Spectral spaces of countable abelian lattice-ordered groups

TL;DR

This paper characterizes spectral spaces that correspond to countable abelian lattice-ordered groups and explores the lattice-theoretic properties related to their spectra, revealing limitations in their logical definability.

Contribution

It provides a topological characterization of spectra of countable abelian ll-groups and shows the non-definability of ll-representability in arbitrary cardinalities.

Findings

Spectral spaces correspond to spectra of countable abelian ll-groups.

A lattice-theoretic condition characterizes ll-representability.

There are limitations in characterizing ll-representability via inite logic.

Abstract

A compact topological space X is spectral if it is sober (i.e., every irreducible closed set is the closure of a unique singleton) and the compact open subsets of X form a basis of the topology of X, closed under finite intersections. Theorem. A topological space X is homeomorphic to the spectrum of some countable Abelian {\ell}-group with unit (resp., MV-algebra) iff X is spectral, has a countable basis of open sets, and for any points x and y in the closure of a singleton {z}, either x is in the closure of {y} or y is in the closure of {x}. We establish this result by proving that a countable distributive lattice D with zero is isomorphic to the lattice of all principal ideals of an Abelian {\ell}-group (we say that D is {\ell}-representable) iff for all a, b $\in$ D there are x, y $\in$ D such that a $\lor$ b = a $\lor$ y = b $\lor$ x and x $\land$ y = 0. On the other hand, we construct a non-{\ell}-representable bounded distributive lattice, of cardinality $\aleph$ 1 , with an {\ell}-representable countable L$\infty, \omega$-elementary sublattice. In particular, there is no characterization, of the class of all {\ell}-representable distributive lattices, in arbitrary cardinality, by any class of L$\infty, \omega$ sentences.

Equations

$$x_1\setminus x_2=x_1\setminus (x_1\wedge x_2)=x_1\setminus x_1=0,$$

$$a\setminus b=\bigvee \left(p\setminus p_{\ast }\mid p\in \mathrm{Ji}L,p\le a,p\nleqq b\right),\text{for all }a,b\in L.$$

$$a^{\prime }\setminus b=\bigvee \left(q\setminus q_{\ast }\mid q\in \mathrm{Ji}L,q\le a^{\prime },q\nleqq b\right).$$

$$a_1\setminus b_1$$

$$=(a_1\wedge a_2)\setminus b_1$$

$$\le (a_1\wedge a_2)\setminus (a_1\wedge b_2)$$

$$=(a_1\wedge a_2)\setminus b_2$$

$$\le a_2\setminus b_2$$

$$a{\ominus }_{D}b\underset{\mathrm{def}}{=}\left\{x\in D\mid a\le b\vee x\right\}.$$

$$a_1{\setminus }_{S}b$$

$$\displaystyle=(a_{1}\mathbin{\smallsetminus}_{S}a_{2})\vee\bigl{(}{(a_{1}\wedge a_{2})\mathbin{\smallsetminus}_{S}b}\bigr{)}$$

$$(a_{1}\mathbin{\smallsetminus}_{S}b)\wedge(a_{2}\mathbin{\smallsetminus}_{S}b)\leq\bigl{(}{(a_{1}\mathbin{\smallsetminus}_{S}a_{2})\wedge(a_{2}\mathbin{\smallsetminus}_{S}a_{1})}\bigr{)}\vee\bigl{(}{(a_{1}\wedge a_{2})\mathbin{\smallsetminus}_{S}b}\bigr{)}\,.$$

$$(a_1{\setminus }_{S}b)\wedge (a_2{\setminus }_{S}b)\le (a_1\wedge a_2){\setminus }_{S}b.$$

$$(a_1{\setminus }_S{b}_1)\wedge (a_2{\setminus }_S{b}_2)$$

$$=(a_1\wedge a_2){\setminus }_{S}b$$

$$=0$$

$$D\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}$$

$$\displaystyle g\bigl{(}{x\mathbin{\smallsetminus}_{E}(y_{1}\vee y_{2})}\bigr{)}$$

$$\displaystyle\leq\bigl{(}{g(x)\mathbin{\smallsetminus}_{S}g(y_{1})}\bigr{)}\wedge\bigl{(}{g(x)\mathbin{\smallsetminus}_{S}g(y_{2})}\bigr{)}\qquad(\text{because }y_{1},y_{2}\in D_{x})$$

$$\displaystyle=g(x)\mathbin{\smallsetminus}_{S}\bigl{(}{g(y_{1})\vee g(y_{2})}\bigr{)}$$

$$(\text{because }g[D]\text{ is consonant in }S\text{ and by Lemma refL:sdSLatHom})$$

$$=g(x){\setminus }_{S}g(y_1\vee y_2)(\text{because }g\text{ is a join-ho-mo-mor-phism}),$$

$$g(x,y,z)\underset{\mathrm{def}}{=}\bigl{(}{f(x)\wedge a}\bigr{)}\vee\bigl{(}{f(y)\wedge b}\bigr{)}\vee f(z)\,,\qquad\text{for all }(x,y,z)\in D\ast\mathsf{J}_{2}\,.$$

$$f_{\ast }(t)$$

$$f^{\ast }(t)$$

$$f(x){\setminus }_{L}f(y)=\bigvee \left(f(p){\setminus }_{L}f(p_{\ast })\mid p\in \mathrm{Ji}D,p\le x,p\nleqq y\right),\text{for all }x,y\in D.$$

$$g\bigl{(}{(x\wedge a)\vee(y\wedge b)\vee z}\bigr{)}=(f(x)\wedge\alpha)\vee(f(y)\wedge\beta)\vee f(z)\,,\quad\text{for all }x,y,z\in D\,.$$

$$u\le (x\wedge a)\vee (y\wedge b)\vee z$$

$$u\wedge a\le (x\wedge a)\vee (y\wedge b)\vee z$$

$$u\wedge b\le (x\wedge a)\vee (y\wedge b)\vee z$$

$$u$$

$$f(u)$$

$$f(u)\wedge \alpha \le (f(x)\wedge \alpha )\vee f(z)\le (f(x)\wedge \alpha )\vee (f(y)\wedge \beta )\vee f(z)$$

$$d(x,y,z)=(x\wedge a)\vee(y\wedge b)\vee z\text{ and }\delta(x,y,z)=\bigl{(}{f(x)\wedge a}\bigr{)}\vee\bigl{(}{f(y)\wedge b}\bigr{)}\vee f(z)$$

Full text

We obtain this result by proving that a countable distributive lattice $D$ with zero is isomorphic to the Stone dual of some $\ell$-spectrum (we say that $D$ is $\ell$-representable) iff for all $a,b\in D$ there are $x,y\in D$ such that $a\vee b=a\vee y=b\vee x$ and $x\wedge y=0$. On the other hand, we construct a non-$\ell$-representable bounded distributive lattice, of cardinality $\aleph_{1}$, with an $\ell$-representable countable $\mathbin{\mathscr{L}}_{\infty,\omega}$-elementary sublattice. In particular, there is no characterization, of the class of all $\ell$-representable distributive lattices, by any class of $\mathbin{\mathscr{L}}_{\infty,\omega}$ sentences.

1. Introduction

A lattice-ordered group, or $\ell$-group for short, is a group $G$ endowed with a translation-invariant lattice ordering. An $\ell$-ideal of $G$ is an order-convex, normal $\ell$-subgroup $I$ of $G$. We say that $I$ is prime if $I\neq G$ and $x\wedge y\in I$ implies that either $x\in I$ or $y\in I$, for all $x,y\in G$. We define the $\ell$-spectrum of $G$ as the set $\operatorname{Spec_{\ell}}{G}$ of all prime $\ell$-ideals of $G$, endowed with the “hull-kernel” topology, whose closed sets are exactly the sets $\left\{{P\in\operatorname{Spec_{\ell}}{G}}\mid{X\subseteq P}\right\}$ for $X\subseteq G$. Characterizing the topological spaces $\operatorname{Spec_{\ell}}{G}$, for Abelian $\ell$-groups $G$, is a long-standing open problem, which we shall call the $\ell$-spectrum problem.

A topological space $X$ is generalized spectral if it is sober (i.e., every irreducible closed set is the closure of a unique singleton) and the collection of all compact open subsets of $X$ forms a basis of the topology of $X$, closed under intersections of any two members. If, in addition, $X$ is compact, we say that it is spectral. It is well known that the $\ell$-spectrum of any Abelian $\ell$-group is a generalized spectral space; in addition, this space is completely normal, that is, for any points $x$ and $y$ in the closure of a singleton $\left\{z\right\}$, either $x$ is in the closure of $\left\{y\right\}$ or $y$ is in the closure of $\left\{x\right\}$ (cf. Bigard, Keimel, and Wolfenstein [22, Ch. 10]). Delzell and Madden found in [10] an example of a completely normal spectral space which is not an $\ell$-spectrum. However, their example is not second countable. The main aim of the present paper is proving that there is no such counterexample in the second countable case (cf. Theorem 9.1). We also prove, in Section 10, that the class of all Stone dual lattices of $\ell$-spectra is neither closed under products nor under homomorphic images. We also prove that they have no $\mathbin{\mathscr{L}}_{\infty,\omega}$-characterization.

For further background on the $\ell$-spectrum problem and related problems, we refer the reader to Mundici [31, Problem 2] (where the $\ell$-spectrum problem is stated in terms of MV-algebras), Marra and Mundici [26, 27], Cignoli and Torrens [7], Di Nola and Grigolia [12], Cignoli, Gluschankof, and Lucas [6], Iberkleid, Martínez, and McGovern [19], Delzell and Madden [10, 11], Keimel [23]. Our main reference on $\ell$-groups will be Bigard, Keimel, and Wolfenstein [5], of which we will mostly follow the notation and terminology. All our $\ell$-groups will be written additively. For background on lattice theory, we refer to Grätzer [17, 18]. As customary, we denote by $\rightarrow$, or $\rightarrow_{D}$ if $D$ needs to be specified, the Heyting implication in a Heyting algebra $D$ (cf. Johnstone [20]): hence $a\rightarrow_{D}b$ is the largest $x\in D$ such that $a\wedge x\leq b$.

2. Strategy of the proof

2.1. Reduction to a lattice-theoretical problem; consonance

Recall the classical Stone duality (cf. Stone [35]), between distributive lattices with zero and [math]-lattice homomorphisms with cofinal111A subset $X$ in a poset $P$ is cofinal if every element of $P$ lies below some element of $X$. range on the one hand, generalized spectral spaces and spectral222A map between generalized spectral spaces is spectral if the inverse image of any compact open set is compact open. maps on the other hand. This duality sends every distributive lattice $D$ with zero to the set $\operatorname{Spec}{D}$ of all its (proper) prime ideals, endowed with the usual hull-kernel topology (cf. Grätzer [18, § 2.5], Johnstone [20, § II.3]); in the other direction, it sends every generalized spectral space $X$ to the lattice ${\overset{\circ}{\mathcal{K}}}(X)$ of all its compact open subsets.

Characterizing all $\ell$-spectra of Abelian $\ell$-groups amounts to characterizing all their Stone duals, which are distributive lattices with zero.

Now for every Abelian $\ell$-group $G$, the Stone dual ${\overset{\circ}{\mathcal{K}}}(\operatorname{Spec_{\ell}}{G})$ of $\operatorname{Spec_{\ell}}{G}$ is isomorphic to the (distributive) lattice $\operatorname{Id_{c}}{G}$ of all principal $\ell$-ideals of $G$ (cf. Proposition 1.19, together with Theorem 1.10 and Lemma 1.20, in Keimel [22]); call such lattices $\ell$-representable. Hence, we get

Lemma 2.1**.**

A topological space $X$ is homeomorphic to the $\ell$-spectrum of an Abelian $\ell$-group iff it is generalized spectral and the lattice ${\overset{\circ}{\mathcal{K}}}(X)$ is $\ell$-representable.

In the sequel, we will denote by $\langle{x}\rangle$, or $\langle{x}\rangle_{G}$ if $G$ needs to be specified, the $\ell$-ideal of $G$ generated by any element $x$ of an $\ell$-group $G$.

The lattice-theoretical analogue of complete normality is given as follows.

Definition 2.2**.**

Two elements $a$ and $b$, in a distributive lattice $D$ with zero, are consonant, in notation $a\sim_{D}b$, if there are $x,y\in D$ such that $a\leq b\vee x$, $b\leq a\vee y$, and $x\wedge y=0$. A subset $X$ of $D$ is consonant if every pair of elements in $X$ is consonant. We say that $D$ is completely normal if it is a consonant subset of itself.

The following result is a restatement of Monteiro [30, Théorème V.3.1].

Proposition 2.3**.**

A generalized spectral space $X$ is completely normal iff its lattice ${\overset{\circ}{\mathcal{K}}}(X)$, of all compact open subsets, is completely normal.

The countable333In the present paper, “countable” will always mean “at most countable”. case of the $\ell$-spectrum problem can thus be restated more ambitiously as follows:

Prove that every countable completely normal distributive lattice with zero is $\ell$-representable.

2.2. Closed homomorphisms

Denote by $\operatorname{F}_{\ell}(\omega)$ the free Abelian $\ell$-group on the first infinite ordinal $\omega\underset{\mathrm{def}}{=}\left\{0,1,2,\dots\right\}$. Given a countable, bounded, completely normal distributive lattice $L$, our main goal is to construct a surjective lattice homomorphism $f\colon\operatorname{Id_{c}}{\operatorname{F}_{\ell}(\omega)}\twoheadrightarrow L$ which induces an isomorphism \operatorname{Id_{c}}\bigl{(}{\operatorname{F}_{\ell}(\omega)/I}\bigr{)}\cong L for a suitable $\ell$-ideal $I$ of $\operatorname{F}_{\ell}(\omega)$. Our next definition introduces the lattice homomorphisms allowing (a bit more than) the latter step, itself contained in Lemma 2.5.

Definition 2.4**.**

A join-homomorphism $f\colon A\to B$, between join-semilattices $A$ and $B$, is closed if whenever $a_{0},a_{1}\in A$ and $b\in B$, if $f(a_{0})\leq f(a_{1})\vee b$, then there exists $x\in A$ such that $a_{0}\leq a_{1}\vee x$ and $f(x)\leq b$.

Lemma 2.5**.**

Let $G$ be an Abelian $\ell$-group, let $S$ be a distributive lattice with zero, and let $\varphi\colon\operatorname{Id_{c}}G\twoheadrightarrow S$ be a closed surjective join-homomorphism. Then $I\underset{\mathrm{def}}{=}\left\{{x\in G}\mid{\varphi(\langle{x}\rangle_{G})=0}\right\}$ is an $\ell$-ideal of $G$, and there is a unique isomorphism $\psi\colon\operatorname{Id_{c}}(G/I)\to S$ such that $\psi(\langle{x+I}\rangle_{G/I})=\varphi(\langle{x}\rangle_{G})$ for every $x\in G^{+}$.

Proof.

It is straightforward to verify that $I$ is an $\ell$-ideal of $G$ and that there is a unique map $\psi\colon\operatorname{Id_{c}}(G/I)\to S$ such that $\psi(\langle{x+I}\rangle_{G/I})=\varphi(\langle{x}\rangle_{G})$ for every $x\in G^{+}$. Since $\varphi$ is a surjective join-homomorphism, so is $\psi$. It remains to verify that $\psi$ is an order-embedding.

