Independent family of sets (${\bigcap }_{j=1}^m{A}_j^{{\epsilon }_j}\ne \varnothing$)

A family of subsets of $X$ is independent when every finite choice of distinct members and their complements has nonempty intersection. Write $A^1=A$ and $A^0=X\setminus A$ in the displayed condition. Every assignment of zero or one to the family then gives a collection with the finite intersection property, which can be extended to an ultrafilter.

For every infinite set $X$ there is an independent family of sets consisting of $2^{|X|}$ subsets of $X$. One construction uses a set $D$ of pairs $(s,H)$, where $s\subseteq X$ is finite and $H\subseteq \mathcal{P}(s)$. For each $A\subseteq X$, let $I_A=\{(s,H):A\cap s\in H\}$. Given distinct $A_1,\dots ,A_m$, choose a finite $s$ distinguishing all their traces and choose $H$ to contain precisely the traces prescribed to have value one. This realizes every finite Boolean choice. With the axiom of choice, $|D|=|X|$, so a bijection transports this family to $X$. Its independent assignments give $2^{2^{|X|}}$ distinct ultrafilters.