If $R$ is a symmetric and anti-symmetric relation on $A$, and $D$ is a diagonal in $A\times A$, then $R\subset D$, and $R\neq D$ in some cases.

Question

We know that if D is a diagonal in $A\times A$, that $D=\{(a,a)\in A\times A\}$. And $R$ is a relation on $A$, so $R = \{(a,b)\in A\times A}$. And since $R$ is symmetric and antisymmetric, then $aRb$ and $bRa$ and $a=b$, right?

I don't know if the proof is as easy as it seems, but can't you just say that $R = \{(a,a)\in A\times A\}$ because $a = b$?

So what I'm stuck on is proving that $R \neq D$ as well and that $R = D$ doesn't have to be true for all cases, which is different than what my simple (yet probably incorrect) proof. So ya, I'm confused. Please help me.

Answer 1

You are right in saying that if $R$ is symmetric and anti-symmetric on $A$, it doesn't follow that $R = D$.
That would be the case if $R$ were also reflexive.

As a minimal counterexample, let $A = \{0,1\}$ and $R = \{ (0,0) \}$.
Then $R$ is symmetric and anti-symmetric.