Stone Space of Boolean Lattice is Topological Space
Theorem
Let $\struct {B, \preceq, \wedge, \vee}$ be a Boolean lattice.
Let $\struct {U, \tau}$ be the Stone space of $B$.
Then $\struct {U, \tau}$ is a topological space.
Proof
The topology of the Stone space is defined as the topology generated by the basis $Q$ consisting of all sets of the form
- $\set {x \in U: b \in x}$
for some $b \in B$, where $U$ is the set of all ultrafilters on $B$.
By Union from Synthetic Basis is Topology, it suffices to show that $Q$ is a synthetic basis.
First, we must show that $Q$ is a cover of $U$.
Let $x_0 \in U$ be arbitrary.
By filter axiom $\paren 1$, $x_0 \ne \O$.
Thus, there is some $b_0 \in x_0$.
Then, $x_0 \in \set {x \in U: b_0 \in x} \in Q$.
As $x_0$ was arbitrary, it follows that $Q$ is a cover of $U$.
Hence, synthetic basis axiom $\paren {\text B 1}$ holds for $Q$.
$\Box$
Let $P, Q \in Q$ be arbitrary.
Then for some $b, c \in B$:
- $P = \set {x \in U: b \in x}$
- $Q = \set {y \in U: c \in y}$
Let $R = \set {z \in U : b \wedge c \in z}$
We will show that:
- $P \cap Q = R$
Let $z \in P \cap Q$ be arbitrary.
Then by the definitions of $P$ and $Q$:
- $b \in z$
- $c \in z$
By definition of lattice, $\wedge$ denotes the meet operator.
Therefore:
- $b \wedge c \preceq b$
- $b \wedge c \preceq c$
- $\forall d: (d \preceq b) \land (d \preceq c) \implies (d \preceq b \wedge c)$
Therefore, by filter axioms $\paren 2$ and $\paren 3$, $b \wedge c \in z$.
As $z \in P \cap Q$ was arbitrary, it follows that $P \cap Q \subseteq R$.
Now, let $v \in R$ be arbitrary.
Then $b \wedge c \in v$.
As stated above:
- $b \wedge c \preceq b$
- $b \wedge c \preceq c$
Therefore, by filter axiom $\paren 3$:
- $b \in v$
- $c \in v$
Thus $v \in P \cap Q$.
Since this holds for each $v \in R$, $R \subseteq P \cap Q$.
From the above, it follows that $P \cap Q = R$ by the definition of set equality.
Therefore, synthetic basis axiom $\paren {\text B 2}$ holds for $Q$ by taking $\AA = \set {P \cap Q}$.
$\blacksquare$