Suppose $R \subseteq A \times A$. Let $B = \{X \in \mathscr P(A) \mid X ≠ \emptyset\}$, where $\mathscr P(A)$ is a power set; and define relation $S$ on $B$ as follows $$ S = \{(X,Y) \in B \times B \mid \forall x \in X \forall y \in Y (xRy)\}$$ 1. Prove that if $R$ is transitive, then $S$ is transitive as well.
2.Why did the empty set have to to be excluded from the set $B$ to make this proof work?
My attempt:
1.
Suppose $R$ is transitive.
Suppose we have some $(X,Y),(Y,Z)$ such that both are in $S$.
Take arbitrary $x \in X$ and $z \in Z$.
Take arbitrary $y \in Y$.
By definition of $S$, we know that $(x,y) \in R$ and $(y,z) \in R$. Since $R$ is transitive, we conclude that $(x,z) \in R$. Since $x$ and $z$ both were arbitrary, we conclude that for all $x \in X$ and for all $z \in Z$, we will have $xRz$. Which implies that $(X,Z) \in S$. And since $(X,Y)$ and $(Y,Z)$ were arbitrary, we conclude that $S$ is transitive. $\Box$
2.
I have doubts, but:
Suppose empty set is not excluded. Then there will be two non-empty subsets $X$ and $Y$, such that $(X,\emptyset) \in S$ and $(\emptyset, Y) \in S$, but $(X,Y)$ is not necessarily in $S$.