Given a Banach lattice $X$, it is well know that every finite subset $A$ of $X$ has a supremum and an infimum. For example, if $\{ x_1, \ldots, x_n\}$ is a finite subset of $X$, then we have $$ sup (\{ x_1, \ldots, x_n\}) = \bigvee^n_{i=1} x_i \, \,, \quad \text{and} \quad inf (\{ x_1, \ldots, x_n\}) = \bigwedge^n_{i=1} x_i \, \,. $$
Now we consider an uncountable subset $A = \{ a_i \colon i \in I \} \subset X$ (where $I$ is an index set). Also we assume that there is an element $\bar{a} \in X$ such that $a_i \leq \bar{a}$ for each $i \in I$ and there is an element $\underline{a} \in X$ such that $\underline{a} \leq a_i$ for each $i \in I$.
I am curious that under which kind of conditions, the existence of a supremum and an infimum of the set $A$ can be guaranteed?
Any idea or suggestions are really much appreciated! Thank you so much in advance!
The standard example of a Banach lattice is $C(K)$ for compact Hausdorff space $K$. Typically, bounded infinite sets here might not have a supremum or infimum. For example, if $s \in K$ is not isolated, $\{f \in C(K):\; f \ge 0, \; f(s) = 1\}$ has no infimum.
EDIT: Using the Tietze extension theorem, for any $t \ne s$ there is $f \in C(K)$ with $f \ge 0$, $f(s)=1$ and $f(t)=0$. Thus if an infimum $g$ existed, it would have to be $$ g(x) = \cases{1 & if $x=s$\cr 0 & otherwise}$$ But that's not continuous unless $\{s\}$ is open, i.e. $s$ is an isolated point.