Proposition. For every set $X$ and preorder $\to$ on $X$, exists a set $S$ and function $s : X \to 2^S$ such that $x \to y$ iff $s(x) \subseteq s(y)$ Proof. Take $(X, \to)$. For $x \in X$, let $s(x)$ denote the set of values $y \in X$ with $y \to x$. Note that $s$ is of type $s : X \to \mathcal P(X) = 2^X$, as promised. I claim that $x \to y \iff s(x) \subseteq s(y)$: $(\Rightarrow)$ Take $x \to y$ and $z \in s(x)$. Then $z \to x$ so by transitivity $z \to y$ so $z \in s(y)$. Thus $s(x) \subseteq s(y)$. $(\Leftarrow)$ Take $s(x) \subseteq s(y)$. Note $x \to x$ so $x \in s(x)$ so $x \in s(y)$ so $x \to y$. #onwards Referenced by: