Relations properties prove

Question

Given a relation $r$, define $r^{-1}$ to be the converse relation, and define $r \cdot r$ to be the usual "$x (r \cdot r) y$ if and only if there is $z$ such that $x r z$ and $z r y$".

Is it possible that relation $r$ in set A has this property that:

$r^{-1}\subset r$ and $r^{-1} \neq r$
$r\cdot r=r$ and $\forall x\in A \;\;(\neg\,x\,r\,x)$

I cannot find any example of relation which would have the first or the second property but I don't know how to start deal with it.

In property 1, are you assuming that the relation $r$ is actually invertible? Or is $r^{-1}$ just the converse relation $x,r,y\iff y,r^{-1},x$? — Arthur, Nov 28 '18 at 07:58
Does "$r \cdot r$" mean the relation defined by $x (r \cdot r) y$ if and only if there is $z$ such that $x r z$ and $z r y$? — Patrick Stevens, Nov 28 '18 at 08:00

score 1 · Answer 1 · answered Nov 28 '18 at 08:15

1

is already impossible: taking the inverse of relations is a monotonic operation. If $r^{-1}\subseteq r$, and $a\, r\, b$ then $b\, r^{-1}\, a$, consequently $b\, r\, a$, i.e. $a\, r^{-1}\,b$, meaning $r\subseteq r^{-1}$, which together with the original assumption yields $r=r^{-1}$.
Even if we take hypothesis 1. as only $r^{-1}\subseteq r$, we must have reflexive elements by transitivity and symmetry, unless $r$ is empty.
To see this, let $a\, r\, b$, then by symmetry we have $b\, r\, a$ as well, and by transitivity, $a\, r\, a$ follows.

answered Nov 28 '18 at 08:15

Berci

90,745

Sorry but the first one isn't connected with the second one - they are independent properties. – avan1235 Nov 28 '18 at 08:21
2

@avan1235 In that case, consider the relation $<$ on the real numbers. – Arthur Nov 28 '18 at 08:48

Relations properties prove

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