Proposition 5

Let $X, X' \subset S$ and $w, w' \in W$. If $w(C_X) \cap w'(C_{X'}) \neq \varnothing$, then $X = X'$, $wW_X = w'W_{X'}$ and $w(C_X) = w'(C_{X'})$.

We are reduced immediately to the case $w' = 1$. The proof is by induction on the length $n$ of $w$. If $n = 0$, the assertion is clear. If $l(w) > 0$, there exists $s \in S$ such that $l(sw) = l(w) - 1$ and then (cf. the end of no. 4) $w(C) \subset s(A_s)$, hence $w(\overline{C}) \subset s(\overline{A}_s)$. Since $\overline{C} \subset \overline{A}_s$, it follows that

Then $s(x^*) = x^*$ for all $x^* \in \overline{C} \cap w(\overline{C})$, and a fortiori for all

Consequently, the relation $C_{X'} \cap w(C_X) \neq \varnothing$ implies on the one hand that $C_{X'} \cap H_s \neq \varnothing$, and hence that $s \in X'$, and on the other hand that $C_{X'} \cap sw(C_X) \neq \varnothing$. The induction hypothesis then implies that $X = X'$ and $swW_X = W_{X'} = W_X$, so $sw \in W_X$ and $w \in W_X$ since $s \in W_X$. It follows that $wW_X = W_{X'}$ and that $w(C_X) = C_X = C_{X'}$.