then ${\mathscr{T}}\cap{\mathscr{L}}_{\lambda,\geq C}$ is a lower order ideal in ${\mathscr{T}}_B$. \[cor\_lower\_order\_ideal\]

It suffices to prove the second statement. Let $T,T'$ be such that $S(B,T)\subset S(B,T')$ and $T' \in {\mathscr{L}}_{\lambda,\geq C}$. We want to show that $T\in{\mathscr{L}}_{\lambda,\geq C}$. As ${\mathscr{T}}_B$ is a graded poset and the intersection ${\mathscr{T}}\cap{\mathscr{L}}_{\geq C}$ is a lower order ideal in ${\mathscr{T}}_B$ (see Remark \[rmk\_star\_convex\]), we know that $T\in{\mathscr{T}}\cap{\mathscr{L}}_{\geq C}$, and it suffices to treat the case where $S(T',B)-S(T,B)$ is a singleton. Let $f$ be the single index of $S(B,T')-S(B,T)$. As $B,T'\in{\mathscr{T}}\cap{\mathscr{L}}_{\geq C}$, we have $f\in E(C)$ by Lemma \[lemma\_star\](ii), so $B\subset H^+_{\alpha_f}$. As $f \in S(B,T')-S(B,T)$, we have $T'\subset H^-_{\alpha_f}$ and $T\subset H^+_{\alpha_f}$. Also, as $f \in E(C)$, we have $(\lambda,\alpha_f) \geq 0$. So we may apply Proposition \[prop\_crossing\_wall\], and we obtain that $T\in{\mathscr{L}}_{\lambda}$.

Suppose that the arrangement ${\mathcal{H}}$ satisfies condition \[(A)\]. Then the complex $\Sigma({\mathscr{L}}_\lambda)$ is shellable. Moreover, there exists a shelling order on its chambers which is an initial shelling of $\Sigma({\mathscr{L}})$. In particular, if $\lambda\not=0$ then $\Sigma({\mathscr{L}}_\lambda)$ is a shellable $PL$ ball of dimension $\dim(V/V_0)-1$. \[cor\_shelling\_order\]

If $\lambda=0$, then ${\mathscr{L}}_\lambda={\mathscr{L}}$, so $\Sigma({\mathscr{L}}_\lambda)=\Sigma({\mathscr{L}})$, and the corollary is just Theorem \[thm\_shelling\_order\].

We now assume that $\lambda\not=0$. By Theorem \[thm\_shelling\_order\] and Corollary \[cor\_lower\_order\_ideal\], it suffices to 