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Let \(f: S \rightarrow T,\) let \(A\) and \(B\) be subsets of \(S,\) and let \(C\) and \(D\) be subsets of \(T\). For \(x \in S\) and \(y \in T\), carefully explain what it means to say that (a) \(y \in f(A \cap B)\) (b) \(y \in f(A \cup B)\) (c) \(y \in f(A) \cap f(B)\) (d) \(y \in f(A) \cup f(B)\) (e) \(x \in f^{-1}(C \cap D)\) (f) \(x \in f^{-1}(C \cup D)\) (g) \(x \in f^{-1}(C) \cap f^{-1}(D)\) (h) \(x \in f^{-1}(C) \cup f^{-1}(D)\)

Short Answer

Expert verified
(a) \(y \in f(A \cap B)\) means there exists \(x \in A \cap B\) such that \(f(x) = y\). (b) \(y \in f(A \cup B)\) means there exists \(x \in A \cup B\) such that \(f(x) = y\). (c) \(y \in f(A) \cap f(B)\) means there exist \(x_1 \in A\) and \(x_2 \in B\) such that \(f(x_1) = y\) and \(f(x_2) = y\). (d) \(y \in f(A) \cup f(B)\) means there exists \(x \in A\) or \(B\) such that \(f(x) = y\). (e) \(x \in f^{-1}(C \cap D)\) means \(f(x) \in C\) and \(f(x) \in D\). (f) \(x \in f^{-1}(C \cup D)\) means \(f(x) \in C\) or \(f(x) \in D\). (g) \(x \in f^{-1}(C) \cap f^{-1}(D)\) means \(f(x) \in C\) and \(f(x) \in D\). (h) \(x \in f^{-1}(C) \cup f^{-1}(D)\) means \(f(x) \in C\) or \(f(x) \in D\).

Step by step solution

01

(a) \(y \in f(A \cap B)\)

To say \(y \in f(A \cap B)\) means that \(y\) is an element of the image of the intersection of sets \(A\) and \(B\) under the function \(f\). In other words, there exists an element \(x \in A \cap B\), such that \(f(x) = y\).
02

(b) \(y \in f(A \cup B)\)

To say \(y \in f(A \cup B)\) means that \(y\) is an element of the image of the union of sets \(A\) and \(B\) under the function \(f\). In other words, there exists an element \(x \in A \cup B\), such that \(f(x) = y\).
03

(c) \(y \in f(A) \cap f(B)\)

To say \(y \in f(A) \cap f(B)\) means that \(y\) is an element of the intersection of the images of sets \(A\) and \(B\) under the function \(f\). In other words, there exist elements \(x_1 \in A\) and \(x_2 \in B\) such that \(f(x_1) = y\) and \(f(x_2) = y\).
04

(d) \(y \in f(A) \cup f(B)\)

To say \(y \in f(A) \cup f(B)\) means that \(y\) is an element of the union of the images of sets \(A\) and \(B\) under the function \(f\). In other words, there exists an element \(x \in A\) or \(B\) such that \(f(x)=y\).
05

(e) \(x \in f^{-1}(C \cap D)\)

To say \(x \in f^{-1}(C \cap D)\) means that \(x\) is an element of the inverse image of the intersection of sets \(C\) and \(D\) under the function \(f\). In other words, \(f(x) \in C\) and \(f(x) \in D\).
06

(f) \(x \in f^{-1}(C \cup D)\)

To say \(x \in f^{-1}(C \cup D)\) means that \(x\) is an element of the inverse image of the union of sets \(C\) and \(D\) under the function \(f\). In other words, \(f(x) \in C\) or \(f(x) \in D\).
07

(g) \(x \in f^{-1}(C) \cap f^{-1}(D)\)

To say \(x \in f^{-1}(C) \cap f^{-1}(D)\) means that \(x\) is an element of the intersection of the inverse images of sets \(C\) and \(D\) under the function \(f\). In other words, \(f(x) \in C\) and \(f(x) \in D\).
08

(h) \(x \in f^{-1}(C) \cup f^{-1}(D)\)

To say \(x \in f^{-1}(C) \cup f^{-1}(D)\) means that \(x\) is an element of the union of the inverse images of sets \(C\) and \(D\) under the function \(f\). In other words, \(f(x) \in C\) or \(f(x) \in D\).

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Key Concepts

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

Image of a Set
When discussing functions and set theory, the image of a set is a fundamental concept. For example, if you have a function, denoted by \(f:S \rightarrow T\), mapping elements from set \(S\) to elements in set \(T\), the image of a subset \(A\) of \(S\) under \(f\) is the set of all images of elements of \(A\). This set is denoted by \(f(A)\).

