Power set and equivalence relation Is $\mathcal{P}(X) / \sim \; = \mathcal{P} (X / \sim ) $ with $\sim$ induced by an equivalence relation on $X$ ?
 A: $\def\P{\mathscr P}$Define, as suggested by @dhr84, a equivalence relation $\sim_\P$ on $\P(X)$ by 
$$ U \preceq V \iff \forall u \in U \; \exists v \in V: u \sim v $$
and 
$$ U \sim_\P V \iff (U \preceq V) \land (V \preceq U)  $$
We can rephrase this as (denoting by $[x]_\sim$ the equivalence class of an element $x \in X$) 
$$ U \sim_\P V \iff \forall x \in X : ([x]_\sim \cap U \ne\emptyset) \Leftrightarrow ([x]_\sim \cap V \ne \emptyset) $$
that is $U$ and $V$ intersect exactly the same equivalence classes of $\sim$. Define a map $f \colon \P(X/\sim) \to \P(X)/\sim_\P$ by 
$$ f(A) := \left[\bigcup A\right]_{\sim_\P}, \quad A \subseteq X. $$
Then 


*

*$f$ is one-to-one: Suppose $f(A) = f(B)$, let $[x]_\sim \in A$, then as $\bigcup A \sim_\P \bigcup B$, and $[x]_\sim \cap \bigcup A \ne \emptyset$, we have $[x]_\sim \cap \bigcup B \ne \emptyset$. Since equivalence classes are either disjoint or equal, we must have $[x]_\sim \in B$. So $A \subseteq B$, by symmetry $A = B$.

*$f$ is onto: Let $U \in \P(X)$. Set $A = \{[x]_\sim \mid x \in U\}$, we will show $f(A) = [U]_{\sim_\P}$, that is $U \sim_\P \bigcup A$. If $x \in U$, we have $x \in [x]_\sim \subseteq \bigcup A$, so $U \subseteq \bigcup A$, and therefore $U \preceq \bigcup A$. If $y \in \bigcup A$, there is an $x \in U$ with $y \in [x]_\sim$, hence $y \sim x$. That proves $\bigcup A \preceq U$. So $U \sim_\P \bigcup A$ and we are done.

