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Chapter 5: Problem 5

Prove Theorem \(5.31 .\) Let \(\Lambda\) be a nonempty indexing set, let \(\mathcal{A}=\left\\{A_{\alpha} \mid \alpha \in \Lambda\right\\}\) be an indexed family of sets, and let \(B\) be a set. Then (a) \(B \cap\left(\bigcup_{\alpha \in \Lambda} A_{\alpha}\right)=\bigcup_{\alpha \in \Lambda}\left(B \cap A_{\alpha}\right),\) and (b) \(B \cup\left(\bigcap_{\alpha \in \Lambda} A_{\alpha}\right)=\bigcap_{\alpha \in \Lambda}\left(B \cup A_{\alpha}\right)\).

Short Answer

Expert verified
In conclusion, we have proved Theorem 5.31 with two parts: (a) \(B \cap\left(\bigcup_{\alpha \in \Lambda} A_{\alpha}\right)=\bigcup_{\alpha \in \Lambda}(B \cap A_{\alpha})\) (b) \(B \cup\left(\bigcap_{\alpha \in \Lambda} A_{\alpha}\right)=\bigcap_{\alpha \in \Lambda}(B \cup A_{\alpha})\) To prove each part, we demonstrated that each side of the given equations was a subset of the other side. This was accomplished by a series of set manipulations and a careful analysis of the definitions of set operations.

Step by step solution

01

Prove Part (a) of Theorem 5.31

Suppose \(x \in B \cap\left(\bigcup_{\alpha \in \Lambda} A_{\alpha}\right)\). By the definition of intersection, this means that \(x \in B\) and \(x \in \bigcup_{\alpha \in \Lambda} A_{\alpha}\). The latter part implies that there exists an index \(\alpha_0\) such that \(x \in A_{\alpha_0}\). Since \(x\) belongs to both \(B\) and \(A_{\alpha_0}\), we can conclude that \(x \in B \cap A_{\alpha_0}\). Since \(\alpha_0 \in \Lambda\), we can also say that \(x \in \bigcup_{\alpha \in \Lambda}(B \cap A_{\alpha})\). Therefore, we have \(B \cap\left(\bigcup_{\alpha \in \Lambda} A_{\alpha}\right) \subseteq \bigcup_{\alpha \in \Lambda}(B \cap A_{\alpha})\). Next, suppose \(x \in \bigcup_{\alpha \in \Lambda}(B \cap A_{\alpha})\), which implies that there exists an index \(\alpha_1\) such that \(x \in B \cap A_{\alpha_1}\). This means that \(x\) belongs to both \(B\) and \(A_{\alpha_1}\). Since \(x \in A_{\alpha_1}\), we know that \(x \in \bigcup_{\alpha \in \Lambda} A_{\alpha}\). Thus, \(x\) belongs to both \(B\) and \(\bigcup_{\alpha \in \Lambda} A_{\alpha}\), which means that \(x \in B \cap\left(\bigcup_{\alpha \in \Lambda} A_{\alpha}\right)\). Therefore, we have \(\bigcup_{\alpha \in \Lambda}(B \cap A_{\alpha}) \subseteq B \cap\left(\bigcup_{\alpha \in \Lambda} A_{\alpha}\right)\). Since both subsets are true, we can now conclude that \[B \cap\left(\bigcup_{\alpha \in \Lambda} A_{\alpha}\right)=\bigcup_{\alpha \in \Lambda}(B \cap A_{\alpha})\].
02

