441 lines
13 KiB
Markdown
441 lines
13 KiB
Markdown
Page 512
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**Definition**
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Let $R$ be a relation from $A$ to $B$. Define the inverse relation $R^{-1}$ from
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$B$ to $A$ as follows:
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$$ R^{-1} = \{(y, x) \in B \times A | (x, y) \in R\} $$
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---
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Page 513
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**Definition**
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A **relation on a set** A is a relation from $A$ to $A$.
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---
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Page 514
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**Definition**
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Given sets $A_1, A_2, \dots, A_n$ an **$n$-ary relation** $R$ on
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$A_1 \times A_2 \times \cdots \times A_n$ is a subset of
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$A_1 \times A_2 \times \cdots \times A_n$. The special cases of $2$-ary,
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$3$-ary, and $4$-ary relations are called **binary**, **ternary**, and
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**quarternary relations**, respectively.
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---
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Page 518
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**Definition**
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Let $R$ be a relation on a set $A$.
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1. $R$ is **reflexive** if, and only if, for every $x \in A, x R x$.
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2. $R$ is **symmetric** if, and only if, for every
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$x, y \in A, \text{ if } x R y \text{ then } y R x$.
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3. $R$ is **transitive** if, and only if, for every
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$x, y, z \in A, \text{ if } x R y \text{ and } y R z \text{ then } x R z$.
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---
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Page 523
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**Proof of Reflexivity:**
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Suppose $m$ is a particular but arbitrarily chosen integer. _[We must show that
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$m T m$.]_ Now $m - m = 0$. But $3 | 0$ since $0 = 3 \cdot 0$. Hence
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$3 | (m - m)$. Thus, by definition of $T$, $m T m$ _[as was to be shown]_.
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---
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Page 524
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**Proof of Symmetry:**
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Suppose $m$ and $n$ are particular but arbitrarily chosen integers that satisfy
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the condition $m T n$. _[We must show that $n T m$.]_ By definition of $T$,
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since $m T n$ then $3 | (m - n)$. By definition of "divides", this means that
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$m - n = 3k$, for some integer $k$. Multiplying both sides by $-1$ gives
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$n - m = 3(-k)$. Since $-k$ is an integer, this equation shows that
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$3 | (n - m)$. Hence, by definition of $T$, $n T m$ _[as was to be shown]_.
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---
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Page 524
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**Proof of Transitivity:**
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Suppose $m$, $n$, and $p$ are particular but arbitrarily chosen integers that
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satisfy the condition $m T n$ and $n T p$. _[We must show that $m T p$.]_ By
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definition of $T$, since $m T n$ and $n T p$, then $3 | (m - n)$ and
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$3 | (n - p)$. By definition of "divides", this means that $m - n = 3r$ and
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$n - p = 3s$, for some integers $r$ and $s$. Adding the two equations gives
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$(m - n) + (n - p) = 3r + 3s$, and simplifying gives that $m - p = 3(r + s)$.
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Since $r + s$ is an integer, this equation shows that $3 | (m - p)$. Hence, by
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definition of $T$, $m T p$ _[as was to be shown]_.
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---
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Page 525
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**Definition**
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Let $A$ be a set and $R$ a relation on $A$. The **transitive closure** of $R$ is
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the relation $R^t$ on $A$ that satisfies the following three properties:
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1. $R^t$ is transitive.
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2. $R \subseteq R^t$.
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3. If $S$ is any other transitive relation that contains $R$, then
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$R^t \subseteq S$.
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---
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Page 529
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**Definition**
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Given a partition of a set $A$, the **relation induced by the partition**, $R$,
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is defined on $A$ as follows: For every $x, y \in A$,
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$$ x R y \Leftrightarrow \text{ there is a subset } A_i \text{ of the partition such that both } x \text{ and } y \text{ are in } A_i $$
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---
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Page 530
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**Theorem 8.3.1**
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Let $A$ be a set with a partition and let $R$ be the relation induced by the
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partition. Then $R$ is reflexive, symmetric, and transitive.
