In at least two papers ([here](http://www.math.vanderbilt.edu/~ual/papers/mar-zen-001.pdf) and [here](http://www.math.vanderbilt.edu/~ual/papers/mar-zen-002.pdf)) 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?



"Zorn's Lemma guarantees that all algebraic frames are spatial." Why?
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 $x=\bigwedge\{z\in {\cal M}(L): z\geq x\}$.



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



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


