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Is $\mathcal{P}(X) / \sim \; = \mathcal{P} (X / \sim ) $ with $\sim$ induced by an equivalence relation on $X$ ?

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Have you tried any examples? – Gerry Myerson Dec 7 '12 at 12:50
What is the equivalence relation that $\sim$ induces on $\mathcal{P}(X)$? – Rhys Dec 7 '12 at 12:50
I think this is the right idea: Let $U , V \subset X$ such that for all $ u \in U$ there exists $v \in V $ such that $ u \sim v $. Then $U \sim V $. – dhr84 Dec 7 '12 at 12:53
Let $X = \{ \{1 \} , \{2 \} , \{3 \} \}$. I write $[1,2]$ for the equivalence class containing $1$ and $2$. $\mathcal{P} (X) = \mathcal{P} (\{ \{1 \} , \{2 \} , \{ 3 \} \} $ – dhr84 Dec 7 '12 at 12:57
I think you meant equipotent not equal – Amr Dec 7 '12 at 12:58
up vote 2 down vote accepted

$\def\P{\mathscr P}$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.
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Thanks very much! – dhr84 Dec 7 '12 at 13:40

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