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**Proof:**
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Suppose $A$ is a set with a partition. In order to simplify notation, we assume
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that the partition consists of only a finite number of sets. The proof for an
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infinite partition is identical except for notation. Denote the partition
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subsets by
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$$ A_1, A_2, \dots, A_n $$
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Then $A_i \cap A_j = \emptyset$ whenever $i \neq j$, and
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$A_1 \cup A_2 \cup A_3 \cdots \cup A_n = A$. The relation $R$ induced by the
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partition is defined as follows: For every $x, y \in A$,
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$$ x R y \Leftrightarrow \text{ there is a set } A_i \text{ of the partition such that } x \in A_i \text{ and } y \in A_i $$
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_[**Idea for the proof of reflexivity:** For $R$ to be reflexive means that each
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element of $a$ is related by $R$ to itself. But by definition of $R$, for an
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element $x$ to be related to itself means that $x$ is in the same subset of the
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partition itself. Well, if $x$ is in some subset of the partition, then it is
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certainly in the same subset as itself. And $x$ is in some subset of the
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partition because the union of the subsets of the partition is all of $A$. This
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reasoning is formalized as follows.]_
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**Proof that $R$ is reflexive:**
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Suppose $x \in A$. Since $A_1, A_2, \dots A_n$ is a partition of $A$, it follows
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that $x \in A_i$, for for some $i$, and so the statement
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there is a set $A_i$ of the partition such that $x \in A_i$ and $x \in A_i$
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is true. Thus by definition of $R$, $x R x$.
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_[**Idea for the proof of symmetry:** For $R$ to be symmetric means that any
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time one element is related to a second, then the second is related to the
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first. Now for one element $x$ to be related to a second element $y$ means that
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$x$ and $y$ are in the same subset of the partition. But if this is the case,
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then $y$ is in the same subset of the partition as $x$, so $y$ is related to $x$
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by definition of $R$. This reasoning is formalized as follows.]_
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**Proof that $R$ is symmetric:**
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Suppose $x$ and $y$ are elements of $A$ such that $x R y$. Then there is a
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subset $A_i$ of the partition such that $x \in A_i$ and $y \in A_i$ by
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definition of $R$. It follows that the statement
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there is a subset $A_i$ of the partition such that $y \in A_i$ and $x \in A_i$
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is also true. Hence, by definition of $R$, $y R x$.
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_[**Idea for the proof of transitivity:** For $R$ to be transitive means that
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any time one element of $A$ is related by $R$ to a second and that second is
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related to a third, then the first element is related to the third. But for one
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element to be related to another means that there is a subset of the partition
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that contains both. So suppose $x$, $y$, and $z$ are elements such that $x$ is
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in the same subset as $y$ and $y$ is in the same subset as $z$. Must $x$ be in
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the same subset as $z$? Yes, because the subsets 9f the partition are mutually
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disjoint. Since the subset that contains $x$ and $y$ has an element in common
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with the subset that contains $y$ and $z$ (namely, $y$), the two subsets are
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equal. But this means that $x$, $y$, and $z$ are all in the same subset, and so,
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in particular, $x$ and $z$ are in the same subset. Hence $x$ is related by $R$
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to $z$. This reasoning is formalized as follows.]_
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**Proof that $R$ is transitive:**
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Suppose $x$, $y$, and $z$ are in $A$ and $x R y$ and $y R z$. By definition of
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$R$, there are subsets $A_i$ and $A_j$ of the partition such that
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$$ x \text{ and } y \text{ are in } A_i \quad \text{ and } \quad y \text{ and } z \text{ are in } A_j $$
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Suppose $A_i \neq A_j$. _[We will deduce a contradiction.]_ Then
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$A_i \cap A_j = \emptyset$ since $\{A_1, A_2, A_3, \dots, A_n\}$ is a partition
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of $A$. But $y$ is in $A_i$ and $y$ is in $A_j$ also. Hence
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$A_i \cap A_j \neq \emptyset$. _[This contradicts the statement that
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$A_i \cap A_j = \emptyset$.]_ Thus $A_i = A_j$. It follows that $x$, $y$, and
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$z$ are all in $A_i$, and so, in particular,
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$$ x \text{ and } z \text{ are in } A_i $$
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Thus $x R z$ by definition of $R$.
