What is the intuitive motivation for defining equivalence relations with the properties of reflexivity, symmetry, and transitivity?

Question

I am trying to understand why equivalence relations are defined using the three properties of reflexivity, symmetry, and transitivity.

Using an example set of $S = \{1,2,3,4,5,6,7,8,9\}$

It seems to me like a good first intuition is the following:

Reflexive

An equivalence relation's reflexive property basically ensures that every element in the set under examination can be partitioned (i.e. occupy its own equivalence class).

For example, if we had $R_1$ be a relation on $S$ defined as $x-y$ is divisible by $10$, then, because $(1,1)$ , $(2,2)$ , $(3,3)$ , $etc$ are order pairs that will all satisfy $R_1$, we will end up with 10 different equivalence classes ($e.g. [1],[2],[3], etc)$. Therefore, all elements are partitioned...which makes sense because each of these elements $mod(10)$ have a unique remainder...namely $1$, $2$, $3$, $etc$

Transitive

An equivalence relation's transitive property basically allows you to "unidirectionally" (my meaning will be understood shortly) link all elements belonging to the same equivalence class.

For example, if we had $R_2$ be a relation on $S$ defined as $x-y$ is divisible by $2$, then the following ordered pairs (non-exhaustive) are certainly in the set: $(2,4)$, $(4,6)$, $(6,8)$. Pretend I do not know any other ordered pairs.

Now, if I choose as my representative element $[2]$, I can build my equivalence class by starting with $2\ R_2\ n$. Well, $(2,4)$ is in $R_2$ therefore, $4$ belongs to the same equivalence class as $2$. What works with $4\ R_2\ n$? Well $(4,6)$ is in $R_2$ and therefore $6$ is in the same equivalence class as $2$ and $4$.

However, let's say that I start with $[8]$ instead of $[2]$. Which of the elements in set $S$ are in the same equivalence class? Namely, for which value of $n$ is $8\ R_2\ n$ true? Well, I know I have the ordered pair $(6,8)$...but that is not the same form as $8\ R_2\ 6$. If only I knew that $(8,6)$ was also in the set describing $R_2$...and this is where symmetry comes into play.

Symmetric

An equivalence relation's symmetry property, in conjunction with the transitive property, allows you to bidirectionally link all elements belonging to an equivalence class (regardless of which 'starting element' you choose to represent your equivalence class).

For example, incorporating the symmetry property with the above transitivity example, I now know that if $(2,4)$, $(4,6)$, and $(6,8)$ are in my $R_2$ set, then I also know that $(4,2)$,$(6,4)$, and $(8,6)$ are in my $R_2$ set. Consequently, if someone asks me what are the other elements that belong to $[8]$, without hesitation I can say $2$,$4$, and $6$.

Are these the correct ways (at a very basic level) of intuitively understanding the motivation behind using these 3 properties to define equivalence relations?

Answer 1

1. Your thoughts about reflexivity, symmetrivity, transitivity and partitions (especially in the comments) are basically correct and on the right track.
   We can, however, consider even more view points (alternative definitions, if you like) for 'what is an equivalence relation'.

   For example, a relation $R\subseteq A\times A$ is an equivalence relation if and only if
   a) $R$ is left Euclidean: $a\, R\,c$ and $b\, R\, c$ implies $a\, R\, b$, and
   b) $R$ is serial: the domain of $R$ is whole of $A$, i.e. for all $a\in A$ there exists a $b\in A$ such that $a\, R\, b$.

   And if it was taught so primarily, you would be now asking what are the importance of these two properties. If we studied them separately, we would arrive to very different answers.

   ---

2. As written in the comments, reflexivity, symmetry and transitivity together (or even the two conditions above) capture the most fundamental properties of the equality relation $=$.
   The basic concept behind equivalence relations is that of sameness.
   Informally speaking, $R$ is an equivalence relation on $A$ iff there is a 'type of comparison', regarding to elements $a$ and $b$ are compared to be the same iff $a\, R\, b$.

   We can make it formal by introducing a function that somehow 'measures' the attribute under comparison. (For example, you can think about colour as an attribute and the corresponding equivalence relation 'to have the same colour'.)

   A relation $R$ on a set $A$ is an equivalence relation if and only if there is a function $f:A\to X$ such that $a\, R\, b\iff f(a)=f(b)$.

Answer 2

Say $ M $ is a set and $ \sim $ is a relation on it ( Recall that a relation $ R $ on a set $ S $ is simply a subset $ R \subseteq S \times S $, and that $ (a,b) \in R $ is also written as $ aRb $ ). Now $ \sim $ immediately gives rise to a collection $ C := \{ [x] : x \in M \} $ of subsets of $ M $, where $ [x] := \{ y \in M : x \sim y \} $.

In general the collection $ C $ has no specific structure. We can ask ourselves the following question : Under what kinds of $ \sim $ do we get that $ C $ is a partition of $ M $, and that " $ \sim $ can be naturally reconstructed from knowing the set $ C $ alone " i.e. $ \sim = \sim' $ where $ a \sim' b \stackrel{\text{def}}{\iff} ( a, b \text{ belong to same element of } C ) $ ?

[ By the way recall a partition of a set $ S $ is just a collection of pairwise disjoint non-empty sets whose union is $ S $ ]

It's clear any such $ \sim $ should be reflexive, symmetric and transitive, i.e. reflexive symmetric transitive relations (on $ M $) are the only potential candidates for $ \sim $ . Turns out any reflexive symmetric transitive relation will do : Let $ R $ be a reflexive symmetric transitive relation on $ M $. This gives a collection $ C $ of classes $ [x] = \{ y \in M : xRy \} $. As $ x \in [x] $, elements of $ C $ are non-empty and have union $ M $. From symmetry and transitivity of $ R $ we get that "If $ [x] \cap [y] \neq \phi $, then $ [x] = [y] $", i.e. that elements of $ C $ are pairwise-disjoint. Also $ R = R' $ where $ aR'b \stackrel{\text{def}}{\iff} ( a, b \text{ belong to same element of } C ) $.

Hence our question is settled, relations which are reflexive symmetric and transitive are precisely the ones we were looking for. These are also called "equivalence relations".

Edit : Saying "$\sim$ is equal to the natural grouping relation arising from $C$" instead of "$\sim$ can be naturally reconstructed from knowing $C$ alone", as pointed out in the comments, would've been more appropriate.