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Chapter 2: Problem 8

Show that \(A \cup(A \cap B)=A\) for any sets \(A\) and \(B\).

Short Answer

Expert verified
The set identity \(A \cup (A \cap B) = A\) holds because \(A \cap B\) is a subset of \(A\), so the union doesn't add new elements.

Step by step solution

01

Understand the Exercise

We need to show that the union of set \(A\) with the intersection of sets \(A\) and \(B\) results in set \(A\) itself. Mathematically, we want to prove \(A \cup (A \cap B) = A\). We will first analyze the two parts of the union operation individually.
02

Analyze \(A \cap B\)

The intersection \(A \cap B\) represents all elements that are in both set \(A\) and set \(B\). Therefore, \(A \cap B\) is a subset of \(A\) because every element in \(A \cap B\) is also in \(A\).
03

Analyze the Union \(A \cup (A \cap B)\)

The union \(A \cup (A \cap B)\) combines all elements in \(A\) with all elements in \(A \cap B\). Since every element in \(A \cap B\) is already an element in \(A\), the union does not introduce any new elements beyond those already in \(A\). Thus, \(A \cup (A \cap B) = A\).
04

Verify Subset Relations

For two sets \(X\) and \(Y\), if \(X\) is contained within \(Y\), we write \(X \subseteq Y\). From step 2, we know \(A \cap B \subseteq A\), and thus \(A \cup (A \cap B) \subseteq A\). Since \(A\) is always contained within itself by definition, \(A \subseteq A \cup (A \cap B)\). Therefore, \(A = A \cup (A \cap B)\).

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Key Concepts

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

Union of sets
In set theory, the **union** of two sets combines all the elements from both sets without duplicating any of them. When we denote a union, we use the symbol \( \cup \). For example, if we have set \( A = \{1, 2, 3\} \) and set \( B = \{3, 4, 5\} \), the union \( A \cup B \) would be \( \{1, 2, 3, 4, 5\} \). In this operation:
  • Every unique element from both sets appears in the result.
  • Common elements are not repeated.

In the exercise, we calculated \( A \cup (A \cap B) \), showing that this operation didn鈥檛 add new elements beyond those already in \( A \). This is because any element in \( A \cap B \) is also in \( A \), so combining them with \( A \) keeps only elements of \( A \).
Intersection of sets
The **intersection** of sets helps us identify elements that are common to all involved sets. This is denoted by the symbol \( \cap \). Consider two sets: \( A = \{1, 2, 3\} \) and \( B = \{2, 3, 4\} \). Their intersection \( A \cap B \) would be \( \{2, 3\} \) because these are the elements that both sets share.
  • The intersection provides a way to find shared characteristics or commonalities between groups銆�
  • It results in elements that exist in every single set being intersected.

In our original exercise, we saw that \( A \cap B \) is a subset of \( A \), meaning every element of \( A \cap B \) is also an element of \( A \). This is a crucial part of understanding intersection and its effect when performing further operations like union.
Subset relations
A **subset** is a set where every element is also part of another set. This relationship is denoted by the symbol \( \subseteq \). For example, if \( A = \{1, 2, 3\} \) and \( B = \{1, 2\} \), we say \( B \subseteq A \) because all elements of \( B \) are in \( A \).
  • The subset relationship is essential in understanding how sets relate to each other.
  • Through this, we know that a set is always a subset of itself.

In our exercise, we used the subset concept to show \( A \cap B \subseteq A \), which means that the intersection didn't introduce any new elements not already in \( A \). Also, since \( A \) is itself a subset of \( A \cup (A \cap B) \), it further reassures us that showing these subset relations ensures \( A = A \cup (A \cap B) \).

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Most popular questions from this chapter

Suppose that \(f: A \rightarrow B\) and \(g: B \rightarrow C\), and suppose that the composition \(g \circ f\) is an onto function. a) Prove that \(g\) is an onto function. b) Find a specific example that shows that \(f\) is not necessarily onto.

Use induction to prove the following generalized DeMorgan's Law for set theory: For any natural number \(n \geq 2\) and for any sets \(X_{1}, X_{2}, \ldots, X_{n}\), $$ \overline{X_{1} \cap X_{2} \cap \cdots \cap X_{n}}=\overline{X_{1}} \cup \overline{X_{2}} \cup \cdots \cup \overline{X_{n}} $$

This question assumes that you know how to add binary numbers. Suppose \(x\) and \(y\) are binary numbers. Under what circumstances will the binary numbers \(x+y\) and \(x \mid y\) be the same?

Let \(\sim\) be the relation on \(\mathbb{R}\), the set of real numbers, such that for \(x\) and \(y\) in \(\mathbb{R}, x \sim y\) if and only if \(x-y \in \mathbb{Z} .\) For example, \(\sqrt{2}-1 \sim \sqrt{2}+17\) because the difference, \((\sqrt{2}-1)-(\sqrt{2}+17)\), is \(-18\), which is an integer. Show that \(\sim\) is an equivalence relation. Show that each equivalence class \([x]_{\sim}\) contains exactly one number \(a\) which satisfies \(0 \leq a<1\). (Thus, the set of equivalence classes under \(\sim\) is in one-to-one correspondence with the half-open interval \([0,1) .)\)

Let \(M\) be any natural number, and let \(P(n)\) be a predicate whose domain of discourse includes all natural numbers greater than or equal to \(M\). Suppose that \(P(M)\) is true, and suppose that \(P(k) \rightarrow P(k+1)\) for all \(k \geq M\). Show that \(P(n)\) is true for all \(n \geq M\)
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