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Chapter 1: Problem 5

Prove \(A \cup \emptyset=A\) and \(A \cap \emptyset=\emptyset\).

Short Answer

Expert verified
Question: Prove that for any set A, we have \(A \cup \emptyset = A\) and \(A \cap \emptyset = \emptyset\). Answer: We showed two proofs: 1. To prove \(A \cup \emptyset = A\), we demonstrated that if an element \(x\) belongs to \(A \cup \emptyset\), then it must also belong to \(A\), and vice versa. This double containment establishes the equality. 2. To prove \(A \cap \emptyset = \emptyset\), we showed that it is impossible for an element to belong to both \(A\) and \(\emptyset\) due to the definition of an intersection and the empty set. Therefore, \(A \cap \emptyset\) has no elements and is equal to \(\emptyset\).

Step by step solution

01

Prove \(A \cup \emptyset = A\)

We want to show that any element \(x\) that belongs to the set \(A \cup \emptyset\) also belongs to the set \(A\), and vice versa. Proof: Let \(x\) be an arbitrary element. (鈬�) Suppose \(x \in A \cup \emptyset\). By the definition of the union, this means that \(x\) must belong to either \(A\) or \(\emptyset\) (or both). However, since \(\emptyset\) has no elements, \(x\) cannot be in \(\emptyset\). Therefore, \(x\) must be in \(A\). (鈬�) Suppose \(x \in A\). By the definition of the union, this means that \(x\) is in at least one of the sets \(A\) or \(\emptyset\). Thus, \(x \in A \cup \emptyset\). Since we have shown the double containment, we conclude that \(A \cup \emptyset = A\).
02

Prove \(A \cap \emptyset = \emptyset\)

We want to show that any element \(x\) that belongs to the set \(A \cap \emptyset\) also belongs to the set \(\emptyset\), and vice versa. Proof: Let \(x\) be an arbitrary element. (鈬�) Suppose \(x \in A \cap \emptyset\). By the definition of intersection, this means that \(x\) must belong to both sets \(A\) and \(\emptyset\). However, since \(\emptyset\) has no elements, there is no element that belongs to both \(A\) and \(\emptyset\). Therefore, it's impossible for \(x\) to be in \(A \cap \emptyset\). This implies that \(A \cap \emptyset\) has no elements, and thus \(A \cap \emptyset = \emptyset\). (鈬�) Suppose \(x \in \emptyset\). By definition of intersection, \(x\) cannot be in \(A \cap \emptyset\), since there is no element in the empty set. Since both directions of containment have been shown, we conclude that \(A \cap \emptyset = \emptyset\).

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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 concept that helps us combine elements from different sets into a new set. When we talk about the union of two sets, say, set A and set B, we define it as a set that includes all the elements that are in A, in B, or in both. Mathematically, this is expressed as:
  • The union of A and B is denoted by \( A \cup B \).
  • An element \( x \) is in \( A \cup B \) if and only if \( x \in A \) or \( x \in B \).
For example, if \( A = \{ 1, 3, 5 \} \) and \( B = \{ 2, 3, 4 \} \), then \( A \cup B = \{ 1, 2, 3, 4, 5 \} \). This idea becomes particularly interesting when one of the sets is the empty set, \( \emptyset \). The union of any set with the empty set is simply the original set:
  • Since the empty set has no elements, it contributes nothing new to the union.
  • Therefore, \( A \cup \emptyset = A \).
Intersection of Sets
The intersection of sets focuses on the commonality between sets, only containing elements that are in all the sets involved. For two sets A and B, the intersection is defined as:
  • The intersection of A and B is denoted by \( A \cap B \).
  • An item \( x \) is in \( A \cap B \) if and only if \( x \in A \) and \( x \in B \).
Consider the sets \( A = \{ 1, 2, 3 \} \) and \( B = \{ 2, 3, 4 \} \). Their intersection, \( A \cap B = \{ 2, 3 \} \), includes only the elements present in both sets. When intersecting a set with the empty set, we find a unique outcome:
  • Because the empty set has no elements, there are no elements common to both sets.
  • Hence, \( A \cap \emptyset = \emptyset \).
Empty Set
The empty set, symbolized as \( \emptyset \), is a foundational concept in set theory. By definition, an empty set is a set without any elements. It plays a crucial role in understanding and defining operations such as union and intersection:
  • When a set is combined with the empty set through union, no new elements are added, maintaining the original set as \( A \cup \emptyset = A \).
  • In terms of intersection, since there are no elements to share between the empty set and any other set, the result is always the empty set: \( A \cap \emptyset = \emptyset \).
Despite its simplicity, the empty set is indispensable in mathematical proofs and logic, as it provides a standard baseline for comparing and operating on other sets. It also illustrates the idea that sets in mathematics can exist as abstract entities, even without any physical elements.

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Most popular questions from this chapter

Show that an \(m \times n\) matrix gives rise to a well-defined map from \(\mathbb{R}^{n}\) to \(\mathbb{R}^{m}\).

Which of the following relations \(f: \mathbb{Q} \rightarrow \mathbb{Q}\) define a mapping? In each case, supply a reason why \(f\) is or is not a mapping. (a) \(f(p / q)=\frac{p+1}{p-2}\) (c) \(f(p / q)=\frac{p+q}{q^{2}}\) (b) \(f(p / q)=\frac{3 p}{3 q}\) (d) \(f(p / q)=\frac{3 p^{2}}{7 q^{2}}-\frac{p}{q}\)

Prove the relation defined on \(\mathbb{R}^{2}\) by \(\left(x_{1}, y_{1}\right) \sim\left(x_{2}, y_{2}\right)\) if \(x_{1}^{2}+y_{1}^{2}=x_{2}^{2}+y_{2}^{2}\) is an equivalence relation.

Prove \((A \cap B) \backslash B=\emptyset\).

Prove \(A \cup(B \cap C)=(A \cup B) \cap(A \cup C)\).
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