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Give an example of a function \(f: A \rightarrow B\) and \(A_{1}, A_{2} \subseteq A\) for which \(f\left(A_{1} \cap A_{2}\right) \neq f\left(A_{1}\right) \cap f\left(A_{2}\right)\). [Thus the inclusion in Theorem 5.2(b) may be proper.]

Short Answer

Expert verified
An example for which \(f(A1 \cap A2) \neq f(A1) \cap f(A2)\) is a function \(f\) from a set \(A = {1, 2}\) to a set \(B = {3, 4}\) defined as \(f(x) = 3\) for \(x\) in \(A\), and subsets \(A1 = {1}\) and \(A2 = {2}\). The function and these subsets satisfy the proposed condition.

Step by step solution

01

Defining Sets and Function

Let \(A = {1, 2}\), \(B = {3, 4}\), \(A_1 = {1}\), \(A_2 = {2}\) and function \(f: A \rightarrow B\) be defined as \(f(x) = 3\) for all \(x\) in set \(A\). Here \(f\) is a constant function, assigning all elements in set A to a single element (3) in set B.
02

Using Function on Intersection of Sets

We calculate the image of the intersection \(A_1 \cap A_2\) under function \(f\). Since \(A_1\) and \(A_2\) have no elements in common, their intersection is an empty set, i.e., \(A_1 \cap A_2 = \emptyset\). The image of the empty set under any function is always an empty set, i.e., \(f(A_1 \cap A_2) = \emptyset\).
03

Using Function on Individual Sets

We calculate the image of the individual sets \(A_1\) and \(A_2\) under the function \(f\). For both \(A_1\) and \(A_2\), function \(f\) maps their elements to 3, i.e., \(f(A_1)= f(A_2) = {3}\). The intersection of the images of \(A_1\) and \(A_2\) is \(f(A_1) \cap f(A_2) = {3}\).
04

Comparing the Results

We see that the image of the intersection of the sets under the function \(f(A_1 \cap A_2) = \emptyset\), does not equal the intersection of the individual images of the sets under the function \(f(A_1) \cap f(A_2) = {3}\). Therefore, our function and subsets of its domain are examples of an instance where the image of the intersection of the sets under function \(f\) does not equal the intersection of their individual images under function \(f\). Thus the inclusion in Theorem 5.2(b) may be proper.

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Key Concepts

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

Functions
In mathematics, a function is a special kind of relation between two sets. It connects each element in one set, known as the domain, to exactly one element in another set, known as the codomain. The key feature of a function is that each input is related to one and only one output. This unique relationship is what differentiates functions from general relations.
Functions are denoted with names such as \(f\), \(g\), or \(h\). For example, if \(f: A \rightarrow B\), we say that function \(f\) maps elements from set \(A\) (domain) to set \(B\) (codomain). The output for an input \(x\) in the domain \(A\) is denoted as \(f(x)\).
When dealing with functions, it's crucial to understand:
  • The domain: the set of all possible inputs.
  • The codomain: the set where all outputs reside.
  • The image: actual outputs corresponding to the inputs from the domain.
Functions can be used in diverse areas of mathematics and science to model relationships and transformations.
Intersection of Sets
The intersection of sets is a fundamental concept in set theory, and it involves determining the common elements between two or more sets. If we have two sets \(A_1\) and \(A_2\), their intersection, written as \(A_1 \cap A_2\), consists of elements that are present in both \(A_1\) and \(A_2\).
For instance, if \(A_1 = \{1\}\) and \(A_2 = \{2\}\), the intersection \(A_1 \cap A_2\) is an empty set since there are no elements common to both sets. This is denoted as \( \emptyset \).
Some key points about intersections include:
  • If two sets have no common elements, their intersection is an empty set, or \( \emptyset \).
  • Intersections are commutative, meaning \(A_1 \cap A_2 = A_2 \cap A_1\).
  • Intersections are associative, so \((A_1 \cap A_2) \cap A_3 = A_1 \cap (A_2 \cap A_3)\).
Intersections are used extensively in probability, logic, and algebra, wherever relationships and commonalities between sets need to be identified.
Constant Function
A constant function is a simple but special type of function in mathematics. It assigns the same single value in the codomain to every element in the domain.
For a constant function \(f: A \rightarrow B\), every input from set \(A\) is mapped to a fixed output in set \(B\). For example, if \(f(x) = c\) for all \(x\) in \(A\), then \(c\) is a constant value in \(B\). In the case presented, the function \(f(x) = 3\) maps every element in the set \(A = \{1, 2\}\) to the constant value \(3\).
Key characteristics of constant functions include:
  • Their graphs are straight horizontal lines when plotted on a coordinate system.
  • They do not change values, hence are very predictable.
  • For any two points in the domain, \(f(x_1) = f(x_2)\).
Despite their simplicity, constant functions play an important role in many areas of mathematics, such as defining scalar fields and serving as solutions to differential equations under specific boundary conditions.

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Most popular questions from this chapter

Mrs. Blasi has five sons (Michael, Rick, David, Kenneth, and Donald) who enjoy reading books about sports. With Christmas approaching, she visits a bookstore where she finds 12 different books on sports. a) In how many ways can she select nine of these books? b) Having made her purchase, in how many ways can she distribute the books among her sons so that each of them gets at least one book? c) Two of the nine books Mrs. Blasi purchased deal with basketball, Donald's favorite sport. In how many ways can she distribute the books among her sons so that Donald gets at least the two books on basketball?

At St. Xavier High School ten candidates \(C_{1}, C_{2}, \ldots, C_{10}\), run for senior class president. a) How many outcomes are possible where (i) there are no ties (that is, no two, or more, candidates receive the same number of votes? (ii) ties are permitted? [Here we may have an outcome such as \(\left\\{C_{2}, C_{3}, C_{7}\right\\},\left\\{C_{1}, C_{4}, C_{9}, C_{10}\right\\}\), \(\left\\{C_{5}\right\\},\left\\{C_{6}, C_{8}\right\\}\), where \(C_{2}, C_{3}, C_{7}\) tie for first place, \(C_{1}, C_{4}, C_{9}, C_{10}\) tie for fourth place, \(C_{5}\) is in eighth place, and \(C_{6}, C_{8}\) are tied for ninth place.] (iii) three candidates tie for first place (and other ties are permitted)? b) How many of the outcomes in section (iii) of part (a) have \(C_{3}\) as one of the first-place candidates? c) How many outcomes have \(C_{3}\) in first place (alone, or tied with others)?

. Let \(f, g, h: \mathbf{Z}^{+} \rightarrow \mathbf{R}\) where \(f \in O(g)\) and \(g \in O(h)\). Prove that \(f \in O(h)\).

. Let \(f, g: \mathbf{R} \rightarrow \mathbf{R}\) where \(f(x)=a x+b\) and \(g(x)=c x+d\) for all \(x \in \mathbf{R}\), with \(a, b, c, d\) real constants. What relationship(s) must be satisfied by \(a, b, c, d\) if \((f \circ g)(x)=(g \circ f)(x)\) for all \(x \in \mathbf{R} ?\)

Let \(A, B \subseteq \mathbf{N}\) with \(1<|A|<|B|\). If there are 262,144 relations from \(A\) to \(B\), determine all possibilities for \(|A|\) and \(|B|\).

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