Let $x,y\in G^{+}$ such that $\psi(\langle{x+I}\rangle_{G/I})\leq\psi(\langle{y+I}\rangle_{G/I})$. This means that $\varphi(\langle{x}\rangle_{G})\leq\varphi(\langle{y}\rangle_{G})$, thus, since $\varphi$ is a closed map, there exists $\boldsymbol{z}\in\operatorname{Id_{c}}G$ such that $\langle{x}\rangle_{G}\subseteq\langle{y}\rangle_{G}\vee\boldsymbol{z}$ and $\varphi(\boldsymbol{z})=0$. Writing $\boldsymbol{z}=\langle{z}\rangle_{G}$, for $z\in G^{+}$, this means that $z\in I$ and $x\leq ny+nz$ for some positive integer $n$. Therefore, $x+I\leq n(y+I)$, so $\langle{x+I}\rangle_{G/I}\subseteq\langle{y+I}\rangle_{G/I}$. ∎

Although this fact will not be used further in the paper, we record here that much of the relevance of closed maps is contained in the following easy result.

Proposition 2.6**.**

Let $G$ and $H$ be Abelian $\ell$-groups and let $f\colon G\to H$ be an $\ell$-homomorphism. Then the map $\operatorname{Id_{c}}{f}\colon\operatorname{Id_{c}}{G}\to\operatorname{Id_{c}}{H}$, $\langle{x}\rangle_{G}\mapsto\langle{f(x)}\rangle_{H}$ is a closed [math]-lattice homomorphism.

Proof.

It is obvious that the map $\boldsymbol{f}\underset{\mathrm{def}}{=}\operatorname{Id_{c}}{f}$ is a [math]-lattice homomorphism. Let $\boldsymbol{a}_{0},\boldsymbol{a}_{1}\in\operatorname{Id_{c}}G$ and let $\boldsymbol{b}\in\operatorname{Id_{c}}H$ such that $\boldsymbol{f}(\boldsymbol{a}_{0})\subseteq\boldsymbol{f}(\boldsymbol{a}_{1})\vee\boldsymbol{b}$. Pick $a_{0},a_{1}\in G^{+}$, $b\in H^{+}$ such that each $\boldsymbol{a}_{i}=\langle{a_{i}}\rangle_{G}$ and $\boldsymbol{b}=\langle{b}\rangle_{H}$. Then the assumption $\boldsymbol{f}(\boldsymbol{a}_{0})\subseteq\boldsymbol{f}(\boldsymbol{a}_{1})\vee\boldsymbol{b}$ means that there exists a positive integer $n$ such that $f(a_{0})\leq n(f(a_{1})+b)$, which, since $b\geq 0$, is equivalent to $(f(a_{0})-nf(a_{1}))^{+}\leq nb$, that is, since $f$ is an $\ell$-homomorphism, f\bigl{(}{(a_{0}-na_{1})^{+}}\bigr{)}\leq nb. Therefore, setting $\boldsymbol{x}\underset{\mathrm{def}}{=}\langle{(a_{0}-na_{1})^{+}}\rangle_{G}$, we get $\boldsymbol{a}_{0}\subseteq\boldsymbol{a}_{1}\vee\boldsymbol{x}$ and $\boldsymbol{f}(\boldsymbol{x})\subseteq\boldsymbol{b}$. ∎

Example 2.7**.**

Using Proposition 2.6, it is easy to construct examples of non-$\ell$-representable $0,1$-lattice homomorphisms between $\ell$-representable finite distributive lattices: for example, consider the unique zero-separating map $\boldsymbol{f}\colon\mathbf{3}\twoheadrightarrow\mathbf{2}$ (where $\mathbf{2}\underset{\mathrm{def}}{=}\left\{0,1\right\}$ and $\mathbf{3}\underset{\mathrm{def}}{=}\left\{0,1,2\right\}$ with their natural orderings).

2.3. Elementary blocks: the lattices $\operatorname{Op}^{-}({\mathcal{H}})$

Our construction of a closed surjective lattice homomorphism $f\colon\operatorname{Id_{c}}{\operatorname{F}_{\ell}(\omega)}\twoheadrightarrow D$ will be performed stepwise, by expressing $\operatorname{Id_{c}}{\operatorname{F}_{\ell}(\omega)}$ as a countable ascending union $\bigcup_{n<\omega}E_{n}$, for suitable finite sublattices $E_{n}$ (the “elementary blocks” of the construction) and homomorphisms $f_{n}\colon E_{n}\to L$, then extending each $f_{n}$ to $f_{n+1}$. Each step of the construction will be one of the following:

1. (1)

extend the domain of $f_{n}$ — in order to get the final map $f$ defined on all of $\operatorname{Id_{c}}{\operatorname{F}_{\ell}(\omega)}$; this will be done in Section 6, via a lattice-theoretical homomorphism extension result (Lemma 4.2) established in Section 4; 2. (2)

correct “closure defects” of $f_{n}$ (i.e., $f_{n}(a_{0})\leq f_{n}(a_{1})\vee b$ with no $x$ such that $a_{0}\leq a_{1}\vee x$ and $f_{n}(x)\leq b$) — in order to get $f$ closed (Section 7); 3. (3)

add elements to the range of $f_{n}$ — in order to get $f$ surjective (Section 8).

Elaborating on the final example in Di Nola and Grigolia [12], it can be seen that not all the $E_{n}$ can be taken completely normal. Our $E_{n}$ will be defined as sublattices, of the powerset lattice of an infinite-dimensional vector space $\mathbb{R}^{(\omega)}$, generated by open half-spaces arising from finite collections of hyperplanes. Those lattices will be denoted in the form $\operatorname{Op}^{-}({\mathcal{H}})$ (cf. Notation 5.3 and Lemma 5.7). This will be made possible by the Baker-Beynon duality.

While Steps (1) and (2) above require relatively complex arguments, they remain valid with $\mathbb{R}^{(\omega)}$ replaced by $\mathbb{R}^{d}$ for any positive integer $d$, and in fact any topological vector space. On the other hand, while the argument handling Step (3) is noticeably easier, it requires an infinite-dimensional ambient space.

3. Difference operations

The present section consists of a few technical lattice-theoretical results, mostly aimed at Lemmas 4.2 and 7.1, describing how the concept of a difference operation (Definition 3.2) works in the presence of consonance.

We denote by $\operatorname{Ji}L$ (resp., $\operatorname{Mi}L$) the set of all join-irreducible (resp., meet-irreducible) elements in a lattice $L$. For $p\in L$, we denote by $p_{*}$ the largest element of $L$ smaller than $p$ — also called the lower cover of $p$ (cf. Grätzer [18, p. 102]). If $L$ is finite, then $p_{*}$ exists iff $p\in\operatorname{Ji}{L}$. We first state a preparatory lemma.

Lemma 3.1** (folklore; see Exercises 8.5 and 8.6 in Davey and Priestley [9]).**

Let $D$ be a finite distributive lattice. Then every join-irreducible element $p$ of $D$ is join-prime, that is, it is nonzero and $p\leq x\vee y$ implies that $p\leq x$ or $p\leq y$, for all $x,y\in D$. Moreover, the subset $\left\{{x\in D}\mid{p\nleq x}\right\}$ has a largest element $p^{\dagger}$. The assignment $p\mapsto p^{\dagger}$ defines an order-isomorphism from $\operatorname{Ji}D$ onto $\operatorname{Mi}D$.

We now introduce one of our main lattice-theoretical concepts.

Definition 3.2**.**

Let $L$ be a lattice and let $S$ be a $(\vee,0)$-semilattice. A map $L\times L\to S$, $(x,y)\mapsto x\mathbin{\smallsetminus}y$ is an $S$-valued difference operation on $L$ if the following statements hold:

• (D0)

$x\mathbin{\smallsetminus}x=0$, for all $x\in L$.

• (D1)

$x\mathbin{\smallsetminus}z=(x\mathbin{\smallsetminus}y)\vee(y\mathbin{\smallsetminus}z)$, for all $x,y,z\in L$ such that $x\geq y\geq z$.

• (D2)

$x\mathbin{\smallsetminus}y=(x\vee y)\mathbin{\smallsetminus}y=x\mathbin{\smallsetminus}(x\wedge y)$, for all $x,y\in L$.

Although we will need the following lemma only in case $L$ is distributive, we found it worth noticing that it holds in full generality.

Lemma 3.3**.**

Let $L$ be a lattice, let $S$ be a $(\vee,0)$-semilattice, and let $\mathbin{\smallsetminus}$ be an $S$-valued difference operation on $L$. Then $x\mathbin{\smallsetminus}z\leq(x\mathbin{\smallsetminus}y)\vee(y\mathbin{\smallsetminus}z)$, for all $x,y,z\in S$ (triangle inequality). Furthermore, the map $(x,y)\mapsto x\mathbin{\smallsetminus}y$ is order-preserving in $x$ and order-reversing in $y$.

Proof.

As in [36], denote by $\Delta(x,y)$, for $x\leq y$ in $L$, the canonical generators of the dimension monoid $\operatorname{Dim}L$ of $L$. By the universal property defining $\operatorname{Dim}L$, our axioms (D0)–(D2) ensure that there is a unique monoid homomorphism $\mu\colon\operatorname{Dim}L\to\nobreak S$ such that $\mu(\Delta(x,y))=y\mathbin{\smallsetminus}x$ for all $x\leq y$ in $L$. Set $\Delta^{+}(x,y)\underset{\mathrm{def}}{=}\Delta(x\wedge y,x)$, now for all $x,y\in L$.

Denoting by $\leq$ the algebraic preordering of $\operatorname{Dim}L$ (i.e., $\alpha\leq\beta$ if there exists $\gamma$ such that $\beta=\alpha+\gamma$), we established in [36, Prop. 1.9] that $\Delta^{+}(x,z)\leq\Delta^{+}(x,y)+\Delta^{+}(y,z)$, for all $x,y,z\in L$. By taking the image of that inequality under the monoid homomorphism $\mu$ and using (D2), the triangle inequality follows.

Now let $x_{1},x_{2},y\in L$ with $x_{1}\leq x_{2}$. From (D2) and (D0) it follows that

[TABLE]

thus, by the triangle inequality, $x_{1}\mathbin{\smallsetminus}y\leq(x_{1}\mathbin{\smallsetminus}x_{2})\vee(x_{2}\mathbin{\smallsetminus}y)=x_{2}\mathbin{\smallsetminus}y$. The proof, that $y_{1}\leq y_{2}$ implies $x\mathbin{\smallsetminus}y_{2}\leq x\mathbin{\smallsetminus}y_{1}$, is similar. ∎

Lemma 3.4**.**

Let $L$ be a finite lattice, let $S$ be a $(\vee,0)$-semilattice, and let $\mathbin{\smallsetminus}$ be an $S$-valued difference operation on $L$. Then the following statement holds:

[TABLE]

Proof.

Since neither side of (3.1) is affected by changing the pair $(a,b)$ to $(a,a\wedge b)$, we may assume that $a\geq b$, and then prove (3.1) by induction on $a$. The result is trivial for $a=b$ (use (D0)). Dealing with the induction step, suppose that $a>b$. Pick $a^{\prime}\in L$ such that $b\leq a^{\prime}$ and $a^{\prime}$ is a lower cover of $a$. The set $\left\{{x\in L}\mid{x\leq a\text{ and }x\nleq a^{\prime}}\right\}$ has a minimal element $p$. Necessarily, $p$ is join-irreducible and $p_{*}\leq a^{\prime}$, so $a=p\vee a^{\prime}$ and $p_{*}=p\wedge a^{\prime}$. By (D2), $p\mathbin{\smallsetminus}p_{*}=a\mathbin{\smallsetminus}a^{\prime}$. Moreover, by the induction hypothesis,

[TABLE]

Using (D1), we get $a\mathbin{\smallsetminus}b=(a\mathbin{\smallsetminus}a^{\prime})\vee(a^{\prime}\mathbin{\smallsetminus}b)\geq\bigvee\left({q\mathbin{\smallsetminus}q_{*}}\mid{q\in\operatorname{Ji}L\,,\ q\leq a\,,\ q\nleq b}\right)$. For the converse inequality, let $q\in\operatorname{Ji}L$ such that $q\leq a$ and $q\nleq b$. Observing that $q\wedge b\leq q_{*}<q$ and $b<q\vee b\leq a$, we obtain, by using (D2) together with the second statement of Lemma 3.3, $q\mathbin{\smallsetminus}q_{*}\leq q\mathbin{\smallsetminus}(q\wedge b)=(q\vee b)\mathbin{\smallsetminus}b\leq a\mathbin{\smallsetminus}b$. ∎

Lemma 3.5**.**

Let $D$ be a distributive lattice, let $S$ be a $(\vee,0)$-semilattice, and let $\mathbin{\smallsetminus}$ be an $S$-valued difference operation on $D$. Then for all $a_{1},b_{1},a_{2},b_{2}\in D$, if $a_{1}\leq b_{1}\vee a_{2}$ and $a_{1}\wedge b_{2}\leq b_{1}$ (within $D$), then $a_{1}\mathbin{\smallsetminus}b_{1}\leq a_{2}\mathbin{\smallsetminus}b_{2}$ (within $S$).

Proof.

From $a_{1}\leq b_{1}\vee a_{2}$ if follows that $a_{1}\leq b_{1}\vee(a_{1}\wedge a_{2})$, thus

[TABLE]

For any elements $a$ and $b$ in a distributive lattice $D$, we shall set

[TABLE]

Moreover, we shall denote by $a\mathbin{\smallsetminus}_{D}b$ the least element of $a\ominus_{D}b$ if it exists, and then call it the pseudo-difference of $a$ and $b$ relatively to $D$. Further, we shall say that $D$ is a generalized dual Heyting algebra if $a\mathbin{\smallsetminus}_{D}b$ exists for all $a,b\in D$.

The following lemma will be a crucial source of difference operations throughout the paper. The proof is straightforward and we leave it to the reader.

Lemma 3.6**.**

For any generalized dual Heyting algebra $S$, the operation $\mathbin{\smallsetminus}_{S}$ is an $S$-valued difference operation on $S$.

The two following lemmas state that the pseudo-difference operation behaves especially well in the presence of consonance.

Lemma 3.7**.**

The following statements hold, for every generalized dual Heyting algebra $S$ and all $a_{1},a_{2},a,b_{1},b_{2},b\in S$:

1. (1)

$(a_{1}\vee a_{2})\mathbin{\smallsetminus}_{S}b=(a_{1}\mathbin{\smallsetminus}_{S}b)\vee(a_{2}\mathbin{\smallsetminus}_{S}b)$. 2. (2)

$a\mathbin{\smallsetminus}_{S}(b_{1}\wedge b_{2})=(a\mathbin{\smallsetminus}_{S}b_{1})\vee(a\mathbin{\smallsetminus}_{S}b_{2})$. 3. (3)

If $a_{1}\sim_{S}a_{2}$, then $(a_{1}\wedge a_{2})\mathbin{\smallsetminus}_{S}b=(a_{1}\mathbin{\smallsetminus}_{S}b)\wedge(a_{2}\mathbin{\smallsetminus}_{S}b)$. 4. (4)

If $b_{1}\sim_{S}b_{2}$, then $a\mathbin{\smallsetminus}_{S}(b_{1}\vee b_{2})=(a\mathbin{\smallsetminus}_{S}b_{1})\wedge(a\mathbin{\smallsetminus}_{S}b_{2})$.

