Exercise 3, § 5

This solution was written by laguna-s-2.1-free and judged by a machine. It is not Bourbaki's and it has not been checked by a person.

Let $x$ be an element of $\bigcap_{\mathrm{H} \in \mathfrak{F}_k} \mathrm{P_H}$ when $k \leqslant \frac{1}{2}(n+1)$. For every $\mathrm{H} \in \mathfrak{F}_k$, the element $x$ belongs to $\mathrm{P_H} = \bigcup_{\iota \in \mathrm{H}} \mathrm{X_i}$, so there exists $\iota \in \mathrm{H}$ such that $x \in \mathrm{X_{i}}$. Define $\mathrm{T} = \{ i \in [1,n] \mid x \in \mathrm{X_i} \}$. For every $\mathrm{H} \in \mathfrak{F}_k$ we must have $\mathrm{H} \cap \mathrm{T} \neq \varnothing$, for otherwise $x \in \mathrm{P_H}$ would fail. Hence, no subset of $[1,n]\setminus \mathrm{T}$ can contain a subset $\mathrm{H} \in \mathfrak{F}_k$, implying $|[1,n]\setminus \mathrm{T}| < k$. Therefore, $|\mathrm{T}| \geqslant n - (k-1)$. Since $k \leqslant \frac{1}{2}(n+1)$, this gives $|\mathrm{T}| \geqslant n - (k-1) \geqslant k$, so $\mathrm{T}$ contains a subset $\mathrm{H}_0 \in \mathfrak{F}_k$. Then $x \in \mathrm{Q_{H_0}}$, hence $x \in \bigcup_{\mathrm{H} \in \mathfrak{F}_k} \mathrm{Q_H}$.

Conversely, suppose $k \geqslant \frac{1}{2}(n+1)$ and let $x \in \bigcup_{\mathrm{H} \in \mathfrak{F}_k} \mathrm{Q_H}$. There exists $\mathrm{H}_0 \in \mathfrak{F}_k$ such that $x \in \mathrm{Q_{H_0}}$, meaning $x \in \mathrm{X_i}$ for all $i \in \mathrm{H}_0$. For any $\mathrm{H} \in \mathfrak{F}_k$, the two subsets $\mathrm{H}_0$ and $\mathrm{H}$ of $[1,n]$ satisfy $|\mathrm{H}_0 \cap \mathrm{H}| \geqslant 2k - n$, which is positive because $k \geqslant \frac{1}{2}(n+1)$ implies $2k - n \geqslant 1$. Thus, there exists $i \in \mathrm{H}_0 \cap \mathrm{H}$ such that $x \in \mathrm{X_i}$. Since $i \in \mathrm{H}$, it follows that $x \in \mathrm{P_H}$. Hence, $x \in \bigcap_{\mathrm{H} \in \mathfrak{F}_k} \mathrm{P_H}$.