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---
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Page 531
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**Definition**
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Let $A$ be a set and $R$ a relation on $A$. $R$ is an **equivalence relation**
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if, and only if, $R$ is reflexive, symmetric, and transitive.
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---
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Page 533
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**Definition**
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Suppose $A$ is a set and $R$ is an equivalence relation on $A$. For each element
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$a$ in $A$, the **equivalence class of $a$**, denoted $[a]$ and called the
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**class of $a$** for short, is the set of all elements $x$ in $A$ such that $x$
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is related to $a$ by $R$.
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In symbols:
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$$ [a] = \{x \in A | x R a\} $$
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---
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Page 536
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**Lemma 8.3.2**
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Suppose $A$ is a set, $R$ is an equivalence relation on $A$, and $a$ and $b$ are
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elements of $A$. If $a R b$, then $[a] = [b]$.
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---
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Page 536
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**Proof of Lemma 8.3.2**
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Let $A$ be a set, let $R$ be an equivalence relation on $A$, and suppose
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$$ a \text{ and } b \text{ are elements of } A \text{ such that } a R b $$
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_[We must show that $[a] = [b]$.]_
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**Proof that $[a] \subseteq [b]$:**
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Let $x \in [a]$. _[We must show that $x \in [b]$.]_
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Since
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$$ x \in [a] $$
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then
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$$ x R a $$
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by definition of class. But
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$$ a R b $$
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by hypothesis. Thus, by transitivity of $R$,
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$$ x R b $$
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Hence
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$$ x \in [b] $$
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by definition of class. _[This is what was to be shown.]_
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**Proof that $[b] \subseteq [a]$.
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Let $x \in [b]$. _[We must show that $x \in [a]$.]_
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Since
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$$ x \in [b] $$
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then
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$$ x R b $$
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by definition of class. Now
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$$ a R b $$
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by hypothesis. Thus, since $R$ is symmetric,
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$$ b R a $$
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also. Then, since $R$ is transitive and $x R b$ and $b R a$,
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$$ x R a $$
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Hence,
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$$ x \in [a] $$
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by definition of class. _[This is what was to be shown.]_
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Since $[a] \subseteq [b]$ and $[b] \subseteq [a]$, it follows that $[a] = [b]$
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by definition of set equality.
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---
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Page 537
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**Lemma 8.3.3**
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If $A$ is a set, $R$ is an equivalence relation on $A$, and $a$ and $b$ are
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elements of $A$, then
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$$ \text{either } [a] \cap [b] = \emptyset \quad \text{ or } \quad [a] = [b] $$
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---
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Page 537
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**Proof of Lemma 8.3.3**
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Suppose $A$ is a set, $R$ is an equivalence relation on $A$, $a$ and $b$ are
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elements of $A$, and
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$$ [a] \cap [b] \neq \emptyset $$
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_[We must show that $[a] = [b]$.]_
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Since $[a] \cap [b] \neq \emptyset$, there exists an element $x$ in $A$ such
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that $x \in [a] \cap [b]$. By definition of intersection,
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$$ x \in [a] \quad \text{ and } \quad x \in [b]$$
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and so
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$$ x R a \quad \text{ and } \quad x R b $$
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by definition of class. Since $R$ is symmetric _[being an equivalence relation]_
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and $x R a$, then $a R x$. But $R$ is also transitive _[since it is an
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equivalence relation]_, and so, since $a R x$ and $x R b$,
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$$ a R b $$
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Now $A$ and $b$ satisfy the hypothesis of Lemma 8.3.2. Hence, by that lemma,
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$$ [a] = [b] $$
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_[as was to be shown]._
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---
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Page 537
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**Theorem 8.3.4 The Partition Induced by an Equivalence Relation**
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If $A$ is a set and $R$ is an equivalence relation on $A$, then the distinct
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equivalence classes of $R$ form a partition of $A$; that is, the union of the
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equivalence classes is all of $A$, and the intersection of any two distinct
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classes is empty.
