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Chapter 5: Problem 25

For all sets \(A, B\), and \(C\), $$ (A \cap B) \cup C=(A \cup C) \cap(B \cup C) . $$

Short Answer

Expert verified
Using the double inclusion method, we prove that (A ∩ B) ∪ C ⊆ (A ∪ C) ∩ (B ∪ C) and (A ∪ C) ∩ (B ∪ C) ⊆ (A ∩ B) ∪ C. In both cases, elements X and Y from each side of the equation are either in C or in A and B. Therefore, (A ∩ B) ∪ C = (A ∪ C) ∩ (B ∪ C).

Step by step solution

01

Prove (A ∩ B) ∪ C ⊆ (A ∪ C) ∩ (B ∪ C)

Let X be an element in (A ∩ B) ∪ C. This means that X is either in (A ∩ B) or in C. If X is in C, then X is in A ∪ C and X is in B ∪ C, so X is in (A ∪ C) ∩ (B ∪ C). If X is in A ∩ B, then X is in both A and B. This means that X is in A ∪ C (since it's in A) and X is in B ∪ C (since it's in B). Thus, X is also in (A ∪ C) ∩ (B ∪ C). In both cases, every element X in (A ∩ B) ∪ C is in (A ∪ C) ∩ (B ∪ C). Therefore, (A ∩ B) ∪ C ⊆ (A ∪ C) ∩ (B ∪ C).
02

Prove (A ∪ C) ∩ (B ∪ C) ⊆ (A ∩ B) ∪ C

Let Y be an element in (A ∪ C) ∩ (B ∪ C). This means that Y is in A ∪ C and Y is in B ∪ C. First, we consider two cases: Case 1: If Y is in C, then Y is in (A ∩ B) ∪ C. Case 2: If Y is not in C, then Y must be in A and Y must be in B (since Y is in A ∪ C and Y is in B ∪ C). This means that Y should be in A ∩ B. Thus, Y is in (A ∩ B) ∪ C. In both cases, every element Y in (A ∪ C) ∩ (B ∪ C) is in (A ∩ B) ∪ C. Therefore, (A ∪ C) ∩ (B ∪ C) ⊆ (A ∩ B) ∪ C.
03

Conclusion

Since (A ∩ B) ∪ C ⊆ (A ∪ C) ∩ (B ∪ C) and (A ∪ C) ∩ (B ∪ C) ⊆ (A ∩ B) ∪ C, we can conclude that (A ∩ B) ∪ C = (A ∪ C) ∩ (B ∪ C), proving the equality.

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Key Concepts

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

Union of Sets
The union of sets is a fundamental concept in set theory. It represents the combination of different sets, where any element is included if it belongs to at least one of the sets. For example, if we have two sets, say set A and set B, their union, denoted as \(A \cup B\), consists of all elements from both sets without repetition.A few important things to remember when dealing with union of sets:
  • The union is commutative, meaning \(A \cup B = B \cup A\).
  • The union is associative, which implies \((A \cup B) \cup C = A \cup (B \cup C)\).
  • The set itself and its union with an empty set is the set, i.e., \(A \cup \emptyset = A\).
With the example from our exercise, understanding that adding another set like \(C\) into a union involves including all distinct elements from the original union \((A \cap B)\) and the set \(C\) helps grasp this concept clearly.
Intersection of Sets
The intersection of sets finds common elements shared between sets. If we consider two sets, A and B, the intersection, represented as \(A \cap B\), includes only those elements that are present in both A and B.Key characteristics of the intersection of sets are:
  • The intersection is commutative: \(A \cap B = B \cap A\).
  • The intersection is associative: \((A \cap B) \cap C = A \cap (B \cap C)\).
  • If two sets have no elements in common, their intersection is an empty set, that is, \(A \cap B = \emptyset\) for disjoint sets.
In the context of our exercise, understanding how the intersection \(A \cap B\) filters common elements from sets A and B before these elements interact further with set C is essential. Recognizing the order of operations in terms of brackets also plays a critical role when working through these set equations.
Subset Proofs
Proofs involving subsets are foundational in demonstrating the equality of sets. A subset is defined such that if every element of set A is also an element of another set B, then A is a subset of B, written as \(A \subseteq B\).In these proofs, the goal is often to show that two sets are equal by proving that each set is a subset of the other. This involves:
  • Showing \((A \cap B) \cup C \subseteq (A \cup C) \cap (B \cup C)\).
  • Reproving \((A \cup C) \cap (B \cup C) \subseteq (A \cap B) \cup C\), completing the equality demonstration.
By following logical steps to show that every element of one set belongs to another, we complete the proof. This logical quality ensures that complex set relations, like the one in our exercise, can be succinctly expressed and confirmed through subset proofs, ensuring both sides of the equation hold true.

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Most popular questions from this chapter

$$ \text { For all sets } A \text { and } B \text {, if } B \subseteq A^{c} \text { then } A \cap B=\emptyset $$

Find the mistake in the following "proof." "Theorem:" For all sets \(A\) and \(B, A^{c} \cup B^{c} \subseteq(A \cup B)^{c}\). "Proof: Suppose \(A\) and \(B\) are sets, and \(x \in A^{c} \cup B^{c}\). Then \(x \in A^{c}\) or \(x \in B^{c}\) by definition of union. It follows that \(x \notin A\) or \(x \notin B\) by definition of complement, and so \(x \notin A \cup B\) by definition of union. Thus \(x \in(A \cup B)^{c}\) by definition of complement, and hence \(A^{c} \cup B^{c} \subseteq(A \cup B)^{c} . "\)

$$ \text { For all sets } A, B, \text { and } C, \text { if } A \subseteq B \text { then } A \cap C \subseteq B \cap C \text {. } $$

Write in words how to read each of the following out loud. Then write the shorthand notation for each set. a. \(\\{x \in U \mid x \in A\) and \(x \in B\\}\) b. \(\\{x \in U \mid x \in A\) or \(x \in B\\}\) c. \(\\{x \in U \mid x \in A\) and \(x \notin B\\}\) d. \(\\{x \in U \mid x \notin A\\}\)

$$ \text { Prove that for all sets } A \text { and } B, B-A=B \cap A^{c} \text {. } $$
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