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Let \(U=\\{1,2,3,4,5,6,7,8,9,10\\}\) \(A=\\{1,3,5,7,9\\}, B=\\{2,4,6,8,10\\}\), and \(C=\\{1,2,4\) \(5,8,9\\}\). List the elements of each set. a. \(A^{c} \cap\left(B \cap C^{c}\right)\) b. \(\left(A \cup B^{c}\right) \cup\left(B \cap C^{c}\right)\) c. \((A \cup B)^{c} \cap C^{c}\)

Short Answer

Expert verified
a. \(A^{c} \cap(B \cap C^{c}) = \{6, 10\}\) b. \((A \cup B^{c}) \cup(B \cap C^{c}) = \{1, 3, 5, 6, 7, 9, 10\}\) c. \((A \cup B)^{c} \cap C^{c} = \emptyset\)

Step by step solution

01

To find the complement of a set, we need to list all the elements in the universal set U that are not present in the given set. So, • \(A^{c} = \{2, 4, 6, 8, 10\}\) • \(B^{c} = \{1, 3, 5, 7, 9\}\) • \(C^{c} = \{3, 6, 7, 10\}\) #Step 2: Finding a. \(A^{c} \cap\left(B \cap C^{c}\right)\)#

To find this, first, calculate the intersection of B and \(C^{c}\) and then find the intersection of the result with \(A^{c}\). • \(B \cap C^{c} = \{2, 4, 6, 8, 10\} \cap \{3, 6, 7, 10\} = \{6, 10\}\) • Now, \(A^{c} \cap\left(B \cap C^{c}\right) = \{2, 4, 6, 8, 10\} \cap \{6, 10\} = \{6, 10\}\) So, the elements of the set in a. are \(\{6, 10\}\). #Step 3: Finding b. \(\left(A \cup B^{c}\right) \cup\left(B \cap C^{c}\right)\)#
02

First, calculate the union of A and \(B^{c}\) and then find the union of the result with the intersection of B and \(C^{c}\). • \(A \cup B^{c} = \{1, 3, 5, 7, 9\} \cup \{1, 3, 5, 7, 9\} = \{1, 3, 5, 7, 9\}\) • Now, \(\left(A \cup B^{c}\right) \cup\left(B \cap C^{c}\right) = \{1, 3, 5, 7, 9\} \cup \{6, 10\} = \{1, 3, 5, 6, 7, 9, 10\}\) So, the elements of the set in b. are \(\{1, 3, 5, 6, 7, 9, 10\}\). #Step 4: Finding c. \((A \cup B)^{c} \cap C^{c}\)#

First, calculate the union of A and B, then find the complement of the result and finally find the intersection of this complement with \(C^{c}\). • \(A \cup B = \{1, 3, 5, 7, 9\} \cup \{2, 4, 6, 8, 10\} = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}\) • Now, \((A \cup B)^{c} = U \setminus (A \cup B) = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\} \setminus \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\} = \emptyset\) • Finally, \((A \cup B)^{c} \cap C^{c} = \emptyset \cap \{3, 6, 7, 10\} = \emptyset\) So, the elements of the set in c. are \(\emptyset\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Complement of a Set
In set theory, the complement of a set contains everything that is not in the original set but is present in the universal set. It's like the 'opposite' of having a membership in the original set. To find the complement of a set, denoted as 'Ac' or sometimes 'A'', you would list all elements that are in the universal set 'U' but not in set 'A'.

For instance, if you have a universal set U that includes the numbers from 1 to 10, and a set A that contains the odd numbers from 1 to 9, then the complement of A is the set containing all the even numbers from 2 to 10, as these are the numbers in U not present in A. The steps given in the exercise show this process clearly. Understanding how to correctly list this 'opposite' set is crucial in set theory exercises.
Union of Sets
The union of two sets combines all the unique elements from both sets. It's symbolized by the cup-like symbol '∪'. If you picture two circles overlapping, the union is everything inside both circles combined, without duplicating any elements.

As seen in the solution steps, when we take the union of sets A and B's complement, denoted as 'A ∪ Bc', we gather all unique elements that belong to either A or B's complement into one new set. This is particularly useful when trying to understand how different groups can be combined to encompass every member from each group without repetition. Remember, the key to forming a union is to ensure that every element appears once and only once, regardless of how many times it's found across the original sets.
Intersection of Sets
The intersection of sets refers to the common elements that two or more sets share. This is indicated by the symbol '∩'. Imagine two circles that overlap; the area they share, or the overlap itself, represents the intersection. When determining the intersection, we only list elements that appear in all the sets we are considering.

In the example given, to find the intersection 'Ac ∩ (B ∩ Cc)', we first identify the common elements between set B and the complement of set C, and then we see which of those are also in the complement of set A. The result is a set composed solely of elements that are shared between these sets. Clear understanding of intersections helps in analyzing the common traits or numbers in different groups.
Set Elements Listing
Listing set elements is a fundamental skill in set theory. It involves writing out all the unique members of a set in a clear and precise manner, typically enclosed within curly braces '{}'. This method is not just about knowing what's in the set but also about presenting it in a standard mathematical format.

In exercises like the ones here, being able to accurately list set elements is crucial. Each solution involves a different operation (complement, union, or intersection), and the final answer requires you to list the resulting set elements. Whether the outcome is a non-empty set or the empty set '∅', indicating that there are no common elements between the sets under intersection, every set should be clearly represented by its elements. This clarity is essential for students to confirm that their solutions match the mathematical notation required in set theory.

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