My question concerns the left-to-right implication of the following:
Theorem A frame $L$ is compact and zero-dimensional iff it is isomorphic to the lattice of ideals $\mathcal{I}(B)$ of some Boolean algebra $B$.
A frame $L$ is a complete and bounded lattice which satisfies: $x\wedge \bigvee_{i\in I}a_i=\bigvee_{i\in I} x\wedge s_i$ (it is a complete Heyting algebra). $a\in L$ is complemented iff there is a unique $a^\ast$ such that $a\wedge a^\ast=0$ and $a\vee a^\ast=1$. We take $Z(L)$ to be the set of all complemented elements of $L$. $Z(L)$ is a Boolean algebra, and is usually called the center of $L$.
$L$ is zero-dimensional iff for every $a\in L$ there exists $S\subseteq Z(L)$ such that $a=\bigvee S$. Finally $L$ is compact iff it's compact as a lattice.
So, in a proof of the theorem we take the lattice of ideals of the center of $L$ and define $h\colon L\rightarrow \mathcal{I}(Z(L))$ as $h(a)=(a]\cap Z(L)$, where $(a]=\{x\in L\mid x\leqslant a\}$. Of course, $h(a)$ must be an ideal, but I fail to see how to prove that it is a downward set (the other two properties of ideals are rather obvious: $0\in h(a)$ and $h(a)$ must closed for finite joins since $Z(L)$ is a sublattice of $L$). But taking $x\in h(a)$ and $y\leqslant x$, I cannot see how to prove that $y\in Z(L)$.
Could you please help me with this?
Best Answer
You want to prove that $h(a)$ is an ideal of $Z(L)$. Thus, it has to be downward closed in $Z(L)$, not in $L$.
Showing that $h(a)$ is downward closed in $Z(L)$ is very easy because $h(a)$ is the intersection of a downward closed subset of $L$ with $Z(L)$.