Create $n$ subsets with a special union property Suppose we have a set $Α$ with $m$ elements (you may choose $m$ as large as you like).
$A= \{b_1,b_2,...,b_m\}$.
I would like to create $n$ subsets $A_1,…,A_n$ such that:


*

*Every union of more than  $\geq \frac{n}{2}$ of $A_i$’s gives $A$

*Every union of less $<\frac{n}{2}$ of $A_i$'s is strictly contained in $Α$.


Cases $m=n=2,m=n=3$ are easy. Is there an algorithm for the general case? 
 A: Let $N = \lfloor \frac{n}{2}-\frac{1}{4} \rfloor$, the largest integer strictly less than $\frac{n}{2}$. Set $m = {n \choose N}$, and enumerate the $m$ subsets of $A=\{1, \ldots, n\}$ of size $N$ as $\{{\cal B}_1, \ldots, {\cal B}_m\}$. Define $A_i = \{j \in \{1\ldots m\} | i \notin {\cal B}_j\}$.
Let $S$ be any collection of $A_i$'s with $|S| <= N$; then there exists at least one $x \in \{1 \ldots m\}$ such that $i \in {\cal B}_x$ for all $i$ with $A_i \in S$. Thus $x \notin A_i$ for all $A_i \in S$, and the union of the $A_i$ in $S$ is strictly contained in $A$.
Let $S$ be any collection of $A_i$'s with $|S| > N$. Every element $x \in \{1 \ldots m\}$ is {\em not} contained in exactly $N$ of the $A_i$'s, therefore $x$ must be contained in some $A_i \in S$; hence the union of the $A_i \in S$ must be all of $A$.
The first interesting case seems to be $n=5$, where $N=2$. I get the sets
$A_1 = \{5,6,7,8,9,10\}$
$A_2 = \{2,3,4,8,9,10\}$
$A_3 = \{1,3,4,6,7,10\}$
$A_4 = \{1,2,4,5,7,9\}$
$A_5 = \{1,2,3,5,6,8\}$
I think it is an exercise to prove that for fixed $n$, the value of $m$ given is the smallest value that can work.
