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Chapter 5: Problem 34

Use the properties in Theorem \(5.2 .2\) to simplify the given expression. $$A \cap\left(\left(B \cup A^{c}\right) \cap B^{c}\right)$$

Short Answer

Expert verified
The final simplified expression is \(A \cap \left( B^{c}\cap A^{c}\right)\).

Step by step solution

01

Identify properties relevant to the given expression

The properties in Theorem 5.2.2 that will be useful to simplify this expression are: 1. Intersection distributes over union: \(A \cap (B \cup C) = (A \cap B) \cup (A \cap C)\) 2. Union distributes over intersection: \(A \cup (B \cap C) = (A \cup B) \cap (A \cup C)\) 3. Complement laws: \(A \cup A^{c} = U\), \(A \cap A^{c} = \emptyset\) 4. De Morgan's laws: \((A \cup B)^{c} = A^{c} \cap B^{c}\), \((A \cap B)^{c} = A^{c} \cup B^{c}\)
02

Apply properties to the given expression

We are given the expression as: \(A \cap\left(\left(B \cup A^{c}\right) \cap B^{c}\right)\) Now, we'll apply the properties step by step: 1. First, use the intersection distributes over union property on the inside bracket expression \(B\cup A^{c}\): \(A \cap\left(\left( B^{c}\cap B\right) \cup \left( B^{c}\cap A^{c}\right)\right)\) 2. Apply the complement law, as \(A \cap A^{c} = \emptyset\) and \(B \cap B^{c} =\emptyset\): \(A \cap\left(\emptyset \cup \left( B^{c}\cap A^{c}\right)\right)\) 3. As a union with an empty set does not change the set itself, we have: \(A \cap \left( B^{c}\cap A^{c}\right)\)
03

Final simplified expression

After applying the relevant properties in Theorem 5.2.2 to the given expression, we have simplified the expression as follows: \(A \cap \left( B^{c}\cap A^{c}\right)\) This is the final simplified expression.

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Key Concepts

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

De Morgan's Laws
Understanding De Morgan's Laws is crucial for simplifying complex set expressions. These laws concern the complement of the union and intersection of two sets. Specifically, the laws state that the complement of the union of two sets is the intersection of their complements, \( (A \cup B)^c = A^c \cap B^c \), and the complement of the intersection is the union of their complements, \( (A \cap B)^c = A^c \cup B^c \).

By applying De Morgan's Laws, you can transform a seemingly intricate set expression into a more manageable form. This was demonstrated in the exercise where De Morgan鈥檚 Laws were not directly applied but were essential to understanding how the set operations interact, particularly how the complement operation affects the outcome when combined with intersection and union.
Complement Laws
Complement laws are simple yet powerful tools in set theory. They assert that a set combined with its complement yields the universal set, \( A \cup A^{c} = U \), and the intersection of a set with its complement is the empty set, \( A \cap A^{c} = \emptyset \).

These laws were pivotal in the problem's Step 2 of the solution, where the intersection of a set with its complement was used to simplify the inner expression to the empty set, \( \emptyset \). When encountering \( A \cap A^{c} \) or \( A \cup A^{c} \) in a set expression, you can instantly simplify it to \( \emptyset \) or the universal set \( U \) respectively. This strategy assists in vastly reducing the complexity of set equations.
Intersection Distribution
The property known as intersection distribution allows you to expand the intersection of a set with a union of sets. Formally, it's expressed as \( A \cap (B \cup C) = (A \cap B) \cup (A \cap C) \).

This distributive property plays a crucial role in breaking down complex expressions into simpler components, which can then be individually addressed. During Step 1 of our exercise, this property was used to distribute the set A across the union contained within the parentheses, leading to an intersection between A and each set inside the union. By understanding how to distribute intersections, you unravel the given set expression, isolating each piece for further simplification.
Union Distribution
Conversely, union distribution deals with spreading a union operation over an intersection. The formal representation is \( A \cup (B \cap C) = (A \cup B) \cap (A \cup C) \).

