Let $\ell$ be an integer parameter. I want to ask the existence of the following design: There is a universal constant $\beta < 1$ such that for all sufficiently large $\ell$, the following holds:

• The universe size $d = O(\ell)$.
• There are $\ell^2$ sets in the design, $S_1, \dots, S_{\ell^2} \subseteq [d]$.
• $(\forall i \in [\ell^2])\ |S_i| = \ell$.
• For every $t+1 \in [\ell^2]$, there exist $2$ disjoint subsets of $S_{t+1}$: $F_1, F_2$ (these are "forbidden" subsets), such that:
  1. $|F_1|, |F_2| \ge \beta \cdot \ell$.
  2. $S_1, \dots, S_t$ can be divided into $2$ consecutive chunks, $S_1, \dots, S_{i}$; $S_{i+1}, \dots, S_t$ such that for chunk $j \in [2]$, its intersection with $F_j$ is empty. In other words, for the first chunk, $\left( \bigcup_{j=1}^{i} S_j \right) \cap F_1 = \emptyset$, and so on.

btw. do we have a name for such block designs?

Some motivations. If we do not insist on the size of each set, then one could choose all the $2$-subsets of $[d]$, and then order them in, for example, reverse lexicographic order. Then for $S_{t+1} = \{b_1 < b_2\}$, all the sets that are lexicographically larger than it can be divided into two chunks, one that does not have $b_1$, and one that does not have $b_2$. The question I have is whether such a phenomenon still exists once we enlarge the set size while keeping (asymptotically) the same number of sets in the design.

Construction 1 with large universe. If $d = O(\ell^2)$, then there is an easy construction as well. Group $d$ into $O(\ell)$ blocks each with $\ell/2$ elements, then use the choose $2$ trick mentioned above, we have each set with $\ell$ elements, yet, the sets before any set can be split into two chunks that the first chunk does not know the first block of $\ell/2$ elements, and the second chunk that does not know the second block of $\ell/2$ elements.

Construction 2 with smaller universe but more chunks. Let $d = O(\ell\log \ell)$. We could again group $d$ into $O(\log\ell)$ blocks each with $O(\ell)$ elements. Therefore, but choose all $O(\log \ell)$-subsets, we can generate any $\ell^{O(1)}$ sets, each of $\log\ell$ blocks with each block $\ell$ elements. Sort these sets in reverse lexicographic order according to block number, therefore we have $\log\ell$ chunks, each not covered some $\ell$ elements. Note that this works for any polynomial number of sets, but the number of chunks fails the requirement.

Misc. Comments I have a feeling that the above constructions are not optimal, though they are certainly nontrivial. The reason is that for a fixed the number of sets to generate, intuitively it holds that as we permit more chunks, the size of the forbidden sets (measured by the fraction of elements) should also go up. However, in our first construction, we have $2$ chunks, each has half elements uncovered, whereas in the second construction, the number of chunks goes to $\log\log m$, yet each now only has a $1/\log\log m$ fraction of uncovered elements. Another sign that the second construction is not optimal is that that setting of parameters simultaneously handle all $\ell^{O(1)}$ sets.

Thank you for the attention.