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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?

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  • $\begingroup$ 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 $R\subseteq A\times A$. $\endgroup$ 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

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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 $=$.

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

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