Similar to its intersection counterpart, this property lets you dissect an imposing set operation into more digestible parts. Although union distribution wasn't directly applied in the provided exercise, it's essential to recognize the inverse relationship between intersection and union distribution. It embodies the flexibility of set operations, allowing various routes to simplify to the most concise form suitable for the context.

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Most popular questions from this chapter

Refer to the definition of symmetric difference given above. Prove each of \(40-45\), assuming that \(A, B\), and \(C\) are all subsets of a universal set \(U\). $$A \Delta B=B \Delta A$$

The following problem, devised by Ginger Bolton, appeared in the January 1989 issue of the College Mathematics Journal (Vol. 20, No. 1, p. 68): Given a positive integer \(n \geq 2\), let \(S\) be the set of all nonempty subsets of \(\\{2,3, \ldots, n\\}\). For each \(S_{i} \in S\), let \(P_{i}\) be the product of the elements of \(S_{i}\). Prove or disprove that $$ \sum_{i=1}^{2^{n-1}-1} P_{i}=\frac{(n+1) !}{2}-1 $$ In 23 and 24 supply a reason for each step in the derivation.

Prove each statement that is true and find a counterexample for each statement that is false. Assume all sets are subsets of a universal set \(U\). For all sets \(A, B\), and \(C, A-(B-C)=(A-B)-C\).

Let the universal set be the set \(\mathbf{R}\) of all real numbers and let \(A=\\{x \in \mathbf{R} \mid 0

The following is a proof that for any sets \(A, B\), and \(C\), \(A \cap(B \cup C)=(A \cap B) \cup(A \cap C)\). Fill in the blanks. Proof: Suppose \(A, B\), and \(C\) are any sets. (1) \(A \cap(B \cup C) \subseteq(A \cap B) \cup(A \cap C):\) Let \(x \in A \cap(B \cup C)\). [We must show that \(x \in \underline{\text { (a) }}\).] By definition of intersection, \(x \in \underline{\text { (b) }}\) and \(x \in \frac{\text { (c) }}{\text { Thus }}\) \(x \in A\) and, by definition of union, \(x \in B\) or \((\mathrm{d})\) Case \(1(x \in A\) and \(x \in B):\) In this case, by definition of intersection, \(x \in \underline{(e)}\), and so, by definition of union, \(x \in(A \cap B) \cup(A \cap C) .\) Case \(2(x \in A\) and \(x \in C):\) In this case, (f) Hence in either case, \(x \in(A \cap B) \cup(A \cap C)\) [as was to be shown]. \([\) So \(A \cap(B \cup C) \subseteq(A \cap B) \cup(A \cap C)\) by definition of subset. (2) \((A \cap B) \cup(A \cap C) \subseteq A \cap(B \cup C):\) Let \(x \in(A \cap B) \cup(A \cap C)\). [We must show that \(\underline{\text { (a) }}\).] By definition of union, \(x \in \stackrel{\text { (b) }}{\text { or } x \in \stackrel{\text { (c) }}{\text { Let } x \in} \text {. }}\) Case \(1(x \in A \cap B):\) In this case, by definition of intersection, \(\frac{(\mathrm{d})}{}\) and \((\mathrm{e})\). Since \(x \in B\), then by definition of union, \(x \in B \cup C\). Hence \(x \in A\) and \(x \in B \cup C\), and so, by definition of intersection, \(x \in \underline{\text { (f) }}\). Case \(2(x \in A \cap C):\) In this case, \((g) .\) In either case, \(x \in A \cap(B \cup C)\) [as was to be shown]. [Thus \((A \cap B) \cup(A \cap C) \subseteq A \cap(B \cup C)\) by definition of subset. [Since both subset relations have been proved, it follows, by definition of set equality, that \(\stackrel{(\mathrm{h})}{-}\).]
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