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For a relation \(R\) on a set \(X\), let \(R^{*}\) denote the reflexive and transitive closure of \(R\). (a) For any relation \(R\) on a set \(X\), define a relation \(\sim\) on \(X\) as follows: \(x \sim y\) if and only if \(x R^{*} y\) and \(y R^{*} x .\) Prove that \(\sim\) is an equivalence relation. (b) Let \(x_{1} \sim x_{2}\) and \(y_{1} \sim y_{2} .\) Show that \(x_{1} R^{*} y_{1}\) if and only if \(x_{2} R^{*} y_{2}\).

Short Answer

Expert verified
(a) \(\sim\) is an equivalence relation. (b) \(x_1 R^* y_1\) if and only if \(x_2 R^* y_2\).

Step by step solution

01

Understanding Reflexive and Transitive Closure

The reflexive and transitive closure of a relation \(R\) on a set \(X\) is the smallest relation \(R^*\) that includes \(R\) and is both reflexive and transitive. This means every element is related to itself (reflexive), and if an element \(a\) is related to \(b\), and \(b\) to \(c\), then \(a\) is related to \(c\) (transitive).
02

Defining the Relation \(\sim\)

The relation \(x \sim y\) is defined such that \(x \sim y\) if and only if both \(x R^* y\) and \(y R^* x\). This means that \(x\) and \(y\) are mutually reachable via the reflexive and transitive closure of \(R\).
03

Proving \(\sim\) is Reflexive

To show that \(\sim\) is reflexive, for any element \(x\) in \(X\), \(x R^* x\) holds because \(R^*\) is reflexive. Thus, \(x \sim x\).
04

Proving \(\sim\) is Symmetric

To show that \(\sim\) is symmetric, assume \(x \sim y\), meaning \(x R^* y\) and \(y R^* x\). Then, \(y R^* x\) and \(x R^* y\), implying \(y \sim x\).
05

Proving \(\sim\) is Transitive

To show that \(\sim\) is transitive, assume \(x \sim y\) and \(y \sim z\). This means \(x R^* y\), \(y R^* x\), \(y R^* z\), and \(z R^* y\). By transitivity of \(R^*\), \(x R^* z\) and \(z R^* x\), implying \(x \sim z\). Thus, \(\sim\) is an equivalence relation.
06

Showing \(x_1 R^* y_1\) implies \(x_2 R^* y_2\)

Given \(x_1 \sim x_2\) and \(y_1 \sim y_2\), we have \(x_1 R^* x_2\) and \(y_1 R^* y_2\). If \(x_1 R^* y_1\), then by the properties of \(R^*\) and the equivalence class behavior, \(x_2 R^* x_1 R^* y_1 R^* y_2\), ultimately implying \(x_2 R^* y_2\).
07

Converse: Showing \(x_2 R^* y_2\) implies \(x_1 R^* y_1\)

Assume \(x_2 R^* y_2\). Because \(x_1 R^* x_2\) and \(y_1 R^* y_2\), then \(x_1 R^* x_2 R^* y_2 R^* y_1\), confirming \(x_1 R^* y_1\). This completes the proof that \(x_1 R^* y_1\) if and only if \(x_2 R^* y_2\).

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Key Concepts

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

Reflexive and Transitive Closure
The concept of reflexive and transitive closure is key when seeking to extend a relation to include all possible connections without adding unnecessary elements. Imagine you have a set of points, and some are connected by lines. A reflexive closure ensures that each point is connected to itself. Meanwhile, a transitive closure adds connections wherever possible based on existing ones, such that if point A is connected to point B, and point B to point C, point A should also connect to point C.
This creates a network where the shortest path between any two points is utilized without adding extra connections that are not necessary. Therefore, the reflexive and transitive closure, denoted as \( R^* \), is a crucial foundation whenever the underlying structure of the data requires processing in its complete connected form. The reflexive property ensures that every element is self-related, while the transitive property builds indirect pathways, optimizing the relational graph.
Relation on a Set
A relation on a set is essentially a way to describe how the elements within that set are connected or related. A set \( X \) can have many elements, and when we talk about a relation \( R \) on \( X \), it means there are pairs of elements \( (a, b) \) such that \( a \) is related to \( b \) by \( R \).
These relations can have properties like being reflexive, symmetric, and transitive, making them more comprehensively defined. Reflexive means each element is related to itself, symmetric means if \( a \) is related to \( b \), then \( b \) is also related to \( a \), and transitive means if \( a \) is related to \( b \), and \( b \) to \( c \), then \( a \) is related to \( c \).
Defining a relation like \( x \sim y \), which only holds when both \( x R^* y \) and \( y R^* x \) are true, creates a specific kind of connection called an equivalence relation. This type of relation structures the elements into equivalence classes, grouping together mutually reachable elements.
Mutual Reachability
Mutual reachability is an interesting concept, particularly when dealing with closure properties of relations. This idea centers around the bidirectional ability to traverse between two elements within the structure of the set. When we say elements \( x \) and \( y \) are mutually reachable, it means there exists a way to go from \( x \) to \( y \) and vice versa, using the set of relations.
The relationship \( x \sim y \) is built on mutual reachability, implying that starting from \( x \), you can reach \( y \) and from \( y \) you can also travel back to \( x \).
This is a strong condition because it doesn't just involve indirect connections (like in transitive closures); it fulfills a symmetric requirement, strengthening the importance of both directional travels. Consequently, \( x \sim y \) epitomizes the essence of mutual connectivity within a closed network, reflecting the interplay of reflexivity, symmetry, and transitivity in the combined closure of any initial relation \( R \).

