Subfield: Combinatorics


Question: Let $\alpha<\frac{3-\sqrt{5}}{2}$ be a positive real number. Prove that there exist positive integers $n$ and $p>\alpha \cdot 2^{n}$ for which one can select $2 p$ pairwise distinct subsets $S_{1}, \ldots, S_{p}, T_{1}, \ldots, T_{p}$ of the set $\{1,2, \ldots, n\}$ such that $S_{i} \cap T_{j} \neq \varnothing$ for all $1 \leq i, j \leq p$.

Solution:
Let $k$ and $m$ be positive integers (to be determined later) and set $n=k m$. Decompose the set $\{1,2, \ldots, n\}$ into $k$ disjoint subsets, each of size $m$; denote these subsets by $A_{1}, \ldots, A_{k}$. Define the following families of sets:
$$
\begin{aligned}
\mathcal{S} & =\left\{S \subset\{1,2, \ldots, n\}: \forall i S \cap A_{i} \neq \varnothing\right\} \\
\mathcal{T}_{1} & =\left\{T \subset\{1,2, \ldots, n\}: \exists i A_{i} \subset T\right\}, \quad \mathcal{T}=\mathcal{T}_{1} \backslash \mathcal{S} .
\end{aligned}
$$
For each set $T \in \mathcal{T} \subset \mathcal{T}_{1}$, there exists an index $1 \leq i \leq k$ such that $A_{i} \subset T$. Then for all $S \in \mathcal{S}$, $S \cap T \supset S \cap A_{i} \neq \varnothing$. Hence, each $S \in \mathcal{S}$ and each $T \in \mathcal{T}$ have at least one common element.

Below we show that the numbers $m$ and $k$ can be chosen such that $|\mathcal{S}|,|\mathcal{T}|>\alpha \cdot 2^{n}$. Then, choosing $p=\min \{|\mathcal{S}|,|\mathcal{T}|\}$, one can select the desired $2 p$ sets $S_{1}, \ldots, S_{p}$ and $T_{1}, \ldots, T_{p}$ from families $\mathcal{S}$ and $\mathcal{T}$, respectively. Since families $\mathcal{S}$ and $\mathcal{T}$ are disjoint, sets $S_{i}$ and $T_{j}$ will be pairwise distinct.

To count the sets $S \in \mathcal{S}$, observe that each $A_{i}$ has $2^{m}-1$ nonempty subsets so we have $2^{m}-1$ choices for $S \cap A_{i}$. These intersections uniquely determine set $S$, so
$$
|\mathcal{S}|=\left(2^{m}-1\right)^{k}
$$
Similarly, if a set $H \subset\{1,2, \ldots, n\}$ does not contain a certain set $A_{i}$ then we have $2^{m}-1$ choices for $H \cap A_{i}$ : all subsets of $A_{i}$, except $A_{i}$ itself. Therefore, the complement of $\mathcal{T}_{1}$ contains $\left(2^{m}-1\right)^{k}$ sets and
$$
\left|\mathcal{T}_{1}\right|=2^{k m}-\left(2^{m}-1\right)^{k} .
$$
Next consider the family $\mathcal{S} \backslash \mathcal{T}_{1}$. If a set $S$ intersects all $A_{i}$ but does not contain any of them, then there exists $2^{m}-2$ possible values for each $S \cap A_{i}$ : all subsets of $A_{i}$ except $\varnothing$ and $A_{i}$. Therefore the number of such sets $S$ is $\left(2^{m}-2\right)^{k}$, so
$$
\left|\mathcal{S} \backslash \mathcal{T}_{1}\right|=\left(2^{m}-2\right)^{k}
$$
From (1), (2), and (3) we obtain
$$
|\mathcal{T}|=\left|\mathcal{T}_{1}\right|-\left|\mathcal{S} \cap \mathcal{T}_{1}\right|=\left|\mathcal{T}_{1}\right|-\left(|\mathcal{S}|-\left|\mathcal{S} \backslash \mathcal{T}_{1}\right|\right)=2^{k m}-2\left(2^{m}-1\right)^{k}+\left(2^{m}-2\right)^{k}
$$
Let $\delta=\frac{3-\sqrt{5}}{2}$ and $k=k(m)=\left[2^{m} \log \frac{1}{\delta}\right]$. Then
$$
\lim _{m \rightarrow \infty} \frac{|\mathcal{S}|}{2^{k m}}=\lim _{m \rightarrow \infty}\left(1-\frac{1}{2^{m}}\right)^{k}=\exp \left(-\lim _{m \rightarrow \infty} \frac{k}{2^{m}}\right)=\delta
$$
and similarly
$$
\lim _{m \rightarrow \infty} \frac{|\mathcal{T}|}{2^{k m}}=1-2 \lim _{m \rightarrow \infty}\left(1-\frac{1}{2^{m}}\right)^{k}+\lim _{m \rightarrow \infty}\left(1-\frac{2}{2^{m}}\right)^{k}=1-2 \delta+\delta^{2}=\delta
$$
Hence, if $m$ is sufficiently large then $\frac{|\mathcal{S}|}{2^{m k}}$ and $\frac{|\mathcal{T}|}{2^{m k}}$ are greater than $\alpha$ (since $\alpha<\delta$ ). So $|\mathcal{S}|,|\mathcal{T}|>\alpha \cdot 2^{m k}=\alpha \cdot 2^{n}$.