Let $A$ be a complete Boolean algebra, and let $\kappa$ be a cardinal. We will say that $B\subset A$ is $\kappa$-broad, if $|B|\ge\kappa$ and for any $B'\subset B$ with $|B'|\ge\kappa$ we have $\bigwedge B'=0$ and $\bigvee B'=\bigvee B$.
Example 1: If $A$ is an atomless probability measure algebra, it contains an $\aleph_{0}$-broad set. Indeed, take a sequence of independent events of probability $\frac{1}{2}$.
Example 2: If $A$ is the measure algebra corresponding to Lebesgue measure on $[0,1]$ we were able to show existence of a $\aleph_{1}$-broad set under CH. I think this can be generalized to arbitrary finite measure algebra using Maharam's classification.
Have this property been considered? What are sufficient conditions for existence of $\aleph_{1}$-broad sets?
Now here is the case which actually came up in my work. Let $A$ be an atomless Maharam algebra, i.e. an atomless complete CCC Boolean algebra, endowed with a Maharam submeasure, i.e. an order preserving $\rho:A\to [0,1]$ such that $\rho(a)=0$ iff $a=0$, $\rho(a\vee b)\le \rho(a)+\rho(b)$, for every $a,b\in A$, and whenever $a_{p}\downarrow 0$ we have $\rho(a_{p})\downarrow 0$.
Theorem (Velickovic, 05): Every atomless Maharam algebra contains an $\aleph_{0}$-broad set.
The main question is as follows.
Does every atomless Maharam algebra contain an $\aleph_{1}$-broad set $B$ such that $0\notin B$, but for every $\varepsilon>0$ there is $b\in B$ with $\rho(b)<\varepsilon$?
Note that our construction in Example 2 does in fact yield this second condition as well.
