Equality sign and equivalence relations

[psie]

Homework Statement

Let $A$ be a non-empty set and define the two relations $R$ and $S$ on $A$ by $aRb$ for all $a$ and $b$ in $A$ and $aSb$ if $a=b$. Show that $R$ and $S$ are equivalence relations and find their equivalence classes.

Relevant Equations

An equivalence relation is reflexive, symmetric and transitive, and these properties need to be checked.

I solved this exercise a long time ago, and I am now reviewing my solution. I think it is correct, but more importantly, when I look at this exercise now, I have a hard time understanding it and my solution. Hence I've got some questions.

$S$ is reflexive since $\forall a\in A$ we have $a=a\implies aSa$. $S$ is symmetric since if $a=b$, then $b=a$, i.e. $aSb\implies bSa$. $S$ is transitive since if $a=b$ and $b=c$, then $a=b=c$, i.e. $aSb\wedge bSc\implies aSc$. The equivalence classes are $\{a\}$ for all $a\in A$.

If this is correct, why does "if $a=b$, then $b=a$" hold? The text has not really introduced the equality sign yet as some...I don't know what; I believe this exercise is the point at which it is introduced as an equivalence relation (the text is an introductory course on algebra). But how do you prove symmetry for example? My solution to transitivity also already seems to assume that something is transitive. I feel silly for asking, but I feel I am doubting something fundamental.

My remaining solution is:

$R$ is reflexive because $b$ can equal $a$. Since $aRb$ holds for all $a,b\in A$, we can first choose $b$ and then $a$ and obtain $aRb\iff bRa$. Transitivity follows by a similar argument. The equivalence class is $A$.

 

[PeroK]

You seem to have simply stated that each property holds without explaining why.

For the reflexivity of R, you could say:

As $\forall a,b \in A: aRb$, in particular $\forall a \in A: aRa$. Hence R is reflexive.

 

[psie]

You seem to have simply stated that each property holds without explaining why.

I think for $R$ the solution is OK; yes, one could be more descriptive, but the problem I see in my solution is that I used some symmetry and transitive property of the equals sign, the very thing I need to prove. It feels circular.

 

[PeroK]

I think for $R$ the solution is OK; yes, one could be more descriptive, but the problem I see in my solution is that I used some symmetry and transitive property of the equals sign, the very thing I need to prove. It feels circular.

That's a good point. I'm not sure how to prove that equals is an equivalence relation. You can state that equality has those properties. But, a proof suggests to me starting with something more fundamental. I don't know what that would be.

 

[pasmith]

That's a good point. I'm not sure how to prove that equals is an equivalence relation. You can state that equality has those properties. But, a proof suggests to me starting with something more fundamental. I don't know what that would be.

There's nothing more fundamental than equality (with the possible exception of set inclusion; you could say that two objects are equal iff they are members of the same singleton set, but how you determine that a set is a singleton?).

I don't think it's circular to state that equality is by definition reflexive, symmetric and transitive, and is therefore an equivalence relation. Some basic exerises in the application of definitions are that simple.

 

[FactChecker]

I think for $R$ the solution is OK; yes, one could be more descriptive, but the problem I see in my solution is that I used some symmetry and transitive property of the equals sign, the very thing I need to prove. It feels circular.

The problem did not ask you to prove those properties of '=', so it is not circular. The properties of '=' may not be formally addressed in a typical algebra book. I think you are safe to assume the reflexive, transitive, and symmetry properties of '='.

 

[Orodruin]

The problem did not ask you to prove those properties of '=', so it is not circular.

It kind of did though. It defines $S$ as $a S b$ if $a = b$. This makes $S$ equivalent to $=$.

 

[FactChecker]

It kind of did though. It defines $S$ as $a S b$ if $a = b$. This makes $S$ equivalent to $=$.

I saw this part as just a routine run through the properties defining an equivalence relation. Maybe I was wrong.