1
$\begingroup$

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?

$\endgroup$
3
  • $\begingroup$ It's just equality. There are no pairs $a\neq b$ in $A\times A$ such that $aRb$. $\endgroup$
    – lulu
    Aug 14, 2020 at 19:13
  • $\begingroup$ I have never seen your notation $R=(A,A,R)$. I think you mean $R\subseteq A\times A$. $\endgroup$
    – gen-ℤ ready to perish
    Aug 14, 2020 at 19:44
  • $\begingroup$ 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. $\endgroup$
    – Juju9708
    Aug 14, 2020 at 20:00

2 Answers 2

Reset to default
0
$\begingroup$

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 $=$.

$\endgroup$
0
$\begingroup$

Hint: Assume that $aRb$ and use the definition of symmetry and then the definiton of anti-symmetry to deduce an equality.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.