Consider a net of weak order units in a Riesz space converging in order to a weak order unit. Is there a tail whose infimum is a weak order unit?

Question

Let $X$ be an extremally disconnected (the closure of an open set is open) compact Hausdorff space, and consider the Riesz space $C^\infty(X)$ of continuous functions from $X$ to the extended real number line $\mathbb{R}\cup\{\pm\infty\}$ such that the preimage of $\mathbb{R}$ is dense in $X$. By the Ogasawara theorem, this is a prototypical universally complete Reisz space.

A net $(x_i)_i$ in a Reisz space converges in order to $x$ if there exists a decreasing net $(y_j)_j$ with infimum zero such that for any $j$ there is an $i_0$ with $|x_i-x|\le y_j$ for all $i\ge i_0$.

Suppose $(f_i)_i$ is a net of weak order units (positive invertible functions) in $C^\infty(X)$, that is, for every $f_i$ the set $\{x\in X\colon f_i(x)>0\}$ is dense in $X$, that converges in order to a weak order unit $f$. Is it then true that there is an $i_0$ such that $\inf_{i\ge i_0}f_i$ is a weak order unit?

Answer 1

It is not true. Let $X = \beta \mathbb{N}$, so that $C(X) \cong l^\infty$. For each $i, k\in \mathbb{N}$ let $f_{i,k}$ be the function which is constantly $1$ on $\{1, \ldots, i\}$ and constantly $\frac{1}{k}$ on the rest of $X$. Also let $g_i$ be the function which is constantly $0$ on $\{1, \ldots,i\}$ and constantly $1$ on the rest of $X$.

Order the index set by $(i,k) \leq (i', k')$ if $i\leq i'$ and $k \leq k'$. The sequence $(g_i)$ is decreasing with infimum zero, and for each $i$ we have $|f_{i',k} - 1_X| \leq g_i$ for all $i' \geq i$ and all $k$. So $(f_{i,k})$ converges in order to $1_X$. But for any $i_0$, $k _0$ the infimum of all later $f_{i,k}$ is constantly zero off of $\{1, \ldots, i_0\}$.