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Relations

Binary relations

Definition: a binary relation R between the sets S and T is a subset of the Cartesian product S \times T.

  • If (a,b) \in R then a is in relation R to b, denoted by aRb.
  • The set S is called the domain of the relation R and the set T the codomain.
  • If S=T then R is a relation on S.
  • This definition can be expanded to n-ary relations.

Definition: let R be a relation from a set S to a set T. Then for each element a \in S we define [a]_R to be the set

[a]_R := {b \in T ;|; aRb}.

This set is called the ($R$-) image of a.

For b \in T the set

_R[b] := {a \in S ;|; aRb}

is called the ($R$-) pre-image of B or $R$-fiber of b.


Relations between finite sets can be described using matrices.

Definition: if S = \{s_1, \dots, s_n\} and T = \{t_1, \dots, t_m\} are finite sets and R \subseteq S \times T is a binary relation, then the adjacency matrix A_R of the relation R is the n \times n matrix whose rows are indexed by S and columns by T defined by

A_{s,t} = \begin{cases} 1 &\text{ if } (s,t) \in R, \ 0 &\text{ otherwise}. \end{cases}

For example, the adjacency matrix of relation \leq on the set \{1,2,3,4,5\} is the upper triangular matrix

\begin{pmatrix} 1 & 1 & 1 & 1 & 1 \ 0 & 1 & 1 & 1 & 1 \ 0 & 0 & 1 & 1 & 1 \ 0 & 0 & 0 & 1 & 1 \ 0 & 0 & 0 & 0 & 1\end{pmatrix}


Some relations have special properties

Definitions: let R be a relation on a set S. Then R is called

  • Reflexive if \forall x \in S there is (x,x) \in R.
  • Irreflexive if \forall x \in S there is (x,x) \notin R.
  • Symmetric if \forall x,y \in S there is that xRy \implies yRx.
  • Antisymmetric if \forall x,y \in S there is that xRy \land yRx \implies x = y.
  • Transitive if \forall x,y,z \in S there is that xRy \land yRz \implies xRz.

Equivalence relations

Definition: a relation R on a set S is called an equivalence relation on S if and only if it is reflexive, symmetric and transitive.


Lemma: let R be an equivalence relation on a set S. If b \in [a]_R, then [b]_R = [a]_R.

??? note "Proof:"

Suppose $b \in [a]_R$, therefore $aRb$. If $c \in [b]_R$, then $bRc$ and as $aRb$ there is transitivity $aRc$. In particular $[b]_R \subseteq [a]_R$. By symmetry of $R$, $aRb \implies bRa$ and hence $a \in [b]_R$, obtaining $[a]_R \subseteq [b]_R$.

Definition: let R be an equivalence relation on a set S. Then the sets [s]_R where s \in S are called the $R$-equivalence classes on S. The set of $R$-equivalence classes is denoted by S/R.


Theorem: let R be an equivalence relation on a set S. Then the set S/R of $R$-equivalence classes partitions the set S.

??? note "Proof:"

Let $\Pi_R$ be the set of $R$-equivalence classes. Then by reflexivity of $R$ we find that each element $a \in S$ is inside the class $[a]_R$ of $\Pi_R$. If an element $a \in S$ is in the classes $[b]_R$ and $[c]_R$ of $\Pi_R$, then by the previous lemma we find $[b]_R = [a]_R$ and $[c]_R = [a]_R$. Then $[b]_R = [c]_R$, therefore each element $a \in S$ is inside a unique member of $\Pi_R$, which therefore is a partition of $S$.

Composition of relations

If R_1 and R_2 are two relations between a set S and T, new relations can be formed between S and T by taking the intersection R_1 \cap R_2, the union R_1 \cup R_2 or the complement R_1 \backslash R_2. Furthermore a relation R^\top from T to S can be considered as the relation \{(t,s) \in T \times S \;|\; (s,t) \in R\} and the identity relation from T to S is given by I = \{(s, t) \in S \times T \;|\; s = t\}

Another way of making new relations out of existing ones is by taking the composition.

Definition: if R_1 is a relation between S and T and R_2 is a relation between T and U then the composition R = R_1;R_2 is the relation between S and U defined by sRu for s \in S and u \in U, if and only if there is a t \in T with sR_1t and tR_2u.


Proposition: suppose R_1 is relation from S to T, R_2 a relation from T to U and R_3 a relation from U to V. Then R_1;(R_2;R_3) = (R_1;R_2);R_3. Composing relations is associative.

