Let R be a reflexive relation on a nonempty set X. The asymmetric part of R is defined as the relation $P_r$ on X as $xP_ry$ iff $xRy$ but not $yRx$. The relation $I_r$ = $R\setminus P_r$ on X is then called the symmetric part of R.
- Show that if R is transitive , so are $P_r$ and $I_r$.
( Question from maths for economists by Efe A. Ok)
My attempt : I tried using the definition of transitivity i.e. if $xRy$ and $yRz$ then $xRz$
However to use it properly I had to use cases where $zRx$ belonged to R and also a case where it did not belong to R.
However I could not prove it even when using this approach.
Please let me know how to proceed.
Best Answer
Transitivity of $I_r$ is easy to check: you can verify it by just following the definition of $I_r$ and transitivity of $R$. (In fact, $I_r$ is an equivalence relation.)
To prove that $P_r$ is transitive, examine how to derive the transitivity of strict order relation $<$ from that of the usual order $\le$. It goes as follows: if $x<y$ and $y<z$ then $x\le z$. If $z\le x$, then we have $x<y$ and $y<z\le x$, which is impossible.
We are going to mimic the proof: if $(x,y),\, (y,z)\in P_r$ then $(x,z)\in R$. On the contrary assume that $(z,x)\in R$. Since $R$ is transitive and $(y,z)\in R$, $(y,x)\in R$. However $(x,y)\in P_r$ implies $(y,x)\not\in R$ by definition, a contradiction. Therefore $(z,x)\not\in R$.