Proof.

The proofs of (1) and (2) are easy exercises.

Ad (3). We first compute as follows:

[TABLE]

Symmetrically, a_{2}\mathbin{\smallsetminus}_{S}b\leq(a_{2}\mathbin{\smallsetminus}_{S}a_{1})\vee\bigl{(}{(a_{1}\wedge a_{2})\mathbin{\smallsetminus}_{S}b}\bigr{)}. By meeting the two inequalities, we obtain, by using the distributivity of $S$, the following inequality:

[TABLE]

Now our assumption $a_{1}\sim_{S}a_{2}$ means that $(a_{1}\mathbin{\smallsetminus}_{S}a_{2})\wedge(a_{2}\mathbin{\smallsetminus}_{S}a_{1})=0$, so we obtain the following inequality:

[TABLE]

The converse inequality is trivial.

The proof of (4) is similar to the one of (3). ∎

Lemma 3.8**.**

Let $S$ be a generalized dual Heyting algebra and let $a_{1},a_{2},b_{1},b_{2}\in S$. If $a_{1}\sim_{S}a_{2}$ and $a_{1}\wedge a_{2}\leq b_{1}\wedge b_{2}$, then $(a_{1}\mathbin{\smallsetminus}_{S}b_{1})\wedge(a_{2}\mathbin{\smallsetminus}_{S}b_{2})=0$.

Proof.

Set $b\underset{\mathrm{def}}{=}b_{1}\wedge b_{2}$. We compute as follows:

[TABLE]

Lemma 3.9**.**

Let $D$ and $L$ be distributive lattices, let $E$ and $S$ be generalized dual Heyting algebras, and let $g\colon E\to L$ be a lattice homomorphism. We assume that $D$ is a sublattice of $E$, $S$ is a sublattice of $L$, and $g[D]$ is a consonant subset of $S$. Let $\Sigma$ be a subset of $D$, generating $D$ as a lattice. If $g(x\mathbin{\smallsetminus}_{E}y)\leq g(x)\mathbin{\smallsetminus}_{S}g(y)$ for all $x,y\in\Sigma$, then $g(x\mathbin{\smallsetminus}_{E}y)\leq g(x)\mathbin{\smallsetminus}_{S}g(y)$ for all $x,y\in D$.

The situation in Lemma 3.9 is partly illustrated in Figure 3.1.

Proof.

Let $x\in\Sigma$. We claim that the set $D_{x}\underset{\mathrm{def}}{=}\left\{{y\in D}\mid{g(x\mathbin{\smallsetminus}_{E}y)\leq g(x)\mathbin{\smallsetminus}_{S}g(y)}\right\}$ is equal to $D$. Indeed, it follows from our assumptions that $\Sigma\subseteq D_{x}$. For all $y_{1},y_{2}\in D_{x}$,

[TABLE]

that is, $y_{1}\vee y_{2}\in D_{x}$. The proof that $y_{1}\wedge y_{2}\in D_{x}$ is similar, although easier since it does not require any consonance assumption. Hence, $D_{x}$ is a sublattice of $D$. Since it contains $\Sigma$, it contains $D$; whence $D_{x}=D$.

This holds for all $x\in\Sigma$, which means that for all $y\in D$, the set $D^{\prime}_{y}\underset{\mathrm{def}}{=}\left\{{x\in D}\mid{g(x\mathbin{\smallsetminus}_{E}y)\leq g(x)\mathbin{\smallsetminus}_{S}g(y)}\right\}$ contains $\Sigma$. Moreover, by an argument similar to the one used in the paragraph above, $D^{\prime}_{y}$ is a sublattice of $D$. Hence, $D^{\prime}_{y}=D$. This holds for all $y\in D$, as required. ∎

4. The Main Extension Lemma for distributive lattices

The key idea, of our proof of Theorem 9.1, is the possibility of extending certain lattice homomorphisms $f\colon D\to L$, where $D$ and $L$ are distributive [math]-lattices with $D$ finite and $L$ completely normal, to finite, or countable, distributive extensions of $D$. The present section is mostly devoted to the required technical lattice-theoretical extension result (Lemma 4.2).

We first state a preparatory result. For any lattice $D$, denote by $D\ast\mathsf{J}_{2}$ the sublattice of $D^{3}$ consisting of all triples $(x,y,z)$ such that $z\leq x$ and $z\leq y$. The following lemma means that, as the notation suggests, $D\ast\mathsf{J}_{2}$ is, in the bounded case, the free distributive product (cf. Grätzer [17, Thm. 12.5]) of $D$ with the second entry $\mathsf{J}_{2}$ of Jaskowsky’s sequence, represented in Figure 4.1. For $x,y,z\in D$, the triple $(x\vee z,y\vee z,z)\in D\ast\mathsf{J}_{2}$ can then be identified with $(x\wedge\mathsf{a})\vee(y\wedge\mathsf{b})\vee z$. The proof of Lemma 4.1 is a straightforward exercise.

Lemma 4.1**.**

Let $D$ and $E$ be bounded distributive lattices and let $a,b\in E$ such that $a\wedge b=0$. Then for every $0,1$-lattice homomorphism $f\colon D\to E$, there exists a unique $0,1$-lattice homomorphism $g\colon D\ast\mathsf{J}_{2}\to E$ such that $g(1,0,0)=a$, $g(0,1,0)=\nobreak b$, and $g(x,x,x)=x$ whenever $x\in D$. This map is defined by the rule

[TABLE]

Lemma 4.2** (Main Extension Lemma).**

Let $E$ be a finite distributive lattice, let $D$ be a $0,1$-sublattice of $E$, and let $a,b\in E$ such that the following conditions hold:

1. (1)

$E$* is generated, as a lattice, by $D\cup\left\{a,b\right\}$.* 2. (2)

$D$* is a Heyting subalgebra of $E$.* 3. (3)

$a\wedge b=0$. 4. (4)

For all $p\in\operatorname{Ji}{D}$, if $p\leq p_{*}\vee a\vee b$, then either $p\leq p_{*}\vee a$ or $p\leq p_{*}\vee b$. 5. (5)

For all $p,q\in\operatorname{Ji}{D}$, if $p\leq p_{*}\vee a$ and $q\leq q_{*}\vee b$, then $p$ and $q$ are incomparable.

Let $L$ be a generalized dual Heyting algebra and let $f\colon D\to L$ be a consonant [math]-lattice homomorphism. For every $t\in E$, we set

[TABLE]

Call a pair $(\alpha,\beta)$ of elements of $L$ $f$-admissible if there exists a (necessarily unique) lattice homomorphism $g\colon E\to L$, extending $f$, such that $g(a)=\alpha$ and $g(b)=\beta$. Then the following statements hold:

1. (i)

$f_{*}(t)\leq f^{*}(t)$* for every $t\in E$.* 2. (ii)

$f_{*}(a)\wedge f_{*}(b)=0$. 3. (iii)

The $f$-admissible pairs are exactly the pairs $(\alpha,\beta)$ satisfying the inequalities $f_{*}(a)\leq\alpha\leq f^{*}(a)$, $f_{*}(b)\leq\beta\leq f^{*}(b)$, and $\alpha\wedge\beta=0$.

Note*.*

Although the proof of our main result (viz. Theorem 9.1) will require only the consideration of $(\alpha,\beta)=(f_{*}(a),f_{*}(b))$, we keep the more general formulation, due to possible relevance to further extensions of the present work. The proof of Lemma 4.2 is mostly unaffected by that increase in generality.

Proof.

The uniqueness statement on $g$ follows immediately from Assumption (1), so we need to deal only with the existence statement.

Since $f$ is consonant, the assignment $(x,y)\mapsto f(x)\mathbin{\smallsetminus}_{L}f(y)$ defines an $L$-valued difference operation on $D$ (use Lemma 3.6). By Lemma 3.4, it follows that

[TABLE]

The remainder of our proof consists mainly of a series of claims.

Claim 1**.**

For all $x,y\in D$ and all $t\in E$, $x\leq y\vee t$ implies that $f(x)\leq f(y)\vee f_{*}(t)$.

Proof of Claim..

Let $x,y\in D$ such that $x\leq y\vee t$, and let $p\in\operatorname{Ji}{D}$ such that $p\leq x$ and $p\nleq y$. The latter relation means that $y\leq p^{\dagger}$. Hence, from $x\leq y\vee t$ it follows that $p\leq p^{\dagger}\vee t$, or, equivalently, $p\leq p_{*}\vee t$. Therefore, by the definition of $f_{*}(t)$, we get $f(p)\mathbin{\smallsetminus}_{L}f(p_{*})\leq f_{*}(t)$. Joining those inequalities over all possible values of $p$ and invoking (4.1), we get $f(x)\mathbin{\smallsetminus}_{L}f(y)\leq f_{*}(t)$ and we are done. ∎ Claim 1.

Claim 2**.**

Let $x,y\in D$ such that $x\leq y\vee a\vee b$. Then $f(x)\leq f(y)\vee f_{*}(a)\vee f_{*}(b)$.

Proof of Claim..

We must prove that $f(x)\mathbin{\smallsetminus}_{L}f(y)\leq f_{*}(a)\vee f_{*}(b)$. By (4.1), it suffices to prove that $f(p)\mathbin{\smallsetminus}_{L}f(p_{*})\leq f_{*}(a)\vee f_{*}(b)$, for every $p\in\operatorname{Ji}{D}$ such that $p\leq x$ and $p\nleq y$. The latter relation means that $y\leq p^{\dagger}$, thus, for any such $p$, the inequality $p\leq p^{\dagger}\vee a\vee b$, or, equivalently, $p\leq p_{*}\vee a\vee b$, holds. By Assumption (4), this implies that either $p\leq p_{*}\vee a$ or $p\leq p_{*}\vee b$. By the definition of $f_{*}(a)$ and $f_{*}(b)$, this implies that either $f(p)\mathbin{\smallsetminus}_{L}f(p_{*})\leq f_{*}(a)$ or $f(p)\mathbin{\smallsetminus}_{L}f(p_{*})\leq f_{*}(b)$. In both cases, $f(p)\mathbin{\smallsetminus}_{L}f(p_{*})\leq f_{*}(a)\vee f_{*}(b)$. ∎ Claim 2.

Claim 3**.**

$f_{*}(t)\leq f^{*}(t)$, for every $t\in E$.

Proof of Claim..

We must prove that for all $p\in\operatorname{Ji}{D}$ and all $x\in D$ such that $p\leq p_{*}\vee t$ and $t\leq x$, the inequality $f(p)\mathbin{\smallsetminus}_{L}f(p_{*})\leq f(x)$ holds. Since obviously, $p\leq p_{*}\vee x$, it follows, since $p$ is join-prime in $D$, that $p\leq x$, so we obtain the inequalities $f(p)\mathbin{\smallsetminus}_{L}f(p_{*})\leq f(p)\leq f(x)$. ∎ Claim 3.

Claim 4**.**

$f_{*}(a)\wedge f_{*}(b)=0$.

Proof of Claim..

It suffices to prove that for all $p,q\in\operatorname{Ji}{D}$ with $p\leq p_{*}\vee a$ and $q\leq q_{*}\vee b$, the relation $(f(p)\mathbin{\smallsetminus}_{L}f(p_{*}))\wedge(f(q)\mathbin{\smallsetminus}_{L}f(q_{*}))=0$ holds. By Lemma 3.8, it suffices to prove that $f(p)\wedge f(q)\leq f(p_{*})\wedge f(q_{*})$. Since $f$ is a meet-homomorphism, it suffices to prove that $p\wedge q\leq p_{*}\wedge q_{*}$. However, it follows from (5) that $p$ and $q$ are incomparable, so this is obvious. ∎ Claim 4.

It is clear that every $f$-admissible pair $(\alpha,\beta)$ satisfies $f_{*}(a)\leq\alpha\leq f^{*}(a)$, $f_{*}(b)\leq\beta\leq f^{*}(b)$, and $\alpha\wedge\beta=0$. It thus remains to prove that conversely, every such pair $(\alpha,\beta)$ is $f$-admissible.

Claim 5**.**

There exists a unique map $g\colon E\to L$ such that

[TABLE]

Moreover, $g(a)=\alpha$, $g(b)=\beta$, and $g$ is a join-homomorphism extending $f$.

Proof of Claim..

By Assumptions (1) and (3), every element $t$ of $E$ has the form $(x\wedge a)\vee(y\wedge b)\vee z$, where $x,y,z\in D$. This implies the uniqueness statement on $g$, and says that all we need to do is to verify that the right hand side of (4.2) depends only on $t$; the map $g$ thus defined, via (4.2), would then automatically be a join-homomorphism extending $f$, satisfying, by virtue of the relations $f(0)=0$, $\alpha\leq f(1)$, and $\beta\leq f(1)$, the equations $g(a)=\alpha$ and $g(b)=\beta$. Hence, we only need to verify that the following implications hold, for every $u\in D$:

[TABLE]

for all $u,x,y,z\in D$. Since $E$ is distributive, the premise of (4.3) is equivalent to the conjunction of the following inequalities:

[TABLE]

Since $f$ is a join-homomorphism and by Claims 1 and 2, together with the inequalities $f_{*}(a)\leq\alpha$ and $f_{*}(b)\leq\beta$, those inequalities imply the following inequalities:

[TABLE]

Since $L$ is distributive, this implies, by reversing the argument above, the inequality $f(u)\leq(f(x)\wedge\alpha)\vee(f(y)\wedge\beta)\vee f(z)$, thus completing the proof of (4.3).

Further, since $E$ is distributive and since $a\wedge b=0$, the premise of (4.4) is equivalent to the inequality $u\wedge a\leq(x\wedge a)\vee z$, thus to the inequality $u\wedge a\leq x\vee z$, which can be written $a\leq(u\rightarrow_{E}(x\vee z))$. By Assumption (2), this is equivalent to $a\leq v$, where we set $v\underset{\mathrm{def}}{=}(u\rightarrow_{D}(x\vee z))$. Since $\alpha\leq f^{*}(a)$, this implies that $\alpha\leq f(v)$. Hence, $f(u)\wedge\alpha\leq f(u)\wedge f(v)=f(u\wedge v)\leq f(x\vee z)=f(x)\vee f(z)$. Since $L$ is distributive, this implies in turn that

[TABLE]

thus completing the proof of (4.4). The proof of (4.5) is symmetric. ∎ Claim 5.

In order to conclude the proof of Lemma 4.2, it is sufficient to prove that $g$ is a meet-homomorphism. By Assumption (3) and since $\alpha\wedge\beta=0$, respectively, it follows from Lemma 4.1 that there are unique lattice homomorphisms $d\colon D\ast\mathsf{J}_{2}\to E$ and $\delta\colon D\ast\mathsf{J}_{2}\to L$ such that

[TABLE]

for all $(x,y,z)\in D\ast\mathsf{J}_{2}$. Then Claim 5 implies that $\delta=g\circ d$. Moreover, it follows from Assumptions (1) and (3) that $d$ is surjective. Now any two elements of $E$ have the form $d(t_{1})$ and $d(t_{2})$, where $t_{1},t_{2}\in D\ast\mathsf{J}_{2}$, and

[TABLE]

The converse inequality $g(d(t_{1})\wedge d(t_{2}))\leq g(d(t_{1}))\wedge g(d(t_{2}))$ is trivial. ∎

5. Lattices of convex open polyhedral cones

Throughout this section we shall fix a real topological vector space $\mathbb{E}$. Denote by $\operatorname{int}(A)$ and $\operatorname{cl}(A)$ the interior and closure of a subset $A$, respectively. We begin with two preparatory lemmas.

