Are two relations equal if they are both equivalence relations

Question

If R and S are both equivalence relations on a non empty set A, then does R=S?

That was the question on my assignment, I think they are because equivalence relations have to be reflexive, symmetric and transitive.

Take A = {a,b,c}, if they are equivalence relations, then they both have to be :

{(a,a),(b,b),(c,c),(a,b),(b,a),(a,c),(c,b),(b,c),(c,a)} in order for them to all the properties of equivalence relations, then R cannot be different from S because then otherwise one of them may not be equivalent.

Is my logic correct here? And if it is, how can I go about proving it, because I don't think the above is a proof, and if it isn't correct, then can you guys point me in the right direction?

Thanks

Answer 1

It is certainly possible that the two are equal, however several counter examples exist.

Some basic examples of equivalence relations that you should be aware of are the "trivial" equivalence relations. The first of which where every element is related only to itself. The second of which is where every element is related to every other element.

Take for counterexample:

$A=\{1,2\}$

$S = \{(1,1),(2,2)\}$

$R = \{(1,1),(1,2),(2,1),(2,2)\}$

Note: both of these do count as equivalence relations. $S$ is indeed symmetric (despite not having any examples of pairs of nonequal elements). And both are indeed transitive.

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Additional information: the set of equivalence relations on $A$ are in direct bijection with the set of partitions of $A$.

Given an equivalence relation, $R$, one can define a partition of $A$ using the equivalence classes of $R$.

Conversely, given a partition of $A$, say $A=A_1\cup A_2\cup\dots\cup A_n$, one can define an equivalence relation $R=A_1^2\cup A_2^2\cup\dots\cup A_n^2$. (here $A_i^2$ represents $A_i\times A_i$, the cartesian product of $A_i$ with itself)

This lets us know that $A=\{a,b,c\}$ has five different equivalence relations:

$\begin{array}{|l|l|}\hline\text{partition}&\text{relation}\\ \hline\{a,b,c\} & \{(a,a),(a,b),(a,c),(b,a),(b,b),(b,c),(c,a),(c,b),(c,c)\}\\ \hline\{a,b\},\{c\}& \{(a,a),(a,b),(b,a),(b,b),(c,c)\}\\ \hline\{a,c\},\{b\}& \{(a,a),(a,c),(b,b),(c,a),(c,c)\}\\ \hline\{a\},\{b,c\} & \{(a,a),(b,b),(b,c),(c,b),(c,c)\}\\ \hline\{a\},\{b\},\{c\}& \{(a,a),(b,b),(c,c)\}\\ \hline\end{array}$

Answer 2

The equality relation is also an equivalence relation.

Answer 3

For any nonempty set $X$, the sets $$ \Delta_X=\{(x,x):x\in X\} $$ and $X\times X$ are both equivalence relations. They are the same only in a very special case.

The special case is when $X$ has a single element.