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---
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Page 538
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**Proof of Theorem 8.3.4**
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Suppose $A$ is a set and $R$ is an equivalence relation on $A$. For notational
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simplicity, we assume that $R$ has only a finite number of distinct equivalence
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classes, which we denote
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$$ A_1, A_2, \dots, A_n $$
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where $n$ is a positive integer. (When the number of classes is infinite, the
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proof is identical except for notation.)
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**Proof that $A = A_1 \cup A_2 \cup \cdots \cup A_n$:**
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_[We must show that $A \subseteq A_1 \cup A_2 \cup \cdots \cup A_n$ and that
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$A_1 \cup A_2 \cup \cdots \cup A_n \subseteq A$.]_
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To show that $A \subseteq A_1 \cup A_2 \cup \cdots \cup A_n$, suppose $x$ is any
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element of $A$. _[We must show that $x \in A_1 \cup A_2 \cup \cdots A_n$.]_ By
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reflexivity of $R$, $x R x$. And this implies that $x \in [x]$ by definition of
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class. Since $x$ is in _some_ equivalence class, it must be in one of the
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distinct equivalence classes $A_1, A_2, \dots$, or $A_n$. Thus $x \in A_i$ for
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some index $i$, and hence $x \in A_1 \cup A_2 \cup \cdots \cup A_n$ by
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definition of union _[as was to be shown]_.
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To show that $A_1 \cup A_2 \cup \cdots \cup A_n \subseteq A$, suppose
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$x \in A_1 \cup A_2 \cup \cdots \cup A_n$. _[We must show that $x \in A$.]_ Then
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$x \in A_i$ for some $i = 1, 2, \dots, n$, by definition of union. Now each
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$A_i$ is an equivalence class of $R$, and equivalence classes are subsets of
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$A$. Hence $A_i \subseteq A$ and so $x \in A$ _[as was to be shown]._
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Since $A \subseteq A_1 \cup A_2 \cup \cdots A_n$ and
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$A_1 \cup A_2 \cup \cdots \cup A_n \subseteq A$, then by definition of set
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equality, $A = A_1 \cup A_2 \cup \cdots \cup A_n$.
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**Proof that the distinct classes of $R$ are mutually disjoint:**
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Suppose that $A_i$ and $A_j$ are any two distinct equivalence classes of $R$.
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_[We must show that $A_i$ and $A_j$ are disjoint.]_ Since $A_i$ and $A_j$ are
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distinct, then $A_i \neq A_j$. And since $A_i$ and $A_j$ are equivalence classes
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of $R$, there must exist elements $a$ and $b$ in $A$ such that $A_i = [a]$ and
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$A_j = [b]$.
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By Lemma 8.3.3,
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$$ \text{either } [a] \cap [b] = \emptyset \quad \text{ or } \quad [a] = [b]$$
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Now $[a] \neq [b]$ because $A_i \neq A_j$, and hence $[a] \cap [b] = \emptyset$.
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Thus $A_i \cap A_j = \emptyset$, and so $A_i$ and $A_j$ are disjoint _[as was to
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be shown]._
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---
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Page 540
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**Definition**
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Suppose $R$ is an equivalence relation on a set $A$ and $S$ is an equivalence
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class of $R$. A **representative** of the class $S$ is any element $a$ such that
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$[a] = S$.
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--
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Page 541
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**Definition**
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Let $m$ and $n$ be integers and let $d$ be a positive integer. We say that **$m$
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is congruent to $n$ modulo $d$** and write
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$$ m = n (\mod d) $$
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if, and only if,
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$$ d | (m - n) $$
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Symbolically:
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$$ m \equiv n(\mod d) \Leftrightarrow d | (m - n) $$
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