Lemma 5.1**.**

Let $A$ and $F$ be convex subsets in $\mathbb{E}$, with $F$ closed and $F\cap\operatorname{int}(A)\neq\varnothing$. Then $\operatorname{cl}(F\cap A)=F\cap\operatorname{cl}(A)$.

Proof.

Fix $u\in F\cap\operatorname{int}(A)$, and let $p\in F\cap\operatorname{cl}(A)$. Since $F$ and $A$ are both convex, $(1-\lambda)p+\lambda u\in F\cap A$ for each $\lambda\in\left({0,1}\right]$. Since $(1-\lambda)p+\lambda u$ converges to $p$, as $\lambda$ goes to [math] and $\lambda>0$, it follows that $p\in\operatorname{cl}(F\cap A)$. We have thus proved that $F\cap\operatorname{cl}(A)\subseteq\operatorname{cl}(F\cap A)$. The converse containment is trivial. ∎

Lemma 5.2**.**

Let $F$ be the union of finitely many closed subspaces in $\mathbb{E}$ and let $Q$ be a convex subset of $\mathbb{E}$. Then either $Q\subseteq F$ or $Q\cap F$ is nowhere dense in $Q$.

Proof.

We first deal with the case where $F$ is a closed subspace of $\mathbb{E}$. Suppose that $Q\cap F$ is not nowhere dense in $Q$. Since $F$ is a closed subspace of $\mathbb{E}$, $Q\cap F$ is relatively closed in $Q$, thus the relative interior $U$ of $Q\cap F$ in $Q$ is nonempty. Fix $u\in U$ and let $q\in Q$. Since $Q$ is convex, $(1-\lambda)u+\lambda q\in Q$ for every $\lambda\in[0,1]$. Since $U$ is a relative neighborhood of $u$ in $Q$, it follows that $(1-\lambda)u+\lambda q$ belongs to $U$, thus to $F$, for some $\lambda\in\left({0,1}\right]$. Since $\left\{u,(1-\lambda)u+\lambda q\right\}\subseteq F$ with $\lambda>0$, it follows that $q\in F$, therefore completing the proof that $Q\subseteq F$.

In the general case, $F=\bigcup_{i=1}^{n}F_{i}$, where each $F_{i}$ is a closed subspace of $\mathbb{E}$. If $Q\not\subseteq F$, then $Q\not\subseteq F_{i}$, thus, by the paragraph above, $Q\cap F_{i}$ is nowhere dense in $Q$ whenever $1\leq i\leq n$; whence $Q\cap F=\bigcup_{i=1}^{n}(Q\cap F_{i})$ is also nowhere dense in $Q$. ∎

From now on, for any closed hyperplane $H$ of $\mathbb{E}$, we shall denote by $H^{+}$ and $H^{-}$ the open half-spaces with boundary $H$, with associated closed half-spaces $\overline{H}^{+}\underset{\mathrm{def}}{=}\operatorname{cl}(H^{+})$ and $\overline{H}^{-}\underset{\mathrm{def}}{=}\operatorname{cl}(H^{-})$.

Notation 5.3**.**

For a set ${\mathcal{H}}$ of closed hyperplanes of $\mathbb{E}$, we will set

[TABLE]

Furthermore, we will denote by $\operatorname{Bool}({\mathcal{H}})$ the Boolean algebra of subsets of $\mathbb{E}$ generated by $\Sigma_{{\mathcal{H}}}$ (equivalently, by $\overline{\Sigma}_{{\mathcal{H}}}$), and by $\operatorname{Clos}({\mathcal{H}})$ (resp., $\operatorname{Op}({\mathcal{H}})$) the lattice of all closed (resp., open) members of $\operatorname{Bool}({\mathcal{H}})$.

Trivially, $\operatorname{Clos}({\mathcal{H}})$ and $\operatorname{Op}({\mathcal{H}})$ are both $0,1$-sublattices of $\operatorname{Bool}({\mathcal{H}})$, which is a $0,1$-sublattice of the powerset lattice of $\mathbb{E}$. For the remainder of this section we shall fix a nonempty444Lemma 5.4 does not extend to ${\mathcal{H}}=\varnothing$, for $\operatorname{Op}(\varnothing)=\left\{\varnothing,\mathbb{E}\right\}$ while $\Sigma_{\varnothing}=\varnothing$. set ${\mathcal{H}}$ of closed hyperplanes of $\mathbb{E}$ through the origin.

Lemma 5.4**.**

For every $X\in\operatorname{Bool}({\mathcal{H}})$, the subsets $\operatorname{cl}(X)$ and $\operatorname{int}(X)$ both belong to $\operatorname{Bool}({\mathcal{H}})$. Moreover, $\operatorname{Op}({\mathcal{H}})$ is generated, as a lattice, by $\Sigma_{{\mathcal{H}}}\cup\left\{\mathbb{E}\right\}$, and it is Heyting subalgebra of the Heyting algebra ${\mathcal{O}}(\mathbb{E})$ of all open subsets of $\mathbb{E}$.

Proof.

For the duration of the proof, we shall denote by $\operatorname{Clos}^{\prime}({\mathcal{H}})$ (resp., $\operatorname{Op}^{\prime}({\mathcal{H}})$) the sublattice of $\operatorname{Bool}({\mathcal{H}})$ generated by $\overline{\Sigma}_{{\mathcal{H}}}\cup\left\{\varnothing\right\}$ (resp., $\Sigma_{{\mathcal{H}}}\cup\left\{\mathbb{E}\right\}$).

We first prove that the closure of any member of $\operatorname{Bool}({\mathcal{H}})$ belongs to $\operatorname{Clos}^{\prime}({\mathcal{H}})$. Writing the elements of $\operatorname{Bool}({\mathcal{H}})$ in disjunctive normal form, we see that every element of $\operatorname{Bool}({\mathcal{H}})$ is a finite union of finite intersections of open half-spaces and closed half-spaces with boundaries in ${\mathcal{H}}$. Since $H^{\sigma}=\overline{H}^{\sigma}\setminus H$ for all $H\in{\mathcal{H}}$ and all $\sigma\in\left\{+,-\right\}$, it follows that every element of $\operatorname{Bool}({\mathcal{H}})$ is a finite union of sets of the form $Q\setminus F$, where $Q$ is a finite intersection of closed half-spaces with boundaries in ${\mathcal{H}}$ and $F$ is a finite union of members of ${\mathcal{H}}$. Since the closure operator commutes with finite unions, the first statement of Lemma 5.4 thus reduces to verifying that $\operatorname{cl}(Q\setminus F)$ belongs to $\operatorname{Clos}^{\prime}({\mathcal{H}})$, for any $Q$ and $F$ as above. Now this follows from Lemma 5.2: if $Q\subseteq F$ then $\operatorname{cl}(Q\setminus F)=\varnothing$, and if $Q\not\subseteq F$, then $Q\cap F$ is nowhere dense in $Q$, thus $\operatorname{cl}(Q\setminus F)=\operatorname{cl}(Q)=Q$. The statement about the closure follows; in particular, $\operatorname{Clos}^{\prime}({\mathcal{H}})=\operatorname{Clos}({\mathcal{H}})$. By taking complements, the statement about the interior follows; in particular, $\operatorname{Op}^{\prime}({\mathcal{H}})=\operatorname{Op}({\mathcal{H}})$.

For all $X,Y\in\operatorname{Op}({\mathcal{H}})$, the Heyting residue $X\rightarrow Y$, evaluated within the lattice ${\mathcal{O}}(\mathbb{E})$ of all open subsets of $\mathbb{E}$, is equal to \operatorname{int}\bigl{(}{(\complement X)\cup Y}\bigr{)} (where $\complement$ denotes the complement in $\mathbb{E}$), thus, as $(\complement X)\cup Y$ belongs to $\operatorname{Bool}({\mathcal{H}})$ and by the paragraph above, it belongs to $\operatorname{Op}({\mathcal{H}})$. ∎

In particular, the members of $\operatorname{Op}({\mathcal{H}})$ are open polyhedral cones, that is, finite unions of finite intersections of open half-spaces of $\mathbb{E}$. Lemma 5.4 also says that the topology on $\mathbb{E}$ could be, in principle, omitted from the study of $\operatorname{Bool}({\mathcal{H}})$ and $\operatorname{Op}({\mathcal{H}})$.

Define a basic open member of $\operatorname{Op}({\mathcal{H}})$ as a nonempty finite intersection of open half-spaces with boundaries in ${\mathcal{H}}$. In particular, the intersection of the empty collection yields the basic open set $\mathbb{E}$. Since every element of $\operatorname{Op}({\mathcal{H}})$ is a finite union of basic open sets, we obtain the following.

Corollary 5.5**.**

Every join-irreducible element of $\operatorname{Op}({\mathcal{H}})$ is basic open. In particular, it is convex.

It is easy to find examples showing that the converse of Corollary 5.5 does not hold: a basic open member of $\operatorname{Op}({\mathcal{H}})$ may not be join-irreducible.

Corollary 5.6**.**

Let $H$ be a closed hyperplane of $\mathbb{E}$, with associated open half spaces $H^{+}$ and $H^{-}$. Then the members of $\operatorname{Op}({\mathcal{H}}\cup\left\{H\right\})$ are exactly the sets of the form $(X\cap H^{+})\cup(Y\cap H^{-})\cup Z$, where $X,Y,Z\in\operatorname{Op}({\mathcal{H}})$; moreover, one can take $Z\subseteq X$ and $Z\subseteq Y$.

Proof.

For every basic open set $U$ in $\operatorname{Op}({\mathcal{H}}\cup\left\{H\right\})$, there is a basic open set $T$ in $\operatorname{Op}({\mathcal{H}})$ such that $U=T\cap H^{+}$ or $U=T\cap H^{-}$ or $U=T$. By Lemma 5.4, every element of $\operatorname{Op}({\mathcal{H}}\cup\left\{H\right\})$ is a finite union of basic open sets, thus it has the given form. Moreover, changing $X$ to $X\cup Z$ and $Y$ to $Y\cup Z$ does not affect the value of $(X\cap H^{+})\cup(Y\cap H^{-})\cup Z$.∎

Lemma 5.7**.**

The top element of $\operatorname{Op}({\mathcal{H}})$ (viz. $\mathbb{E}$) is join-irreducible in $\operatorname{Op}({\mathcal{H}})$. Consequently, the subset $\operatorname{Op}^{-}({\mathcal{H}})\underset{\mathrm{def}}{=}\operatorname{Op}({\mathcal{H}})\setminus\left\{\mathbb{E}\right\}$ is a [math]-sublattice of $\operatorname{Op}({\mathcal{H}})$. It is generated, as a lattice, by $\Sigma_{{\mathcal{H}}}$.

Proof.

Any basic open member of $\operatorname{Op}({\mathcal{H}})$, distinct from $\mathbb{E}$, omits the origin. Hence, any member of $\operatorname{Op}({\mathcal{H}})$, distinct from $\mathbb{E}$, omits the origin, and so the union of any two such sets is distinct from $\mathbb{E}$. This proves that $\mathbb{E}$ is join-irreducible in $\operatorname{Op}({\mathcal{H}})$. The verifications of the other statements of Lemma 5.7 are straightforward. ∎

Remark 5.8*.*

Let ${\mathcal{H}}$ be finite. Then the unit of $\operatorname{Op}^{-}({\mathcal{H}})$ is equal to $\complement\bigcap{\mathcal{H}}$, which is distinct from the unit of $\operatorname{Op}({\mathcal{H}})$, which is equal to $\mathbb{E}$. In particular, $\operatorname{Op}^{-}({\mathcal{H}})$ is not a Heyting subalgebra of $\operatorname{Op}({\mathcal{H}})$.

6. The Main Extension Lemma for lattices $\operatorname{Op}({\mathcal{H}})$

Throughout this section we shall fix a real topological vector space $\mathbb{E}$. Our main goal is to show that Lemma 4.2 can be applied to lattices of the form $\operatorname{Op}({\mathcal{H}})$ (cf. Lemma 6.6). This goal will be achieved via a convenient description of the join-irreducible members of $\operatorname{Op}({\mathcal{H}})$ (cf. Lemma 6.4), involving an operator that we will denote by $\nabla_{{\mathcal{H}}}$ (cf. Notation 6.3).

For any subset $X$ in $\mathbb{E}$, we denote by $\operatorname{conv}(X)$ the convex hull of $X$, and by $\operatorname{cone}(X)\underset{\mathrm{def}}{=}\mathbb{R}^{+}\cdot\operatorname{conv}(X)$ the closed convex cone generated by $X$. For a set $X$, a poset $P$, and maps $f,g\colon X\to P$, we shall set $[\![{f\leq g}]\!]\underset{\mathrm{def}}{=}\left\{{x\in X}\mid{f(x)\leq g(x)}\right\}$, and similarly for $[\![{f<g}]\!]$, $[\![{f>g}]\!]$, and so on. We first state two preparatory lemmas.

Lemma 6.1**.**

Let $n$ be a nonnegative integer and let $b_{1}$, …, $b_{n}$, $c$ be linear functionals on $\mathbb{E}$. Then $\bigcap_{i=1}^{n}[\![{b_{i}\geq 0}]\!]\subseteq[\![{c\geq 0}]\!]$ iff $c\in\operatorname{cone}(\left\{b_{1},\dots,b_{n}\right\})$.

Proof.

By working in the quotient space $\mathbb{E}/{\bigcap_{i=1}^{n}\ker(b_{i})}$, the problem is reduced to the classical finite-dimensional case (cf. Schrijver [34, Thm. 7.1]). ∎

Lemma 6.2**.**

Suppose that $\mathbb{E}$ is Hausdorff. Then $\operatorname{cone}(X)$ is a closed subset of $\mathbb{E}$, for every finite subset $X$ of $\mathbb{E}$.

Proof.

The subspace $F$ of $\mathbb{E}$ generated by $X$ is finite-dimensional, thus (since $\mathbb{E}$ is Hausdorff) closed. This reduces the problem to the case where $\mathbb{E}=\mathbb{R}^{d}$ for some nonnegative integer $d$. By the Farkas-Minkowski-Weyl Theorem (cf. Schrijver [33, Cor. 7.1a]), $\operatorname{cone}(X)$ is then a finite intersection of closed half-spaces of $\mathbb{E}$. ∎

Until the end of this section, we will fix a nonempty finite set ${\mathcal{H}}$ of closed hyperplanes of $\mathbb{E}$ through the origin.

Notation 6.3**.**

For every $U\in\operatorname{Op}({\mathcal{H}})$, we set ${\mathcal{H}}_{U}\underset{\mathrm{def}}{=}\left\{{H\in{\mathcal{H}}}\mid{H\cap U\neq\varnothing}\right\}$. The intersection $\nabla_{{\mathcal{H}}}{U}$ of all members of ${\mathcal{H}}_{U}$ is a closed subspace of $\mathbb{E}$.

