Consider a set $X$ and a partial order $\preceq$ on the power set $2^X$ of $X$. We assume that $\preceq$ extends the usual subset relation $\subseteq$, i.e. whenever $A\subseteq B\subseteq X$ then $A\preceq B$.
Is it always possible to find a set of partial orders $\mathcal{P}$ so that for all $A,B\subseteq X$ we have $A\preceq B$ if and only if for every $\leq \, \in \mathcal{P}$ and $a\in A$ there is some $b\in B$ so that $ a \leq b$?
No. Suppose $X$ has at least $2$ elements; $a,b\in X$, $a\ne b$. Extend the partial order $\subseteq$ to a partial order $\preceq$ on $2^X$ so that $\{a\}\preceq\{b\}$. Suppose there is a set $\mathcal P$ of partial orders of $X$ with the property you want. Since $\{a\}\preceq\{b\}$, we must have $a\le b$ for every $\le\in\mathcal P$. But then, for every $\le\in\mathcal P$ and every $x\in\{a,b\}$, we have $x\le y$ for some $y\in\{b\}$, although $\{a,b\}\not\preceq\{b\}$.
