The question is in the title: what property does a lattice need to have such that for every element of the lattice $x$, there exists a set of ultrafilters in the lattice such that the intersection of that set of ultrafilters is identical to the principal filter of $x$?
The above is at least true for finite lattices. Is this property the same as being atomistic?
Best Answer
Assuming that by "ultrafilter" you mean "maximal proper filter", this property of a bounded lattice $L$ is equivalent to the statement that for all $a,b\in L$ with $a\not\leq b$, there exists a nonzero $c\leq a$ such that $b\wedge c=0$.
Indeed, assuming that statement, then given a principal filter $F$ generated by an element $a$ and any $b\not\in F$, there is a nonzero $c\leq a$ such that $b\wedge c=0$. Then an ultrafilter containing the filter generated by $c$ is an ultrafilter which contains $F$ but does not contain $b$. So, $b$ is not in the intersection of all ultrafilters containing $F$. Since $b\not\in F$ was arbitrary, this means $F$ is the intersection of all the ultrafilters that contain it.
Conversely, assuming your property, let $a,b\in L$ with $a\not\leq b$. There must exist an ultrafilter $U$ containing the principal filter generated by $a$ such that $b\not\in U$. Since $U$ is maximal, this means there exists $d\in U$ such that $b\wedge d=0$. Then $c=d\wedge a$ is nonzero (since $a,d\in U$) and satisfies $c\leq a$ and $b\wedge c=0$.
This property holds for an atomistic lattice and the converse is true for finite lattices. Indeed, if $L$ is atomistic and $a,b\in L$ satisfy $a\not\leq b$, then there exists some atom $c$ such that $c\leq a$ and $c\not\leq b$, and then $b\wedge c=0$. Conversely, if $L$ is finite and satisfies your property, let $a\in L$ and let $b$ be the join of all the atoms below $a$. Then if $a\not=b$, then $a\leq b$, so there exists a nonzero $c\leq a$ such that $b\wedge c=0$. There then exists an atom $d\leq c$, and we have $d\leq a$ but $b\wedge d=0$. This contradicts the definition of $b$. Thus $a=b$, and $L$ is atomistic.
(In particular, this property does not hold in every finite lattice, since not every finite lattice is atomistic. For instance, consider a chain with more than 2 elements.)
For infinite lattices, though, this condition is much weaker than being atomistic. For instance, every Boolean algebra satisfies this condition, taking $c=a\wedge \neg b$.