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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?

S.C.
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    I would say intuitively, that these properties mimic the properties of equality. – saulspatz Jul 21 '19 at 19:42
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    As a consequence of the reflexive, symmetric, and transitive properties, any equivalence relation provides a partition of the underlying set into disjoint equivalence classes – J. W. Tanner Jul 21 '19 at 19:47
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    The idea is for the relation to capture the notion of "same" in some specified sense. Certainly any object is the "same" as itself (so reflexive). If $A$ is the "same" as $B$, then by the concept of "sameness", $B$ should be the "same" as $A$ (so symmetric). Similarly, if $A$ is the "same" as $B$, and $B$ is the "same" as $C$, then by the concept of "sameness", $A$ should be the "same" as $C$ (so transitive). – quasi Jul 21 '19 at 19:47
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    Are there only 3 ways of fundamentally describing equality? Clearly these 3 properties must be "more special" than other ways of defining equality. – S.C. Jul 21 '19 at 19:49
  • A better way of phrasing my confusion is "why do I need all 3". What is left out if my relation only has the properties of symmetric and transitive? What is left out if my relation only has the properties of transitive and reflexive? (etc) – S.C. Jul 21 '19 at 19:51
  • The $\le$ relation on the integers is relexive and transitive, but not symmetric. So for example, you have $3\le 5$, but you wouldn't want to regard $3$ as "equivalent" to $5$. – quasi Jul 21 '19 at 19:56
  • Sure, I understand these "literal" examples. But I am not seeing the 'abstract' consequences. If I have a relation, say the one that you just used, that is not symmetric, the resulting partitions will NOT be equivalence classes? Is that the real issue? Or is the issue that I won't even be able to partition my set. – S.C. Jul 21 '19 at 19:58
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    The partitionability is a consequence of the definition, not the main motivation. Review the first comment of saulspatz and my first comment. Equivalence is a kind of "equals". That's the main motivation. – quasi Jul 21 '19 at 20:00
  • Cool. That's certainly something that is interesting to know. As I final issue, though, are there really only 3 ways of abstractly defining equivalence? Clearly each one of these 'equivalence properties' brings something 'different' to the table. Are there absolutely no other properties that mimic 'equality' that could also be brought to the table? – S.C. Jul 21 '19 at 20:02
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    Those $3$ properties capture everything about "abstract" equality, in the sense that anything else that holds for the abstract concept is implies by those $3$ properties. – quasi Jul 21 '19 at 20:05
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    Is there a proof of that? Or is it just something that is axiomatically accepted? i.e. is there a way of proving that these 3 criteria 'completely' describe the concept of equality? Or is it the case that these 3 chosen criteria have "worked so well" that there is no need to find new or different criteria. – S.C. Jul 21 '19 at 20:07
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    Modeling a concept by axioms doesn't require proof, just agreement that the concept is sensibly modeled. But as a test of your potential agreement, can you think of any property that is always true of abstract equals that is not implied by those $3$ properties? – quasi Jul 21 '19 at 20:09
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    Further motivation: if we wish to consider the consequences of forcing some elements to be equal (equivalent) then this amounts to considering the smallest equivalence relation containing the forced equalities. Often we desire the equivalence relation to further be compatible with ambient algebraic operations, yielding what's known as a congruence. – Bill Dubuque Jul 21 '19 at 20:10

2 Answers2

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  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 $a\, R\, b\iff f(a)=f(b)$.

Berci
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    Ohhh, I really like the function perspective on this. That makes the ‘equivalence class’ structure much more intuitive. – S.C. Jul 22 '19 at 00:33
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    And most importantly, purely brings in the equality relation (on another set). – Berci Jul 22 '19 at 00:38
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    @S.Cramer See also this answer for more on such kernel equivalence relations (whose classes are known as fibers, level sets / curves, etc) and, more genrally, equalizers. – Bill Dubuque Jul 22 '19 at 13:31
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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.

Venkata Karthik Bandaru
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  • Does the union of C produce R? (i.e. $\bigcup C = R$). (also...I think you meant $\emptyset$ instead of $\phi$) – S.C. Dec 02 '20 at 19:11
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    Union of the family C of sets produces M (not R). And yeah I guess any of the variants $ \emptyset $, $ \varnothing $, $ \phi $, ${}$, etc can be safely used to denote the null set. – Venkata Karthik Bandaru Dec 02 '20 at 19:49
  • Why exactly does $R=R'$ require transitivity, reflexivity, and symmetry? I interpreted you comments as: "If two relationships produce the same $C$ and both of these relations are reflexive, symmetric, and transitive, then the two relations are equivalent." However, I do not know see why you need to stipulate that the relationships are reflexive, symmetric, and transitive (as I was able to prove that $R=R'$ without invoking those properties) – S.C. Dec 06 '20 at 01:34
  • I think you're having in mind this situation (correct me if I am wrong) : R is a general relation on M. Now set C, relation R' arise as defined above. You're saying R=R' holds even here. Here is a quick counterexample. Taking M = {1,2} and R = {(1,2), (2,1)} gives C = { {1} ( = [2] ) , {2} ( = [1] ) } from which R' = { (1,1), (2,2) }. – Venkata Karthik Bandaru Dec 06 '20 at 08:34
  • hmmm. Perhaps I misunderstand what you are trying to say in the statement, "$\sim$ can be naturally reconstructed from knowing the set $C$ alone". I interpreted this as meaning: For a given set $C$ (defined using your above notation) there is only one unique relation, call it $\sim$, that can be used to produce it...i.e. if another relation $\sim'$ can also be used to produce that same $C$, then $\sim'=\sim$. – S.C. Dec 06 '20 at 19:40
  • I think I see the issue in my proof, where I wrongly claimed that you do not need transitivity, reflexivity, and symmetry to prove R=R'. Thank you for the help! – S.C. Dec 06 '20 at 21:07
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    Just in case you were wondering: I want to show that an arbitrary $\langle a,b \rangle \in R$ is also in $R'$ (and without loss of generality, show that an arbitrary $\langle a,b\rangle \in R'$ is also in $R$). So let $\langle a,b \rangle \in R$. By definition, $b \in [a]_R$. We know from assumption that $C=C'$ which implies that $[a]_R \in C'$. If $[a]_R \in C' \land b \in [a]_R$, then $\exists x \exists y (y \in R' \land x \in M \land y=\langle x,b \rangle)$. Now, to show that this $x$ is precisely $a$ (which would mean that $\langle a,b \rangle \in R'$), we need trans, symm, and reflex. – S.C. Dec 06 '20 at 21:19