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Let $(\Omega_1,\mathcal {F_1})$ and $(\Omega_2,\mathcal {F_2})$ be two measurable spaces.we call $\mathcal{F_1}$ and $\mathcal {F_2}$ are isomorphic if there exist a measurable map $f: (\Omega_1,\mathcal {F_1}) \to (\Omega_2,\mathcal {F_2})$ such that the natural map $f_{*}: \mathcal {F_2} \to \mathcal {F_1}$ is bijective and preserves arbitrary union and complements.

Let $(\Omega,\mathcal {F})$ $( \vert \mathcal {F} \vert < \infty)$ and $(\{0,1\}^k,\mathcal {P}(\{0,1\}^k))$ be measurable spaces. Does there always exist a $k$ such that $\mathcal {F}$ is isomorphic to $\mathcal {P}(\{0,1\}^k)$?

I know that cardinality of $\mathcal {F}$ is $2^n$ for some $n$.Any ideas to proceed?

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  • $\begingroup$ In other words, must exist two sigma algebras of the same cardinality that preserve union and complement operations for sets of different cardinality. $\endgroup$
    – Masacroso
    Sep 22, 2016 at 19:28
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    $\begingroup$ Consider the 'atoms' of the space i.e. the equivalence classes given by $x \sim y$ iff ($x \in A$ iff $y \in A$) for each $A \in \Omega$. Give the set of classes the power set algebra. The measurable map that sends each $x \in F$ to its equivalence class induces a bijection on the algebras. $\endgroup$
    – basket
    Sep 22, 2016 at 19:34
  • $\begingroup$ @basket: Did you mean $x \in \Omega$ instead of $\mathcal {F}$? $\endgroup$
    – Math Lover
    Sep 22, 2016 at 19:42
  • $\begingroup$ I meant $F$ as in the set being measured, not the algebra of subsets. $\endgroup$
    – basket
    Sep 22, 2016 at 19:47
  • $\begingroup$ @basket But $f$ should be defined on $\Omega$ right to induce a map between sigma algebras? $\endgroup$
    – Math Lover
    Sep 22, 2016 at 19:50

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