Is there a relation which is neither symmetric nor antisymmetric?

Question

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?

Answer 1

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.

Answer 2

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?

Answer 3

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 $bRc$ and $cRb$.

Answer 4

To say that a relation $R$ on a set $A$ is not symmetric is equivalent to saying that there exist elements $a$ and $b$ in $A$ such that $aRb$ and $\require{cancel}b\cancel{R}a$. To say that a relation $R$ on a set $A$ is not antisymmetric is equivalent to saying that there exists an element $a\in A$ and an element $b\in A$ such that $a\ne b$, $aRb$, and $bRa.$ Consider the relation $R = \{\ (a,b)\ |\ ab^{2}\ \gt\ 0\}$ on the set of all integers $\mathbb Z$. Since $2\cdot (-1)^{2} = 2\gt 0$, the ordered pair $(2, -1)\in R$. However, since $(-1)\cdot 2^{2} = -4 \not\gt 0$, $(-1, 2)\not\in R$, thus $R$ is not symmetric. Similarly, we can show that $R$ is not antisymmetric by noting that the inequality $ab^{2}\gt0$ will hold for any two positive integers $a$ and $b$. Thus, there exists a distinct pair of integers $a$ and $b$ such that $aRb$ and $bRa$. Hence, $R$ cannot be antisymmetric.