Stone Space of Boolean Lattice is Topological Space

Theorem
Let $\left({B,\preceq,\wedge,\vee}\right)$ be a non-empty Boolean algebra.

Let $\left({U,\mathcal T}\right)$ be the Stone space of $B$.

Then $\left({U,\mathcal T}\right)$ is a topological space.

Proof
$U$ is non-empty by the Ultrafilter Lemma.

The topology of the Stone space is defined as the topology generated by the basis $Q$ consisting of all sets of the form
 * $\left\{{x \in U): b \in x}\right\}$

for some $b \in B$.

We must show that $Q$ is in fact a basis.

$Q$ is trivially a subset of the power set of $U$.

Suppose that $x \in U$.

Then by the definition of ultrafilter, $x \ne \varnothing$.

Thus $x$ has some element $b$. Then $x \in \left\{{x \in U): b \in x}\right\} \in Q$.

So $Q$ covers $U$.

Let $P,Q \in Q$. Then for some $b,c \in B$,
 * $P = \left\{{x \in U): b \in x}\right\}$ and
 * $Q = \left\{{y \in U): c \in y}\right\}$.

Let $z \in P \cap Q$.

Then by the definitions of $P$ and $Q$,
 * $b \in z$ and $c \in z$. So $z$ is an ultrafilter on $B$ containing $b$ and $c$.

Since $z$ is a filter containing $b$ and $c$, it also contains $b \wedge c$.

Let $R = \left\{{w \in U): b \wedge c \in w}\right\}$.

$R$ is clearly an element of $Q$ containing $z$.

Suppose that $v \in R$.

Then $b \wedge c \in v$.

Since $v$ is a filter,

$b \in v$ and $c \in v$.

Thus $v \in P \cap Q$.

Since this holds for each $v \in R$, $R \subseteq P \cap Q$