find a family of signs $(\varepsilon_e) \in \{\pm 1\}^E$ such that:

-   for every $e \in E$, we have $(\lambda,\varepsilon_e\alpha_e)\geq 0$;

-   there exists a chamber $B\in{\mathscr{L}}_\lambda$ with $B\subset
    H^+_{\varepsilon_e\alpha_e}$ for every $e \in E$.

Indeed, Corollary \[cor\_lower\_order\_ideal\] will then imply that ${\mathscr{T}}\cap{\mathscr{L}}_\lambda$ is a lower order ideal in ${\mathscr{T}}_B$, so it will be an initial segment for at least one linear extension of $\preceq_B$.

Let $F = \{e\in E : (\lambda,\alpha_e)\not=0\}$. For every $e\in F$, we choose $\varepsilon_e \in \{\pm 1\}$ such that $(\lambda,\varepsilon_e\alpha_e)>0$. Let $x_0$ be a point in $V$ not on any hyperplane of ${\mathcal{H}}$, that is, $x_0\in V - \bigcup_{e \in E} H_{\alpha_e}$. Then for every $e \in F$, the inner product $(x_0 + c \cdot \lambda,\varepsilon_e\alpha_e)
= (x_0,\varepsilon_e\alpha_e)+c \cdot (\lambda,\varepsilon_e\alpha_e)
$ tends to $+\infty$ as $c$ tends to $+\infty$, so it is positive for $c$ large enough. Similarly, the inner product $(x_0+c \cdot \lambda,\lambda)=(x_0,\lambda)+c \cdot (\lambda,\lambda)$ is positive for $c$ large enough. On the other hand, if $e\in E -  F$, then $(x_0+c \cdot \lambda,\alpha_e)=(x_0,\alpha_e)\not=0$ for every $c\in{{\mathbb{R}}}$. So, if $c\in{{\mathbb{R}}}$ is large enough, then $x=x_0+c \cdot \lambda\in V - 
\bigcup_{e \in E} H_{\alpha_e}$, and $x$ is in $H^+_{\varepsilon_e\alpha_e}$ for every $e\in F$ and in 