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Chapter 2: Problem 32

Suppose that \(A \times B=\emptyset,\) where \(A\) and \(B\) are sets. What can you conclude?

Short Answer

Expert verified
At least one of the sets \(A\) or \(B\) must be empty.

Step by step solution

01

Understand the Cartesian Product

The Cartesian product of two sets, denoted by \(A \times B\), is the set of all ordered pairs \((a, b)\) where \(a \in A\) and \(b \in B\).
02

Analyze the Given Condition

In this exercise, it is given that \(A \times B = \emptyset\). This means that there are no ordered pairs \((a, b)\) such that \(a \in A\) and \(b \in B\).
03

Conclude Based on the Definition

For the Cartesian product \(A \times B\) to be empty, at least one of the sets \(A\) or \(B\) must be empty. If either set contained at least one element, there would be at least one ordered pair, and thus the Cartesian product would not be empty.

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Key Concepts

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

Empty Set
In set theory, the empty set is a fundamental concept. The empty set, denoted by \(\backslashemptyset\), is a set that contains no elements. It's the unique set with zero elements, playing a crucial role in various mathematical theories and concepts.

Here are some important points about the empty set:
  • The empty set is a subset of every set.
  • It's denoted by a pair of curly braces with nothing between them: \(\backslash{\backslash}\).
  • The cardinality (or size) of the empty set is 0.
In the context of this exercise, understanding the empty set helps us see why a Cartesian product would be empty if one of the sets involved has no elements.
Ordered Pairs
An ordered pair is a fundamental concept in set theory and is crucial for understanding Cartesian products. Ordered pairs consist of two elements arranged in a specific order, typically written as \((a, b)\). The order in which these elements are positioned matters; thus, \((a, b)\) is generally distinct from \((b, a)\).

Here are more details about ordered pairs:
  • The first element is known as the first coordinate, and the second element is the second coordinate.
  • Ordered pairs are central to defining Cartesian products of sets.
  • They help in mapping elements from one set to another in a structured manner.
For the given exercise, we see that the Cartesian product contains no ordered pairs if either set is empty.
Set Theory
Set theory is the branch of mathematical logic that studies sets, which are collections of objects. It forms the foundation for various areas of mathematics. Here are some key concepts within set theory:
  • Sets can be finite or infinite, and their elements are distinct.
  • Sets are typically denoted by capital letters like \(\backslashmathbb{A}\) and \(\backslashmathbb{B}\).
  • Important operations include union, intersection, and difference of sets.
  • Sets can include various kinds of elements, including numbers, letters, or other sets.
In our exercise, we are particularly interested in the Cartesian product of sets. The Cartesian product, denoted by \(A \backslashes times B\), forms a new set comprising all ordered pairs from two sets \(A\) and \(B\). However, if one or both of these sets are empty, the product will also be empty. This exercise reinforces these basic but essential principles of set theory.

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Most popular questions from this chapter

Let \(g(x)=\lfloor x\rfloor .\) Find a) \(g^{-1}(\\{0\\})\) b) \(g^{-1}(\\{-1,0,1\\})\) c) \(g^{-1}(\\{x | 0 < x < 1\\})\)

Show that if \(x\) is a real number and \(n\) is an integer, then a) \(x < n\) if and only if \(\lfloor x\rfloor < n .\) b) \(n < x\) if and only if \(n < \lceil x\rceil\)

Assume that \(a \in A,\) where \(A\) is a set. Which of these statements are true and which are false, where all sets shown are ordinary sets, and not multisets. Explain each answer. a) \(\\{a, a\\} \cup\\{a, a, a\\}=\\{a, a, a, a, a\\}\) b) \(\\{a, a\\} \cup\\{a, a, a\\}=\\{a\\}\) c) \(\\{a, a\\} \cap\\{a, a, a\\}=\\{a, a\\}\) d) \(\\{a, a\\} \cap\\{a, a, a\\}=\\{a\\}\) e) \(\\{a, a, a\\}-\\{a, a\\}=\\{a\\}\)

In this exercise we show that the meet and join operations are commutative. Let \(A\) and \(B\) be \(m \times n\) zero-one matrices. Show that $$ \begin{array}{lll}{\text { a) } \mathbf{A} \vee \mathbf{B}=\mathbf{B} \vee \mathbf{A}} & {\text { b) } \mathbf{B} \wedge \mathbf{A}=\mathbf{A} \wedge \mathbf{B}}\end{array} $$

Let \(f\) be the function from \(R\) to \(R\) defined by \(f(x)=x^{2} .\) Find $$\begin{array}{ll}{\text { a) } f^{-1}(\\{1\\},} & {\text { b) } f^{-1}(\\{x | 0 < x < 1\\}} \\ {\text { c) } f^{-1}(\\{x | x>4\\}}\end{array}$$
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