Simple question on predense set in a boolean algebra

Question

Let B a complete boolean algebra and D a subsets of B. Then D is predense below $ b\in B $, i.e. the downward closure of D is dense below b, iff $b\le \bigvee D$.Proving this equivalence seemed like it would be an easy exercise, but I think I'm missing something. Can someone help me, please?

Answer 1

HINT:

• If $b\not\le\bigvee D$, let $x=b\land\left(\neg\bigvee D\right)$, and use $x$ to show that $D$ is not predense below $b$.

• Show that if $D$ is not predense below $b$, there is an $x\le b$ such that $x\land d=0$ for all $d\in D$, and deduce that $b\not\le\bigvee D$.

You may find it helpful to start by proving the special case $b=1$: $D$ is predense in $B$ iff $\bigvee D=1$. It uses the same ideas, but with slightly less clutter.