Need a clarification of the proof that the prime ideal space of a distributive bounded lattice is compact

11.19 Theorem, from B. A. Davey, H. A. Priestley, Introduction to lattices and order, Let $L$ be a bounded distributive lattice, then the prime ideal space $\langle \mathcal{I}_p(L); \tau \rangle$ is compact. $\tau$ is the topology whose basis is $\mathcal{B}= \{X_b \cap(X\setminus X_c) : b,c \in L\}$, with $X_a = \{I \in \mathcal{I_p}(L) : a \notin I\}$.

We want to prove that the subbasis $\mathcal{S} = \{X_b : b \in L\}\cup \{X \setminus X_c : c \in L\}$ satisfies Alexander's Lemma. Let $\mathcal{U} := \{X_b : b\in A_0\}\cup\{X \setminus X_c : c \in A_1\}$, a open cover of $\mathcal{I_p}(L)$.

Let $J$ be the ideal generated by $A_0$ and $G$ the filter generated by $A_1$. It is easy to prove that $J \cap G \neq \emptyset$. So $J \cap G \neq \emptyset$ and let $a \in J \cap G$; if $A_0$ and $A_1$ are both non-empty, there exist $b_1, \dots, b_j \in A_0$ and $c_1, \dots , c_k \in A_1\:$ s.t. $\:c_1 \wedge \dots \wedge c_k \le a \le b_1 \vee \dots \vee b_j$, whence $X = X_1 = X_{b_1}\cup \dots \cup X_{b_j} \cup (X \setminus X_{c_1}) \cup \dots \cup (X \setminus X_{c_k})$.

What escapes me is why we can write $1$ in that way if $J \cap G \neq \emptyset$?

Solution 1:

To show that $$X = X_{b_1} \cup \cdots \cup X_{b_j} \cup (X \setminus X_{c_1}) \cup \cdots \cup (X \setminus X_{c_k}),$$ pick a prime ideal $I$ of $L$.

Given that $$c_1 \wedge \cdots \wedge c_k \leq b_1 \vee \cdots \vee b_j,$$ then either $$b_1 \vee \cdots \vee b_j \notin I$$ or $$c_1 \wedge \cdots \wedge c_k \in I.$$

In the first case, there is $i_0$ with $1\leq i_0\leq j$ such that $b_{i_0} \notin I$ (because $I$ is an ideal), whence $I \in X_{b_{i_0}}$.
In the second case, there is $i_1$ such that $1 \leq i_1 \leq k$ and $c_{i_1} \in I$ (because $I$ is prime), whence $I \notin X_{c_{i_1}}$ and therefore $I \in X \setminus X_{c_{i_1}}$.