Simple question about relations Yea, I hate myself when I ask this question. But I got a question below:
Consider the following relations on the set $\{a, b, c\}$.
Which of the relations are equivalence relations?
It's from a multiple choice. One of the correct answers is:
$\{(a, a),(b, b),(c, c)\}$
I understand all the other answers but I just have to convince my self about this is true. Here is my ideas:


*

*It's reflective because $aRa$.


Let $a=b$ then


*

*It's s symmetric because $aRb$ and $bRa$ where $a=b$.


Let $a=b=c$ then 


*

*It's transitive because $aRb$ and $bRc$ then $aRc$ where $a=b=c$.


Am I correct? If not, how can I convince my self it's true?
 A: Your step-by-step method and the logic for each step is entirely correct. This correct method should convince you that it is true. That's really all that is there to this.
A: I'm going to have to disagree with the first answer about this solution being correct. Your relation is explicitly on the set $\{a,b,c\}$. That is, $a,b,c$ are fixed objects. You can't "let $a = b$" in your reasoning, any more than you can "let $2=5$" if you were investigating a relation on the integers.
It is reflexive because every member of the set is related to itself.
The arguments for symmetry and reflexivity are essentially vacuous and degenerate cases of those properties. Your arguments do have the right idea behind them, but you need to express these ideas with different symbols (IE, let $xRy$, then $x=y$. Then...). You can't declare explicit members of your set to be equal.
Notice that this is the identity relation on your set. Equivalence relations are essentially a generalization and abstraction of this idea in the same way that a ring is a generalization of the integers or a metric is a generalization of distance.
A: The axioms defining an equivalence relation $R$ over a set $A$ use universal quantification, and those defining symmetry and transitivity also use implication – and this is important.
The reflexivity axiom states that


For every $a \in A$ we have that $aRa$


Clearly, if $a = \{ a,b,c \}$ this holds for our choice of $R$.
The symmetry axiom states that


For every $a,b \in R$, if $aRb$ then also $bRa$


To check that this axiom holds for our choice of $R$, we need to check that for any choice of $a$ and $b$, then whenever $aRb$ then also $bRa$. We only have that $aRa$, $bRb$ and $cRc$, and clearly if $aRa$, then $aRa$ -- the cases for $bRb$ and $cRc$ are similar. All other cases are vacuously true; for instance if $aRc$ (which is not the case) then also $cRa$.
The case of transitivity is similar; the axiom here states that


For all $a,b,c \in A$, if $aRb$ and $bRc$ then also $aRc$.


Here, the only cases that must be checked are that 


*

*if $aRa$ and $aRa$ then also $aRa$

*if $bRb$ and $bRb* *then* also $bRb$
 -  *if* $cRc$ and $cRc$ *then* also $cRc$

