I'm dealing with Stone's representation theorem for finite Boolean algebras, i.e. if $B$ is a finite Boolean algebra and $A$ is the set of its atoms then
\begin{align} \phi: B &\to P(A)\\ b &\mapsto \{a \in A \,.\, a \leq b\} \end{align}
is a Boolean algebra-isomorphism.
For an atomic Boolean algebra $x \leq y$ if and only if $\phi(x) \subseteq \phi(y)$:
the implication $\Rightarrow$ is trivial (transitivity) while the other follows from $$x = (x \wedge y) \vee (x \wedge (\neg y))$$ and the fact that $x \wedge (\neg y) = 0$, since otherwise we would have an atom $a \leq x \wedge (\neg y)$ therefore $a \leq x$ and $a > y$ against hypotesis, so $x = x \wedge y$ and $x \leq y$.
With that we have
- $\phi$ is injective since if $\phi(b_1) = \phi(b_2)$ then $b_1 = \sup(\phi(b_1)) = \sup(\phi(b_2)) = b_2$;
Questions:
do I need $B$ to be finite for having that $\phi(\sup(X)) = X$, i.e. $\{a \in A \,.\, a \leq \sup(X)\} \subseteq X$? Or is this theorem true for atomic BA?
I can prove that $\phi$ preserve the structure $(0, 1, \wedge, \vee, \neg)$, but it seems a bit unclean(gosh, at least two equalities!); is it enough if $\phi$ preserve $0$, $1$ and is strictly monic (so it follows from the first observation and injectivity) to be a BA-morphism?