The separability of superextensions

Question

The superextension $\lambda X$ of a compact Hausdorff space $X$ is the space of maximal linked systems of closed subsets of $X$, endowed with the Vietoris topology inherited from the double hyperspace of $X$.

It is easy to show that the density $d(\lambda X)$ of the superextension of a compact Hausdorff space $X$ does not exceed the density $d(X)$ of $X$. Is the converse true?

Question. Is $d(\lambda X)=d(X)$ for every compact Hausdorff space $X$?

I recall that the density $d(X)$ of a topological space $X$ is the smallest cardinality of a dense set in $X$.

Answer 1

There are non-separable zero-dimensional compact Hausdorff spaces with a separable superextension. Because of zero-dimensionality one obtains $\lambda X$ as the space of all maximal linked systems of clopen sets. In order that $\lambda X$ be separable it is necessary and sufficient that the Boolean algebra of clopen sets is $\sigma$-linked: extend each of the countaly many linked systems to a maximal one, the resulting countable set is dense in $\lambda X$. In order that $X$ itself is not separable it is necessary and sufficient that the Boolean algebra of clopen sets is not $\sigma$-centered.

The measure algebra on $[0,1]$ is $\sigma$-linked, but not $\sigma$-centered. Hence its Stone space is a counterexample.

Other examples were constructed by Murray Bell in Two Boolean Algebras with Extreme Cellular and Compactness Properties (Canadian Journal of Mathematics, Volume 35, Issue 5, 01 October 1983, pp. 824-838). For example a Boolean algebra that is $\sigma$-linked, but not $\sigma$-$3$-linked. These have the advantage that they are subalgebras of $\mathcal{P}(\omega)/\mathit{fin}$; hence their Stone spaces provide examples that are even remainders in compactifications of $\mathbb{N}$. The measure algebra need not be embeddable in that algebra. Thanks to Jan van Mill for this reference.