Mathematically, this can be expressed as \(f(A) = \{ f(x) | x \in A \}\). If you encounter an expression like \(y \in f(A)\), it simply means that \(y\) is one of the outputs produced by the function \(f\) for some input \(x\) from the set \(A\).

The concept becomes even more intriguing when we talk about the image of a combination of sets, such as the intersection \(A \cap B\), where the image \(f(A \cap B)\) consists of all elements that result from applying \(f\) on all elements that are both in \(A\) and \(B\). Similarly, the image of the union \(A \cup B\) combines the outcomes from both \(A\) and \(B\) without duplication.
Inverse Image of a Set
The inverse image, or preimage, is a complimentary concept to the image of a set. Given a function from set \(S\) to set \(T\), denoted \(f:S \rightarrow T\), and a subset \(C\) of \(T\), the inverse image of \(C\) under \(f\) is the set of elements in \(S\) that are mapped to \(C\). This set is represented by \(f^{-1}(C)\).

The formal definition is \(f^{-1}(C) = \{ x | x \in S, f(x) \in C \}\). So, when you see \(x \in f^{-1}(C)\), it is saying that \(x\) is an element in \(S\) that, when fed into \(f\), yields an output belonging to \(C\).

It's important to note that the function does not necessarily have to be inversible for the inverse image to exist. The term 'inverse image' merely refers to the reversal of the 'mapping direction' when considering the function's effects on sets. It's a key tool for understanding how functions relate elements across domains.
Function Intersection and Union
In set theory, the concepts of intersection and union are often used to combine sets in various ways. When combined with functions, the intersection \(\cap\) and union \(\cup\) can be applied to both images and inverse images.

For instance, the intersection of images, \(f(A) \cap f(B)\), contains all elements that are images of elements from both \(A\) and \(B\) independently. However, this doesn't mean that the intersection of \(A\) and \(B\) is necessary for an element to be in \(f(A) \cap f(B)\). On the contrary, the union of images, \(f(A) \cup f(B)\), includes all elements that are images of elements from \(A\), \(B\), or both.

The intersection and union of inverse images follow the same logic but in the opposite 'direction'. The set \(f^{-1}(C \cap D)\) consists of elements from \(S\) that map to both \(C\) and \(D\), and \(f^{-1}(C \cup D)\) contains elements from \(S\) that map to either \(C\), \(D\), or both.
Mathematical Reasoning
Mathematical reasoning is the backbone of understanding concepts like function mapping and operations on sets. It involves logical deduction and induction to infer properties, relationships, and patterns. When you work through problems involving images and inverse images of sets, you are employing mathematical reasoning to make sense of how elements and sets are transformed and interrelated through functions.

Part of this reasoning entails understanding how different operations—such as intersection and union—modify the relationships between sets. For example, reasoning about whether an element belongs to the intersection of two images requires you to consider the individual images separately before assessing the shared elements. Applying logical steps methodically can deepen your comprehension of set theory and function mappings.
Set Theory
Set theory is a fundamental language of mathematics that deals with the study of sets, which are collections of objects. It serves as the foundation for various mathematical disciplines, including the study of functions and their properties. The actions of a function, whether describing its image or inverse image, are rooted in set-theoretic concepts.

By understanding the basics of sets, including notions like subsets, intersections, unions, and the empty set, you can better navigate the operations performed by functions on these sets. Knowing that a function can be seen as a set of ordered pairs also connects set theory directly to understanding function concepts. This connection amplifies the importance of solid set-theoretic knowledge as a prelude to more advanced mathematical explorations.

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Most popular questions from this chapter

