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The composition \(\mathcal{R}_{2} \circ \mathcal{R}_{1}\) of the relations \(\mathcal{R}_{1}\) and \(\mathcal{R}_{2}\) is defined as follows: $$ \mathcal{R}_{2} \circ \mathcal{R}_{1}:=\left\\{(x, z) \mid \exists y\left(x \mathcal{R}_{1} y \wedge y \mathcal{R}_{2} z\right)\right\\} $$ In particular, if \(\mathcal{R}_{1} \subset X \times Y\) and \(\mathcal{R}_{2} \subset Y \times Z\), then \(\mathcal{R}=\mathcal{R}_{2} \circ \mathcal{R}_{1} \subset X \times Z\), and $$ x \mathcal{R} z:=\exists y\left((y \in Y) \wedge\left(x \mathcal{R}_{1} y\right) \wedge\left(y \mathcal{R}_{2} z\right)\right) $$ a) Let \(\Delta_{X}\) be the diagonal of \(X^{2}\) and \(\Delta_{Y}\) the diagonal of \(Y^{2}\). Show that if the relations \(\mathcal{R}_{1} \subset X \times Y\) and \(\mathcal{R}_{2} \subset Y \times X\) are such that \(\left(\mathcal{R}_{2} \circ \mathcal{R}_{1}=\Delta_{X}\right) \wedge\left(\mathcal{R}_{1} \circ \mathcal{R}_{2}=\right.\) \(\left.\Delta_{Y}\right)\), then both relations are functional and define mutually inverse mappings of \(X\) and \(Y\). b) Let \(\mathcal{R} \subset X^{2}\). Show that the condition of transitivity of the relation \(\mathcal{R}\) is equivalent to the condition \(\mathcal{R} \circ \mathcal{R} \subset \mathcal{R}\). c) The relation \(\mathcal{R}^{\prime} \subset Y \times X\) is called the transpose of the relation \(\mathcal{R} \subset X \times Y\) if \(\left(y \mathcal{R}^{\prime} x\right) \Leftrightarrow(x \mathcal{R} y)\) Show that a relation \(\mathcal{R} \subset X^{2}\) is antisymmetric if and only if \(\mathcal{R} \cap \mathcal{R}^{\prime} \subset \Delta_{X}\) d) Verify that any two elements of \(X\) are connected (in some order) by the relation \(\mathcal{R} \subset X^{2}\) if and only if \(\mathcal{R} \cup \mathcal{R}^{\prime}=X^{2}\)

Step by step solution

01

a) Proof that both relations are functional and define mutually inverse mappings

To show that both \(\mathcal{R}_1\) and \(\mathcal{R}_2\) are functional, we need to prove that for each \(x \in X\), there exists a unique \(y \in Y\) such that \((x, y) \in \mathcal{R}_1\), and similarly for each \(y \in Y\), there exists a unique \(x \in X\) such that \((y, x) \in \mathcal{R}_2\). Given that \((\mathcal{R}_2 \circ \mathcal{R}_1) = \Delta_X\) and \((\mathcal{R}_1 \circ \mathcal{R}_2)=\Delta_Y\), for any \(x \in X\), there is a unique \(y \in Y\) such that \((x, y) \in \mathcal{R}_1\) and \((y, x) \in \mathcal{R}_2\). Then, both \(\mathcal{R}_1\) and \(\mathcal{R}_2\) are functional. Now, we need to show that they are mutually inverse mappings of \(X\) and \(Y\). Let \(f:X \to Y\) be the function defined by \(\mathcal{R}_1\) and \(g:Y \to X\) be the function defined by \(\mathcal{R}_2\). 1) We have to show that \(f(g(y)) = y\) for all \(y \in Y\). If \((x, y) \in \mathcal{R}_1\) and \((y, x) \in \mathcal{R}_2\), then \((g(y), y) \in \mathcal{R}_1\) and \((y, g(y)) \in \mathcal{R}_2\). Thus, \(f(g(y)) = y\) for all \(y \in Y\) since \((\mathcal{R}_1 \circ \mathcal{R}_2) = \Delta_Y\). 2) We also have to show that \(g(f(x)) = x\) for all \(x \in X\). If \((x, y) \in \mathcal{R}_1\) and \((y, x) \in \mathcal{R}_2\), then \((x, f(x)) \in \mathcal{R}_2\) and \((f(x), x) \in \mathcal{R}_1\). Thus, \(g(f(x)) = x\) for all \(x \in X\) since \(( \mathcal{R}_2 \circ \mathcal{R}_1 ) = \Delta_X\). So, \(\mathcal{R}_1\) and \(\mathcal{R}_2\) define mutually inverse mappings of \(X\) and \(Y\).

