Relations

Section3.1Relations

Definition3.1.1Relation

For sets $A,B\text{,}$ a relation from $A$ to $B$ is a subset $R$ of the Cartesian product $A \times B\text{.}$ Elements $a \in A$ and $b \in B$ are said to be related if $(a,b) \in R\text{.}$ Moreover, we often write a relation with infix notation, as $a\,R\,b\text{.}$

When $A=B\text{,}$ which is perhaps more common, we just call $R$ a relation on $A$ .

Note that any set of ordered pairs from $A$ and $B$ is a relation. Most of them we don't find particularly interesting, but others are incredibly interesting. Soon we will study some properties of interesting relations.

Example3.1.2

Let $A = \{1,2,3\}$ and $B = \{1,2,3,4,5\}$ and consider the relation $R$ from $A$ to $B$ given by

\begin{equation*} R = \{ (1,2), (1,4), (2,3), (2,5), (3,2), (3,1) \}. \end{equation*}

Then, $2\,R\,3\text{,}$ $3\,R\,2\text{,}$ $1\,R\,2$ but $2\,\not R\,1\text{.}$ Moreover, $1\,R\,4\text{,}$ but saying $4\,R\,1$ or $4\,\not R\,1$ doesn't even make sense since $4 \notin A$ and hence $(4,1) \notin A \times B\text{.}$

Example3.1.3

The “less than” relation $\lt$ is a relation on $\mathbb{R}\text{,}$ with which you are undoubtedly familiar. You probably aren't used to thinking of this as a subset of the plane $\mathbb{R}^2\text{,}$ but it can be. In particular, $x \lt y$ if and only if the point $(x,y)$ lies above the line $y = x$ in the plane. The reason we need to recognize things like less than as a set is because we are using set theory as the foundation of mathematics.

Example3.1.4

Here is an example of a relation on $\mathbb{Z}\text{,}$ which is called the divides relation. Given $a,b \in \mathbb{Z}\text{,}$ we say $a$ divides $b$ and write $a \mid b\text{,}$ if there exists some integer $k$ so that $b = ak\text{.}$ Note, although technically $\mid$ is a subset of $\mathbb{Z}^2\text{,}$ we almost never think of it that way, and it would look kind of weird to write $(2,6) \in \mid$ as opposed to $2 \mid 6\text{.}$

Definition3.1.5Domain, Range

The domain of a relation $R$ from $A$ to $B$ is the subset of $A$ given by

$\begin{equation*} \dom(R) = \{ x \in A \mid (\exists y \in B) (x\,R\,y) \}. \end{equation*}$

Similarly, the range of $R$ is the subset of $B$ given by

$\begin{equation*} \range(R) = \{ y \in B \mid (\exists x \in A) (x\,R\,y) \}. \end{equation*}$

Definition3.1.6Identity relation

The identity relation on a set $A$ is the relation $I_A := \{ (x,x) \mid x \in A \}\text{.}$ In other words, a pair of elements from $A$ are related if and only if they are equal. This could also be called the equality relation.

Definition3.1.7Inverse relation

If $R$ is a relation from $A$ to $B\text{,}$ the inverse relation $R^{-1}$ is the relation from $B$ to $A$ given by

$\begin{equation*} R^{-1} = \{ (b,a) \in B \times A \mid a\,R\,b \}. \end{equation*}$

Note that $(R^{-1})^{-1} = R.$

Exercise3.1.8

What is the inverse of the less than relation on $\mathbb{R}\text{?}$

Exercise3.1.9

Consider a relation $R$ on $\mathbb{R}\text{.}$ Since $R \subseteq \mathbb{R}^2\text{,}$ we can think of this as some geometric subset of the plane. Describe how $R$ and $R^{-1}$ are related geometrically.

Hint

If you are having trouble, consider a few very simple examples along with the less than relation. Remember, whatever the geometric transformation is, applying it twice should give you the original set back since $(R^{-1})^{-1} = R.$

Theorem3.1.10

Given a relation $R\text{,}$ we have

$\dom(R^{-1}) = \range(R)\text{;}$
$\range(R^{-1}) = \dom(R)\text{.}$

Definition3.1.11Relation Composition

Given relations $R,S$ from $A$ to $B$ and from $B$ to $C\text{,}$ respectively, we can form the composite relation $S \circ R$ from $A$ to $C$ as

$\begin{equation*} \{ (a,c) \in A \times C \mid (\exists b \in B)( a\,R\,b \wedge b\,R\,c ) \}. \end{equation*}$

Theorem3.1.12

Let $A,B,C,D$ be sets with relations $R,S,T$ from $A$ to $B\text{,}$ $B$ to $C\text{,}$ and $C$ to $D\text{,}$ respectively. Then

$(R^{-1})^{-1} = R\text{;}$
$T \circ (S \circ R) = (T \circ R) \circ S\text{;}$
$I_B \circ R = R$ and $R \circ I_A = R\text{;}$
$(S \circ R)^{-1} = R^{-1} \circ S^{-1}\text{.}$

Proof

Left as an exercise for the reader. These follow directly from the definitions.