The Inverse Sine Function. We have seen that in order to obtain an inverse function, it is sometimes necessary to restrict the domain (or the codomain) of a function. (a) Let \(f: \mathbb{R} \rightarrow \mathbb{R}\) be defined by \(f(x)=\sin x\). Explain why the inverse of the function \(f\) is not a function. (A graph may be helpful.) Notice that if we use the ordered pair representation, then the sine function can be represented as $$ f=\\{(x, y) \in \mathbb{R} \times \mathbb{R} \mid y=\sin x\\} $$ If we denote the inverse of the sine function by \(\sin ^{-1}\), then $$ f^{-1}=\\{(y, x) \in \mathbb{R} \times \mathbb{R} \mid y=\sin x\\} $$ Part (14a) proves that \(f^{-1}\) is not a function. However, in previous mathematics courses, we frequently used the "inverse sine function." This is not really the inverse of the sine function as defined in Part (14a) but, rather, it is the inverse of the sine function restricted to the domain \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\). (b) Explain why the function \(F:\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \rightarrow[-1,1]\) defined by \(F(x)=\) \(\sin x\) is a bijection. The inverse of the function in Part ( \(14 \mathrm{~b}\) ) is itself a function and is called the inverse sine function (or sometimes the arcsine function). (c) What is the domain of the inverse sine function? What are the range and codomain of the inverse sine function? Let us now use \(F(x)=\operatorname{Sin}(x)\) to represent the restricted sine function in Part (14b). Therefore, \(F^{-1}(x)=\operatorname{Sin}^{-1}(x)\) can be used to represent the inverse sine function. Observe that $$ F:\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \rightarrow[-1,1] \text { and } F^{-1}:[-1,1] \rightarrow\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] $$ (d) Using this notation, explain why \(\operatorname{Sin}^{-1} y=x\) if and only if \(\left[y=\sin x\right.\) and \(\left.-\frac{\pi}{2} \leq x \leq \frac{\pi}{2}\right]\) \(\operatorname{Sin}\left(\operatorname{Sin}^{-1}(y)\right)=y\) for all \(y \in[-1,1] ;\) and $$ \operatorname{Sin}^{-1}(\operatorname{Sin}(x))=x \text { for all } x \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] $$

Let \(g: \mathbb{N} \times \mathbb{N} \rightarrow \mathbb{N}\) by \(g(m, n)=2^{m} 3^{n},\) let \(A=\\{1,2,3\\},\) and let \(C=\) \(\\{1,4,6,9,12,16,18\\}\) . Find (a) \(g(A \times A)\) (b) \(g^{-1}(C)\) (c) \(g^{-1}(g(A \times A))\) (d) \(g\left(g^{-1}(C)\right)\)

In our definition of the composition of two functions, \(f\) and \(g\), we required that the domain of \(g\) be equal to the codomain of \(f\). However, it is sometimes possible to form the composite function \(g \circ f\) even though \(\operatorname{dom}(g) \neq \operatorname{codom}(f) .\) For example, let $$ f: \mathbb{R} \rightarrow \mathbb{R} $$ be defined by \(\quad f(x)=x^{2}+1,\) and let \(g: \mathbb{R}-\\{0\\} \rightarrow \mathbb{R} \quad\) be defined by \(\quad g(x)=\frac{1}{x}\) (a) Is it possible to determine \((g \circ f)(x)\) for all \(x \in \mathbb{R} ?\) Explain. (b) In general, let \(f: A \rightarrow T\) and \(g: B \rightarrow C\). Find a condition on the domain of \(g\) (other than \(B=T\) ) that results in a meaningful definition of the composite function \(g \circ f: A \rightarrow C\).

Let \(f:(\mathbb{R}-\\{0\\}) \rightarrow \mathbb{R}\) by \(f(x)=\frac{x^{3}+5 x}{x}\) and let \(g: \mathbb{R} \rightarrow \mathbb{R}\) by \(g(x)=x^{2}+5\) (a) Calculate \(f(2), f(-2), f(3),\) and \(f(\sqrt{2})\) (b) Calculate \(g(0), g(2), g(-2), g(3),\) and \(g(\sqrt{2})\) (c) Is the function \(f\) equal to the function \(g ?\) Explain. (d) Now let \(h:(\mathbb{R}-\\{0\\}) \rightarrow \mathbb{R}\) by \(h(x)=x^{2}+5\). Is the function \(f\) equal to the function \(h\) ? Explain.

Piecewise Defined Functions. We often say that a function is a piecewise defined function if it has different rules for determining the output for different parts of its domain. For example, we can define a function \(f: \mathbb{R} \rightarrow \mathbb{R}\) by giving a rule for calculating \(f(x)\) when \(x \geq 0\) and giving a rule for calculating \(f(x)\) when \(x<0\) as follows: $$f(x)=\left\\{\begin{array}{ll} x^{2}+1, & \text { if } x \geq 0 \\ x-1 & \text { if } x<0 \end{array}\right.$$ (a) Sketch a graph of the function \(f .\) Is the function \(f\) an injection? Is the function \(f\) a surjection? Justify your conclusions. For each of the following functions, determine if the function is an injection and determine if the function is a surjection. Justify all conclusions. (b) \(g:[0,1] \rightarrow(0,1)\) by (c) \(h: \mathbb{Z} \rightarrow\\{0,1\\}\) by $$g(x)=\left\\{\begin{array}{ll} 0.8, & \text { if } x=0 \\ 0.5 x, & \text { if } 0

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