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Let $B$ be a Boolean algebra. There are two ways to turn $B$ into a topological space.

  1. View $B$ as a discrete space and take the Stone-Cech compactification resulting in $\beta(X)$. This is the same as the Stone space associated to the Boolean algebra $\mathcal{P}(X)$ of powersets of $X$.
  2. Take the Stone space of $\mathcal{B}$, $\mathrm{Ult}(B)$.

What is the relationship between $\mathrm{Ult}(B)$ and $\beta(X)$?

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    $\begingroup$ Definition 1 doesn't use the Boolean algebra structure of $B$, so the answer is "not very much". Consider the fact that there are $2^{\aleph_0}$ non-isomorphic countable Boolean algebras. $\endgroup$
    – Robert Furber
    Commented Jan 20, 2023 at 12:35
  • $\begingroup$ The Stone space of a Boolean algebra has as underlying set its set of ultrafilters, but it's not its powerset; indeed, it's a compact Hausdorff space, and as such it coincides with its Stone-Čech compactification. $\endgroup$
    – amrsa
    Commented Jan 20, 2023 at 12:42
  • $\begingroup$ @RobertFurber I have a follow-up question. Now let $B$ be a Boolean algebra of sets, so a subset of the powerset of some finite or infinite set $X$ such that $B$ forms a topological basis and let the space generated be $Y$. It seems more likely that there is a nice relationship between the Stone-Cech compactification of $Y$ and $Ult(B)$ since they both utilize the Boolean algebra structure. Do you have any guesses about this? $\endgroup$
    – IllogicalUser
    Commented Jan 20, 2023 at 13:27
  • $\begingroup$ @T-Biconditional In your question, do you mind saying what $X$ is? You have not specified it. $\endgroup$
    – PatrickR
    Commented Jan 21, 2023 at 6:01

1 Answer 1

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The Stone space of a product of a family of Boolean algebras is homeomorphic to the Stone-Cech compactification of the union of the Stone spaces of the members of the family, see S. Koppelberg, Handbook of Boolean Algebras Vol 1, Theorem 3.8.9.

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