Types of relations

Let $A$ be a set and let $R$ be a relation on $A$, meaning $R \subseteq A \times A$. Write $aRb$ (equivalently $(a,b) \in R$) to say "$a$ is related to $b$." Three properties classify how $R$ behaves.

• Reflexive — every element relates to itself. Formally, for all $a \in A$, $(a,a) \in R$.
• Symmetric — the relation never goes one way only. Formally, $(a,b) \in R \implies (b,a) \in R$ for all $a,b \in A$.
• Transitive — the relation chains. Formally, $(a,b) \in R$ and $(b,c) \in R \implies (a,c) \in R$ for all $a,b,c \in A$.

A relation satisfying all three simultaneously is an equivalence relation:

where: $A$ is the underlying set, $R \subseteq A \times A$ is the relation, and each implication ($\implies$) must hold for every choice of the elements named, not just some. When $R$ is an equivalence relation, each $a \in A$ generates an equivalence class

where: $[a]$ is the set of all elements related to $a$. These classes have a beautiful structural property: they partition $A$ — every element of $A$ lies in exactly one class, so the classes are non-empty, pairwise disjoint, and their union is all of $A$. Intuitively, $R$ groups $A$ into clusters of mutually related elements, and $[a]$ is simply the cluster containing $a$. A practical habit worth building early: to disprove one of the three properties, a single counterexample suffices — you never need to check every pair, only find the one pair that breaks the rule.

Proving that the equivalence classes of an equivalence relation $R$ on $A$ partition $A$ into non-empty, pairwise disjoint pieces whose union is all of $A$ is what turns "reflexive, symmetric, transitive" from three isolated checks into one powerful structural guarantee, and every step below uses exactly one of the three defining properties, so it doubles as a demonstration of why all three are needed together.

Step 1. Every class is non-empty and covers $A$. Take any $a \in A$. By reflexivity, $(a,a) \in R$, so $a \in [a] = \{x \in A : (x,a) \in R\}$. Hence $a$ belongs to its own class, $[a]$ is never empty, and since this holds for every $a \in A$, the union of all the classes is all of $A$.

Step 2. If two classes share an element, they are identical. Suppose $[a]$ and $[b]$ share some element $c$, so $c \in [a]$ and $c \in [b]$, meaning $(c,a) \in R$ and $(c,b) \in R$. By symmetry, $(c,a) \in R$ gives $(a,c) \in R$. By transitivity applied to $(a,c) \in R$ and $(c,b) \in R$, we get $(a,b) \in R$.

Step 3. Show $[a] = [b]$ using that $(a,b) \in R$. Take any $x \in [a]$, so $(x,a) \in R$. Combined with $(a,b) \in R$ from Step 2, transitivity gives $(x,b) \in R$, i.e. $x \in [b]$. So $[a] \subseteq [b]$. The identical argument with $a$ and $b$ swapped (using symmetry to get $(b,a) \in R$) gives $[b] \subseteq [a]$. Hence $[a] = [b]$.

Step 4. Conclude the partition. Steps 1–3 together say: every element lies in some class (non-empty, covering), and any two classes are either identical or disjoint (since sharing even one element forces them equal). That is precisely the definition of a partition of $A$ — and it required reflexivity for coverage, symmetry to flip the shared relation, and transitivity to chain it into equality.

Let $A = \{1,2,3,4\}$ and $R = \{(1,1),(2,2),(3,3),(4,4),(1,2),(2,1)\}$. Check whether $R$ is reflexive, symmetric, and transitive, and if it is an equivalence relation, find its equivalence classes.

Step 1. Reflexive? We need $(a,a) \in R$ for every $a \in A = \{1,2,3,4\}$. Scanning $R$: $(1,1), (2,2), (3,3), (4,4)$ are all present. So $R$ is reflexive.

Step 2. Symmetric? We need: whenever $(a,b) \in R$, also $(b,a) \in R$. The only non-diagonal pair is $(1,2)$, and its reverse $(2,1)$ is also in $R$. Every diagonal pair $(a,a)$ trivially reverses to itself. So $R$ is symmetric.

Step 3. Transitive? Check every chain. From $(1,2) \in R$ and $(2,1) \in R$, transitivity demands $(1,1) \in R$ — true. From $(2,1) \in R$ and $(1,2) \in R$, it demands $(2,2) \in R$ — true. All remaining chains only involve pairs of the form $(a,a)$, which close up trivially. So $R$ is transitive.

Step 4. Conclude. $R$ is reflexive, symmetric, and transitive, so it is an equivalence relation on $A$. Applying the partition proof above: $[1] = [2] = \{1,2\}$ (since $1$ and $2$ are mutually related), while $3$ and $4$ relate to nothing but themselves, so $[3] = \{3\}$ and $[4] = \{4\}$. The equivalence classes are $\{1,2\}, \{3\}, \{4\}$ — three disjoint sets whose union is $A$.

Competitive stretch

On $\mathbb{Z}$, define $aRb$ iff $5 \mid (a-b)$. Reflexive: $a-a=0=5\times 0$. Symmetric: $a-b=5k \implies b-a=5(-k)$. Transitive: $a-b=5k, b-c=5m \implies a-c=5(k+m)$. So $R$ is an equivalence relation, and $[2] = \{\ldots,-8,-3,2,7,12,17,\ldots\}$ — exactly the five classes $[0],[1],[2],[3],[4]$ partition $\mathbb{Z}$ into remainders mod 5, matching the partition proof above elements-for-elements: coverage (every integer has a remainder), and disjointness (no integer has two different remainders mod 5).

• Treating reflexivity as "obviously true" without checking it — verify $(a,a) \in R$ explicitly for every element of the stated set $A$, not just a couple.
• Confusing "looks symmetric informally" with the actual pair check — $(1,2) \in R$ does not make $R$ symmetric unless $(2,1) \in R$ is also listed.
• Forgetting that transitivity is tested only where a chain exists — if no two pairs in a finite relation actually chain together, transitivity holds vacuously (trivially true), which students often mistake for "not transitive."
• Assuming symmetric-and-transitive automatically implies reflexive — this fails whenever some element of $A$ never appears as a coordinate of any pair in $R$, a favourite exam trick.

Ask the tutor: Want to work through a trickier relation defined by a formula (like "$a R b$ if $a^2 = b^2$") to practice spotting whether it's reflexive, symmetric, and transitive?