The Timpson Lyceum · Discrete Math (2025)

Relations - Lecture 7 (Discrete Math)

Relations – Lecture 7

Overview

This lecture introduces relations as the basic set-theoretic objects that underlie familiar ideas like "less than or equal to," "divides," and "congruent to." A relation is nothing more than a set of ordered pairs, and once we adopt that view we can write notation for membership, define what it means for a relation to live "on" or "between" sets, form the inverse of a relation, and classify relations by five standard properties. These tools let us treat everyday comparisons with full rigor and set the stage for studying orderings and equivalences. By the end you should be able to test whether a set is a relation, take inverses, and decide which of the five properties any given relation has.

Key Concepts

Relation. A relation is simply a set of ordered pairs. To check whether a set is a relation, inspect its elements: if every element is an ordered pair, the set is a relation.

$\{(\text{a},\text{b}),(\text{c},\text{d})\}$ is a relation (both elements are ordered pairs).
The empty set $\varnothing$ is a relation vacuously — there is no element to violate the condition.

Membership notation. If a relation is named $R$ , we write $x\,R\,y$ to mean $(x,y)\in R$ , read " $x$ is related to $y$ ." To deny it, strike the symbol: $x\,\not\!R\,y$ means $(x,y)\notin R$ . Order matters, so $(1,2)\in R$ does not imply $(2,1)\in R$ .

Defining a relation by a rule. A relation can be specified by a condition rather than listed. For example, $Q=\{(x,y): x,y\in\mathbb{Z}\ \text{and}\ y-x\in\mathbb{N}\}.$ Here $\mathbb{Z}$ is the integers and $\mathbb{N}=\{0,1,2,3,\dots\}$ the naturals. Since $y-x\ge 0$ exactly when $x\le y$ , the relation $Q$ is the usual $\le$ on the integers — a familiar comparison rewritten as an explicit set of ordered pairs.

Relation on a set / from one set to another. Let $A$ and $B$ be sets.

$R$ is a relation on $A$ if $R\subseteq A\times A$ (every left and right entry comes from $A$ ).
$R$ is a relation from $A$ to $B$ if $R\subseteq A\times B$ (left entries from $A$ , right entries from $B$ ).

A single relation may satisfy several, one, or none of these descriptions, depending on its pairs and the sets involved. The empty relation qualifies on/from any sets, vacuously.

Inverse of a relation. The inverse $R^{-1}$ is obtained by swapping every ordered pair: $R^{-1}=\{(y,x):(x,y)\in R\}.$ The " $-1$ " is notation, not an exponent or reciprocal. Some relations equal their own inverse (e.g. $\{(1,1),(2,2)\}$ ). The inverse of $\le$ is $\ge$ . Taking the inverse twice returns the original: $(R^{-1})^{-1}=R$ .

Five properties of a relation $R$ on a set $A$ .

Reflexive — for all $x\in A$ , $x\,R\,x$ (every element related to itself).
Irreflexive — for all $x\in A$ , $x\,\not\!R\,x$ (no element related to itself). Reflexive and irreflexive are not opposites: a relation that relates some but not all elements to themselves is neither.
Symmetric — for all $x,y$ , if $x\,R\,y$ then $y\,R\,x$ .
Antisymmetric — for all $x,y$ , if $x\,R\,y$ and $y\,R\,x$ then $x=y$ .
Transitive — for all $x,y,z$ , if $x\,R\,y$ and $y\,R\,z$ then $x\,R\,z$ .

Sample classifications (intuition builders):

$=$ : reflexive, symmetric, antisymmetric, transitive.
$\le$ : reflexive, antisymmetric, transitive (not symmetric).
$<$ : irreflexive, antisymmetric, transitive (antisymmetric vacuously — you can never have both $x<y$ and $y<x$ ).
divides ( $\mid$ ) on $\mathbb{Z}$ : reflexive, transitive.
"same last name" : reflexive, symmetric, transitive.
"sibling of" : symmetric but not transitive (chaining can return to the same person).
"ancestor of," "parent of" : irreflexive; antisymmetric (vacuously for ancestor); "parent of" is not transitive.
$\varnothing$ : irreflexive, symmetric, antisymmetric, transitive — all the conditional properties hold vacuously.

Worked Example

Claim. For any relation $R$ , $(R^{-1})^{-1}=R$ .

We prove the two sets contain exactly the same ordered pairs, using the inverse rule $(a,b)\in S^{-1}\iff(b,a)\in S$ at each step.

\begin{aligned} (x,y)\in R \;&\Longrightarrow\; (y,x)\in R^{-1} \;\Longrightarrow\; (x,y)\in (R^{-1})^{-1}, \\ (x,y)\in (R^{-1})^{-1} \;&\Longrightarrow\; (y,x)\in R^{-1} \;\Longrightarrow\; (x,y)\in R. \end{aligned}

The first line shows $R\subseteq(R^{-1})^{-1}$ ; the second shows $(R^{-1})^{-1}\subseteq R$ . Since each set contains the other, $(R^{-1})^{-1}=R$ . $\qquad\blacksquare$

Summary

A relation is a set of ordered pairs; check membership of pairs to test the definition. The empty set is a relation vacuously.
$x\,R\,y$ means $(x,y)\in R$ ; because pairs are ordered, $x\,R\,y$ and $y\,R\,x$ are different claims.
$R$ is a relation on $A$ when $R\subseteq A\times A$ , and from $A$ to $B$ when $R\subseteq A\times B$ ; familiar comparisons like $\le$ are just relations defined by a rule.
The inverse $R^{-1}$ swaps every pair; the " $-1$ " is notation only, the inverse of $\le$ is $\ge$ , and $(R^{-1})^{-1}=R$ .
Every relation on a set can be tested for five properties: reflexive, irreflexive, symmetric, antisymmetric, transitive — and reflexive/irreflexive are not negations of each other.
Conditional properties (symmetric, antisymmetric, transitive) hold vacuously when no pairs trigger their hypothesis, which is why $\varnothing$ and strict orders satisfy several of them automatically.