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.

Answer 2

You seem to be interested in the notion of join-prime element in a lattice.
In any lattice, a join-prime element is also join-irreducible, and the converse holds for distributive lattices.
Also, in any lattice, an element is join-irreducible iff it covers exactly one element.

In case you are allowing the set $A$ to be infinite (for infinite lattices, of course), then the notion is the one of completely join-prime element.
Again, every completely join-prime element is also completely join-irreducible, and the converse holds for lattices which satisfy the Join Infinite Distributive law: \begin{equation} x \wedge \bigvee_{I\in I}y_i = \bigvee_{I\in I}x \wedge y_i \tag{JID} \end{equation} where $I$ is any set. Of course every lattice satisfying (JID) is distributive, but the converse doesn't hold.
Again, an element is completely join-irreducible iff it covers exactly one element.

There are dual notions of (completely) meet-prime elements, with dual results connecting them to (completely) meet-irreducible ones, and the dual notion of (MID).