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Chapter 2: Problem 12

Suppose that \(\mathcal{R}\) is a reflexive, symmetric binary relation on a set \(A\). Show that the transitive closure \(\mathcal{R}^{*}\) is an equivalence relation.

Short Answer

Expert verified
Since \(\mathcal{R}^{*}\) is reflexive, symmetric, and transitive, it is an equivalence relation.

Step by step solution

01

Understand the Definitions

An equivalence relation is a binary relation that is reflexive, symmetric, and transitive. Given that \(\mathcal{R}\) is reflexive and symmetric, we need to show that \(\mathcal{R}^{*}\) satisfies all three properties to be an equivalence relation.
02

Establish Reflexivity of \(\mathcal{R}^{*}\)

Since \(\mathcal{R}\) is reflexive, for every element \(a \in A\), \((a, a) \in \mathcal{R}\). In a transitive closure, reflexivity is automatically preserved, thus \(\forall a \in A, (a, a) \in \mathcal{R}^{*}\). Therefore, \(\mathcal{R}^{*}\) is reflexive.
03

Establish Symmetry of \(\mathcal{R}^{*}\)

Since \(\mathcal{R}\) is symmetric, for any \((a, b) \in \mathcal{R}\), \((b, a) \in \mathcal{R}\). In the transitive closure, symmetric pairs will be preserved, so for all pairs \((a, b) \in \mathcal{R}^{*}\), \((b, a) \in \mathcal{R}^{*}\) too, which means \(\mathcal{R}^{*}\) is symmetric.
04

Establish Transitivity of \(\mathcal{R}^{*}\)

The transitive closure \(\mathcal{R}^{*}\) is defined as the smallest transitive relation that contains \(\mathcal{R}\). By definition, \(\mathcal{R}^{*}\) is transitive because any sequence of related elements in \(\mathcal{R}\) will have their closure in \(\mathcal{R}^{*}\). Hence, for any \((a, b)\) and \((b, c)\) in \(\mathcal{R}^{*}\), we have \((a, c)\) in \(\mathcal{R}^{*}\).
05

Conclude that \(\mathcal{R}^{*}\) is an Equivalence Relation

Having proven that \(\mathcal{R}^{*}\) satisfies reflexivity, symmetry, and transitivity, it follows that \(\mathcal{R}^{*}\) is indeed an equivalence relation.

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Key Concepts

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

Transitive Closure
The transitive closure of a relation is an essential concept when dealing with equivalence relations. If you have a relation \( \mathcal{R} \) on a set \( A \), the transitive closure, denoted \( \mathcal{R}^* \), encompasses \( \mathcal{R} \) and ensures that it is transitive. In simpler terms, transitivity means if there is a direct path from element \( a \) to \( b \) and from \( b \) to \( c \), then there should also be a direct path from \( a \) to \( c \).
Here鈥檚 how it works:
  • Start with the original pairs in \( \mathcal{R} \).
  • Add pairs to make the relation transitive: if \( (a, b) \) and \( (b, c) \) are in \( \mathcal{R} \), then \( (a, c) \) should also be in \( \mathcal{R}^* \).
  • Continue this process until no more pairs can be added.
Transitive closure is about completing the chain of relations, making it a crucial step in proving if a relation is an equivalence relation.
Reflexive Relation
A reflexive relation is a straightforward yet important concept in the study of relations. For a relation \( \mathcal{R} \) on a set \( A \) to be reflexive, every element must be related to itself. This means that for every \( a \) in \( A \), the pair \( (a, a) \) should exist in the relation.
Key features of reflexivity include:
  • Every element has a loop back onto itself.
  • This property is inherently preserved in the transitive closure of a relation.
Why does reflexivity matter? In the context of equivalence relations, reflexivity ensures that all elements are comparable to themselves, forming the foundation needed to develop further properties such as symmetry and transitivity. Without reflexivity, a relation can't qualify as an equivalence relation.
Symmetric Relation
Symmetry in relations adds a level of reciprocal respect between the elements of a set. A relation \( \mathcal{R} \) on a set \( A \) is symmetric if whenever an element \( a \) is related to an element \( b \), \( b \) is also related to \( a \). In more formal terms, for all \( (a, b) \) in \( \mathcal{R} \), \( (b, a) \) should also be in \( \mathcal{R} \).
Characteristics of symmetric relations include:
  • For each connection in one direction, there must be a corresponding connection in the opposite direction.
  • This quality must be maintained in the transitive closure to ensure symmetry in \( \mathcal{R}^* \).
Maintaining symmetry is crucial in the structure of equivalence relations as it ensures that if one element can relate to another, the relationship is mutual. Symmetry upholds the balance of relationships within the set, essential for defining a full equivalence scenario.

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Most popular questions from this chapter

Consider the functions from \(\mathbb{Z}\) to \(\mathbb{Z}\) which are defined by the following formulas. Decide whether each function is onto and whether it is one-to-one; prove your answers. a) \(f(n)=2 n\) b) \(g(n)=n+1\) c) \(h(n)=n^{2}+n+1\) d) \(s(n)=\left\\{\begin{array}{ll}\mathrm{n} / 2, & \text { if } n \text { is even } \\ (\mathrm{n}+1) / 2, & \text { if } n \text { is odd }\end{array}\right.\)

Let \(A, B\), and \(C\) be sets. Simplify each of the following expressions. In your answer, the complement operator should only be applied to the individual sets \(A, B\), and \(C\). a) \(\overline{A \cup B \cup C}\) b) \(\overline{A \cup B \cap C}\) c) \(\overline{\overline{A \cup B}}\) d) \(\overrightarrow{B \cap \bar{C}}\) e) \(\overrightarrow{A \cap \overline{B \cap \bar{C}}}\) f) \(A \cap \overline{A \cup B}\)

Suppose that \(A, B\), and \(C\) are finite sets which are pairwise disjoint. (That is, \(A \cap B=A \cap C=B \cap C=\emptyset .\) ) Express the cardinality of each of the following sets in terms of \(|A|,|B|\), and \(|C|\). Which of your answers depend on the fact that the sets are pairwise disjoint? a) \(\mathcal{P}(A \cup B)\) b) \(A \times\left(B^{C}\right)\) c) \(\mathcal{P}(A) \times \mathcal{P}(C)\) d) \(A^{B \times C}\) e) \((A \times B)^{C}\) f) \(\mathcal{P}\left(A^{B}\right)\) g) \((A \cup B)^{C}\) h) \((A \cup B) \times A\) i) \(A \times A \times B \times B\)

Use induction on the cardinality of \(B\) to show that for any finite sets \(A\) and \(B,\left|A^{B}\right|=|A|^{|B|}\). (Hint: For the case where \(B \neq \emptyset\), choose \(x \in B\), and divide \(A^{B}\) into classes according to the value of \(f(x) .\)

Recall that the symbol \(\oplus\) denotes the logical exclusive or operation. If \(A\) and \(B\) sets, define the set \(A \triangle B\) by \(A \Delta B=\\{x \mid(x \in A) \oplus(x \in B)\\} .\) Show that \(A \triangle B=(A \backslash B) \cup(B \backslash A) .(A \triangle B\) is known as the symmetric difference of \(A\) and \(B\).)
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