How many sets is in the $\sigma$- field generated by $n$ distinct sets

Question

Given a set $\Omega$, let $A_1,\ldots, A_n$ be distinct subsets of $\Omega$. How many sets do you have in the $\sigma$-field generated by $\{A_1, A_2,\ldots, A_n\}$.

Answer 1

Let $[n]=\{1,2,\ldots,n\}$. For every $\omega$ in $\Omega$, let $\alpha(\omega)=\{i\in[n]\mid \omega\in A_i\}$. For every $I\subseteq[n]$, let $A_I=\alpha^{-1}(I)=\{\omega\in\Omega\mid \alpha(\omega)=I\}$. Finally, let $$ \mathcal P=\{A_I\mid I\subseteq[n]\}. $$ "In general position", $\mathcal P$ is a partition of $\Omega$ and the sigma-field $\mathcal A$ generated by $\{A_i\mid i\in[n]\}$ is also generated by $\mathcal P$. The partition $\mathcal P$ has $2^n$ elements hence the sigma-field $\mathcal F$ has $2^{2^n}$ elements.

If some subsets $A_I$ are empty, $\mathcal P$ has only $1\leqslant p\leqslant 2^n-1$ nonempty elements which still form a partition of $\Omega$ and still generate $\mathcal A$ hence $\mathcal A$ has $2^{p}$ elements.

Edit: For $n$ disjoint sets $A_i$ whose union is not the full space $\Omega$, $A_I=\varnothing$ for every $I$ of size at least $2$ hence the only nonempty sets $A_I$ are $A_\varnothing=\Omega\setminus\bigcup\limits_{i=1}^nA_i$ and each $A_{\{i\}}=A_i$. Hence $p=n+1$ and $\mathcal A$ has $2^{n+1}$ elements. For $n$ disjoint sets $A_i$ whose union is the full space $\Omega$, the only nonempty sets $A_I$ are the sets $A_{\{i\}}=A_i$, hence $p=n$ and $\mathcal A$ has $2^{n}$ elements.