Problem 9 - A Gentle Introduction to the Art of Mathematics

Question

Prove that \(A \cup(B \cap C)=(A \cup B) \cap(A \cup C)\) by showing that \(A \cup(B \cap C) \subseteq\) \((A \cup B) \cap(A \cup C)\) and \((A \cup B) \cap(A \cup C) \subseteq A \cup(B \cap C)\).

Step-by-Step Solution

Answer

We proved \(A \cup (B \cap C) = (A \cup B) \cap (A \cup C)\) by showing that each side is a subset of the other.

Step 1: Showing that \(A \cup (B \cap C) \subseteq (A \cup B) \cap (A \cup C)\)

To prove this, start by assuming an arbitrary element \(x\) in \(A \cup (B \cap C)\). This means that \(x \in A\) or \(x \in (B \cap C)\). If \(x \in A\), then \(x \in A \cup B\) and also \(x \in A \cup C\), so \(x \in (A \cup B) \cap (A \cup C)\). If \(x \in (B \cap C)\), then \(x \in B\) and \(x \in C\). Therefore, \(x \in A \cup B\) and \(x \in A \cup C\), so \(x \in (A \cup B) \cap (A \cup C)\). Hence, \(A \cup (B \cap C) \subseteq (A \cup B) \cap (A \cup C)\).

Step 2: Showing that \((A \cup B) \cap (A \cup C) \subseteq A \cup (B \cap C)\)

To prove this, start by assuming an arbitrary element \(x\) in \((A \cup B) \cap (A \cup C)\). This means that \(x \in (A \cup B)\) and \(x \in (A \cup C)\). If \(x \in A\), then \(x \in A \cup (B \cap C)\). If \(x otin A\), then \(x \in B\) (because \(x \in A \cup B\)) and \(x \in C\) (because \(x \in A \cup C\)). Therefore, \(x \in B \cap C\), and hence \(x \in A \cup (B \cap C)\). Thus, \((A \cup B) \cap (A \cup C) \subseteq A \cup (B \cap C)\).

Step 3: Conclusion

Since both \(A \cup (B \cap C) \subseteq (A \cup B) \cap (A \cup C)\) and \((A \cup B) \cap (A \cup C) \subseteq A \cup (B \cap C)\), we have shown that \(A \cup (B \cap C) = (A \cup B) \cap (A \cup C)\).

Key Concepts

Set Operations

Set operations are mathematical techniques used to combine, relate, and compare different sets. Sets are collections of distinct elements grouped by some rule or criterion. The primary operations include union, intersection, and complement. In this exercise, we focus on union and intersection. The union of two sets, denoted as \(A \cup B\), includes all elements that are in either set A, set B, or both. The intersection of two sets, \(A \cap B\), includes only the elements that are in both sets A and B. Set operations are fundamental to various branches of mathematics and are used to solve numerous problems. Understanding these operations can simplify complex relationships between sets and help in proving various properties.

Subset Proof

A subset proof involves showing that every element of one set is also an element of another set. To prove \(A \subseteq B\), we demonstrate that if an element \(x\) belongs to set A, then \(x\) must also belong to set B. This type of proof is essential in set theory to establish inclusion relationships. In this exercise, we prove two subset relationships:

• \(A \cup (B \cap C) \subseteq (A \cup B) \cap (A \cup C)\)
• \((A \cup B) \cap (A \cup C) \subseteq A \cup (B \cap C)\)

By proving these two relationships, we show that the sets are equal.

Union and Intersection of Sets

The union and intersection are two primary operations used to combine sets. The union, \(A \cup B\), consists of any element that is in set A, set B, or both. The intersection, \(A \cap B\), consists of elements that are only in both sets A and B. Understanding how these operations work is crucial to solving problems involving sets. In our proof, we apply these operations to demonstrate set equality. By understanding that the intersection picks common elements and that the union combines all elements, you can see how the given sets relate to one another.

Element-Wise Argument

An element-wise argument involves examining individual elements of sets to prove a statement. This type of argument is used to show that one set is a subset of another. For instance, to show \(A \cup (B \cap C) \subseteq (A \cup B) \cap (A \cup C)\), we take an arbitrary element \(x\) in \(A \cup (B \cap C)\) and demonstrate it must also be in \((A \cup B) \cap (A \cup C)\). The method relies on structured logical reasoning for each possibility:

• If \(x\) is in A
• If \(x\) is in both B and C but not in A

This meticulous verification ensures that the subset relationship or equality is solidly established.