/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 5 Prove Part (2) of Theorem 6.34. ... [FREE SOLUTION] | 91影视

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Chapter 6: Problem 5

Prove Part (2) of Theorem 6.34. Let \(f: S \rightarrow T\) be a function and let \(A\) and \(B\) be subsets of \(S\). Then \(f(A \cup\) \(B)=f(A) \cup f(B)\)

Short Answer

Expert verified
To prove \(f(A \cup B) = f(A) \cup f(B)\), we showed that \(f(A \cup B) \subseteq f(A) \cup f(B)\) and \(f(A) \cup f(B) \subseteq f(A \cup B)\). By demonstrating both directions of the inclusion, we concluded that \(f(A \cup B) = f(A) \cup f(B)\).

Step by step solution

01

Prove \(f(A \cup B) \subseteq f(A) \cup f(B)\)

Let \(y \in f(A \cup B)\). This means that there exists an element \(x \in A \cup B\) such that \(y = f(x)\). Since \(x \in A \cup B\), we have two possibilities: \(x \in A\) or \(x \in B\). If \(x \in A\), then \(y = f(x) \in f(A)\), so \(y \in f(A) \cup f(B)\). If \(x \in B\), then \(y = f(x) \in f(B)\), so \(y \in f(A) \cup f(B)\). In both cases, we have \(y \in f(A) \cup f(B)\). Since this holds for any \(y \in f(A \cup B)\), we have proved that \(f(A \cup B) \subseteq f(A) \cup f(B)\).
02

Prove \(f(A) \cup f(B) \subseteq f(A \cup B)\)

Let \(y \in f(A) \cup f(B)\). This means that either \(y \in f(A)\) or \(y \in f(B)\). If \(y \in f(A)\), then there exists an element \(x \in A\) such that \(y = f(x)\). Since \(A \subseteq A \cup B\), it follows that \(x \in A \cup B\), and thus \(y = f(x) \in f(A \cup B)\). If \(y \in f(B)\), then there exists an element \(x \in B\) such that \(y = f(x)\). Since \(B \subseteq A \cup B\), it follows that \(x \in A \cup B\), and thus \(y = f(x) \in f(A \cup B)\). In both cases, we have \(y \in f(A \cup B)\). Since this holds for any \(y \in f(A) \cup f(B)\), we have proved that \(f(A) \cup f(B) \subseteq f(A \cup B)\). Since we have proved both \(f(A \cup B) \subseteq f(A) \cup f(B)\) and \(f(A) \cup f(B) \subseteq f(A \cup B)\), we can conclude that \(f(A \cup B) = f(A) \cup f(B)\).

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Key Concepts

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

Mathematical Functions
A mathematical function is a rule that assigns each element in one set, called the domain, to exactly one element in another set, called the codomain. This relationship is often denoted by the symbol \( f: S \rightarrow T \), where \( f \) is the function, \( S \) is the domain, and \( T \) is the codomain.
Functions are essential in connecting elements of different sets in a structured and predictable way. For example:
  • The function \( f(x) = x^2 \) takes a number \( x \) and maps it to its square.
  • The function \( f: S \rightarrow T \) maps elements from set \( S \) to set \( T \).
Functions help in defining relationships and transformations between various mathematical objects, making them crucial in both theoretical and applied mathematics.
Set Theory
Set theory is a branch of mathematical logic that deals with the study of sets, which are collections of objects. It is fundamental to modern mathematics, as it provides the underlying framework for defining and manipulating sets and their related operations.
Within set theory:
  • The union of sets \( A \) and \( B \), denoted \( A \cup B \), is the set containing all the elements of \( A \), \( B \), or both.
  • The intersection, represented as \( A \cap B \), contains elements common to both sets \( A \) and \( B \).
  • Set differences and complements are also essential operations, providing more ways to relate and compare sets.
These operations are not just theoretical. They have practical applications in various fields like computer science, probability, and data analysis. They let us model and solve problems involving collections of objects.
Subsets
In set theory, a subset is defined as a set where every element is also an element of another set. If set \( A \) is a subset of set \( B \), it is written as \( A \subseteq B \).
This relationship tells us that all elements of \( A \) are contained within \( B \), but \( B \) may contain elements not in \( A \). For example:
  • If \( A = \{1, 2\} \) and \( B = \{1, 2, 3\} \), then \( A \subseteq B \).
  • An important case is when a set \( A \) is the empty set, denoted by \( \emptyset \), which is a subset of every possible set.
Understanding subsets is crucial as it forms the basis for many other operations in set theory. In the context of functions, subsets allow us to examine how functions behave over smaller sections of their domain, leading to properties such as injections, surjections, and bijections.

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Most popular questions from this chapter

(a) Draw an arrow diagram that represents a function that is an injection but is not a surjection. (b) Draw an arrow diagram that represents a function that is an injection and is a surjection. (c) Draw an arrow diagram that represents a function that is not an injection and is not a surjection. (d) Draw an arrow diagram that represents a function that is not an injection but is a surjection. (e) Draw an arrow diagram that represents a function that is not a bijection.

Prove Part (2) of Corollary 6.28. Let \(A\) and \(B\) be nonempty sets and let \(f: A \rightarrow B\) be a bijection. Then for every \(y\) in \(B,\left(f \circ f^{-1}\right)(y)=y\).

Let \(f: \mathbb{Z} \times \mathbb{Z} \rightarrow \mathbb{Z}\) be defined by \(f(m, n)=m+3 n\) (a) Calculate \(f(-3,4)\) and \(f(-2,-7)\). (b) Determine the set of all the preimages of 4 by using set builder notation to describe the set of all \((m, n) \in \mathbb{Z} \times \mathbb{Z}\) such that \(f(m, n)=4\).

Let \(R_{5}=\\{0,1,2,3,4\\} .\) Define \(f: R_{5} \rightarrow R_{5}\) by \(f(x)=x^{2}+4(\bmod 5)\), and define \(g: R_{5} \rightarrow R_{5}\) by \(g(x)=(x+1)(x+4)(\bmod 5)\). (a) Calculate \(f(0), f(1), f(2), f(3),\) and \(f(4)\). (b) Calculate \(g(0), g(1), g(2), g(3),\) and \(g(4)\). (c) Is the function \(f\) equal to the function \(g\) ? Explain.

(a) Define \(f: \mathbb{R} \rightarrow \mathbb{R}\) by \(f(x)=e^{-x^{2}}\). Is the inverse of \(f\) a function? Justify your conclusion. (b) Let \(\mathbb{R}^{*}=\\{x \in \mathbb{R} \mid x \geq 0\\} .\) Define \(g: \mathbb{R}^{*} \rightarrow(0,1]\) by \(g(x)=e^{-x^{2}}\). Is the inverse of \(g\) a function? Justify your conclusion.
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