# Equivalence of Definitions of Equivalence Relation

## Theorem

Let $\RR$ be a relation on a set $S$.

The following definitions of the concept of Equivalence Relation are equivalent:

### Definition 1

Let $\RR$ be:

$(1): \quad$ reflexive
$(2): \quad$ symmetric
$(3): \quad$ transitive

Then $\RR$ is an equivalence relation on $S$.

### Definition 2

$\RR$ is an equivalence relation if and only if:

$\Delta_S \cup \RR^{-1} \cup \RR \circ \RR \subseteq \RR$

where:

$\Delta_S$ denotes the diagonal relation on $S$
$\RR^{-1}$ denotes the inverse relation
$\circ$ denotes composition of relations

## Proof

### Definition 1 implies Definition 2

Let $\RR$ be an equivalence relation by definition 1.

By definition, $\RR$ is reflexive, symmetric and transitive.

$\Delta_S \subseteq \RR$
$\RR^{-1} = \RR$

and so by definition of set equality:

$\RR^{-1} \subseteq \RR$
$\RR \circ \RR \subseteq \RR$
$\Delta_S \cup \RR^{-1} \cup \RR \circ \RR \subseteq \RR$

Thus $\RR$ is an equivalence relation by definition 2.

$\Box$

### Definition 2 implies Definition 1

Let $\RR$ be an equivalence relation by definition 2.

That is:

$\Delta_S \cup \RR^{-1} \cup \RR \circ \RR \subseteq \RR$
$(1): \quad \Delta_S \subseteq \RR$
$(2): \quad \RR^{-1} \subseteq \RR$
$(3): \quad \RR \circ \RR \subseteq \RR$

From Relation Contains Diagonal Relation iff Reflexive, $(1)$ gives directly that $\RR$ is reflexive.

From Inverse Relation Equal iff Subset, $(2)$ gives that $\RR^{-1} = \RR$.

So from Relation equals Inverse iff Symmetric $\RR$ is symmetric.

From Relation contains Composite with Self iff Transitive, $(3)$ gives that $\RR$ is transitive.

So $\RR$ has been shown to be reflexive, symmetric and transitive.

Thus $\RR$ is an equivalence relation by definition 1.

$\blacksquare$