"Zorn's Lemma guarantees that all algebraic frames are spatial." Why?

Question

In at least two papers (here and here) Jorge Martínez and Eric R. Zenk say that Zorn's Lemma implies that all algebraic frames are spatial. However, I haven't been able to find an actual explanation or proof of this claim anywhere, nor can I think of one myself. Does anyone have an idea or a reference on proving this?

Answer 1

Let $L$ be a complete lattice with top element $1$. Let ${\cal M}(L)$ be the collection of meet-irreducible elements of $L$.

Recall that a frame is said to be spatial if for all $x\in L$ with $x<1$ we have $x=\bigwedge\{z\in {\cal M}(L): z\geq x\}$.

Lemma: Every complete, algebraic lattice $L$ contains a meet-irreducible element that is stricly smaller than the top-element.

Proof. Let $0$ be the bottom element, and $1$ be the top element of $L$. Since $L$ is compact, $1$ is the join of all compact elements of $L$. Let $c>0$ be a compact element and set $$K = \{k\in L: c\not\leq k\}.$$ Clearly $0\in K$ and the join of any directed subset of $K$ is in $K$ since $c$ is compact. By Zorn's Lemma, $K$ contains a maximal element $m\in K$. By definition, any $x>m$ must contain $c$, so $m\vee c$ is the unique least element of $L$ that properly contains $m$, and so $m$ is meet-irreducible.

Theorem: If $L$ is complete, algebraic, then for all $x\in L$ with $x<1$ we have $x=\bigwedge\{z\in {\cal M}(L): z\geq x\}$.

Proof. Suppose $x < m:=\bigwedge\{z\in {\cal M}(L): z\geq x\}$. Then $[x,m]=\{a\in L: x\leq a \leq m\}$ is a complete algebraic lattice without meet-irreducible elements other than the top-element, contradicting the Lemma.