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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.

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up vote 2 down vote accepted

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.

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Much appreciated. – Bojan Kwitek Nov 26 '12 at 16:37
The proof is similar to the standard proof that the free algebra and/or its regular completion (forcing for $|X|$ Cohen reals) is ccc. It is also easy to see that ccc implies there is no chain whose order type includes $\omega_1$ or $\omega_1^*$, but I don’t know how to reduce to ccc the case where the chain is order-isomorphic to an uncountable set of reals. – Emil Jeřábek Nov 26 '12 at 17:02
(Or any uncountable order type not including $\omega_1$ or $\omega_1^*$, for that matter.) – Emil Jeřábek Nov 26 '12 at 17:15
$\beta N$ has uncountable chains of clopen sets while it is separable (and hence c.c.c.), so I think there is no obvious relation. – Bojan Kwitek Nov 26 '12 at 17:23
Yes, the regular completion of the free algebra also has uncountable chains. What I was thinking about was (loosely speaking) a proof using some simple properties of the free algebra and ccc, but not involving directly any non-obvious combinatorics (such as the $\Delta$-system lemma). – Emil Jeřábek Nov 26 '12 at 17:51

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