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I've proved that there are relations which are both symmetric and antisymmetric ($\forall a \forall b (aRb \rightarrow (a=b))$) and now I'm trying to prove that there are relations which are neither symmetric nor antisymmetric. I got stuck! Any ideas?

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Think of a set that contains a couple of elements. Come up with a relation on that set such that for some pairs of elements (x, y), $x R y$ and $\lnot (y R x)$; but for other pairs of elements (x, y), $x R y$ and $y R x$. –  Tanner Swett Feb 25 '12 at 22:17

3 Answers 3

up vote 6 down vote accepted

If every pair satisfies $aRb\rightarrow bRa$ then the relation is symmetric. If there is at least one pair which fails to satisfy that then it is not symmetric.

Similarly if there is at least one pair which has $(aRb\rightarrow bRa)\land a\neq b$ then antisymmetry is also not satisfied.

We can therefore take the following relation: $\{a,b,c\}$ would be our universe and $R=\{\langle a,b\rangle,\langle b,a\rangle,\langle a,c\rangle\}$.

The fact that $aRc\land\lnot cRa$ shows that the relation is not symmetric, but $a\neq b$ and both $aRb$ and $bRa$ hold.

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Remember that a relation on a set $A$ is just a subset of $A\times A$. So, you can just pick a convenient subset $R \subset A \times A$ so that only for SOME elements $a,b$ of $A$(I.e. not all), both $(a,b)$ and $(b,a)$ are in $R$. Can you take it from here?

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Suppose $aRb$ and $bRc$ and $cRb$. And that's as far as $R$ goes. It's not symmetric since $(\text{not }bRa)$ and it's not antisymmetric since both $aRb$ and $bRa$.

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