What does being smaller that to a join means in a distributive lattice

Question

This is a follow up question to my previous question with more restrictions. It was answered negatively for arbitrary lattices, but mentioned that the result holds "only" in distributive lattices. I would like to know whether being distributive is also sufficient, or not.
So, the question is:

Consider a subset $A$ of a distributive lattice, and some $x\lt\bigvee A.$ Dose there exist $a\in A$ with $x\le a$?

Answer 1

My original answer used an infinite example, which per bof's comment above was totally unnecessary.

In fact my claim mentioned in the OP that distributivity is sufficient if we restrict attention to two-element $A$s was blatantly false. Working in the lattice $L$ of subsets of $\{1,2,3\}$, take $A=\{\{1,2\},\{3\}\}$ and $x=\{1,3\}$. Now $L$ is about as nice as it is possible for a lattice to be - e.g. it's a Boolean algebra, so a fortiori distributive.

In fact, abstracting from this we have the following:

Suppose $L$ is a lattice with elements $a,b,c$ such that $a\vee b\not\ge c$ and $a\vee c\not\ge b$. Then setting $A=\{a\vee b, c\}$ and $x=a\vee c$ we have $x<\bigvee A$ but $x\not<$ any element of $A$.

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What if we modify things further, and simply ask that $x$ be below some finite join of elements of $A$?

Here we obviously need an infinite example, and $\mathcal{P}(\mathbb{N})$ provides one: take $A$ to be the set of finite sets and $x$ to be the set of even numbers.