# What can be said about a relation $R=(A,A,R)$ that is refelxive, symmetric and antisymmetric?

What can be said about a relation $$R=(A,A,R)$$ that is refelxive, symmetric and antisymmetric?

I know the definitions:

• Reflexive: $$xRx$$ $$\forall$$ $$x \in A$$
• Symmetric: if $$\forall a,b \in A, aRb \Rightarrow bRa$$
• Antisymmetric: $$aRb$$ and $$bRa \Rightarrow a=b$$

Can I get some hints?

• It's just equality. There are no pairs $a\neq b$ in $A\times A$ such that $aRb$.
– lulu
Aug 14, 2020 at 19:13
• I have never seen your notation $R=(A,A,R)$. I think you mean $R\subseteq A\times A$. Aug 14, 2020 at 19:44
• Yes, I was about to clarify that, my college has a book and its notation is like that, but it's the same as you wrote. Aug 14, 2020 at 20:00

Note that if $$aRb$$ or $$bRa$$, then $$a = b$$; this follows since either of $$aRb$$, $$bRa$$ implies the other via the symmetry condition, and then the antisymmetry condition yields $$a = b$$. This in turn implies that $$R$$ is transitive, since $$aRb$$ and $$bRc$$ forces $$a = b = c$$, and $$aRa$$ by reflexivity gives us $$aRc$$, so transitivity follows. Thus $$R$$, being reflexive, symmetric, and transitive, satisfies the definition of an equivalence relation. The equivalence classes are all singletons; thus the assertions "$$a = b$$" and "$$aRb$$" are logically equivalent, and $$R$$ functions for all the world exactly as does $$=$$.
Hint: Assume that $$aRb$$ and use the definition of symmetry and then the definiton of anti-symmetry to deduce an equality.