,E,FA, B, C, D, E, F. The task is to compute various set operations based on these sets. Specifically, we need to find:
(a) ABA \cup B, ABA \cap B, UcU^c, ABA \setminus B, BAB \setminus A, DED \setminus E, FDF \setminus D
(b) AΔBA \Delta B, CΔDC \Delta D, EΔFE \Delta F
(c) A(BE)A \cap (B \cup E), (AB)c(A \setminus B)^c, (AD)B(A \cap D) \setminus B, (BF)(CE)(B \cap F) \cup (C \cap E), B×FB \times F

2. Solution Steps

First, let's explicitly list the elements of each set:
U={1,2,3,4,5,6,7,8,9}U = \{1, 2, 3, 4, 5, 6, 7, 8, 9\}
A={1,2,3,4,5}A = \{1, 2, 3, 4, 5\}
B={4,5,6,7}B = \{4, 5, 6, 7\}
C={5,6,7,8,9}C = \{5, 6, 7, 8, 9\}
D={1,3,5,7,9}D = \{1, 3, 5, 7, 9\} (odd numbers in UU)
E={2,4,6,8}E = \{2, 4, 6, 8\} (even numbers)
F={1,5,9}F = \{1, 5, 9\}
(a)
AB={1,2,3,4,5,6,7}A \cup B = \{1, 2, 3, 4, 5, 6, 7\}
AB={4,5}A \cap B = \{4, 5\}
Uc=U^c = \emptyset (complement of UU with respect to UU is empty)
AB={1,2,3}A \setminus B = \{1, 2, 3\}
BA={6,7}B \setminus A = \{6, 7\}
DE={1,3,5,7,9}D \setminus E = \{1, 3, 5, 7, 9\} (DE=DD \setminus E = D since DD and EE are disjoint)
FD=F \setminus D = \emptyset
(b)
AΔB=(AB)(BA)={1,2,3}{6,7}={1,2,3,6,7}A \Delta B = (A \setminus B) \cup (B \setminus A) = \{1, 2, 3\} \cup \{6, 7\} = \{1, 2, 3, 6, 7\}
CΔD=(CD)(DC)={6,8}{1,3}={1,3,6,8}C \Delta D = (C \setminus D) \cup (D \setminus C) = \{6, 8\} \cup \{1, 3\} = \{1, 3, 6, 8\}
EΔF=(EF)(FE)={2,4,6,8}{1,5,9}={1,2,4,5,6,8,9}E \Delta F = (E \setminus F) \cup (F \setminus E) = \{2, 4, 6, 8\} \cup \{1, 5, 9\} = \{1, 2, 4, 5, 6, 8, 9\}
(c)
BE={2,4,5,6,7,8}B \cup E = \{2, 4, 5, 6, 7, 8\}
A(BE)={1,2,3,4,5}{2,4,5,6,7,8}={2,4,5}A \cap (B \cup E) = \{1, 2, 3, 4, 5\} \cap \{2, 4, 5, 6, 7, 8\} = \{2, 4, 5\}
AB={1,2,3}A \setminus B = \{1, 2, 3\}
(AB)c=U(AB)={1,2,3,4,5,6,7,8,9}{1,2,3}={4,5,6,7,8,9}(A \setminus B)^c = U \setminus (A \setminus B) = \{1, 2, 3, 4, 5, 6, 7, 8, 9\} \setminus \{1, 2, 3\} = \{4, 5, 6, 7, 8, 9\}
AD={1,2,3,4,5}{1,3,5,7,9}={1,3,5}A \cap D = \{1, 2, 3, 4, 5\} \cap \{1, 3, 5, 7, 9\} = \{1, 3, 5\}
(AD)B={1,3,5}{4,5,6,7}={1,3}(A \cap D) \setminus B = \{1, 3, 5\} \setminus \{4, 5, 6, 7\} = \{1, 3\}
BF={4,5,6,7}{1,5,9}={5}B \cap F = \{4, 5, 6, 7\} \cap \{1, 5, 9\} = \{5\}
CE={5,6,7,8,9}{2,4,6,8}={6,8}C \cap E = \{5, 6, 7, 8, 9\} \cap \{2, 4, 6, 8\} = \{6, 8\}
(BF)(CE)={5}{6,8}={5,6,8}(B \cap F) \cup (C \cap E) = \{5\} \cup \{6, 8\} = \{5, 6, 8\}
B×F={(4,1),(4,5),(4,9),(5,1),(5,5),(5,9),(6,1),(6,5),(6,9),(7,1),(7,5),(7,9)}B \times F = \{(4,1), (4,5), (4,9), (5,1), (5,5), (5,9), (6,1), (6,5), (6,9), (7,1), (7,5), (7,9)\}

3. Final Answer

(a) AB={1,2,3,4,5,6,7}A \cup B = \{1, 2, 3, 4, 5, 6, 7\}, AB={4,5}A \cap B = \{4, 5\}, Uc=U^c = \emptyset, AB={1,2,3}A \setminus B = \{1, 2, 3\}, BA={6,7}B \setminus A = \{6, 7\}, DE={1,3,5,7,9}D \setminus E = \{1, 3, 5, 7, 9\}, FD=F \setminus D = \emptyset
(b) AΔB={1,2,3,6,7}A \Delta B = \{1, 2, 3, 6, 7\}, CΔD={1,3,6,8}C \Delta D = \{1, 3, 6, 8\}, EΔF={1,2,4,5,6,8,9}E \Delta F = \{1, 2, 4, 5, 6, 8, 9\}
(c) A(BE)={2,4,5}A \cap (B \cup E) = \{2, 4, 5\}, (AB)c={4,5,6,7,8,9}(A \setminus B)^c = \{4, 5, 6, 7, 8, 9\}, (AD)B={1,3}(A \cap D) \setminus B = \{1, 3\}, (BF)(CE)={5,6,8}(B \cap F) \cup (C \cap E) = \{5, 6, 8\}, B×F={(4,1),(4,5),(4,9),(5,1),(5,5),(5,9),(6,1),(6,5),(6,9),(7,1),(7,5),(7,9)}B \times F = \{(4,1), (4,5), (4,9), (5,1), (5,5), (5,9), (6,1), (6,5), (6,9), (7,1), (7,5), (7,9)\}

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