Recall (cf. Lemma 3.1) that for a join-irreducible member $P$ of $\operatorname{Op}({\mathcal{H}})$, $P^{\dagger}$ denotes the largest element of $\operatorname{Op}({\mathcal{H}})$ not containing $P$.

Lemma 6.4**.**

A nonempty, convex member $P$ of $\operatorname{Op}({\mathcal{H}})$ is join-irreducible, within the lattice $\operatorname{Op}({\mathcal{H}})$, iff $P\cap\nabla_{{\mathcal{H}}}{P}$ is nonempty. Moreover, in that case, the lower cover $P_{*}$ of $P$, in $\operatorname{Op}({\mathcal{H}})$, is equal to $P\setminus\nabla_{{\mathcal{H}}}{P}$, and $P^{\dagger}=\complement(\operatorname{cl}(P)\cap\nabla_{{\mathcal{H}}}{P})$.

Proof.

Suppose first that $P$ is join-irreducible. Suppose, by way of contradiction, that $P\cap\nabla_{{\mathcal{H}}}{P}=\varnothing$, that is, $P\subseteq\bigcup_{H\in{\mathcal{H}}_{P}}\complement H$. Since $P$ is join-irreducible in the distributive lattice $\operatorname{Op}({\mathcal{H}})$, it is join-prime in that lattice (cf. Lemma 3.1), thus there exists $H\in{\mathcal{H}}_{P}$ such that $P\subseteq\complement H$; in contradiction with $H\in{\mathcal{H}}_{P}$.

Suppose, conversely, that $P\cap\nabla_{{\mathcal{H}}}{P}\neq\varnothing$. The subset $P\setminus\nabla_{{\mathcal{H}}}{P}$ belongs to $\operatorname{Op}({\mathcal{H}})$ and it is a proper subset of $P$, thus we only need to prove that every proper subset $X$ of $P$, belonging to $\operatorname{Op}({\mathcal{H}})$, is contained in $P\setminus\nabla_{{\mathcal{H}}}{P}$. It suffices to consider the case where $X$ is basic open. There are a subset ${\mathcal{X}}$ of ${\mathcal{H}}$ and a family $\left({\varepsilon_{H}}\mid{H\in{\mathcal{X}}}\right)$ of elements of $\left\{+,-\right\}$ such that $X=\bigcap_{H\in{\mathcal{X}}}H^{\varepsilon_{H}}$. Since $P\not\subseteq X$, there exists $H\in{\mathcal{X}}$ such that $P\not\subseteq H^{\varepsilon_{H}}$. Hence,

[TABLE]

If $P\subseteq H^{-\varepsilon_{H}}$, then $X\subseteq H^{-\varepsilon_{H}}$, thus, since $X\subseteq H^{\varepsilon_{H}}$, we get $X=\varnothing$, a contradiction. Hence, $P\not\subseteq H^{-\varepsilon_{H}}$, that is,

[TABLE]

By (6.1) and (6.2), and since $P$ is convex, it follows that $P\cap H\neq\varnothing$, that is, $H\in{\mathcal{H}}_{P}$. Hence, $\nabla_{{\mathcal{H}}}{P}\subseteq H$. Since $X\cap H=\varnothing$, it follows that $X\cap\nabla_{{\mathcal{H}}}{P}=\varnothing$, that is, $X\subseteq P\setminus\nabla_{{\mathcal{H}}}P$, thus completing the proof of the join-irreducibility of $P$.

Finally, it follows from Lemma 5.4 that the set $U\underset{\mathrm{def}}{=}\operatorname{int}\complement(P\cap\nabla_{{\mathcal{H}}}{P})$ belongs to $\operatorname{Op}({\mathcal{H}})$. Moreover, $U=\complement\operatorname{cl}(P\cap\nabla_{{\mathcal{H}}}{P})$. Since $P\cap\nabla_{{\mathcal{H}}}{P}\neq\varnothing$ and by Lemma 5.1, we get $U=\complement(\operatorname{cl}(P)\cap\nabla_{{\mathcal{H}}}{P})$. For every $V\in\operatorname{Op}({\mathcal{H}})$, $P\not\subseteq V$ iff $P\cap V\subsetneqq P$, iff $P\cap V\subseteq P_{*}$, iff $P\cap V\cap\nabla_{{\mathcal{H}}}{P}=\varnothing$, iff $V\subseteq\complement(P\cap\nabla_{{\mathcal{H}}}{P})$. Since $V$ is open, this is equivalent to $V\subseteq U$. Therefore, $U=P^{\dagger}$. ∎

Proposition 6.5**.**

Let $P$ and $Q$ be join-irreducible elements in $\operatorname{Op}({\mathcal{H}})$. If $P\subsetneqq Q$, then $\nabla_{{\mathcal{H}}}{Q}\subsetneqq\nabla_{{\mathcal{H}}}{P}$.

Proof.

By definition, ${\mathcal{H}}_{P}\subseteq{\mathcal{H}}_{Q}$, thus $\nabla_{{\mathcal{H}}}{Q}\subseteq\nabla_{{\mathcal{H}}}{P}$. Since $P\subsetneqq Q$ and by Lemma 6.4, $P$ is contained in $Q_{*}=Q\setminus\nabla_{{\mathcal{H}}}{Q}$, thus $P\cap\nabla_{{\mathcal{H}}}{Q}=\varnothing$. Since $P\cap\nabla_{{\mathcal{H}}}{P}\neq\varnothing$, it follows that $\nabla_{{\mathcal{H}}}{P}\neq\nabla_{{\mathcal{H}}}{Q}$. ∎

Lemma 6.6** (Extension Lemma for $\operatorname{Op}({\mathcal{H}})$).**

Let $H$ be a closed hyperplane of $\mathbb{E}$, let $L$ be a generalized dual Heyting algebra, and let $f\colon\operatorname{Op}({\mathcal{H}})\to L$ be a consonant [math]-lattice homomorphism. Then $f$ extends to a unique lattice homomorphism $g\colon\operatorname{Op}({\mathcal{H}}\cup\left\{H\right\})\to L$ such that $g(H^{+})=f_{*}(H^{+})$ and $g(H^{-})=f_{*}(H^{-})$.

We refer to Lemma 4.2 for the notations $f_{*}(H^{+})$ and $f_{*}(H^{-})$.

Proof.

It suffices to verify that Conditions (1)–(5) of Lemma 4.2 are satisfied, with $D:=\operatorname{Op}({\mathcal{H}})$, $E:=\operatorname{Op}({\mathcal{H}}\cup\left\{H\right\})$, $a:=H^{+}$, and $b:=H^{-}$. Conditions (1) (use Corollary 5.6) and (3) are obvious. By Lemma 5.4, $\operatorname{Op}({\mathcal{H}})$ is a Heyting subalgebra of $\operatorname{Op}({\mathcal{H}}\cup\left\{H\right\})$; Condition (2) follows.

Let $P$ be a join-irreducible element of $\operatorname{Op}({\mathcal{H}})$ such that $P\subseteq P_{*}\cup H^{+}\cup H^{-}$. By Lemma 6.4, this means that $P\cap\nabla_{{\mathcal{H}}}{P}\subseteq H^{+}\cup H^{-}$. Since $P\cap\nabla_{{\mathcal{H}}}{P}$ is convex, this implies that $P\cap\nabla_{{\mathcal{H}}}{P}$ is contained either in $H^{+}$ or in $H^{-}$, thus that $P$ is contained either in $P_{*}\cup H^{+}$ or in $P_{*}\cup H^{-}$. Condition (4) follows.

For Condition (5), let $P,Q\in\operatorname{Ji}\operatorname{Op}({\mathcal{H}})$ such that $P\subseteq P_{*}\cup H^{+}$ and $Q\subseteq Q_{*}\cup H^{-}$. Suppose for example that $P\subseteq Q$. Then $P\cap\nabla_{{\mathcal{H}}}{P}\subseteq H^{+}$, $Q\cap\nabla_{{\mathcal{H}}}{Q}\subseteq H^{-}$, and $P^{\dagger}\subseteq\nobreak Q^{\dagger}$. Thus, by Lemma 6.4, $\operatorname{cl}(Q)\cap\nabla_{{\mathcal{H}}}{Q}\subseteq\operatorname{cl}(P)\cap\nabla_{{\mathcal{H}}}{P}$. It follows that $Q\cap\nabla_{{\mathcal{H}}}{Q}\subseteq\overline{H}^{+}$, hence $Q\cap\nabla_{{\mathcal{H}}}{Q}=\varnothing$, a contradiction. ∎

7. Correcting a closure defect

Throughout this section we shall fix a real topological vector space $\mathbb{E}$, with topological dual $\mathbb{E}^{\prime}$, endowed with the weak-$*$ topology.

Lemma 7.1**.**

Let ${\mathcal{H}}$ be a finite set of closed hyperplanes in $\mathbb{E}$, let $a,b\in\mathbb{E}^{\prime}$ with respective kernels $A$ and $B$, both belonging to ${\mathcal{H}}$. We set

[TABLE]

for any positive integer $m$. Then for all large enough $m$, the following statement holds: for every generalized dual Heyting algebra $L$, every consonant [math]-lattice homomorphism $f\colon\operatorname{Op}({\mathcal{H}})\to L$ extends to a lattice homomorphism $g\colon\operatorname{Op}({\mathcal{H}}_{m})\to L$ such that $g(A^{+}\mathbin{\smallsetminus}_{\operatorname{Op}^{-}({\mathcal{H}}_{m})}B^{+})=f(A^{+})\mathbin{\smallsetminus}_{L}f(B^{+})$.

Note*.*

The notation $A^{+}\mathbin{\smallsetminus}_{\operatorname{Op}^{-}({\mathcal{H}}_{m})}B^{+}$ might look a bit crowded, in particular due to the use of $\operatorname{Op}^{-}({\mathcal{H}}_{m})$ instead of $\operatorname{Op}({\mathcal{H}}_{m})$. In reality, that distinction is immaterial here, because $\operatorname{Op}^{-}({\mathcal{H}}_{m})$ is an ideal of $\operatorname{Op}({\mathcal{H}}_{m})$, thus $U\mathbin{\smallsetminus}_{\operatorname{Op}^{-}({\mathcal{H}}_{m})}V=U\mathbin{\smallsetminus}_{\operatorname{Op}({\mathcal{H}}_{m})}V$ for all $U,V\in\operatorname{Op}^{-}({\mathcal{H}}_{m})$.

Proof.

We begin by stating exactly how large $m$ should be.

Claim 1**.**

There exists a positive integer $m_{0}$ such that for all $m\geq m_{0}$ and all $X\in\operatorname{Op}({\mathcal{H}})$, $C_{m}^{-}\subseteq X$ implies that $B^{+}\subseteq X$.

Proof of Claim..

Every $P\in\operatorname{Ji}\operatorname{Op}({\mathcal{H}})$ is basic open, thus both $\operatorname{cl}(P)$ and $\nabla_{{\mathcal{H}}}{P}$ are intersections of closed half-spaces with boundaries in ${\mathcal{H}}$. Hence, there is a finite subset $\Phi_{P}$ of $\mathbb{E}^{\prime}\setminus\left\{0\right\}$ such that $\operatorname{cl}(P)\cap\nabla_{{\mathcal{H}}}{P}=\bigcap_{x\in\Phi_{P}}[\![{x\geq 0}]\!]$ and $\ker(x)\in{\mathcal{H}}$ for every $x\in\Phi_{P}$. Since $\mathbb{E}^{\prime}$ is Hausdorff, it follows from Lemma 6.2 that the closed convex cone $K_{P}$ generated by $\Phi_{P}$ is a closed subset of $\mathbb{E}^{\prime}$. Hence, setting ${\mathcal{P}}\underset{\mathrm{def}}{=}\left\{{P\in\operatorname{Ji}\operatorname{Op}({\mathcal{H}})}\mid{-b\notin K_{P}}\right\}$, there exists a positive integer $m_{0}$ such that

[TABLE]

It follows from Lemma 6.1 that for every $y\in\mathbb{E}$ and every $P\in\operatorname{Ji}\operatorname{Op}({\mathcal{H}})$, $-y\in\nobreak K_{P}$ iff $\bigcap_{x\in\Phi_{P}}{[\![{x\geq 0}]\!]}\subseteq[\![{y\leq 0}]\!]$, iff $\operatorname{cl}(P)\cap\nabla_{{\mathcal{H}}}{P}\subseteq[\![{y\leq 0}]\!]$, iff $[\![{y>0}]\!]\subseteq P^{\dagger}$ (cf. Lemma 6.4). In particular, $-b\in K_{P}$ iff $B^{+}\subseteq P^{\dagger}$. Similarly, $-b+(1/m)a\in K_{P}$ iff $C_{m}^{-}\subseteq P^{\dagger}$. Hence, (7.1) means that $C_{m}^{-}\subseteq P^{\dagger}$ implies that $B^{+}\subseteq P^{\dagger}$, whenever $m\geq m_{0}$ and $P\in\operatorname{Ji}\operatorname{Op}({\mathcal{H}})$. Now every meet-irreducible element of $\operatorname{Op}({\mathcal{H}})$ has the form $P^{\dagger}$ (cf. Lemma 3.1), and every element of $\operatorname{Op}({\mathcal{H}})$ is an intersection of meet-irreducible elements of $\operatorname{Op}({\mathcal{H}})$. ∎ Claim 1.

We shall prove that every integer $m\geq m_{0}$ has the property stated in Lemma 7.1. Let $L$ be a generalized dual Heyting algebra and let $f\colon\operatorname{Op}({\mathcal{H}})\to L$ be a consonant [math]-lattice homomorphism. We consider the extension $g$ of $f$, to a homomorphism from $\operatorname{Op}({\mathcal{H}}_{m})$ to $L$, given by Lemma 6.6, with $H:=C_{m}$, $H^{+}:=C_{m}^{+}$, $H^{-}:=C_{m}^{-}$. In particular,

[TABLE]

We claim that the following inequality holds:

[TABLE]

Since $L$ is distributive, this amounts to proving the following statement:

[TABLE]

Let $P\in\operatorname{Ji}\operatorname{Op}({\mathcal{H}})$ such that $P\subseteq P_{*}\cup C_{m}^{+}$; that is, $P\cap\nabla_{{\mathcal{H}}}{P}\subseteq C_{m}^{+}$. It follows that $\operatorname{cl}(P)\cap\nabla_{{\mathcal{H}}}{P}=\operatorname{cl}(P\cap\nabla_{{\mathcal{H}}}{P})\subseteq\overline{C}_{m}^{+}$, that is, $C_{m}^{-}\subseteq P^{\dagger}$. It thus follows from the definition of $m_{0}$ (cf. Claim 1) that $B^{+}\subseteq P^{\dagger}$, that is, $P\not\subseteq B^{+}$. Since $B^{+}\in\operatorname{Op}({\mathcal{H}})$, it follows that $P\cap B^{+}\subseteq P_{*}$.

