I am reading Burris & Sankappanavar, Chapter 1 on lattices, and I am doing Exercise 6 in Section §4: If $L$ is an algebraic lattice and $D$ a subset of $L$ such that for each $d_1$, $d_2 \in D$ there is a $d_3 \in D$ with $d_1 \leq d_3$ and $d_2 \leq d_3$ (i.e., $D$ is upward directed) then, for $a \in L$, $$ a \wedge \bigvee D = \bigvee_{d \in D} (a \wedge d). $$
I tried to prove this proposition, but it seems to me (if the proof is correct) that this is not the beautiful proof I was supposed to find. Are there better ways to prove it?
My approach:
We get the inequality from right to left for free. For the other one, use that $L$ is algebraic: thus $a = \bigvee C$, and $C$ consists of compact elements. Suppose $a \leq \bigvee D$, because the other case is easy. Then for each $c \in C$ we have $c \leq \bigvee D$, and $c$ is compact, thus $c \leq \bigvee D_0$ where $D_0 \subseteq_\omega D$. As $D$ is upward directed, we have $c \leq d_0$ for some $d_0 \in D$. Now $c \leq a$, so $c \leq a \wedge d_0$, and thus $$ c \leq \bigvee_{d \in D} (a \wedge d). $$ This can be done for each $c \in C$, so $$ a \wedge \bigvee D \leq a = \bigvee C \leq \bigvee_{d \in D} (a \wedge d), $$ which is what we wanted to prove.
Is this argument OK, please? And can this be done in a nicer or faster way? Thank you!
You're thinking in the right direction, but what about the case when $a$ and $\bigvee D$ are incomparable?
You can avoid considering these cases (and make a proof a little nicer) by letting $b = a \wedge \bigvee D$. Since $L$ is algebraic we know that $b = \bigvee C$ for some set $C$ of compact elements of $L$. For each $c \in C$ we have $c \leq \bigvee C = b = a \wedge \bigvee D$, hence $c \leq a$ and $c \leq \bigvee D$. By the compactness of $c$ there is a finite $D_0 \subseteq D$ with $c \leq \bigvee D_0$. Since $D$ is upward directed there is some $d_0 \in D$ with $c \leq \bigvee D_0 \leq d_0$. We have $c \leq a \wedge d_0$. It follows that $$a \wedge \bigvee D = b = \bigvee C \leq \bigvee_{d \in D}(a \wedge d).$$