Exercise 6, § 7

Let $S$ be a non-empty finite subset of a vector $K$-space $E$, not containing 0. Suppose that there exists an integer $k \geqslant 1$ with the following property:

(*) for all $x \in S$, there is a partition of $S - \{x\}$ into $h$ free subsets with $h \leqslant k$.

For such a partition $(N_i)_{1 \leqslant i \leqslant h}$ and every sequence $(r_j)_{1 \leqslant j \leqslant m}$ of integers of $\{1, h\}$ a subspace $F_{r_1 r_2 \ldots r_m}$ of $E$ is defined by induction on $m$ as follows: $F_{r_1}$ is the subspace generated by $N_{r_1}$, $F_{r_1 \ldots r_p}$ the subspace generated by the intersection of $F_{r_1 \ldots r_{p-1}}$ and $N_{r_p}$. Suppose that:

(**) for all $x \in S$ and every partition $(N_i)_{1 \leqslant i \leqslant h}$ of $S - \{x\}$ into at most $k$ free subsets, there exists at least one sequence $(r_j)_{1 \leqslant j \leqslant m}$ such that

(a) Let $n$ be the least of the integers $m \geqslant 1$ for all possible choices of $x \in S$ and let $(N_i)_{1 \leqslant i \leqslant h}$ and $(r_j)_{1 \leqslant j \leqslant m}$ satisfy conditions () and (*). Show that of necessity $n = 1$. (Argue by reductio ad absurdum by assuming $x \notin F_{r_1 r_2 \ldots r_n}$ and $n > 1$. Then necessarily $x \in F_{r_n}$. For simplicity write $E_p = F_{r_1 r_2 \ldots r_p}$ for $1 \leqslant p \leqslant n$ and $x = \sum_i \alpha_i y_i$ with $\alpha_i \in K$ and $y_i \in N_{r_n}$; there is at least one $y_i$ not belonging to $E_{n-1}$, denote it by $y$; then write $N = N_{r_n} - \{y\}$, $N'_i = N_i$ for $1 \leqslant i \leqslant h$ and $i \neq r_n$, $N'_{r_n} = N \cup \{x\}$, so that $(N'_i)_{1 \leqslant i \leqslant h}$ is a partition of $S - \{y\}$ consisting of free subsets. Define $E'_0 = E$ and for $1 \leqslant p \leqslant n$, $E'_p$ as the subspace generated by the intersection of $E'_{p-1}$ and $N'_{r_p}$. Now $y \notin E_{n-1}$; let $q$ be the least integer such that $y \notin E_q$; show that $E'_q \not\subset E_q$. Let $s$ be the least integer such that $E'_s \not\subset E_s$; show that $r_s = r_n$; deduce finally a contradiction from the relations $y \in E_s$, $E'_{s-1} \subset E_{s-1}$ and $E'_s \not\subset E_s$.)

(b) Conclude from (a) that with the given hypotheses there exists a partition of $S$ into $h$ free subsets for an integer $h \leqslant k$.