Now suppose that $P\subseteq A^{+}$. Since $P\cap B^{+}\subseteq P_{*}$, the inequalities $P\subseteq P_{*}\cup A^{+}$ and $P\cap B^{+}\subseteq P_{*}$ both hold, thus also $f(P)\leq f(P_{*})\vee f(A^{+})$ and $f(P)\wedge f(B^{+})\leq f(P_{*})$. Since $\mathbin{\smallsetminus}_{L}$ is an $L$-valued difference operation on the range of $f$ (cf. Lemma 3.6), it follows from Lemma 3.5 that $f(P)\mathbin{\smallsetminus}_{L}f(P_{*})\leq f(A^{+})\mathbin{\smallsetminus}_{L}f(B^{+})$, which implies (7.4) right away.

It remains to handle the case where $P\not\subseteq A^{+}$. Due to the obvious containment $C_{m}^{+}\subseteq A^{+}\cup B^{-}$, we get $P\subseteq P_{*}\cup A^{+}\cup B^{-}$, thus, since $P$ is join-prime in $\operatorname{Op}({\mathcal{H}})$, we get $P\subseteq B^{-}$, thus $f(P)\mathbin{\smallsetminus}_{L}f(P_{*})\leq f(B^{-})$, and thus, by using the equation $f(B^{+})\wedge f(B^{-})=0$ and the inequality f(A^{+})\leq f(B^{+})\vee\bigl{(}{f(A^{+})\mathbin{\smallsetminus}_{L}f(B^{+})}\bigr{)},

[TABLE]

thus completing the proof of (7.4) in the general case, and therefore of (7.3).

Now $A^{+}\subseteq B^{+}\cup(A^{+}\cap C_{m}^{+})$, thus $A^{+}\mathbin{\smallsetminus}_{\operatorname{Op}^{-}({\mathcal{H}}_{m})}B^{+}\subseteq A^{+}\cap C_{m}^{+}$, and thus

[TABLE]

Since $f(A^{+})\leq f(B^{+})\vee g(A^{+}\mathbin{\smallsetminus}_{\operatorname{Op}^{-}({\mathcal{H}}_{m})}B^{+})$, the converse inequality

[TABLE]

holds, and therefore $f(A^{+})\mathbin{\smallsetminus}_{L}f(B^{+})=g(A^{+}\mathbin{\smallsetminus}_{\operatorname{Op}^{-}({\mathcal{H}}_{m})}B^{+})$. ∎

Lemma 7.1 deals with closure defects of the form $f(A^{+})\leq f(B^{+})\vee\gamma$. A finite iteration of that result will yield our next lemma, which extends it to closure defects of the form $f(U)\leq f(V)\vee\gamma$, for arbitrary $U,V\in\operatorname{Op}^{-}({\mathcal{H}})$.

Lemma 7.2**.**

Let $\Lambda$ be an additive subgroup of $\mathbb{E}^{\prime}$. Let ${\mathcal{H}}$ be a finite subset of ${\mathcal{H}}_{\Lambda}\underset{\mathrm{def}}{=}\left\{{\ker(x)}\mid{x\in\Lambda\setminus\left\{0\right\}}\right\}$, let $L$ be a completely normal distributive lattice with zero, let $f\colon\operatorname{Op}({\mathcal{H}})\to L$ be a [math]-lattice homomorphism, let $U,V\in\operatorname{Op}^{-}({\mathcal{H}})$, and let $\gamma\in L$ such that $f(U)\leq f(V)\vee\gamma$. Then there are a finite subset $\widetilde{{\mathcal{H}}}$ of ${\mathcal{H}}_{\Lambda}$, containing ${\mathcal{H}}$, $W\in\operatorname{Op}^{-}(\widetilde{{\mathcal{H}}})$, and a lattice homomorphism $g\colon\operatorname{Op}(\widetilde{{\mathcal{H}}})\to L$ extending $f$, such that $U\subseteq V\cup W$ and $g(W)\leq\gamma$.

Proof.

We may assume that ${\mathcal{H}}$ is nonempty. Fix an enumeration $(A_{0},B_{0})$, …, $(A_{n-1},B_{n-1})$ of all pairs of open half-spaces with boundary in ${\mathcal{H}}$. Since $L$ is completely normal, there is a finite chain $S_{0}\subseteq S_{1}\subseteq\cdots\subseteq S_{n}$ of finite sublattices of $L$ such that $S_{0}$ contains $f[\operatorname{Op}({\mathcal{H}})]\cup\left\{\gamma\right\}$ and $S_{i}$ is consonant in $S_{i+1}$ whenever $0\leq i<n$. We construct inductively an ascending chain ${\mathcal{H}}={\mathcal{H}}_{0}\subseteq{\mathcal{H}}_{1}\subseteq\cdots\subseteq{\mathcal{H}}_{n}$ of finite subsets of ${\mathcal{H}}_{\Lambda}$, together with an ascending chain of lattice homomorphisms $f_{l}\colon\operatorname{Op}({\mathcal{H}}_{l})\to S_{l}$, for $0\leq l\leq n$, such that $f_{0}=f$ and

[TABLE]

For $k=0$ there is nothing to verify. Suppose having performed the construction up to level $k$, with $0\leq k<n$. By applying Lemma 7.1, with ${\mathcal{H}}_{k}$ in place of ${\mathcal{H}}$, $f_{k}$ in place of $f$, $S_{k+1}$ (which is a finite distributive lattice, thus, a fortiori, a dual Heyting algebra) in place of $L$, and $(A_{k},B_{k})$ in place of $(A^{+},B^{+})$, we get a finite subset ${\mathcal{H}}_{k+1}$ of ${\mathcal{H}}_{\Lambda}$, containing ${\mathcal{H}}_{k}$, together with a lattice homomorphism $f_{k+1}\colon\operatorname{Op}({\mathcal{H}}_{k+1})\to S_{k+1}$, extending $f_{k}$, such that

[TABLE]

Since $S_{k+1}$ contains $S_{1}$, it follows that

[TABLE]

Since $\operatorname{Op}^{-}({\mathcal{H}}_{k})$ is a sublattice of $\operatorname{Op}^{-}({\mathcal{H}}_{k+1})$ and since $f_{k+1}$ extends $f_{k}$, it follows from the induction hypothesis (7.5) (with fixed $k$) that

[TABLE]

and hence, by (7.6),

[TABLE]

therefore completing the verification of the induction step.

At stage $n$, we obtain a finite subset $\widetilde{{\mathcal{H}}}={\mathcal{H}}_{n}$ of ${\mathcal{H}}_{\Lambda}$, containing ${\mathcal{H}}$, together with a homomorphism $g=f_{n}\colon\operatorname{Op}({\mathcal{H}}_{n})\to S_{n}$, extending $f$, such that

[TABLE]

Since the open half-spaces with boundary in ${\mathcal{H}}$ generate $\operatorname{Op}^{-}({\mathcal{H}})$ as a lattice (cf. Lemma 5.7) and since every pair of such half-spaces has the form $(A_{k},B_{k})$, it follows from Lemma 3.9, applied to (7.7) and the commutative square represented in Figure 7.1, that

[TABLE]

In particular, $f_{n}(U\mathbin{\smallsetminus}_{\operatorname{Op}^{-}({\mathcal{H}}_{n})}V)\leq f(U)\mathbin{\smallsetminus}_{S_{1}}f(V)\leq\gamma$. Let $W\underset{\mathrm{def}}{=}U\mathbin{\smallsetminus}_{\operatorname{Op}^{-}({\mathcal{H}}_{n})}\nobreak V$. ∎

8. Enlarging the range of a homomorphism

Until the end of this section we shall fix a set $I$ and consider the vector space $\mathbb{E}=\mathbb{R}^{(I)}$ with basis $I$, endowed with the coarsest topology making all canonical projections $\delta_{i}\colon\mathbb{E}\to\mathbb{R}$ (for $i\in I$) continuous. We denote by $\Lambda$ the additive subgroup of $\mathbb{E}^{\prime}$ generated by $\left\{{\delta_{i}}\mid{i\in I}\right\}$ and we set (using the notation in Lemma 7.2) ${\mathcal{H}}_{\mathbb{Z}}\underset{\mathrm{def}}{=}{\mathcal{H}}_{\Lambda}$, the set of all integral hyperplanes of $\mathbb{E}$. We shall also set $\Delta_{i}\underset{\mathrm{def}}{=}\ker\delta_{i}$, $\Delta_{i}^{+}\underset{\mathrm{def}}{=}[\![{\delta_{i}>0}]\!]$, and $\Delta_{i}^{-}\underset{\mathrm{def}}{=}[\![{\delta_{i}<0}]\!]$. Any hyperplane $H\in{\mathcal{H}}_{\mathbb{Z}}$ is the kernel of a nonzero element $x=\sum_{i\in I}x_{i}\delta_{i}\in\Lambda$, with all $x_{i}\in\mathbb{Z}$ and the support $\operatorname{supp}(x)\underset{\mathrm{def}}{=}\left\{{i\in I}\mid{x_{i}\neq 0}\right\}$ finite. Since $x$ is determined up to a nonzero scalar multiple, $\operatorname{supp}(x)$ depends of $H$ only, so we may denote it by $\operatorname{supp}(H)$. For a set ${\mathcal{H}}$ of integral hyperplanes of $\mathbb{R}^{(I)}$, we shall set $\operatorname{supp}({\mathcal{H}})\underset{\mathrm{def}}{=}\bigcup\left({\operatorname{supp}(H)}\mid{H\in{\mathcal{H}}}\right)$.

For $x\in\mathbb{R}^{(I)}$ and $S\subseteq I$, we shall denote by $x\mathbin{\restriction}_{S}$ the restriction of $x$ to $S$ extended by zero on $I\setminus S$.

Lemma 8.1**.**

Let ${\mathcal{H}}$ be a set of integral hyperplanes of $\mathbb{R}^{(I)}$, with support $S$, and let $Z\in\operatorname{Bool}({\mathcal{H}})$. Then $x\in Z$ iff $x\mathbin{\restriction}_{S}\in Z$, for all $x\in\mathbb{R}^{(I)}$.

Proof.

For each $H\in{\mathcal{H}}$, pick $p_{H}\in\Lambda$ with kernel $H$, and set $H^{+}\underset{\mathrm{def}}{=}[\![{p_{H}>0}]\!]$, $H^{-}\underset{\mathrm{def}}{=}[\![{p_{H}<0}]\!]$. Then for every $x\in\mathbb{R}^{(I)}$, $x\in H^{+}$ iff $p_{H}(x)>0$, iff $p_{H}(x\mathbin{\restriction}_{S})>0$, iff $x\mathbin{\restriction}_{S}\in H^{+}$. The proof for $H^{-}$ is similar. Since the $H^{+}$ and $H^{-}$ generate $\operatorname{Bool}({\mathcal{H}})$ as a Boolean algebra, the general result follows easily. ∎

Lemma 8.2**.**

Let ${\mathcal{H}}$ be a set of integral hyperplanes of $\mathbb{R}^{(I)}$ and let $i\in I\setminus\operatorname{supp}({\mathcal{H}})$. We denote by $\varphi\colon\operatorname{Op}({\mathcal{H}})\hookrightarrow\operatorname{Op}({\mathcal{H}})\ast\mathsf{J}_{2}$ and $\psi\colon\operatorname{Op}({\mathcal{H}})\hookrightarrow\operatorname{Op}({\mathcal{H}}\cup\left\{\Delta_{i}\right\})$ the diagonal embedding and the inclusion map, respectively, and we set $\varepsilon(X,Y,Z)\underset{\mathrm{def}}{=}(X\cap\Delta_{i}^{+})\cup(Y\cap\Delta_{i}^{-})\cup Z$, for all $(X,Y,Z)\in\operatorname{Op}({\mathcal{H}})\ast\mathsf{J}_{2}$. Then $\varepsilon$ is an isomorphism and $\psi=\varepsilon\circ\varphi$.

We illustrate Lemma 8.2 on Figure 8.1.

Proof.

It is obvious that $\varphi$ and $\psi$ are both $0,1$-lattice homomorphisms, that $\varepsilon$ is lattice homomorphism (use Lemma 4.1), and $\psi=\varepsilon\circ\varphi$. Moreover, it follows from Corollary 5.6 that $\varepsilon$ is surjective.

Set $S\underset{\mathrm{def}}{=}\operatorname{supp}({\mathcal{H}})$. In order to prove that $\varepsilon$ is one-to-one, it is sufficient to prove that every triple $(X,Y,Z)\in\operatorname{Op}({\mathcal{H}})\ast\mathsf{J}_{2}$ is determined by the set $T\underset{\mathrm{def}}{=}\varepsilon(X,Y,Z)$. Let $t\in\mathbb{R}^{(I)}$. Then $t\mathbin{\restriction}_{S}\in\Delta_{i}$, thus $t\mathbin{\restriction}_{S}\in T$ iff $t\mathbin{\restriction}_{S}\in Z$, iff $t\in Z$ (cf. Lemma 8.1); hence $T$ determines $Z$. Likewise, $t\mathbin{\restriction}_{S}+\delta_{i}\in\Delta_{i}^{+}$, thus $t\mathbin{\restriction}_{S}+\delta_{i}\in T$ iff $t\mathbin{\restriction}_{S}+\delta_{i}$ belongs to $X\cup Z=X$, iff (using again Lemma 8.1) $t\mathbin{\restriction}_{S}\in X$, iff $t\in X$. Symmetrically, $t\mathbin{\restriction}_{S}-\delta_{i}\in T$ iff $t\in Y$. Therefore, $T$ determines both $X$ and $Y$. ∎

Lemma 8.3**.**

Let ${\mathcal{H}}$ be a set of integral hyperplanes of $\mathbb{R}^{(I)}$ and let $i\in I\setminus\operatorname{supp}({\mathcal{H}})$. Let $L$ be a bounded distributive lattice, and let $a,b\in L$ such that $a\wedge b=0$. Then every $0,1$-lattice homomorphism $f\colon\operatorname{Op}({\mathcal{H}})\to L$ extends to a unique $0,1$-lattice homomorphism $g\colon\operatorname{Op}({\mathcal{H}}\cup\left\{\Delta_{i}\right\})\to L$ such that $a=g(\Delta_{i}^{+})$ and $b=g(\Delta_{i}^{-})$.

Proof.

Keep the notation of Lemma 8.2. A homomorphism $g\colon\operatorname{Op}({\mathcal{H}}\cup\left\{\Delta_{i}\right\})\to L$ satisfies the given conditions iff the homomorphism $h\underset{\mathrm{def}}{=}g\circ\varepsilon\colon\operatorname{Op}({\mathcal{H}})\ast\mathsf{J}_{2}\to L$ satisfies $h(X,X,X)=f(X)$ whenever $X\in\operatorname{Op}({\mathcal{H}})$, $a=h(\mathbb{R}^{(I)},\varnothing,\varnothing)$, and $b=h(\varnothing,\mathbb{R}^{(I)},\varnothing)$. Apply Lemma 4.1. ∎

9. Representing countable completely normal lattices

This section is devoted to a proof of our main theorem (Theorem 9.1), together with a short discussion of some of its corollaries.

Theorem 9.1**.**

Every countable completely normal distributive lattice with zero is isomorphic to $\operatorname{Id_{c}}G$, for some Abelian $\ell$-group $G$.

Proof.

