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Chapter 1: Problem 5

Let \(E\) be the set of all functions from a set \(X\) into a set \(Y\). Let \(b \in X\) and let \(R\) be the subset of \(E \times E\) consisting of those pairs \((f, g)\) such that \(f(b)\) \(=g(b)\). Prove that \(R\) is an equivalence relation. Define a one-one onto function \(e_{b}: E / R \rightarrow Y\).

Short Answer

Expert verified
\( R \) is an equivalence relation; \( e_{b} \) is a bijection onto \( Y \).

Step by step solution

01

Prove Reflexivity of R

To prove reflexivity, we need to show that for every function \( f \in E \), the pair \((f, f)\) is in \( R \). Since the definition of \( R \) is that it contains pairs \((f, g)\) such that \( f(b) = g(b) \), it is evident that \( f(b) = f(b) \) for any function \( f \). Thus, \( (f, f) \in R \), showing that \( R \) is reflexive.
02

Prove Symmetry of R

To prove symmetry, we need to show that if \( (f, g) \in R \), then \( (g, f) \in R \) as well. From the definition of \( R \), if \( (f, g) \in R \), this implies \( f(b) = g(b) \). Symmetry requires that \( g(b) = f(b) \), which is trivially true because equality is symmetric. Hence, \( (g, f) \in R \), and \( R \) is symmetric.
03

Prove Transitivity of R

To prove transitivity, we need to show that if \( (f, g) \in R \) and \( (g, h) \in R \), then \( (f, h) \in R \). This means if \( f(b) = g(b) \) and \( g(b) = h(b) \), then \( f(b) = h(b) \). By the transitive property of equality, \( f(b) = h(b) \) holds true, so \( (f, h) \in R \). Thus, \( R \) is transitive.
04

Define Function e_{b}

We now define a function \( e_{b}: E / R \rightarrow Y \). The set \( E / R \) is the set of equivalence classes of functions under \(R\). For each equivalence class \([f]\), where \( [f] \) denotes all functions equivalent to \( f \), define \( e_{b}([f]) = f(b) \).
05

Prove e_{b} is Well-Defined

To ensure \( e_{b} \) is well-defined, we must show that if \([f] = [g]\) (i.e., \( f \sim g \) under \( R \)), then \( e_{b}([f]) = e_{b}([g]) \). Since \( f \sim g \) means \( f(b) = g(b) \), \( e_{b}([f]) = f(b) = g(b) = e_{b}([g]) \). Therefore, \( e_{b} \) is well-defined.
06

Prove e_{b} is One-One

To prove that \( e_{b} \) is one-one (injective), assume \( e_{b}([f]) = e_{b}([g]) \). This implies \( f(b) = g(b) \), and hence \( f \sim g \) in \( R \), meaning \([f] = [g]\). Thus, \( e_{b} \) is injective.
07

Prove e_{b} is Onto

To prove that \( e_{b} \) is onto (surjective), consider any \( y \in Y \). We need to find a class \([f]\) such that \( e_{b}([f]) = y \). Define a function \( f \) such that \( f(b) = y \). Then \( e_{b}([f]) = f(b) = y \), making \( e_{b} \) surjective.

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Key Concepts

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

Function Mapping
Function mapping between sets is fundamental in mathematics. It involves creating a relationship where each element from a set, known as the domain, is assigned to an element in another set, called the co-domain. In the exercise, every function is a part of the set \( E \), mapping from set \( X \) to set \( Y \).
For example, consider a function \( f: X \rightarrow Y \). This means for every \( x \in X \), there is a unique \( y \in Y \) such that \( f(x) = y \). In our exercise, the value of \( f \) at a specific element \( b \in X \) is used to define relations among functions.
This mapping approach is crucial as it allows us to classify functions based on a single value (\( b \)) and understand their equivalence, leading to the concept of an equivalence relation.
Reflexivity
Reflexivity is a characteristic of an equivalence relation. It means every element is related to itself. Speaking plainly, if we have a set of functions \( E \), reflexivity ensures that for any function \( f \), the pair \((f, f)\) is part of the relation \( R \).
This is because, by definition, \( R \) contains pairs \((f, g)\) such that \( f(b) = g(b) \). Certainly, for any function \( f \), it has to be that \( f(b) = f(b) \), securing that each function is related to itself under \( R \).
Reflexivity helps establish the foundation for an equivalence relation, marking it out as a potential set of relationships that behave in a specific, predictable manner.
Symmetry
Symmetry in the context of relations is a fascinating concept. If a relation \( R \) on a set is symmetric, it means that if one element is related to another, then the reverse must also be true.
For our set of functions, if \( (f, g) \) is in \( R \) because \( f(b) = g(b) \), then symmetry posits that \( (g, f) \) must also be in \( R \) because \( g(b) = f(b) \).
This property makes the relation mutual; if something (like having the same value at a point) applies in one direction, it must apply back the other way as well. Symmetry is crucial, allowing us to assert equivalences without being directional, thereby maintaining equality between objects.
Transitivity
Transitivity is the property that supports chaining relations together. For a relation \( R \) to be transitive, if \( f \) is related to \( g \), and \( g \) is related to \( h \), then \( f \) should also be related to \( h \).
In formal terms, if \( (f, g) \in R \) and \( (g, h) \in R \), then it follows naturally that \( (f, h) \in R \). This is because both \( f(b) = g(b) \) and \( g(b) = h(b) \) imply \( f(b) = h(b) \) directly.
Transitivity is important as it links elements together in a sequence, ensuring that the relation holds not just between direct pairs, but across connections, indicating a consistent and orderly structure among the functions.

