TITLE: A space walls $(S,M)$ is discrete if and only if $M$ is locally finite?
QUESTION [1 upvotes]: A space with walls is simple a set a pair $(S,M)$ is simply a set $S$ together with a collection of subsets $M$ closed under complementation. A space with walls clearly forms a pocset under inclusion. $A < B$ implies $B^* < A^*$.
A pocset is a poset $M$ together with an order reversing involution $A\to A^*$ such that $A\not= A^*$, $A$ and $A^*$ are incomparable
A pocset is said to be locally finite if every interval is finite, i.e. if $A<B$ then $\{C|A<C<B\}$ is a finite set.
A space with walls clearly forms a pocset under inclusion. It is said to be discrete if for any two elements of $a, b \in S$ the collection of subsets in $M$ containing $a$ and not containing $b$ is finite. 
Is that true $(S,M)$ is discrete if and only if $M$ is locally finite?

REPLY [0 votes]: The following seems to be a counterexample. What am I missing?
Let $S$ be the set of all periodic functions $\sigma:\mathbb Z\to\mathbb Z$.
For $x,y\in\mathbb Z$, let
$L_{x,y}=\{\sigma\in S:\sigma(x)\le y\}$ and
$U_{x,y}=\{\sigma\in S:\sigma(x)\ge y\}$.
Let $M=\{L_{x,y}:x,y\in\mathbb Z\}\cup\{U_{x,y}:x,y\in\mathbb Z\}$.
Then $(S,M)$ is a space with walls, and $M$ is locally finite, but $(S,M)$ is not discrete.
$(S,M)$ is a space with walls.
The definition requires $M$ to be closed under complementation. The complement of $L_{x,y}$ is $U_{x,y+1}$, and the complement of $U_{x,y}$ is $L_{x.y-1}$.
$M$ is locally finite.
Suppose $A,B\in M$ and $A\lt B$. There are two cases.
Case 1. $A=L_{x,y}$ and $B=L_{x,z}$ where $y\lt z$. Then
$\{C:A\lt C\lt B\}=\{L_{x,w}:y\lt w\lt z\}$ is finite.
Case 2. $A=U_{x,y}$ and $B=U_{x,z}$ where $y\gt z$. Then
$\{C:A\lt C\lt B\}=\{U_{x,w}:y\gt w\gt z\}$ is finite.
$(S,M)$ is not discrete.
Suppose $\sigma_1,\sigma_2\in S$ and $\sigma_1\ne\sigma_2$. Let $\sigma_i$ be periodic with period $p_i\gt0$; then both $\sigma_1,\sigma_2$ are periodic with period $p=p_1p_2$. Choose $x\in\mathbb Z$ with $\sigma_1(x)\ne\sigma_2(x)$ and let $y_1=\sigma_1(x)$ and $y_2=\sigma_2(x)$. Then $\sigma_1(x+np)=y_1$ and $\sigma_2(x+np)=y_2$ for all $n\in\mathbb Z$. let
$$A_n=\begin{cases}
L_{x+np,y_1}\text{ if }y_1\lt y_2,\\
U_{x+np,y_1}\text{ if }y_1\gt y_2.
\end{cases}$$
Then $\sigma_1\in A_n$ but $\sigma_2\notin A_n$ for all $n\in\mathbb Z$.
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