To solve the problem, we need to find [tex]\( A \cup B \)[/tex], [tex]\( B \cap C \)[/tex], and [tex]\( B^C \)[/tex] given the sets [tex]\( S \)[/tex], [tex]\( A \)[/tex], [tex]\( B \)[/tex], and [tex]\( C \)[/tex].
### Step-by-Step Solution:
1. List the elements in the set [tex]\( A \cup B \)[/tex]:
The union of two sets, [tex]\( A \cup B \)[/tex], contains all elements that are in either set [tex]\( A \)[/tex] or set [tex]\( B \)[/tex] or in both.
Given:
[tex]\[
A = \{1, 2, 5, 6, 7, 10, 12, 15\}
\][/tex]
[tex]\[
B = \{1, 2, 3, 8, 11, 13, 14\}
\][/tex]
Merging all elements from both sets without duplicates:
[tex]\[
A \cup B = \{1, 2, 3, 5, 6, 7, 8, 10, 11, 12, 13, 14, 15\}
\][/tex]
2. List the elements in the set [tex]\( B \cap C \)[/tex]:
The intersection of two sets, [tex]\( B \cap C \)[/tex], contains only the elements that are present in both set [tex]\( B \)[/tex] and set [tex]\( C \)[/tex].
Given:
[tex]\[
B = \{1, 2, 3, 8, 11, 13, 14\}
\][/tex]
[tex]\[
C = \{1, 4, 7, 9, 12, 15\}
\][/tex]
Identify the common elements:
[tex]\[
B \cap C = \{1\}
\][/tex]
3. List the elements in the set [tex]\( B^C \)[/tex]:
The complement of a set [tex]\( B \)[/tex] with respect to the universal set [tex]\( S \)[/tex], denoted as [tex]\( B^C \)[/tex] or [tex]\( \bar{B} \)[/tex], contains all elements that are in [tex]\( S \)[/tex] but not in [tex]\( B \)[/tex].
Given:
[tex]\[
S = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15\}
\][/tex]
[tex]\[
B = \{1, 2, 3, 8, 11, 13, 14\}
\][/tex]
Subtract the elements of B from S:
[tex]\[
B^C = \{4, 5, 6, 7, 9, 10, 12, 15\}
\][/tex]
### Final Results:
a) [tex]\( A \cup B = \{1, 2, 3, 5, 6, 7, 8, 10, 11, 12, 13, 14, 15\} \)[/tex]
b) [tex]\( B \cap C = \{1\} \)[/tex]
c) [tex]\( B^C = \{4, 5, 6, 7, 9, 10, 12, 15\} \)[/tex]
