Can someone kindly confirm whether the ultrafilter lemma for arbitrary (i.e., not necessarily Boolean) bounded lattices is equivalent to Zorn's lemma?


To be precise, if $\mathbf{L} = (L, \leq, \land, \lor, 0, 1)$ is a bounded lattice, then an *$\mathbf{L}$-filter* is a non-empty subset $F \subseteq L$ such that  

(i) for any $x, y \in F$, there exists some $a \in F$ such that $a \leq x \land y$, and  

(ii) if $x \in F$, $z \in L$, and $x \leq z$, then $z \in F$.


A generalized version of the ultrafilter lemma would be that every proper $\mathbf{L}$-filter can be extended to an $\mathbf{L}$-ultrafilter. When $\mathbf{L}$ is a Boolean algebra (such as the power set of a non-empty set equipped with the standard set-theoretic operations), we have the well-known result that the ultrafilter lemma is equivalent to the Boolean prime ideal theorem, the compactness theorem for PL, etc., and that these are strictly weaker than the axiom of choice. But when $\mathbf{L}$ is an arbitrary lattice, I suppose that Krull's theorem and hence Zorn's lemma can be derived from the ultrafilter lemma. Am I mistaken here?



Ultrafilter lemma for arbitrary lattice
It is equivalent to AC.


Consider any collection $A$ of nonempty sets, and let $\newcommand\P{\mathbb{P}}\P$ be the set of partial choice functions, so that $p\in\P$ if and only if $p$ is a partial function on $A$ for which $p(a)\in a$ for every $a\in\text{dom}(p)$. We place the forcing order on $\P$, so that $q\leq p$ if $q$ extends $p$ to a larger domain, or equivalently, $p=q\upharpoonright\text{dom}(p)$. In particular, being lower in the order means having more information, larger domain, and so on. The empty function is at the top, the largest element of $\P$. Let us also add an object $\bot$ to $\P$ below all others.


The motivating idea is that $\P$ is the forcing notion that adds a choice function for $A$, augmented with $\bot$.


This is a nontrivial bounded lattice, because any two partial functions $p$, $q$ have a least upper bound $p\vee q$, which is their common part as functions, and a greatest lower bound, which is their union $p\cup q$ if they are compatible as functions, and otherwise $\bot$.


I assume that ultrafilters for you cannot be the whole lattice (since otherwise the ultrafilter assertion would become trivialized). Every proper filter in $\P$, I claim, gives rise to a unifying limit partial choice function, the union of the all the functions in the filter, since the filter cannot contain $\bot$ and so all elements of it must be compatible as functions. Furthermore, the limit function arising in this way from an ultrafilter must be totally defined on $A$, since otherwise we could extend it by defining the choice function on one more set $a\in A$ in the collection.


So from an ultrafilter in $\P$ we get a choice function on $A$.


Let me add a note about distributivity, since the lattice $\P$ is not generally a distributive lattice, one for which $p\vee(q\wedge r)=(p\vee q)\wedge (p\vee r)$. The reason is that it could be that $q$ and $r$ are incompatible, which would make $q\wedge r=\bot$ and consequently the LHS would be $p$, but if $q$ and $r$ differ from $p$ on some point $a\in\text{dom}(p)$, then the RHS will be strictly above $p$, lacking that point $a$ in its domain.


So a question remains, I suppose, about the strength of the ultrafilter lemma for distributive lattices. [**Update** Keith Kearnes posted an answer with references to Herrlich & Klimosky, showing that it is also equivalent to AC with nontrivial bounded distributive lattices.]


