Not a duplicate of this, this, or this.
This is exercise $4.3.21$ from the book How to Prove it by Velleman $($$2^{nd}$ edition$)$:
Suppose $R$ is a relation on $A$, and define a relation $S$ on $\mathscr P(A)$ as follows:
$$S=\{(X,Y)\in\mathscr P(A)\times \mathscr P(A)|\forall x\in X\exists y\in Y(xRy)\}.$$
For each part, give either a proof or a counterexample to justify your answer.
$(a)$ If $R$ is reflexive, must $S$ be reflexive$?$
$(b)$ If $R$ is symmetric, must $S$ be symmetric$?$
$(c)$ If $R$ is transitive, must $S$ be transitive$?$
Here are my answers:
Part $(a)$ Proof:
Suppose $R$ is reflexive. Let $X$ be an arbitrary element of $\mathscr P(A)$. Let $x$ be an arbitrary element of $X$. Since $X\in\mathscr P(A)$, then $X\subseteq A$. From $X\subseteq A$ and $x\in X$, $x\in A$. Since $R$ is reflexive, $xRx$ and so $\exists y\in X(xRy)$. Ergo if $x\in X$ then $\exists y\in X(xRy)$. Since $x$ is arbitrary, $(X,X)\in S$. Thus if $X\in\mathscr P(A)$ then $(X,X)\in S$. Since $X$ is arbitrary, $S$ is reflexive. Therefore if $R$ is reflexive then $S$ is reflexive. $Q.E.D.$
Part $(b)$ Counterexample:
Suppose $A=\{1,2\}$ and so $\mathscr P(A)= \Bigr\{\emptyset,\{1\},\{2\},\{1,2\}\Bigr\}$. Let $R=\{(1,2),(2,1)\}$ which is symmetric. Then $S=\Biggr\{\Bigr(\{1\},\{2\}\Bigr),\Bigr(\{1\},\{1,2\}\Bigr),\Bigr(\{2\},\{1\}\Bigr),\Bigr(\{2\},\{1,2\}\Bigr),\Bigr(\{1,2\},\{1,2\}\Bigr)\Biggr\}$ which is clearly not symmetric: for example $\Bigr(\{1\},\{1,2\}\Bigr)\in S$ but $\Bigr(\{1,2\},\{1\}\Bigr)\notin S$.
Part $(c)$ Proof:
Suppose $R$ is transitive. Let $X$, $Y$, and $Z$ be arbitrary elements of $\mathscr P(A)$ such that $(X,Y)\in S$ and $(Y,Z)\in S$. Let $x$ be an arbitrary element of $X$. From $(X,Y)\in S$ and $x\in X$, $\exists y\in Y(xRy)$ and so we can choose some $y_0\in Y$ such that $xRy_0$. From $(Y,Z)\in S$ and $y_0\in Y$, $\exists z\in Z(y_0Rz)$ and so we can choose some $z_0\in Z$ such that $y_0Rz_0$. Since $xRy_0$ and $y_0Rz_0$, $xRz_0$ and so $\exists z\in Z(xRz)$. Ergo if $x\in X$ then $\exists z\in Z(xRz)$. Since $x$ is arbitrary, $(X,Z)\in S$. Thus if $(X,Y)\in S$ and $(Y,Z)\in S$ then $(X,Z)\in S$. Since $X$, $Y$, and $Z$ are arbitrary, $S$ is transitive. Therefore if $R$ is transitive then $S$ is transitive. $Q.E.D.$
Are my answers valid$?$
Thanks for your attention.