Question

8. Show that the inverse of an order relation in A is an order relatitón m 11 9. Let G be a relation in A. Show that G is an orderr IA and G oG G relation if and only if G nG hot GH is an equivalence

how to prove this? a book of set theory exercise3.2 number9

Answer 1

G is an order relation in A. Let
A=(r, y) E R
, then let  $G=\begin{Bmatrix} (x,y) \in A|x<=y \end{Bmatrix}$ ( x<=y is the property of order relation), and also

$G^{-1}=\begin{Bmatrix} (x,y) \in A|y<=x \end{Bmatrix}$

its clear that $G^{-1}$ and will have common ordered pairs only for x=y So $G \cap G^{-1}=\begin{Bmatrix} (x,y)\in A|x=y \end{Bmatrix}$

hence .  $G \cap G^{-1}=\begin{Bmatrix} (x,y)\in A|x=y \end{Bmatrix} =I_{A}$

for second part we now that if there are two relation S and R such that $S\subseteq X *Y$ and $R\subseteq Y*Z$ then  $S\circ R=\begin{Bmatrix} (x,z)\in X*Z | \exists y \in Y:(x,y) \in R \Lambda (y,z) \in S \end{Bmatrix}$

based on this definition

$G\circ G=\begin{Bmatrix} (x,y)|(x,y)\in A \end{Bmatrix}$ = G

So we see the above two things are necessary for G to be an order relation.