A space walls $(S,M)$ is discrete if and only if $M$ is locally finite?

Question

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?

Answer 1

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 $n\in\mathbb Z$.