Exercise 15, § 5

Let E be a finite set, let $\mathfrak{L}$ and $\mathfrak{C}$ be two disjoint non-empty subsets of $\mathfrak{P}(\mathrm{E})$, and let $\lambda, h, k, l$ be four integers $\geqslant 1$ with the following properties : (i) for each $\mathrm{A} \in \mathfrak{L}$ and each $\mathrm{B} \in \mathfrak{C}$, $\mathrm{Card}\,(\mathrm{A} \cap \mathrm{B}) \geqslant \lambda$; (ii) for each $\mathrm{A} \in \mathfrak{L}$, $\mathrm{Card}\,(\mathrm{A}) \geqslant h$; (iii) for each $\mathrm{B} \in \mathfrak{C}$, $\mathrm{Card}\,(\mathrm{B}) \leqslant k$; (iv) for each $x \in \mathrm{E}$, the number of elements of $\mathfrak{L} \cup \mathfrak{C}$ which contain $x$ is exactly $l$. Show that $\mathrm{Card}\,(\mathrm{E}) \leqslant hk/\lambda$. (Let $(a_i)_{1 \leqslant i \leqslant n}$ be the sequence of distinct elements of E arranged in some order, and for each $i$ let $r_i$ be the number of elements of $\mathfrak{L}$ to which $a_i$ belongs. Show that, if $\mathrm{Card}\,(\mathfrak{L}) = s$ and $\mathrm{Card}\,(\mathfrak{C}) = t$, then we have

For $\mathrm{Card}\,(\mathrm{E})$ to be equal to $hk/\lambda$ it is necessary and sufficient that for each $\mathrm{A} \in \mathfrak{L}$ and each $\mathrm{B} \in \mathfrak{C}$ we have $\mathrm{Card}\,(\mathrm{A}) = h$, $\mathrm{Card}\,(\mathrm{B}) = k$,

Card $(A \cap B) = \lambda$, and that there exists an $r \leqslant l$ such that for each $x \in E$ the number of elements of $\mathfrak{L}$ to which $x$ belongs is equal to $r$.