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?
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asked Dec 15, 2015 at 19:51
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1$\begingroup$ More generally (under AC) any continuous frame is spatial - see e. g. VII.4.3 in Johnstone's "Stone Spaces" (pp. 310-311) $\endgroup$– მამუკა ჯიბლაძეDec 15, 2015 at 21:25
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$\begingroup$ I don't have access to that book out in the sticks unfortunately. Is there a sketch of the proof available online? $\endgroup$– Anschel Schaffer-CohenDec 17, 2015 at 18:11
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1$\begingroup$ I am not aware of any. If you have access to "Continuous Lattices and Domains" by Gierz, Hofmann, Keimel, Lawson, Mislove and Scott, there on page 126 is Theorem I-4.25 stating that for any $x\not\leqslant y$ in a bounded complete algebraic domain there is a completely irreducible element $p$ with $x\not\leqslant p$ and $y\leqslant p$. $\endgroup$– მამუკა ჯიბლაძეDec 17, 2015 at 19:57
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$\begingroup$ Every distributive continuous lattice is a spatial frame. See "The spectral theory of distributive continuous lattices", Karl H. Hofmann and Jimmie D. Lawson 1978. $\endgroup$– Zhenchao LyuJul 25, 2019 at 14:52
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1 Answer
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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.
answered Jan 22, 2016 at 10:17