The Boolean algebra associated to a Stone space contains only principal ultrafilters

Question

This is a needed fact in order to prove Stone's Representation Theorem, but I'm struggling to prove it. I tried using the fact that every two points on a Stone space can be separated by disjoint clopen sets and the fact that every non-principal ultrafilter contains only infinite sets to construct a contradiction, but I couldn't do it.

Edit: Given a Stone space $T$, define its associated Boolean algebra as the algebra of clopen sets in $T$. On the other side, given a Boolean algebra $B$ define its associated Stone space as the set of Ultrafilters in $B$, in which each basic open set is the collection of all ultrafilters containing a certain element of $B$.

We (firstly) need to prove that every Stone space $T$ is homeomorphic to $SB(T)$ (which is the Stone space associated to the Boolean algebra associated to $T$). The homeomorphism takes each element in $T$ and assigns it to the principal ultrafilter generated by this element. Using the fact that every two distinct points in $T$ can be separated by disjoint clopen sets one can prove that this function is injective. I'm having trouble on the next step.

Answer 1

Let $X$ be a Boolean space, i.e. compact Hausdorff and having a base of clopen sets. You want to see that every ultrafilter $\mathcal{U} \subset \operatorname{Clop}(X)$ is determined by a point,i.e. there is some $p \in X$, such that

$$\mathcal{U} = \{ A \in \operatorname{Clop}(X): p \in A \}$$

Recall that the Boolean Algebra $\operatorname{Clop}(X)$ has $\cap,\cup$ and complementation as its BA operations, and an ultrafilter on a BA is a maximal subset of the BA (by inclusion) which is, non-empty, upwards closed and closed under finite meets and does not contain $0$ (here $\emptyset$).

This immmediately implies that $\cap\{A \in \mathcal{U}\}$ is non-empty, as the members of $\mathcal{U}$ form a family of closed sets with the finite intersection property (see wikipedia) in a compact space (from having finite meets (intersections, here) and $\emptyset \notin \mathcal{U}$) . Suppose that the intersection contains two points $p \neq q$ of $X$. By Hausdorffness and having a base of clopen sets, there are $U_p, U_q$ clopen in $X$ such that $p \in U_p$, $q \in U_q$ and $U_p \cap U_q = \emptyset$. Then as we have an ultrafilter (see here under Boolean algebra) we know that either $U_p \in \mathcal{U}$ or $X \setminus U_p \in \mathcal{U}$. In the former case, $q \notin \cap\{A \in \mathcal{U}\}$ , in the latter $p \notin \cap\{A \in \mathcal{U}\}$, which is contrary to the original assumption that both were in the intersction. So $\cap\{A \in \mathcal{U}\}$ can only have a single point, say $p$. By the same type of argument ,any clopen set $C$ with $p \in C$ is in $\mathcal{U}$ (it contains either $C$ or $X \setminus C$, but $p$ must be in it). So we have the equality from the beginning: $\mathcal{U}$ is determined by $p$.

Now, if $X$ is any Boolean space (as defined above), we define the map $$F : X \rightarrow \operatorname{Stone}(\operatorname{Clop}(X)) \text{ by } F(x) = \{A \in \operatorname{Clop}(X): x \in A\}$$

Then we have just shown above that $F$ is a surjection: any element of $\operatorname{Stone}(\operatorname{Clop}(X))$, which is by definition an ultrafilter in the Boolean Algebra $\operatorname{Clop}(X)$ is in the image of $F$. The 1-1 ness follows from $X$ being Hausdorff again: if $x \neq y$ find separating clopen sets $x \in C_x, y \in C_y, C_x \cap C_y = \emptyset$. Then $C_x \in F(x), C_y \in F(y)$ by definition, and so by disjointness $C_y \notin F(x), C_y \notin F(x)$, showing that $F(x) \neq F(y)$ (we found two points of $\operatorname{Clop}(X)$ to distinguish the two subsets $F(x)$ and $F(y)$ of it).

By compactness and Hausdorffness of $X$ and $\operatorname{Stone}(\operatorname{Clop}(X))$ rspectively, we only need to show continuity of $F$ to get a homeomorphism between them.

This is clear because the definition of the topology on a stone space $\operatorname{Stone}(A)$ of any Boolean Algebra $A$, is that a (clopen in fact)
base for its topology is given by all sets of the form

$$B(a) = \{\mathcal{U} \in \operatorname{Stone}(A): a \in U\}$$

where the $\mathcal{U}$ are ultrafilters on $A$, and $a$ ranges over all elements of $A$. For continuity of $F$, it suffices to check that any basic set $B(C)$ of $\operatorname{Stone}(\operatorname{Clop}(X))$, where $C \subset X$ is clopen (we apply the above to $\operatorname{Clop}(X)$ whose points are clopen subsets of $X$):

$$x \in F^{-1}[B(C)] \leftrightarrow F(x) \in B(C) \leftrightarrow F(x) \in \{\mathcal{U} \in \operatorname{Stone}(\operatorname{Clop}(X)): C \in \mathcal{U}\} \leftrightarrow \\ C \in F(x) \leftrightarrow x \in C$$

So that $F^{-1}[B(C)] = C$ is indeed open, and continuity has been shown.

Answer 2

When you say "the Boolean algebra associated to a Stone space contains only principal ultrafilters", I guess you mean that for every ultrafilter $U$ on $Cl(S)$ (the Boolean algebra of clopen subsets of the Stone space $S$), there is some point $x$ in $S$ such that $U = \{X\in Cl(S)\mid x\in X\}$.

Terminological quibble: I'm not sure I would call such an ultrafilter "principal". I would reserve the term "principal ultrafilter" for an ultrafilter on a powerset algebra (i.e. $\{X\in \mathcal{P}(S)\mid x\in X\}$), or (more generally) for an ultrafilter on an abstract Boolean algebra $B$ which is of the form $(\uparrow b) = \{a\in B\mid b\leq a\}$ when $b$ is an atom. But unless $S$ is finite, $Cl(S)$ is not the full powerset algebra, and for many Stone spaces it will have no atoms, and hence no principal ultrafilters in my sense. For your notion, I would prefer to say that every ultrafilter on $Cl(S)$ is the restriction of a principal ultrafilter on $\mathcal{P}(S)$ along the inclusion $Cl(S)\to \mathcal{P}(S)$.

Now for how to prove it: Let $U$ be an ultrafilter on $Cl(S)$. Use compactness to show that $\bigcap_{X\in U} X$ is nonempty (take the complement of this big intersection...), and use total disconnectedness and the fact that $U$ is an ultrafilter to show that $\bigcap_{X\in U} X$ contains at most one point.

Answer 3

You are confusing two things here. Unless the Stone space in question has isolated points, there will be no principal ultrafilters on the Boolean algebra of clopen sets. Every ultrafilter is the ultrafilter of sets containing given point, yes, but it is not principal.

Under the Stone correspondence, ultrafilters correspond to points. A principal filter corresponds to a clopen set. A principal ultrafilter corresponds to a clopen, i.e. isolated point.

In other words, every point in the Stone space give you an ultrafilter, but it will only be principal if the point is isolated.

The point is that the intersection of every ultrafilter, principal or not, is a single point, or rather, pendantically speaking, the singleton of a single point. It follows from the fact that you have compactness on clopen sets (to have a point in the intersection) and the fact that clopen sets separate points (to have no more than one point in the intersection). But typically, this singleton is not a clopen set, so not a member of the algebra.