Prove Part (b) of Theorem 5.31

Suppose \(x \in B \cup\left(\bigcap_{\alpha \in \Lambda} A_{\alpha}\right)\). This means that either \(x \in B\) or \(x \in \bigcap_{\alpha \in \Lambda} A_{\alpha}\). If \(x \in B\), then \(x \in B \cup A_{\alpha}\) for each \(\alpha \in \Lambda\). Therefore, \(x \in \bigcap_{\alpha \in \Lambda}(B \cup A_{\alpha})\). If \(x \in \bigcap_{\alpha \in \Lambda} A_{\alpha}\), then \(x \in A_{\alpha}\) for every \(\alpha \in \Lambda\). Consequently, \(x \in B \cup A_{\alpha}\) for all \(\alpha \in \Lambda\). Once again, we can say that \(x \in \bigcap_{\alpha \in \Lambda}(B \cup A_{\alpha})\). Thus, we have \(B \cup\left(\bigcap_{\alpha \in \Lambda} A_{\alpha}\right) \subseteq \bigcap_{\alpha \in \Lambda}(B \cup A_{\alpha})\). Now, suppose \(x \in \bigcap_{\alpha \in \Lambda}(B \cup A_{\alpha})\). This means that \(x\) belongs to \(B \cup A_{\alpha}\) for each \(\alpha \in \Lambda\). If \(x \in B\), then \(x \in B \cup (\bigcap_{\alpha \in \Lambda} A_{\alpha})\). If \(x \notin B\), then \(x\) must be in \(A_{\alpha}\) for every \(\alpha \in \Lambda\), because it is in every union \(B \cup A_{\alpha}\). Therefore, \(x \in \bigcap_{\alpha \in \Lambda} A_{\alpha}\) which means \(x \in B \cup (\bigcap_{\alpha \in \Lambda} A_{\alpha})\). So we have \(\bigcap_{\alpha \in \Lambda}(B \cup A_{\alpha}) \subseteq B \cup\left(\bigcap_{\alpha \in \Lambda} A_{\alpha}\right)\). As both subsets are true, we can now conclude that \[B \cup\left(\bigcap_{\alpha \in \Lambda} A_{\alpha}\right)=\bigcap_{\alpha \in \Lambda}(B \cup A_{\alpha})\]. With these two proofs, we have demonstrated both parts (a) and (b) of Theorem 5.31.

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Key Concepts

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

Understanding Set Theory
At the heart of many mathematical disciplines lies set theory, the study of sets, which are simply collections of objects. These objects can be anything from numbers to letters, or even other sets. Understanding set theory is crucial for exploring more advanced mathematical concepts, as it provides a fundamental language and framework.

In set theory, we denote sets by capital letters (e.g., A, B, C) and their elements by lower-case letters. We describe the elements of a set using braces \( \{ \} \) and list the elements or the condition an element must satisfy to be part of the set. For example, \( A = \{ x \mid x > 0 \} \) represents the set of all positive numbers. Operations in set theory include union (\(\cup\)), intersection (\(\cap\)), and complementation. These operations help us understand how sets combine, overlap, and differ from each other.

When we look at intersections and unions, they tell us about commonality and combination, respectively. The intersection of two sets contains only the elements common to both sets, while the union of two sets contains all elements from either set. When grasping these set operations, one can solve complex problems across various mathematical subjects.
Indexed Family of Sets
An indexed family of sets is a way of organizing a collection of sets using an index set. The index set, often denoted by \( \Lambda \), can be thought of as a set of labels that help us reference each individual set within the family. Each member of this family is a set that corresponds to an element in the index set.

For example, we can define an indexed family of sets \( \mathcal{A} = \{A_{\alpha} \mid \alpha \in \Lambda\} \) where each \( A_{\alpha} \) is a set, and \( \alpha \) is an element of the index set \( \Lambda \). This notation helps us to manage and refer to possibly infinite collections of sets in a structured manner, which is particularly useful when dealing with complex proofs and operations involving multiple sets.

Understanding indexed families is critical when solving problems that involve operations over collections of sets, such as intersections and unions across the entire family鈥攁s seen in the exercise in question. The structure provided by the indexed family allows for a systematic approach to these operations.
Intersection and Union of Sets
The concepts of intersection and union of sets are fundamental operations in set theory that help us understand the relationship between different sets. The intersection of a collection of sets includes only those elements that are present in every set of the collection. Mathematically, for a family of sets \( \mathcal{A} \) indexed by \( \Lambda \), the intersection is written as: \[ \bigcap_{\alpha \in \Lambda} A_{\alpha} \.\] It represents the common elements shared among all \( A_{\alpha} \) within the family.

Conversely, the union of a collection of sets contains all elements that appear in at least one of the sets within the collection. For the same indexed family \( \mathcal{A} \) and index set \( \Lambda \) as earlier, the union is denoted by: \[ \bigcup_{\alpha \in \Lambda} A_{\alpha} \.\] This operation creates a set that encompasses all unique elements present in any of the \( A_{\alpha} \) sets.

