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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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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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