02

b) Equivalence of transitivity and R 鈭� R 鈯� R

Let \(\mathcal{R} \subset X^2\). We need to show that transitivity of \(\mathcal{R}\) is equivalent to the condition \(\mathcal{R} \circ \mathcal{R} \subset \mathcal{R}\). 1) Let's assume \(\mathcal{R}\) is transitive. If \((a, b) \in \mathcal{R}\) and \((b, c) \in \mathcal{R}\), then, due to transitivity, \((a, c) \in \mathcal{R}\). It follows that \(\mathcal{R} \circ \mathcal{R} \subset \mathcal{R}\). 2) Now, let's assume \(\mathcal{R} \circ \mathcal{R} \subset \mathcal{R}\). If \((a, b) \in \mathcal{R}\) and \((b, c) \in \mathcal{R}\), then \((a, c) \in \mathcal{R} \circ \mathcal{R}\). Since \(\mathcal{R} \circ \mathcal{R} \subset \mathcal{R}\), it means that \((a, c) \in \mathcal{R}\). So, \(\mathcal{R}\) is transitive. Thus, transitivity of \(\mathcal{R}\) is equivalent to the condition \(\mathcal{R} \circ \mathcal{R} \subset \mathcal{R}\).

03

c) Antisymmetry and R 鈭� R' 鈯� 螖(X)

A relation \(\mathcal{R} \subset X^2\) is called antisymmetric if and only if for all \(x, y \in X\), if \((x, y) \in \mathcal{R}\) and \((y, x) \in \mathcal{R}\), then \(x = y\). 1) Let's assume \(\mathcal{R}\) is antisymmetric. If \((x, y) \in \mathcal{R} \cap \mathcal{R}^{\prime}\), then \((x, y) \in \mathcal{R}\) and \((y, x) \in \mathcal{R}\). Since \(\mathcal{R}\) is antisymmetric, \(x = y\). Therefore, \((x, y) \in \Delta_X\) and \(\mathcal{R} \cap \mathcal{R}^{\prime} \subset \Delta_X\). 2) Now, let's assume \(\mathcal{R} \cap \mathcal{R}^{\prime} \subset \Delta_X\). If \((x, y) \in \mathcal{R}\) and \((y, x) \in \mathcal{R}\) , then \((x, y) \in \mathcal{R} \cap \mathcal{R}^{\prime}\), and thus \((x, y) \in \Delta_X\). It implies \(x = y\). So, \(\mathcal{R}\) is antisymmetric. Thus, a relation \(\mathcal{R} \subset X^2\) is antisymmetric if and only if \(\mathcal{R} \cap \mathcal{R}^{\prime} \subset \Delta_X\).

