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What are the open sets of a Stone space of a Boolean algebra B? According to Wikipedia, they are the ultrafilters on B that contain an element of B. However, this cannot be the case since an ultrafilter is a set of subsets of B and contains no elements of B. Using the alternate definition of the Stone space as the space of homomorphisms from B to {0,1} doesn't offer much help, either.

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An ultrafilter on a Boolean algebra is a subset of $B$, not of the power set of $B$. You may be confused because $B$ itself may be viewed as (part of) a power set of of another set. – JDH Nov 9 '10 at 20:16
up vote 6 down vote accepted

I think you are confused by the two meanings of ultrafilter. In the context of a Boolean algebra $B$, an ultrafilter is just a maximal proper filter, that is, a subset of $B$ that is closed under $\wedge$, upward closed, and does not contain $0$. The Stone space $M(B)$ consists of all such maximal filters, and its usual topology is generated by basic open sets of the form $$ N_p = \{ f \in M(B) : p \in f \} $$ with one such basic open set $N_p$ for each $p \in B$.

There is a second, incompatible meaning of ultrafilter: an ultrafilter on an arbitrary set $S$ is a maximal filter in the above sense on the lattice of subsets of $S$. This is what people mean when they say "an ultrafilter on $\omega$", for example. When $S$ is already being treated as a lattice, the former sense is usually what is meant, but you can only tell by context.

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