Topology on Stone Spaces of Boolean Algebra

Question

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.

Answer 1

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.