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 $\AA = \set {P \cap Q}$.

$\blacksquare$