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Is there an example of an uncountable Boolean algebra $B$ in which every chain is countable and such that $\ell_\infty$ embeds into the Banach space $C(\mbox{Stone }B)$? The latter requirement is not very important, I just want to exclude some trivial cases, like the algebra of finite/cofinite subsets on some uncountable set.

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The second requirement is too strict: it makes the first one impossible.

The commutative von Neumann algebra $\ell^\infty(\mathbb{N})$ has as Gelfand spectrum the Stone-Cech compactification of $\mathbb{N}$ (with the discrete topology). This, in turn, is the Stone space of the Boolean algebra $\mathcal{P}(\mathbb{N})$, the powerset of the natural numbers. So $\ell^\infty(\mathbb{N}) \cong C(\mathop{Stone}(\mathcal{P}(\mathbb{N})))$ embeds in $C(\mathop{Stone}(B))$ if and only if $\mathcal{P}(\mathbb{N})$ embeds in $B$. But the former has uncountable chains. So if $B$ satisfies the second requirement, it has uncountable chains, and cannot satisfy the first requirement.

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$P(\mathbb{N})$ has uncountable chains. Think of Dedekind cuts in $\mathbb{Q}$. – Joel David Hamkins Nov 30 '12 at 0:03
Every well-ordered chain is countable, but there are uncountable chains isomorphic to the real numbers. – Asaf Karagila Nov 30 '12 at 1:04
Good point, thanks! I'll edit the answer. – Chris Heunen Nov 30 '12 at 1:14

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