# Is it true that every finite sigma algebra is 'isomorphic' to some power set sigma algebra?

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?

• In other words, must exist two sigma algebras of the same cardinality that preserve union and complement operations for sets of different cardinality. Sep 22, 2016 at 19:28
• 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. Sep 22, 2016 at 19:34
• @basket: Did you mean $x \in \Omega$ instead of $\mathcal {F}$? Sep 22, 2016 at 19:42
• I meant $F$ as in the set being measured, not the algebra of subsets. Sep 22, 2016 at 19:47
• @basket But $f$ should be defined on $\Omega$ right to induce a map between sigma algebras? Sep 22, 2016 at 19:50