We must represent a countable completely normal distributive lattice $L$ with zero. The lattice $\overline{L}$, obtained from $L$ by adding a new top element, is also completely normal, and $L$ is an ideal of $\overline{L}$. Any representation of $\overline{L}$ as $\operatorname{Id_{c}}\overline{G}$, for an Abelian $\ell$-group $\overline{G}$, yields $L\cong\operatorname{Id_{c}}G$ for the $\ell$-ideal $G\underset{\mathrm{def}}{=}\left\{{x\in\overline{G}}\mid{\langle{x}\rangle_{\overline{G}}\in L}\right\}$ (cf. Bigard, Keimel, and Wolfenstein [5, § 2.3]). Hence, it suffices to consider the case where $L$ is bounded, following the strategy described in Section 2. Fix a generating subset $\left\{{a_{n}}\mid{n\in\omega}\right\}$ of $L$.

As in Section 8, we shall denote by $\Lambda$ the additive subgroup of \bigl{(}{\mathbb{R}^{(\omega)}}\bigr{)}^{\prime} generated by the canonical projections $\delta_{n}\colon\mathbb{R}^{(\omega)}\to\mathbb{R}$ (where $n<\omega$), and we shall denote by ${\mathcal{H}}_{\mathbb{Z}}={\mathcal{H}}_{\Lambda}=\left\{{H_{n}}\mid{n\in\omega}\right\}$ the set of all integral hyperplanes of $\mathbb{R}^{(\omega)}$. Moreover, let $\left\{{(U_{n},V_{n},\gamma_{n})}\mid{n\in\omega}\right\}$ be an enumeration of all triples $(U,V,\gamma)$, where $U,V\in\operatorname{Op}^{-}({\mathcal{H}}_{\mathbb{Z}})$ and $\gamma\in L$.

We construct an ascending chain $\left({{\mathcal{H}}_{n}}\mid{n\in\omega}\right)$ of nonempty finite subsets of ${\mathcal{H}}_{\mathbb{Z}}$, with union ${\mathcal{H}}_{\mathbb{Z}}$, together with an ascending sequence $\left({f_{n}}\mid{n\in\omega}\right)$ of $0,1$-lattice homomorphisms $f_{n}\colon\operatorname{Op}({\mathcal{H}}_{n})\to L$, as follows.

Take ${\mathcal{H}}_{0}\underset{\mathrm{def}}{=}\left\{\Delta_{0}\right\}$ (cf. Section 8); so $\operatorname{Op}({\mathcal{H}}_{0})=\left\{\varnothing,\Delta_{0}^{+},\Delta_{0}^{-},\Delta_{0}^{+}\cup\Delta_{0}^{-},\mathbb{R}^{(\omega)}\right\}$ is isomorphic to $\mathsf{J}_{2}$ (cf. Section 4). Let $f_{0}\colon\operatorname{Op}({\mathcal{H}}_{0})\to\left\{0,a_{0},1\right\}$ be the unique homomorphism such that $f_{0}(\Delta_{0}^{+})=a_{0}$, $f_{0}(\Delta_{0}^{-})=0$, and f_{0}\bigl{(}{\mathbb{R}^{(\omega)}}\bigr{)}=1.

Suppose $f_{n}\colon\operatorname{Op}({\mathcal{H}}_{n})\to L$ already constructed.

Let $n=3m$ for some integer $m$, denote by $k$ the first nonnegative integer outside $\operatorname{supp}({\mathcal{H}}_{n})$, and set ${\mathcal{H}}_{n+1}\underset{\mathrm{def}}{=}{\mathcal{H}}_{n}\cup\left\{\Delta_{k}\right\}$. By Lemma 8.3, there is a unique lattice homomorphism $f_{n+1}\colon\operatorname{Op}({\mathcal{H}}_{n+1})\to L$, extending $f_{n}$, such that $f_{n+1}(\Delta_{k}^{+})=a_{m}$ and $f_{n+1}(\Delta_{k}^{-})=0$. This will take care of the surjectivity of the restriction, to $\operatorname{Op}^{-}({\mathcal{H}}_{\mathbb{Z}})$, of the union of the $f_{n}$.

Let $n=3m+1$ for some integer $m$, and set ${\mathcal{H}}_{n+1}\underset{\mathrm{def}}{=}{\mathcal{H}}_{n}\cup\left\{H_{m}\right\}$. Since $L$ is completely normal and the range of $f_{n}$ is finite, there is a finite sublattice $S$ of $L$ such that the range of $f_{n}$ is consonant in $S$. By Lemma 6.6, $f_{n}$ extends to a lattice homomorphism $f_{n+1}$ from $\operatorname{Op}({\mathcal{H}}_{n+1})$ to $S$, thus to $L$. This will take care of the union of all $f_{n}$ be defined on $\operatorname{Op}({\mathcal{H}}_{\mathbb{Z}})$.

Let, finally, $n=3m+2$ for some integer $m$. By iterating Lemma 7.2 finitely many times, we get a finite subset ${\mathcal{H}}_{n+1}$ of ${\mathcal{H}}_{\mathbb{Z}}$ containing ${\mathcal{H}}_{n}$, together with an extension $f_{n+1}\colon\operatorname{Op}({\mathcal{H}}_{n+1})\to L$, such that for every $k\leq n$, if $\left\{U_{k},V_{k}\right\}\subseteq\operatorname{Op}^{-}({\mathcal{H}}_{n})$ and $f_{n}(U_{k})\leq f_{n}(V_{k})\vee\gamma_{k}$, then $f_{n+1}(U_{k}\mathbin{\smallsetminus}_{\operatorname{Op}^{-}({\mathcal{H}}_{n+1})}V_{k})\leq\gamma_{k}$. This will take care of the union of the $f_{n}$ be closed (cf. Definition 2.4) on $\operatorname{Op}^{-}({\mathcal{H}}_{\mathbb{Z}})$.

The union $f$ of all the $f_{n}$ is a surjective lattice homomorphism from $\operatorname{Op}({\mathcal{H}}_{\mathbb{Z}})$ onto $L$. Furthermore, the restriction $f^{-}$ of $f$ to $\operatorname{Op}^{-}({\mathcal{H}}_{\mathbb{Z}})$ is a closed, surjective lattice homomorphism from $\operatorname{Op}^{-}({\mathcal{H}}_{\mathbb{Z}})$ onto $L$. Now it follows from the Baker-Beynon duality (cf. Lemmas 3.2 and 3.3, and Section 7, in Baker [1]) that $\operatorname{Id_{c}}\operatorname{F}_{\ell}(\omega)$ is isomorphic to the sublattice of $\mathbb{R}^{\mathbb{R}^{(\omega)}}$ generated by $\left\{{[\![{f>0}]\!]}\mid{f\in\Lambda}\right\}$, that is, $\operatorname{Op}^{-}({\mathcal{H}}_{\mathbb{Z}})$ (cf. Lemma 5.7). Hence, the map $f^{-}$ induces a closed, surjective lattice homomorphism $g\colon\operatorname{Id_{c}}\operatorname{F}_{\ell}(\omega)\twoheadrightarrow L$. By Lemma 2.5, this map factors through an isomorphism from $\operatorname{Id_{c}}(\operatorname{F}_{\ell}(\omega)/I)$ onto $L$, for a suitable $\ell$-ideal $I$ of $\operatorname{F}_{\ell}(\omega)$. ∎

Recall that Delzell and Madden’s results in [10] imply that Theorem 9.1 does not extend to the uncountable case.

Corollary 9.2**.**

A second countable generalized spectral space $X$ is homeomorphic to the $\ell$-spectrum of an Abelian $\ell$-group iff it is completely normal.

Proof.

Since $X$ is second countable, an easy application of compactness shows that ${\overset{\circ}{\mathcal{K}}}(X)$ is countable. Apply Theorem 9.1 and Lemma 2.1. ∎

It is well known that the lattice ${\mathcal{C}}(G)$, of all convex $\ell$-subgroups of any $\ell$-group (not necessarily Abelian) $G$, is the ideal lattice a completely normal distributive lattice with zero (see Iberkleid, Martínez, and McGovern [19, § 1.2] for a short overview). Of course, in the Abelian case, ${\mathcal{C}}(G)$ is isomorphic to the ideal lattice of $\operatorname{Id_{c}}{G}$. A direct application of Theorem 9.1 yields the following.

Corollary 9.3**.**

For every countable $\ell$-group $G$, there exists a countable Abelian $\ell$-group $A$ such that ${\mathcal{C}}(G)\cong{\mathcal{C}}(A)$.

The results of Kenoyer [25] and McCleary [28] imply that Corollary 9.3 does not extend to the uncountable case.

The real spectrum $\operatorname{Spec_{r}}{R}$, of any commutative unital ring $R$, is a completely normal spectral space (cf. Coste and Roy [8], Dickmann [13]). A direct application of Corollary 9.2 yields the following.

Corollary 9.4**.**

For every countable, commutative, unital ring $R$, there exists a countable Abelian $\ell$-group $A$ with unit such that $\operatorname{Spec_{r}}{R}\cong\operatorname{Spec_{\ell}}{A}$.

We prove in [38] that Corollary 9.4 does not extend to the uncountable case.

10. Non-$\ell$-representability results

In this section we shall show that the class of $\ell$-representable distributive lattices is neither first-order, nor closed under infinite products (resp., homomorphic images). All our non-$\ell$-representability results will rely on the following concept. We say that a distributive lattice $D$ has countably based differences if for all $a,b\in D$, the filter $a\ominus_{D}b$ (cf. (3.2)) is countably generated. The following result is a restatement, in terms of lattices of principal $\ell$-ideals, of Cignoli, Gluschankof, and Lucas [6, Thm. 2.2]; see also Iberkleid, Martínez, and McGovern [19, Prop. 4.1.2].

Lemma 10.1**.**

Let $G$ be an Abelian $\ell$-group. Then the lattice $\operatorname{Id_{c}}G$ has countably based differences.

Example 10.2**.**

A countable Abelian $\ell$-group $G$, with unit, such that $\operatorname{Id_{c}}{G}$ is not a dual Heyting algebra.

Proof.

Let $G$ consist of all maps $x\colon\omega\to\mathbb{Z}$ such that there are (necessarily unique) $\alpha,\beta\in\mathbb{Z}$ such that $x(n)=\alpha n+\beta$ for all large enough $n$. Then $G$, ordered componentwise, is an $\ell$-subgroup of $\mathbb{Z}^{\omega}$. The constant function $a$, with value $1$, and the identity function $b$ on $\omega$, both belong to $G^{+}$, $a+b$ is a unit of $G$, and there is no least $\boldsymbol{x}\in\operatorname{Id_{c}}G$ such that $\langle{b}\rangle\subseteq\langle{a}\rangle\vee\boldsymbol{x}$. ∎

It is easy to see that the class of all $\ell$-representable distributive lattices is closed under finite cartesian products. We shall now show that this observation does not extend to infinite products.

Proposition 10.3**.**

Let $D$ be a distributive lattice with zero. If $D$ is not a generalized dual Heyting algebra, then $D^{\omega}$ is not $\ell$-representable.

Proof.

Denote by $\varepsilon\colon D\hookrightarrow D^{\omega}$ the diagonal embedding and suppose that $D^{\omega}$ is $\ell$-representable. Since $D$ is isomorphic to an ideal of $D^{\omega}$, it is also $\ell$-representable, thus, by Lemma 10.1, $D$ has countably based differences. On the other hand, since $D$ is not a generalized dual Heyting algebra, there are $a,b\in D$ such that $a\ominus_{D}b$ has no least element. The filter $a\ominus_{D}b$ has a countably basis $\left({c_{n}}\mid{n\in\omega}\right)$, with each $c_{n+1}\leq c_{n}$.

Now by Lemma 10.1, $D^{\omega}$ has countably based differences. In particular, the filter $\varepsilon(a)\ominus_{D^{\omega}}\varepsilon(b)$ has a countable basis $\left({\boldsymbol{e}_{n}}\mid{n\in\omega}\right)$ with each $\boldsymbol{e}_{n+1}\leq\boldsymbol{e}_{n}$. For all $n,k\in\omega$, $a\leq b\vee\boldsymbol{e}_{n}(k)$, thus there exists $f(n,k)\in\omega$ such that $c_{f(n,k)}\leq\boldsymbol{e}_{n}(k)$. Set $\boldsymbol{x}\underset{\mathrm{def}}{=}\left({c_{f(n,n)+1}}\mid{n\in\omega}\right)$. Since $\varepsilon(a)\leq\varepsilon(b)\vee\boldsymbol{x}$, there exists $n\in\omega$ such that $\boldsymbol{e}_{n}\leq\boldsymbol{x}$. It follows that $c_{f(n,n)}\leq\boldsymbol{e}_{n}(n)\leq\boldsymbol{x}(n)=c_{f(n,n)+1}$, a contradiction. ∎

By taking $D\underset{\mathrm{def}}{=}\operatorname{Id_{c}}{G}$, for the $\ell$-group of Example 10.2, we get

Corollary 10.4**.**

The class of all $\ell$-representable bounded distributive lattices is not closed under infinite products.

Our next example involves the infinitary logic $\mathbin{\mathscr{L}}_{\infty,\omega}$, for which we refer the reader to Keisler and Knight [24] (see also Bell [3]), of which we will adopt the terminology, in particular about back-and-forth families. We say that a submodel $M$, of a model $N$, is an $\mathbin{\mathscr{L}}_{\infty,\omega}$-elementary submodel of $N$, if for every $\mathbin{\mathscr{L}}_{\infty,\omega}$ sentence $\varphi$, with (finitely many, by definition of a sentence) parameters from $M$, $M$ satisfies $\varphi$ iff $N$ does. Our example will show that there is no class of $\mathbin{\mathscr{L}}_{\infty,\omega}$ sentences whose class of models is the one of all $\ell$-representable bounded distributive lattices. As customary, we denote by $\omega_{1}$ the first uncountable ordinal.

Example 10.5**.**

A non-$\ell$-representable bounded distributive lattice $\boldsymbol{D}_{\omega_{1}}$, of cardinality $\aleph_{1}$, with a countable $\ell$-representable $\mathbin{\mathscr{L}}_{\infty,\omega}$-elementary sublattice $\boldsymbol{D}_{\omega,\omega_{1}}$.

Proof.

For any sets $I$ and $J$ with $I\subseteq J$, we denote by $[{I}]^{<\omega}$ the set of all finite subsets of $I$, and we set

[TABLE]

(Observe, in particular, that if $J$ is finite, then $\boldsymbol{D}_{J}=\boldsymbol{B}_{J}\times\mathbf{3}$.) We endow $\boldsymbol{D}_{J}$ and $\boldsymbol{D}_{I,J}$ with their componentwise orderings (i.e., $(X,k)\leq(Y,l)$ if $X\subseteq Y$ and $k\leq l$). They are obviously bounded distributive lattices. Further, we set

[TABLE]

For any sets $I$ and $J$ and any bijection $f\colon I\to J$, the map $\overline{f}\colon\boldsymbol{D}_{I}\to\boldsymbol{D}_{J}$, $(X,k)\mapsto(f[X],k)$ is a lattice isomorphism. The following claim states some elementary properties of the maps $\varepsilon_{I,J}$ and $\overline{f}$; its proof is straightforward and we omit it.

