This sounds like a very naïve question, but I couldn't find a correct argument to prove/disprove it rigorously.
Suppose we have a a subset $A$ of a lattice (or any join-semilattice), and some $x\le\bigvee A.$
Dose this means that there is $a\in A$ with $x\le a$?
The answer is clearly "Yes" in a totally ordered set. But unfortunately, I am working in an arbitrary lattice.
In case you need, the definition of the join $\bigvee A$ says that:
- $a\le \bigvee A$ for all $a\in A$;
- if there is some $b$ such that $a\le b$ for all $a\in A,$ then $\bigvee A\le b.$
Actually, the statement isn't true in arbitrary total orders. For example, consider the obvious ordering on $\mathbb{N}\cup\{\infty\}$ and let $A=\mathbb{N}$. Then $\bigvee A=\infty$ but no element of $A$ is $\ge\infty$.
If we restrict attention to finite $A$s then the statement in question is trivially true in total orders, and it is also trivially true for arbitrary $A$s in co-well-founded total orders, but those modifications are not very interesting.
And in fact the statement is never true in non-totally-ordered lattices even for finite $A$s: if $L$ is a non-totally-ordered lattice, let $A=\{a,b\}$ be a pair of $L$-incomparable elements. Then $\bigvee A=a\vee b$ is strictly above all elements of $A$.
Things get a bit more interesting if we replace "$x\le\bigvee A$" with "$x<\bigvee A$." Now for example the statement is true in all total orders.
It's still not true in all lattices, however, even when $A$ has only two elements: consider the lattice $M_3$ and let $A$ be one of the three pairs of incomparable elements. Then $\bigvee A$ is the top element, which lies above the third "intermediate" element of the lattice. A counterexample with $\vert A\vert=2$ also occurs in $N_5$ (exercise), and so we get that the even the modified statement above restricted to two-element $A$s fails in all non-distributive lattices.
In short, we have the following results in terms of which versions of the property in the OP hold in which lattices:
The original property in the OP holds exactly in the co-well-founded total orders.
The variant of that property gotten by restricting attention to finite $A$s holds exactly in the total orders.
The variant of the original property gotten by replacing "$\le$" with "$<$" holds only in distributive lattices.
This leaves open the question of whether there is a distributive lattice in which the variant of the original property with "$\le$" replaced with "$<$" but allowing arbitrary $A$s fails.
I strongly suspect the answer is yes, but I don't immediately see a counterexample.EDIT: I was thinking too hard.All in all, anything along the lines of the property in the OP seems extremely rare.