My question will be very short. 


*Suppose we have a Boolean algebra $B$ which admits an uncountable independent family. Does it follow that there is an uncountable chain of elements in $B$?*


Manifestly, this is the case for (infinite) complete Boolean algebras, although the proofs of existence of uncountable independent families/chains in these algebras seem to have nothing in common. 



Independent families and chains
The free Boolean algebra in any number of generators contains no uncountable chain. This can be seen as follows.


If $X$ is the set of generators, we can identify elements of the algebra with functions $f\colon2^X\to2$ which only depend on finitely many variables. Let $d(f)$ be the set of variables $f$ depends on. Assume for contradiction that $C$ is an uncountable chain in the algebra. By the $\Delta$-system lemma, we may assume that $\{d(f):f\in C\}$ is a $\Delta$-system with kernel $u$. Let $a\in2^u$. For any $f,g\in C$, $f< g$ implies that $f(b)=0$ for every $b\in2^X$ extending $a$, or $g(b)=1$ for every $b$ extending $a$, because $d(f)\cap d(g)=u$. One of these two possibilities has to occur uncountably many often. We can repeat this for all of the finitely many possible choices of $a$. We thus obtain an uncountable chain $C$ such that every $f\in C$ depends only on variables in $u$, but this is impossible, as there are only finitely many such functions.