04

d) Connection of all elements and R 鈭� R' = X虏

Let \(\mathcal{R} \subset X^2\). We need to show that any two elements of \(X\) are connected (in some order) by the relation \(\mathcal{R} \subset X^2\) if and only if \(\mathcal{R} \cup \mathcal{R}^{\prime}=X^2\). 1) Let's assume any two elements of \(X\) are connected by \(\mathcal{R}\). For all \(x, y \in X\), \((x, y) \in \mathcal{R}\) or \((y, x) \in \mathcal{R}\). If \((y, x) \in \mathcal{R}\) , then \((x, y) \in \mathcal{R}^{\prime}\). Therefore, for all \(x, y \in X\), \((x, y) \in \mathcal{R} \cup \mathcal{R}^{\prime}\). It follows that \(\mathcal{R} \cup \mathcal{R}^{\prime} = X^2\). 2) Now, let's assume \(\mathcal{R} \cup \mathcal{R}^{\prime}=X^2\). For all \(x, y \in X\), \((x, y) \in \mathcal{R} \cup \mathcal{R}^{\prime}\). If \((x, y) \in \mathcal{R}^{\prime}\), then \((y, x) \in \mathcal{R}\). Therefore, either \((x, y) \in \mathcal{R}\) or \((y, x) \in \mathcal{R}\). So, any two elements of \(X\) are connected by the relation \(\mathcal{R}\). Thus, any two elements of X are connected by the relation 饾憛 鈯� 饾憢虏 if and only if \(\mathcal{R} \cup \mathcal{R}^{\prime}=X^2\).

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Key Concepts

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

Composition of Relations

The composition of relations is a fundamental concept in set theory where you combine two relations to form a new one. The formal definition states that if you have two relations, \(\mathcal{R}_1\subset X \times Y\) and \(\mathcal{R}_2\subset Y \times Z\), their composition \(\mathcal{R}_2 \circ \mathcal{R}_1\) will consist of ordered pairs \( (x, z) \) such that there exists an element \( y \) in \( Y \) that is related to \( x \) by \(\mathcal{R}_1\) and to \( z \) by \(\mathcal{R}_2\).

This combination process is quite like a relay race, where the baton(y) is passed from a runner(x) to the next(z), ensuring a continuous link. To illustrate its practical application, consider if \(\mathcal{R}_1\) mapped students to their chosen subjects, and \(\mathcal{R}_2\) connected subjects to classrooms. The composition \(\mathcal{R}_2 \circ \mathcal{R}_1\) would tell us which students should be in which classrooms through their choice of subjects.

Functional Relations

Functional relations are a special type of relation that associates each element of the first set with a unique element of the second set. It鈥檚 a relation where no element from the first set is left out or matched with two elements from the second set.

Think of it like a strict one-on-one coffee meeting schedule - each person (from set \( X \) in this case) has a coffee meeting with exactly one other person (from set \( Y \) here). If \(\mathcal{R}_1\) and \(\mathcal{R}_2\) are such that their compositions with each other give the diagonal \(\Delta_X\) and \(\Delta_Y\) respectively, this relationship assures that the functions representing each relation are perfectly reciprocal. This situation mirrors a world where every action has an equal and opposite reaction - for every meeting scheduled, there is a reverse meeting, ensuring that each person's schedule is balanced.

Transitivity

Transitivity is a property of a relation that tells us if a link can be made indirectly through a third element. For a relation \(\mathcal{R}\subset X^2\), it is transitive if whenever an element \(a\) is related to \(b\) and \(b\) is related to \(c\), then \(a\) must also be related to \(c\).

Imagine a chain of friends where if Alice is friends with Bob, and Bob is friends with Carol, then Alice must also be friends with Carol for their friendship circle to be transitive. Mathematically, this translates to \(\mathcal{R} \circ \mathcal{R} \subset \mathcal{R}\), which ensures that the entire group is connected through shared friendships. Transitivity is a crucial concept because it allows us to infer relationships without direct observation, simplifying the understanding of complex networks.

Antisymmetry

Antisymmetry is a characteristic of a relation involving order or hierarchy, which prevents two distinct elements from being mutually related in the opposite direction. For any antisymmetric relation \(\mathcal{R}\subset X^2\), if element \(x\) is related to \(y\), and \(y\) is also related to \(x\), then it must be that \(x\) and \(y\) are, in fact, the same element. This property is crucial in defining mathematical structures like partially ordered sets where the hierarchy or pecking order is significant.

Visualize a simple corporate ladder; if Joe reports to Nancy, and Nancy also reports to Joe, they must actually be the same person (or hold the exact position), defying the corporate hierarchy. This idea in a real-world setting is absurd, showcasing why antisymmetry is important to maintain order and structure.