Claim 1**.**

1. (1)

For any sets $I\subseteq J$, $\boldsymbol{D}_{I,J}$ is a bounded sublattice of $\boldsymbol{D}_{J}$ and $\varepsilon_{I,J}$ defines an isomorphism from $\boldsymbol{D}_{I}$ onto $\boldsymbol{D}_{I,J}$. 2. (2)

The maps $\varepsilon_{I,J}$ form a direct system: that is, $\varepsilon_{I,I}=\mathrm{id}_{\boldsymbol{D}_{I}}$ and $\varepsilon_{I,K}=\varepsilon_{J,K}\circ\varepsilon_{I,J}$ whenever $I\subseteq J\subseteq K$. 3. (3)

For any set $J$, the set $\boldsymbol{D}_{J}$ is the ascending union of all subsets $\boldsymbol{D}_{I,J}$, for $I\in[{J}]^{<\omega}$. 4. (4)

Let $I^{\prime}$, $I^{\prime\prime}$, $J^{\prime}$, $J^{\prime\prime}$ be sets with $I^{\prime}\subseteq I^{\prime\prime}$ and $J^{\prime}\subseteq J^{\prime\prime}$, let $g\colon I^{\prime\prime}\to J^{\prime\prime}$ be a bijection with $g[I^{\prime}]=J^{\prime}$, and let $f$ be the domain-range restriction of $g$ from $I^{\prime}$ onto $J^{\prime}$. Then $\overline{g}\circ\varepsilon_{I^{\prime},I^{\prime\prime}}=\varepsilon_{J^{\prime},J^{\prime\prime}}\circ\overline{f}$.

For any set $K$, we denote by ${\mathcal{L}}_{K}$ the first-order language obtained by adding to the language $(\vee,\wedge,0,1)$, of bounded lattices, a collection of constant symbols indexed by $\boldsymbol{D}_{K}$. Then for every set $I$ containing $K$, the lattice $\boldsymbol{D}_{I}$ is naturally equipped with a structure of model for $\mathbin{\mathscr{L}}_{K}$, by interpreting every $\boldsymbol{a}\in\boldsymbol{D}_{K}$ by $\varepsilon_{K,I}(\boldsymbol{a})$.

For infinite sets $I$ and $J$, a finite subset $K$ of $I\cap J$, and finite sequences $(\boldsymbol{x}_{1},\dots,\boldsymbol{x}_{n})$ of elements of $\boldsymbol{D}_{I}$ and $(\boldsymbol{y}_{1},\dots,\boldsymbol{y}_{n})$ of elements of $\boldsymbol{D}_{J}$, let the statement $(\boldsymbol{x}_{1},\dots,\boldsymbol{x}_{n})\simeq_{K}(\boldsymbol{y}_{1},\dots,\boldsymbol{y}_{n})$ hold if there are $I^{\prime}\in[{I}]^{<\omega}$ and $J^{\prime}\in[{J}]^{<\omega}$ both containing $K$, a bijection $f\colon I^{\prime}\to J^{\prime}$ extending the identity of $K$, and elements $\boldsymbol{x}^{\prime}_{1},\dots,\boldsymbol{x}^{\prime}_{n}\in\boldsymbol{D}_{I^{\prime}}$, such that each $\boldsymbol{x}_{i}=\varepsilon_{I^{\prime},I}(\boldsymbol{x}^{\prime}_{i})$ and each $\boldsymbol{y}_{i}=\varepsilon_{J^{\prime},J}(\overline{f}(\boldsymbol{x}^{\prime}_{i}))$.

Claim 2**.**

The relation $\simeq_{K}$ is a back-and-forth family for $(\boldsymbol{D}_{I},\boldsymbol{D}_{J})$ with respect to the language ${\mathcal{L}}_{K}$.

Proof of Claim..

Trivially, $\varnothing\simeq_{K}\varnothing$. Further, if $(\boldsymbol{x}_{1},\dots,\boldsymbol{x}_{n})\simeq_{K}(\boldsymbol{y}_{1},\dots,\boldsymbol{y}_{n})$ holds via $I^{\prime}$, $J^{\prime}$, and $f$ as above, then $\varepsilon_{J^{\prime},J}\circ\overline{f}\circ\varepsilon_{I^{\prime},I}^{-1}$ is an isomorphism from $\boldsymbol{D}_{I^{\prime},I}$ onto $\boldsymbol{D}_{J^{\prime},J}$, sending each $\boldsymbol{x}_{i}$ to $\boldsymbol{y}_{i}$ and each $\varepsilon_{K,I}(\boldsymbol{z})$, where $\boldsymbol{z}\in\boldsymbol{D}_{K}$, to $\varepsilon_{K,J}(\boldsymbol{z})$; whence $(\boldsymbol{x}_{1},\dots,\boldsymbol{x}_{n})$ and $(\boldsymbol{y}_{1},\dots,\boldsymbol{y}_{n})$ satisfy the same quantifier-free formulas of ${\mathcal{L}}_{K}$.

Now let $(\boldsymbol{x}_{1},\dots,\boldsymbol{x}_{n})\simeq_{K}(\boldsymbol{y}_{1},\dots,\boldsymbol{y}_{n})$, via $I^{\prime}$, $J^{\prime}$, $f\colon I^{\prime}\to J^{\prime}$, and elements $\boldsymbol{x}^{\prime}_{i}\in\boldsymbol{D}_{I^{\prime}}$. Let $\boldsymbol{x}\in\boldsymbol{D}_{I}$. We need to find $\boldsymbol{y}\in\boldsymbol{D}_{J}$ such that $(\boldsymbol{x}_{1},\dots,\boldsymbol{x}_{n},\boldsymbol{x})\simeq_{K}(\boldsymbol{y}_{1},\dots,\boldsymbol{y}_{n},\boldsymbol{y})$. There are a finite set $I^{\prime\prime}$, with $I^{\prime}\subseteq I^{\prime\prime}\subseteq I$, and $\boldsymbol{x}^{\prime\prime}\in\boldsymbol{D}_{I^{\prime\prime}}$, such that $\boldsymbol{x}=\varepsilon_{I^{\prime\prime},I}(\boldsymbol{x}^{\prime\prime})$. We set $\boldsymbol{x}^{\prime\prime}_{i}=\varepsilon_{I^{\prime},I^{\prime\prime}}(\boldsymbol{x}^{\prime}_{i})$ for each $i$. Since $J$ is infinite, we can extend $f$ to a bijection $g\colon I^{\prime\prime}\to J^{\prime\prime}$, with $J^{\prime\prime}\subseteq J$. Then each $\boldsymbol{x}_{i}=\varepsilon_{I^{\prime\prime},I}(\boldsymbol{x}^{\prime\prime}_{i})$ and (using Claim 1) \boldsymbol{y}_{i}=\varepsilon_{J^{\prime\prime},J}\bigl{(}{\overline{g}(\boldsymbol{x}^{\prime\prime}_{i})}\bigr{)}. Hence, setting \boldsymbol{y}\underset{\mathrm{def}}{=}\varepsilon_{J^{\prime\prime},J}\bigl{(}{\overline{g}(\boldsymbol{x}^{\prime\prime})}\bigr{)}, we get the relation

[TABLE]

Symmetrically, for all $\boldsymbol{y}\in\boldsymbol{D}_{J}$, there exists $\boldsymbol{x}\in\boldsymbol{D}_{I}$ such that (10.1) holds. ∎ Claim 2.

By Karp’s Theorem (cf. Karp [21], Barwise [2, Thm. VII.5.3], Keisler and Knight [24, Thm. 1.2.1]), it follows that $\boldsymbol{D}_{I}$ and $\boldsymbol{D}_{J}$ satisfy the same $\mathbin{\mathscr{L}}_{\infty,\omega}$-sentences of the language ${\mathcal{L}}_{K}$. By letting $I\underset{\mathrm{def}}{=}\omega$, $J\underset{\mathrm{def}}{=}\omega_{1}$ and by letting $K$ range over all finite subsets of $\omega$, we thus obtain the following claim.

Claim 3**.**

The lattice $\boldsymbol{D}_{\omega,\omega_{1}}$ $(\cong\boldsymbol{D}_{\omega})$ is an $\mathbin{\mathscr{L}}_{\infty,\omega}$-elementary sublattice of $\boldsymbol{D}_{\omega_{1}}$.

Now we move to $\ell$-representability.

Claim 4**.**

Let $I$ be countably infinite. Then $\boldsymbol{D}_{I}$ is $\ell$-representable.

Proof of Claim..

While Claim 4 trivially follows from Theorem 9.1, it is also easy to verify that $\boldsymbol{D}_{I}\cong\operatorname{Id_{c}}{G}$ where $G$ is the $\ell$-group of Example 10.2. ∎ Claim 4.

Claim 5**.**

The lattice $\boldsymbol{D}_{\omega_{1}}$ does not have countably based differences. In particular, it is not $\ell$-representable.

Proof of Claim..

The elements $\boldsymbol{a}\underset{\mathrm{def}}{=}(\omega_{1},1)$ and $\boldsymbol{b}\underset{\mathrm{def}}{=}(\omega_{1},2)$ both belong to $\boldsymbol{D}_{\omega_{1}}$. Furthermore, the filter $\boldsymbol{b}\ominus_{\boldsymbol{D}_{\omega_{1}}}\boldsymbol{a}=\left\{{(X,2)}\mid{X\subseteq\omega_{1}\text{ cofinite}}\right\}$ is not countably based. The second part of our claim follows from Lemma 10.1. ∎ Claim 5.

This claim finishes the proof of Example 10.5. ∎

Note*.*

Denote by $Z$ the completely normal spectral space constructed by Delzell and Madden in [10, Thm. 2]. Although there is an obvious $0,1$-lattice embedding from $\boldsymbol{D}_{\omega_{1}}$ into ${\overset{\circ}{\mathcal{K}}}(Z)$, it is not hard to see that the two lattices are not isomorphic. Hence, $Z$ is not homeomorphic to the spectrum of $\boldsymbol{D}_{\omega_{1}}$.

Example 10.6**.**

An $\ell$-representable bounded distributive lattice of cardinality $\aleph_{1}$, with a non-$\ell$-representable lattice homomorphic image.

Proof.

The set $\boldsymbol{D}$, of all almost constant maps from $\omega_{1}$ to $\mathbf{3}$, is a $0,1$-sublattice of $\mathbf{3}^{\omega_{1}}$. It is straightforward to verify that $\boldsymbol{D}\cong\operatorname{Id_{c}}{H}$, where $H$ denotes the Abelian $\ell$-group of all almost constant maps from $\omega_{1}$ to the lexicographical product of $\mathbb{Z}$ by itself. Now consider the non-$\ell$-representable lattice $\boldsymbol{D}_{\omega_{1}}$ of Example 10.5. The map $\rho\colon\boldsymbol{D}\to\boldsymbol{D}_{\omega_{1}}$, $\boldsymbol{x}\mapsto(\operatorname{supp}(\boldsymbol{x}),x(\infty))$ is a surjective lattice homomorphism. ∎

By Stone duality, it follows that a spectral subspace of an $\ell$-spectrum may not be an $\ell$-spectrum.

11. Discussion

11.1. Ideal lattices of dimension groups

A partially ordered Abelian group $G$ is a dimension group if $G$ is directed, unperforated (i.e., $mx\geq 0$ implies that $x\geq 0$, whenever $x\in G$ and $m$ is a positive integer), and $G^{+}$ satisfies the Riesz refinement property (cf. Goodearl [15]). The construction $\operatorname{Id_{c}}{G}$, for an Abelian $\ell$-group $G$, extends naturally to arbitrary dimension groups, by replacing “$\ell$-ideal” by “directed convex subgroup” (in short ideal). However, now $\operatorname{Id_{c}}{G}$ is only a $(\vee,0)$-semilattice. This semilattice is always distributive (i.e., it satisfies the Riesz refinement property), but it may not be a lattice. In fact, every countable distributive $(\vee,0)$-semilattice is isomorphic to $\operatorname{Id_{c}}{G}$ for some countable dimension group $G$ (this is stated in Goodearl and Wehrung [16, Thm. 5.2]; it is also implicit in Bergman [4]); moreover, the countable size is optimal (Wehrung [37]).

In particular, it follows from Goodearl and Wehrung [16, Thm. 4.4] that for every distributive lattice $L$ with zero, there exists a dimension group $G$ such that $\operatorname{Id_{c}}{G}\cong L$ (without any restriction on the cardinality of $L$). Attempting to infer, via Theorem 1 of Elliott and Mundici [14], that if $L$ is completely normal, then $G$ is lattice-ordered, would already fail for the lattice $L\underset{\mathrm{def}}{=}\boldsymbol{D}_{\omega_{1}}$ of Example 10.5, simply because $\boldsymbol{D}_{\omega_{1}}$ is not $\ell$-representable. The problem lies in the impossibility to read, on $\operatorname{Id_{c}}{G}$ alone, that every prime quotient of $G$ be totally ordered, as illustrated by the following example (cf. [14, p. 181]): let $G$ be any non totally ordered simple dimension group (e.g., $G=\mathbb{Q}\times\mathbb{Q}$ with positive cone consisting of all $(x,y)$ with either $x=y=0$ or $x>0$ and $y>0$). Then $\operatorname{Id_{c}}{G}\cong\mathbf{2}$, yet $G$ is not totally ordered.

11.2. Lattices of $\ell$-ideals in non-Abelian $\ell$-groups

It is proved in Růžička, Tůma, and Wehrung [32, Thm. 6.3] that every countable distributive $(\vee,0)$-semilattice is isomorphic to $\operatorname{Id_{c}}{G}$ for some $\ell$-group $G$; moreover, this result does not extend to semilattices of cardinality $\aleph_{2}$. The gap at size $\aleph_{1}$ is not filled yet.

11.3. Open problems

Mellor and Tressl proved in [29] that for any infinite cardinal $\lambda$, there is no $\mathbin{\mathscr{L}}_{\infty,\lambda}$ characterization of Stone duals of real spectra of commutative unital rings. Our first open problem calls for an extension of that result to $\ell$-spectra, which would thus also extend the result of Example 10.5 (where we get only $\mathbin{\mathscr{L}}_{\infty,\omega}$).

Problem 1**.**

Is the class of all $\ell$-representable lattices the class of all models of a class of $\mathbin{\mathscr{L}}_{\infty,\lambda}$ sentences, for some infinite cardinal $\lambda$?

Recall from Example 10.6 that a spectral subspace of an $\ell$-spectrum may not be an $\ell$-spectrum. We also extend this result to real spectra in [38]. This suggests the following problem.

Problem 2**.**

Is every retract of an $\ell$-spectrum (resp., real spectrum) also an $\ell$-spectrum (resp., real spectrum)?

The analogy between $\ell$-spectra and real spectra (cf. Delzell and Madden [11]), together with Corollary 9.2, suggests the following problem.

Problem 3**.**

Is every second countable completely normal spectral space homeomorphic to the real spectrum of some commutative, unital ring?

The more general question, of characterizing real spectra of commutative, unital rings, is part of Problem 12 in Keimel’s survey paper [23]. Due to results by Mellor and Tressl [29], this is essentially hopeless without any cardinality restriction.

Acknowledgments

The MV-space problem was introduced to me during a visit, to Università degli studi di Salerno, in September 2016. Excellent conditions provided by the math department are greatly appreciated.

Uncountably many thanks are also due to Estella and Lynn, who stayed at my side during those very hard moments, and also to our family’s cat Nietzsche, who had decided, that fateful November morning, that I shouldn’t leave my desk.