The exercise demonstrates these concepts by showing how they distribute over each other when combined with another set \( B \). When approaching problems such as these, it's important to methodically apply definitions and use logical reasoning to prove the desired equalities. In doing so, we can reveal fundamental properties of intersection and union that hold true in the broader context of set theory.

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Most popular questions from this chapter

Let \(A\) and \(B\) be subsets of some universal set \(U .\) Prove each of the following: (a) \(A \cap B \subseteq A\) (b) \(A \subseteq A \cup B\) (c) \(A \cap A=A\) (d) \(A \cup A=A\) (e) \(A \cap \emptyset=\varnothing\) (f) \(A \cup \emptyset=A\)

Let \(A, B,\) and \(C\) be subsets of some universal set \(U\) (a) Draw two general Venn diagrams for the sets \(A, B,\) and \(C .\) On one, shade the region that represents \(A-(B \cap C),\) and on the other, shade the region that represents \((A-B) \cup(A-C) .\) Based on the Venn diagrams, make a conjecture about the relationship between the sets \(A-(B \cap C)\) and \((A-B) \cup(A-C)\) (b) Use the choose-an-element method to prove the conjecture from Exercise \((5 a)\) (c) Use the algebra of sets to prove the conjecture from Exercise (5a).

Prove Theorem 5.25, Part (4): \((A \cup B) \times C=(A \times C) \cup(B \times C)\).

To help with the proof by induction of Theorem \(5.5,\) we first prove the following lemma. (The idea for the proof of this lemma was illustrated with the discussion of power set after the definition on page \(222 .\) ) Lemma 5.6. Let \(A\) and \(B\) be subsets of some universal set. If \(A=B \cup\\{x\\},\) where \(x \notin B\), then any subset of \(A\) is either a subset of \(B\) or a set of the form \(C \cup\\{x\\},\) where \(C\) is a subset of \(B\) Proof. Let \(A\) and \(B\) be subsets of some universal set, and assume that \(A=\) \(B \cup\\{x\\}\) where \(x \notin B\). Let \(Y\) be a subset of \(A\). We need to show that \(Y\) is a subset of \(B\) or that \(Y=C \cup\\{x\\},\) where \(C\) is some subset of \(B\). There are two cases to consider: (1)\(x\) is not an element of \(Y,\) and (2)\(x\) is an element of \(Y\) Case 1: Assume that \(x \notin Y\). Let \(y \in Y .\) Then \(y \in A\) and \(y \neq x .\) Since $$A=B \cup\\{x\\}.$$ this means that \(y\) must be in \(B\). Therefore, \(Y \subseteq B\). Case 2: Assume that \(x \in Y .\) In this case, let \(C=Y-\\{x\\} .\) Then every element of \(C\) is an element of \(B\). Hence, we can conclude that \(C \subseteq B\) and that \(Y=C \cup\\{x\\}\) Cases (1) and (2) show that if \(Y \subseteq A,\) then \(Y \subseteq B\) or \(Y=C \cup\\{x\\},\) where \(C \subseteq B\). To begin the induction proof of Theorem \(5.5,\) for each nonnegative integer \(n,\) we let \(P(n)\) be, "If a finite set has exactly \(n\) elements, then that set has exactly \(2^{n}\) subsets." (a) Verify that \(P(0)\) is true. (This is the basis step for the induction proof.) (b) Verify that \(P(1)\) and \(P(2)\) are true. (c) Now assume that \(k\) is a nonnegative integer and assume that \(P(k)\) is true. That is, assume that if a set has \(k\) elements, then that set has \(2^{k}\) subsets. (This is the inductive assumption for the induction proof.) Let \(T\) be a subset of the universal set with card \((T)=k+1,\) and let \(x \in T .\) Then the set \(B=T-\\{x\\}\) has \(k\) elements. Now use the inductive assumption to determine how many subsets \(B\) has. Then use Lemma 5.6 to prove that \(T\) has twice as many subsets as \(B\). This should help complete the inductive step for the induction proof.

Let \(A\) be a subset of some universal set \(U\). Prove each of the following (from Theorem 5.20): (a) \( \left(A^{c}\right)^{c}=A\) \((\mathbf{c}) \emptyset^{c}=U\) (b) \(A-\emptyset=A\) (d) \(U^{c}=\emptyset\)
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