Let $\mathcal{F}_1, \ldots, \mathcal{F}_h$ be families of finite sets. Let $\mathcal{G}_{i,j} = \mathcal{F}_i \cap \mathcal{F}_j, 1 \le i \lt j \le h$. We know that $|\mathcal{G}_{i,j}| \ge h$ and $\mathcal{F}_i \cap \mathcal{F}_j \cap \mathcal{F}_k =\emptyset, 1 \le i \lt j \lt k \le h$ We know also that the intersection of any $A_i \in \mathcal{F}_{i}, 1 \le i \le n$, is empty: $\bigcap_{i=1}^h A_i = \emptyset$. Let $\mathcal{U} = \{A \cup B : A,B \in \mathcal{F}_1 \cup \cdots \cup \mathcal{F}_h, A \not= B\}$.
I would like to find a good lower bound for the size of $\mathcal{U}$ as a function of $h$. Ideally, I would like a bound $\approx h$.
My attempt:
Let $\mathcal{H} = \{\mathcal{G}_{i_k,j_k}: 1 \le k \le n\} \subseteq \{\mathcal{G}_{i,j}: 1 \le i \lt j \le h \}$ with the property that $\bigcup_{k=1}^n \{i_k,j_k\} = [h] = \{1, \ldots, h \}$ and $\{i_k,j_k\}, 1 \le k \le n$ are all different. By the conditions in the first paragraph, the intersection of any two families in $\mathcal{H}$ must be empty.
Let $\mathcal{U}_{p,q} = \{A_p \cup A_q: A_p \in \mathcal{G}_{i_p,j_p} \in \mathcal{H}, A_q \in \mathcal{G}_{i_q,j_q} \in \mathcal{H}\}, 1 \le p \lt q \le n$.
I know that the map:
$$\mathcal{G}_{i_1,j_1} \times \cdots \times \mathcal{G}_{i_n,j_n} \rightarrow \mathcal{U}_{1,2} \times \cdots \times \mathcal{U}_{1,n} \times \mathcal{U}_{2,3} \times \cdots \times \mathcal{U}_{2,n} \times \mathcal{U}_{3,4} \times \cdots \times \mathcal{U}_{3,n} \times \cdots \times \mathcal{U}_{n-1,n}, \\(A_1,\ldots,A_n) \rightarrow (A_1 \cup A_2, \ldots, A_1 \cup A_n, A_2 \cup A_3, \ldots, A_2 \cup A_n, A_3 \cup A_4, \ldots, A_4 \cup A_n, \ldots, A_{n-1} \cup A_n)$$
is injective, because if $(A_1 \cup A_2, \ldots, A_{n-1} \cup A_n) = (B_1 \cup B_2, \ldots, B_{n-1} \cup B_n)$ on the right side, then $A_k \setminus B_k \subseteq A_k \cap \bigcap_{1 \le m \le n, m \not= k} B_m = \emptyset, 1 \le k \le n$ and similarly $B_k \setminus A_k \subseteq \emptyset$, thus $A_k = B_k$.
Since $|\mathcal{U}| \ge |\mathcal{U}_{p,q}|, 1 \le p \lt q \le n$ and $|\mathcal{G}_{i,j}| \ge h, 1 \le i \lt j \le h$ then $|\mathcal{U}|^{n(n-1)/2} \ge h^n$, i.e.:
$$|\mathcal{U}| \ge h^\frac{2}{n-1}$$
For $h$ even, we could choose for example $i_k = 2k-1, j_k = 2k, 1 \le k \le n = h/2$. But this doesn't seem much unless $h$ is small. Is there a different construction that can do better than that?
