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?



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?
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 $\{1, \ldots, i\_0\}$.


