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Setting

Let $W = \{w_1, w_2, \dots, w_n \}$ be some set and denote $2^W$ as its power set. For any set $S \subseteq 2^W$, define $F(S)$ as $\{ U \cup U'\mid U\in S, U'\in S \}$.

Example) When $S=\{\{ \}, \{w_1 \}, \{w_2\},\{ w_1, w_3 \} \},\tag{1}$

we have $F(S) = \{\{ \}, \{w_1 \}, \{w_2\},\{ w_1, w_2 \},\{w_1, w_3\}, \{w_1, w_2, w_3\} \},\tag{2}$

where we use the equality $\{w_1\} \cup \{w_1, w_3\} = \{w_1, w_3\}$.

Problem

For any $n$, find the set $S$ with minimum cardinality such that $F(S) = 2^W$.

Any ideas, pointers, references would be appreciated.

Example) For $n=3$ the set where we add $\{w_3\}$ to Eq.(1) is the desired set, and for $n=4$, $S = \{\{ \}, \{w_1 \}, \{w_2\},\{ w_3 \},\{w_4\}, \{w_1, w_2\}, \{w_3, w_4\} \}$ is the desired set.

So far, I had no luck on finding an explicit formula on how to create a set with the above property. Even for small $n$ such as $n=5$, it turned out to be highly non-trivial. Although I am not quite sure if this is the minimal set, $S = \{\{ \}, \{w_1 \}, \{w_2\},\{ w_3 \},\{w_4\}, \{w_5\},\{w_1, w_2\}, \{w_3, w_4\}, \{w_3, w_5\}, \{w_4, w_5\}, \{w_3, w_4, w_5\}\}$ has the property $F(S) = 2^W$, where $|S|=11$.

One simple observation I found so far is that, we need

$|S| \ge \Omega(2^{n/2}), \tag{1}$

since $|F(S)| = |2^W|=2^n$ and $|F(S)| \le |S|^2$. For intuition on the latter inequality, think of a 2-by-2 matrix where the columns and rows are indexed by the elements of $S$, and fill each entry of the matrix by taking the union of elements that the column and row index represents. Then, the set of entries of the matrix must $2^W$ by the property of $S$. There are a lot of duplicate entries in the matrix, but this gives us a rough upper bound for $|F(S)|$.

Note @Test123 gave a more tighter upper bound of $|F(S)| \le (n^2 + n)/2$.

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    Let $[W]^m$ be the set of subsets of $W$ that each have m members. Then $\cup_{m\geq 0} [W]^{m(m+1)/2}$ may not be the smallest possible $S$ (It is not, when $n=4$) but by using Stirling's Formula or Chebychev's Formula (from statistics theory) it seems to be a lot less than $2^{n/2}$ for large $n.$ Note that we only need the set of $m$ such that $0\leq (m^2+m/2)\leq n.$ – 2017-02-06
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    I dk of a standard symbol for $\{U\cup U': U,U'\in S\}$ but $S\times S$ is not preferable as it commonly means the Cartesian product.... Not meant to criticize the Q. It's deceptively simple. – 2017-02-06
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    Following your notation, when you say that $|S\times S|\leq |S|^2$ this is not true. Actually, $|S\times S|\leq ^nC_2+^nC_1$ with $n=|S|$. – 2017-02-06
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    @user254665 Oops sorry, I wasn't aware of that. :p Changed the notation $|S \times S|$ to $F(S)$. – 2017-02-07
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    @Test123 The inequality you gave is definitely tighter. However, I wouldn't say $|F(S)| \le |S|^2$ is "not true", since $nC2 + nC1 \le (n^2 + n)/ 2 \le n^2 \le |S|^2$. – 2017-02-07
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    @user254665 Can you explain why "Note that we only need the set of $m$ such that $0≤(m^2+m/2)≤n$"? It seems to me we need the set of $m$ such that $0 \le m \le n/2$, since to create the largest element $\{w_1, w_2, \dots, w_n\} \in 2^W$, we need at least two sets $U, U'$ in $S$ such that $n/2 \le |U|$ or $n/2 \le |U'|$. – 2017-02-07
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    Because it's $[W]^{(m^2+m)/2},$ not $[W]^m.$ .. E.g if $n=29$ and $m=7$ we have $[W]^{(7^2+7)/2)}=[W]^{28}.$ Take a look at the first few members of the sequence $ F(T_j)$ where $T_0=\{\phi\} $ and $T_{j+1}= T_j\cup [W]^{(j+1)(j+2)/2}. $ – 2017-02-07
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    @user254665 We have that $$\left|\bigcup_{m\geq 0}[W]^{m(m+1)/2}\right|=\sum_{m\geq 0}\binom{n}{\frac{m(m+1)}{2}}$$ $$=\sum_{\displaystyle 0\leq m\leq\left\lfloor\frac{-1+\sqrt{1+8n}}{2}\right\rfloor}\binom{n}{\frac{m(m+1)}{2}}$$ For the first few values of $n$, this gives us $$1, 2, 3, 5, 9, 16, 28, 50, 93, 178, 342, 650$$ which is [A143961](http://oeis.org/A143961) in the OEIS. – 2017-02-07
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    Although the problem setting is a little different, I found the paper "Generating all sets with bounded unions" [link](https://hal.archives-ouvertes.fr/hal-00189041/document) to be very close to my question. Thank you all for the helpful discussions! – 2017-02-07

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