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Most popular questions from this chapter

Identify the equivalence classes of \(\mathrm{N}\) for the following relations: (a) \(\equiv(\bmod 4)\) (b) \(\equiv(\bmod 6)\)

Let \(R\) and \(S\) be equivalence relations on a set \(X\). (a) Show that \(R \cap S\) is an equivalence relation. (b) Show by example that \(R \cup S\) need not be an equivalence relation. (c) Show that \((R \cup S)^{*}\), the reflexive and transitive closure of \(R \cup S\), is the smallest equivalence relation containing both \(R\) and \(S\).

Define a binary relation \(R\) on \(\mathbb{R}\) as \(\\{(x, y) \in \mathbb{R} \times \mathbb{R}: \sin (x)=\sin (y)\\}\). Prove that \(R\) is an equivalence relation. What are its equivalence classes?

(a) For \(k, n_{1}, n_{2}, m_{1}, m_{2} \in \mathbb{N},\) show that if $$ n_{1} \equiv n_{2}(\bmod k) $$ and $$ m_{1}=m_{2}(\bmod k) $$ then $$ n_{1}+m_{1} \equiv n_{2}+m_{2}(\bmod k) $$ and $$ n_{1} \cdot m_{1} \equiv n_{2} \cdot m_{2}(\bmod k) $$ (b) Part (a) says that if we take two equivalence classes \([m]\) and \([n]\), then we can unambiguously define \([m]+[n]\) and \([m] \cdot[n]\). Pick any \(m_{1} \in[m]\) and any \(n_{1} \in[n],\) and define $$ [m]+[n]=\left[m_{1}+n_{1}\right] $$ and $$ [m] \cdot[n] \equiv\left[m_{1} \cdot n_{1}\right] $$ The definition is unambiguous since it doesn't matter which \(m_{1}\) and \(n_{1}\) we pick. Find the addition and multiplication tables for the equivalence classes of \(\equiv(\bmod 4)\) and \(\equiv(\bmod 5) .\) (Hint: For both \(\equiv(\bmod 4)\) and \(\equiv(\bmod 5),\) your answer should include $$ [0]+[0] \equiv[0],[0]+[1] \equiv[1],[0] \cdot[0] \equiv\\{0] $$ and $$ [1] \cdot[1] \equiv[1] $$ but, for \(\equiv(\bmod 4)\). $$ [2]+[2] \equiv[0] $$ whereas, that will be false for \(\equiv(\bmod 5) .\)

Since relations are sets, it is possible to define union, intersection, relative complement, and absolute complement on pairs of relations. A natural question is which properties of the original relations still hold for the resulting new relation. Fill in the following table with \(Y / N,\) representing YES and NO, respectively. If the entry is \(N,\) find an example that shows the property is not preserved under the operation. For instance, enter a \(Y\) in the first row, second column, if the intersection of two reflexive relations is still reflexive; otherwise, enter an \(N\). $$\begin{array}{|l|l|l|l|l|}\hline & \text { Union } & \text { Intersection } & \begin{array}{l}\text { Relative } \\\\\text { Complement }\end{array} & \begin{array}{l}\text { Absolute } \\\\\text { Complement }\end{array} \\\\\hline \text { Reflexive } & & & & \\\\\hline \text { Irrefiexive } & & & & \\\\\hline \text { Symmetric } & & & & \\\\\hline \text { Antisymmetric } & & & & \\\\\hline \text { Transitive } & & & & \\\\\hline\end{array}$$

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