A wall is a union of panels Let $(W,S)$ be a Coxeter system.  Let $A$ be the set of cosets $wW_{S-\{s\}} : w \in W, s \in S$.  And for each $w \in W$, let
$$C_w = \{ wW_{S-\{s\}} : s \in S \}$$
Then $A$ is an apartment with chambers $C_w : w \in W$.  The group acts on $A$ by left translation, and permutes the chambers simply transitively.  A panel is a subset of $A$ obtained by removing one element from a chamber.  
I am trying to do the following exercise in Bourbaki, where walls of $A$ are defined.  From a previous exercise, if $F$ is a subset of a chamber $C_w$, then $j(F)$ is defined to be the intersection of the elements of $F$.  It will be a coset of the form $wW_X$, where $X \subseteq S$, and $\textrm{card }C_w - F = X$.  Thus $F$ is a panel if and only if $j(F) = wW_{S - \{s\}}$ for some $s \in S$.

I have worked out most of this exercise, but I am still trying to understand why a wall $L_t$ should be a union of panels.  From the fact that $F \subseteq L_t$ if and only if $j(F)$ is of the form $wW_{\{s\}}$ with $t = wsw^{-1}$, what should happen is that 
$$L_t = \bigcup\limits_{w \in W, w^{-1}tw \in S} \{ w W_{S - \{s\}} : s \neq w^{-1}tw \}$$
 A: You have to think geometrically. An apartment $A$ is the unit sphere in $R^n$, on which $W$ acts isometrically with the elements $s\in S$ acting as reflections. Walls in $A$ are fixed point sets of reflections in $W$. The components of the complement to the union of walls in $A$ are the open chambers. Closures of the chambers are the fundamental domains for the action of $W$ on $A$; they are called chambers. This cover of $A$ by chambers defines on $A$ the structure of a simplicial complex. The top-dimensional simplices are the chambers. The codimension 1 simplices are the panels. No wall passes through an open chamber (just by the definition). Since walls have dimension $n-2$, each wall $H$ is contained in the union of panels $\pi$ whose intersection with $H$ such that $dim(\pi \cap H)=n-2$. It then remains to observe that every such $\pi$ is contained in $H$. Indeed, $\pi$ is a spherical simplex of dimension $n-2$. Thus, each $\pi$ spans a great sphere $H_\pi\subset A$ (of dimension $n-2$). Since $H_\pi\cap H$ has dimension $n-2$, the great spheres $H_\pi, H$ have to be equal.  Hence, $\pi\subset H$. Therefore, $H$ is the union of panels, as required.  
