# Defining Equivalence relations

So I am not really comfortable with equivalence relations, so this example from Wikipedia gives me trouble. Here is what it says:

Let the set $\{a,b,c\}$ have the equivalence relation $\{(a,a),(b,b),(c,c),(b,c),(c,b)\}$.

So it can also be $\lbrace(a,a), (b,b), (c,c) \rbrace$ if I so choose?

• Yes, that is an equivalence relation, too. In fact, a little meditations shows that an equivalence relation on a set is essentially the same thing as a partition on the set. – Travis Willse May 3 '15 at 4:23
• @Travis So the equivalence classes depend on which equivalence relation I choose? – SalmonKiller May 3 '15 at 4:25
• @SalmonKiller Yes, that is exactly right. – apnorton May 3 '15 at 4:26

Your relation is indeed an equivalence relation:

Reflexivity clearly holds, as symmetry does. Transitivity does hold since

$$\forall x,y,z\in X: xRy \wedge yRz \implies xRz$$

is a true statement: $xRy \wedge yRz$ is true in this case if and only if $x=y=z$, in which case clearly (reflexivity) $xRz$.

EDIT: Simpler argument (without symbolic logic)

You want to prove that the relation $R=\lbrace(a,a), (b,b), (c,c) \rbrace$ defined on the set $X=\{a,b,c\}$ is an equivalence relation, so you must prove three properties:

1. For every $x\in X$, $(x,x)\in R$. (Reflexivity)
2. For every $x,y\in X$, if $(x,y)\in R$ then necessarily $(y,x)\in R$ (Symmetry)
3. For every $x,y,z\in R$, if $(x,y)\in R$ and $(y,z)\in R$, then necessarily $(x,z)\in R$. (Transitivity).

We now prove those properties.

1. Take any element $x$ of $X$, we have only three possibilities, namely $x$ equals $a$, $b$ or $c$, in any case, $(x,x)\in R$ so the relation is reflexive.
2. Take any element $x$ and $y$ of $X$, we must prove that if $(x,y)\in R$, then $(y,x)\in R$. We only need to focus on the cases where $(x,y)\in R$ (if $(x,y)\notin R$ there's nothing to prove!), so the only cases are $x=y=a, b, c$ where the symmetry clearly holds.
3. Here's the point, we must take any three elements $x,y,z$ of $X$ and prove that if $(x,y)\in R$ and $(y,z)\in R$, then $(x,z)\in R$. Again, we only focus on the case $(x,y)\in R$ and $(y,z)\in R$, but this happens only if $x=y=z$ and therefore, clearly, $(x,z)\in R$.

Please notice not every relation you define on $X$ becomes an equivalence relation. For example, if you choose to add (only) $(a,b)$ to $R$, then symmetry is lost.

• Can you please explain that same idea without symbolic logic? I would really appreciate that. – SalmonKiller May 3 '15 at 4:30