Let $ \mathcal{L} = \left(L, \leqslant \right)$ be a bounded distributive lattice, and $\left(Idl \left( \mathcal{L} \right), \subseteq \right)$ be a lattice of ideals of the lattice $\mathcal{L}$.
I met the following problem in my course on Algebraic logic:
$$\text{Prove that} \left(Idl \left( \mathcal{L} \right), \subseteq \right) \text{ is a complete pseudo-complemented lattice.}$$
I have proved that this is a complete lattice, but why it is a pseudo-complemented lattice I cannot understand.
I know that a complete lattice is a pseudo-complemented lattice iff an infinite distributive law is fulfilled in it:
$$x \wedge \bigvee S = \bigvee \{x \wedge y ~ | ~ y \in S \} \text{ for all } x \in Idl \left( \mathcal{L} \right), S \subseteq Idl \left( \mathcal{L} \right).$$
But I don't understand how to prove the inequality from left to right in this case.
Any help or hints will be very helpful. Thanks!
Best Answer
Perhaps you know that, if $\mathbf L$ is a lattice and $A \subseteq L$, then, $(A]$, the ideal of $\mathbf L$ generated by $A$ is given by $$a \in (A] \iff a \leq a_1 \vee \cdots \vee a_n,$$ for some $n \in \omega$ and $a_1, \ldots, a_n \in A$.
Thus, if $S$ is a set of ideals of $\mathbf L$, then $\bigvee S = \left( \bigcup S \right]$ is given by $$a \in \bigvee S \iff a \leq a_1 \vee \cdots \vee a_n,$$ where $a_i \in y_i \in S$.
Now, if $\mathbf L$ is distributive, $x$ is an ideal of $\mathbf L$ and $S$ is a set of ideals of $\mathbf L$, then, taking $b \in x \wedge \bigvee S$, we have $b \in x$ and $b \leq b_1 \vee \cdots \vee b_n$, where $b_i \in y_i \in S$, and so $$b = b \wedge (b_1 \vee \cdots \vee b_n) = (b \wedge b_1) \vee \cdots \vee (b \wedge b_n) \in \bigvee_{y \in S}(x \wedge y).$$ Therefore, $x \wedge \bigvee S \leq \bigvee_{y\in S}(x \wedge y)$, while the converse inequality holds for all lattices.