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On page 81, Set Theory, Jech(2006), to prove the Stone's Representation Theorem, a mapping $\pi$ is defined as

Let $B$ be a Boolean algebra. We let $$S=\{p:p \text{ is an ultrafilter on }B\}.$$ For every $u \in B$, Let $X_u$ be the set of all $p \in S$ such that $u \in p$. Let $$\mathcal{S} =\{X_u : u \in B\}.$$ Let us consider the mapping $\pi(u) = X_u$ from $B$ to $\mathcal{S}$.

It is claimed that $\pi(u \cdot v)=\pi(u) \cap \pi(v)$ and $\pi(u + v)=\pi(u) \cup \pi(v)$, which "follows from the definition of ultrafilter".

I can show the former, but don't know how to deal with the latter. At first, I thought applying the duality argument to the first proof should work. However, it's not the case, unless we replace "ultrafilter" by "prime ideal" in the reasoning.

Addendum: Here's how I show $\pi(u \cdot v)=\pi(u) \cap \pi(v)$. Semantically, $\pi$ maps $u$ to the set of all ultrafilters on $B$ such that $u$ belongs to. Since $u \cdot v \leq u$, we have $\pi(u \cdot v)\subseteq\pi(u)$. Thus $\pi(u \cdot v) \subseteq\pi(u) \cap \pi(v)$.

If $p \in \pi(u) \cap \pi(v)$, then $u \in p$ and $v \in p$. Thus $u \cdot v \in p$. As a consequence, $\pi(u) \cap \pi(v) \subseteq \pi(u \cdot v)$

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up vote 2 down vote accepted

Let $p\in S$. If $u\in p$ or $v\in p$, then certainly $u+v\in p$, so $X_u\cup X_v\subseteq X_{u+v}$. Now suppose that $u+v\in p$. Since $p$ is an ultrafilter, at least one of $u$ and $v$ must belong to $p$, so $p\in X_u\cup X_v$, and $X_{u+v}\subseteq X_u\cup X_v$. Thus, $X_{u+v}=X_u\cup X_v$, as claimed.

In case it isn’t clear why $u+v\in p$ implies that at least one of $u$ and $v$ is in $p$, suppose that $u,v\notin p$. Then since $p$ is an ultrafilter we must have $-u,-v\in p$ and hence $$-(u+v)=(-u)\cdot(-v)\in p\;,$$ contradicting the hypothesis that $u+v\in p$.

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