??? note "Proof:"

Suppose $s \in S$ and $v \in V$ with $sR_1;(R_2;R_3)v$. Then a $t \in T$ with $sR_1t$ and $t(R_2;R_3)v$ can be found. Then there is also a $u \in U$ with $tR_2u$ and $uR_3v$. For this $u$ there is $sR_1;R_2u$ and $uR_3v$ and hence $s(R_1;R_2);R_3v$.

Similarly, if $s \in S$ and $v \in V$ with $s(R_1;R_2);R_3v$. Then a $u \in U$ with $s(R_1;R_2)u$ and $uR_3v$ can be found. Then there is also a $t \in T$ with $sR_1t$ and $tR_2u$. For this $t$ there is $tR_2;R_3u$ and $sR_1t$ and hence $sR_1;(R_2;R_3)v$.

Transitive closure

Lemma: let \ell be a collection of relations R on a set S. If all relations R in \ell are transitive, reflexive or symmetric then the relation \bigcap_{R \in \ell} R is also transitive, reflexive or symmetric respectively.

??? note "Proof:"

Let $\bar R = \bigcap_{R \in \ell} R$. Suppose all members of $\ell$ are transitive. Then for all $a,b,c \in S$ with $a \bar R b$ and $b \bar R c$ there is $aRb$ and $bRc$ for all $R \in \ell$. Thus by transitivity of each $R \in \ell$ there is also $aRc$ for each $R \in \ell$. Thus there is $a \bar R c$. Hence $\bar R$ is also transitive.

Proof for symmetric relation will follow.

Proof for reflexive relation will follow.

The above lemma makes it possible to define the reflexive, symmetric or transitive closure of a relation R on a set S. It is the smallest reflexive, symmetric or transitive relation containing R.

For example suppose R = \{(1,2), (2,2), (2,3), (5,4)\} is a relation on S = \{1, 2, 3, 4, 5\}.

The reflexive closure of R is then the relation

\big{(1,1), (1,2), (2,2), (2,3), (3,3), (4,4), (5,5), (5,4) \big},

the symmetric closure of R is then the relation

\big{ (1,2), (2,1), (2,2), (2,3), (3,2), (4,5), (5,4) \big},

and the transitive clusure of R is then the relation

{(1,2), (1,3), (2,2), (2,3), (5,4)}.

It may be observed that the reflexive closure of R equals the relation I \cup R and the symmetric closure equals R \cup R^\top. For the transitive closure there is:

Proposition: \bigcup_{n > 0} R^n is the transitive closure of the relation R on a set S.

??? note "Proof:"

Define $\bar R = \bigcup_{n>0} R^n$, to show that $\bar R$ is the least transitive relation containing $R$, $\bar R$ must contain $R$, must be transitive and must be the smallest set with both of those properties. 

Since $R \subseteq \bar R$, $\bar R$ contains all of the $R^i, i \in \mathbb{N}$, so in particular $\bar R$ contains $R$.

If $(s_1, s_2), (s_2, s_3) \in \bar R$, then $(s_1, s_2) \in R^j$ and $(s_2, s_3) \in R^k$ for some $j,k$. Since composition is [associative](#composition-of-relations), $R^{j+k} = R^j ; R^k$ and hence $(s_1, s_3) \in R^{j+k} \subseteq \bar R$. 

We claim that if $T$ is any transitive relation containing $R$, then $\bar R \subseteq T$. By taking $R^n \subseteq \bar R \subseteq T \; \forall n \in \mathbb{N}$ . 

:   We first check for $n=1$

    $$
        R^1 = R \subseteq T.
    $$

:   Now suppose that for some $k \in \mathbb{N}$ we have $R^k \subseteq T$. Then by assumption $R^{k+1} \subseteq T$. Let $(s_1, s_3) \in R^{k+1} = R^k ; R$, then $(s_1, s_2) \in R$ and $(s_2, s_3) \in R^k$ for some $s_2$. Hence $(s_1, s_2), (s_2, s_3) \in T$ and by transitivity of $T$, $(s_1, s_3) \in T$. 

Hence if the claim holds for some $k \in \mathbb{N}$ then it also holds for $k+1$. The principle of natural induction implies now that $\forall n \in \mathbb{N}$ we have $R^n \subseteq \bar R \subseteq T$.