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Most popular questions from this chapter

Let \(f: X \rightarrow X\) be a one-one function of a set \(X\) into itself. Define a sequence of functions \(f^{0}, f^{1}, f^{2}, \ldots, f^{n}, \ldots: X \rightarrow X\) by letting \(f^{0}\) be the identity, \(f^{1}=f\), and inductively \(f^{n}(x)=f\left(f^{n-1}(x)\right)\). Prove that each of these functions is one-one. Let \(R\) be the subset of \(X \times X\) consisting of those pairs \((a, b)\) such that \(b=f^{k}(a)\) for some integer \(k\) or \(a=f^{i}(b)\) for some integer \(j\). Prove that \(R\) is an equivalence relation.

Let \(A \subset S, B \subset S\). Prove the following: (a) \(A \subset B\) if and only if \(A \cup B=B\). (b) \(A \subset B\) if and only if \(A \cap B=A\). (c) \(A \subset C(B)\) if and only if \(A \cap B=\emptyset\). (d) \(C(A) \subset B\) if and only if \(A \cup B=S\). (e) \(A \subset B\) if and only if \(C(B) \subset C(A)\). (f) \(A \subset C(B)\) if and only if \(B \subset C(A)\).

Let \(A\) be the set of all functions \(f:[a, b] \rightarrow R\) that are continuous on \([a\), \(b]\). Let \(B\) be the subset of \(A\) consisting of all functions possessing a continuous derivative on \([a, b] .\) Let \(C\) be the subset of \(B\) consisting of all functions whose value at \(a\) is 0 . Let \(d: B \rightarrow A\) be the correspondence that associates with each function in \(B\) its derivative. Is the function \(d: B\) \(\rightarrow A\) invertible? To each \(f \in A\), let \(h(f)\) be the function defined by $$ (h(f))(x)=\int_{a}^{x} f(t) d t, $$ for \(x \in[a, b]\). Verify that \(h: A \rightarrow C\). Find the function \(g: C \rightarrow A\) such that these two functions are inverse functions.

Let \(A=\left\\{a_{1}, a_{2}\right\\}\) and \(B=\left\\{b_{1}, b_{2}\right\\}\) be two sets, each having precisely two distinct elements. Let \(f: A \rightarrow B\) be the constant function such that \(f(a)=\) \(b_{1}\) for each \(a \in A\). (a) Prove that \(f^{-1}\left(f\left(\left\\{a_{1}\right\\}\right)\right) \neq\left\\{a_{1}\right\\}\). [Thus it is usually the case that \(f^{-1}(f(X))\) and \(X\) are not equal.] (b) Prove that \(f\left(f^{-1}(B)\right) \neq B\). [Thus it is usually the case that \(f\left(f^{-1}(B)\right)\) and \(B\) are not equal.] (c) Prove that \(f\left(\left\\{a_{1}\right\\} \cap\left\\{a_{2}\right\\}\right) \neq f\left(\left\\{a_{1}\right\\}\right) \cap f\left(\left\\{a_{2}\right\\}\right)\). [Thus it is usually the case that \(f\left(X \cap X^{\prime}\right)\) are \(f(X) \cap f\left(X^{\prime}\right)\) are not equal.]

Let \(f: X \rightarrow Y\) be a function from a set \(X\) onto a set \(Y\). Let \(R\) be the subset of \(X \times X\) consisting of those pairs \(\left(x, x^{\prime}\right)\) such that \(f(x)=f\left(x^{\prime}\right)\). Prove that \(R\) is an equivalence relation. Let \(\pi: X \rightarrow X / R\) be the projection. Verify that, if \(\alpha \in X / R\) is an equivalence class, to define \(F(\alpha)=f(a)\), whenever \(\alpha=\pi(a)\), establishes a well-defined function \(F: X / R \rightarrow Y\) which is oneone and onto.
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