Show that a relation is equivalent if it is both reflexive and cyclic.

Question

A relation $R$ on set $X$ is called cyclic if whenever both $xRy$ and $yRz$ then $zRx$ where $x,y,z\in X$. Show that a relation on $X$ is an equivalence relation if and only if it is both reflexive and cyclic.

Since an equivalence relation is reflexive, symmetric and transitive and we already know the relation is reflexive I assume you prove that symmetric and transitive is the same as reflexive and cyclic. I honestly don't know where to start though or if my intuition is correct so any help would be great!

Answer 1

1) reflexive + cyclic => equivalence

a) reflexive => reflix

b) reflexive + cyclic => symmetry

xRy => xRy and yRy => yRx.

c) symmetry + cyclic => transitivity

xRy + yRz => zRx => xRz

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2) Equivalence => reflixe + cyclic

a) reflexive => reflexive

b) transitive + symmetry => cyclic

xRy and yRz => xRz => zRx.

Answer 2

If you're clever about the use of the definitions that you're given, you can replace those three variables to prove that the relationship is both symmetric and transitive. However, once you prove symmetry, the rest of the problem is relatively simple. If you'd like more detailed proof you can keep reading.

If a relation R defined on $A$ is both reflexive and circular, then it is equivalent. A relation is circular if $\forall a, b, c\in A, aRb \land bRc \implies cRa$.

Proof:

It is given that the relation is reflexive, so $\forall a \in A, aRa$. It is also given that the relation is circular, so $\forall a, b, c\in A, aRb \land bRc \implies cRa$. By setting $a, b=x, x \in A$ and $c=y,\ y \in A$. If we replace these variables we get $xRx \land xRy \implies yRx$. As mentioned earlier, we know the relation is reflexive, so $xRx$ will always be true. As such, $xRy \implies yRx$. This means that the relation is symmetric. Finally, since $aRb \land bRc \implies cRa$ and $cRa \implies aRc$, $aRb \land bRc \implies aRc$. This means that the relation is transitive. As the relation is reflexive, symmetric, and transitive, it is an equivalence relation $\square$