Almost orthogonality and Hausdorff interval topologies of atomic lattice effect algebras

TL;DR

This paper investigates the topological properties of atomic lattice effect algebras, establishing conditions under which their interval topology is Hausdorff and exploring implications for order convergence and state smearing.

Contribution

It proves that the interval topology of certain atomic lattice effect algebras is Hausdorff under almost orthogonality, and links this to order continuity and compactness, extending the understanding of their topological structure.

Findings

Interval topology is Hausdorff under almost orthogonality.

Order convergence coincides with topological convergence in complete cases.

State smearing theorem applies to these effect algebras.

Abstract

We prove that the interval topology of an Archimedean atomic lattice effect algebra $E$ is Hausdorff whenever the set of all atoms of $E$ is almost orthogonal. In such a case $E$ is order continuous. If moreover $E$ is complete then order convergence of nets of elements of $E$ is topological and hence it coincides with convergence in the order topology and this topology is compact Hausdorff compatible with a uniformity induced by a separating function family on $E$ corresponding to compact and cocompact elements. For block-finite Archimedean atomic lattice effect algebras the equivalence of almost orthogonality and s-compact generation is shown. As the main application we obtain the state smearing theorem for these effect algebras, as well as the continuity of $\oplus$-operation in the order and interval topologies on them.

Equations

$$x=⨁\{k_{\alpha }{a}_{\alpha }\mid \alpha \in \mathcal{E}\}=\bigvee \{k_{\alpha }{a}_{\alpha }\mid \alpha \in \mathcal{E}\}$$

$$x\uparrow u_{\alpha }\le x_{\alpha }\le v_{\alpha }\downarrow x,\alpha \in \mathcal{E}$$

$$x_{\alpha }\stackrel{(o)}{\to }x,\alpha \in \mathcal{E}⟹x_{\alpha }\stackrel{{\tau }_i}{\to }x,\alpha \in \mathcal{E}.$$

$$x_{\alpha }\stackrel{{\tau }_i^F}{\to }x,\alpha \in \mathcal{E}⟺x_{\alpha }\stackrel{{\tau }_i}{\to }x,\alpha \in \mathcal{E}.$$

$$z=⨁\{t_{\gamma }{c}_{\gamma }\mid \gamma \in \mathcal{E}\}=\bigvee \{t_{\gamma }{c}_{\gamma }\mid \gamma \in \mathcal{E}\}.$$

$$x=\bigvee \{u\in \mathcal{U}\mid u\le x\}=\bigwedge \{v\in \mathcal{V}\mid x\le v\}.$$

$$\begin{array}[]{l}x_{\alpha}\stackrel{{\scriptstyle\tau_{\Phi}}}{{\longrightarrow}}x\hbox{ iff }\varphi(x_{\alpha})\to\varphi(x)\hbox{ for any }\varphi\in\Phi.\end{array}$$

$$\begin{array}[]{l}x_{\alpha}\stackrel{{\scriptstyle\tau_{\Phi}}}{{\longrightarrow}}x\ \hbox{ iff }\ \varphi(x_{\alpha})\to\varphi(x)\hbox{ for any }\varphi\in\Phi.\end{array}$$

$$x_{\alpha }\stackrel{(o)}{⟶}x\text{mbox}inE\text{mbox}ifandonlyifx\in D\text{mbox}and{x}_{\alpha }\stackrel{(o)}{⟶}x\text{mbox}inD.$$

Full text

fn1fn1footnotetext: This work was supported by the Slovak Resaerch and Development Agency under the contract No. APVV–0071–06 and the grant VEGA-1/3025/06 of MŠ SR.

1 Introduction, basic definitions and facts

In the study of effect algebras (or more general, quantum structures) as carriers of states and probability measures, an important tool is the study of topologies on them. We can say that topology is practically equivalent with the concept of convergence. From the probability point of view the convergence of nets is the main tool in spite of that convergence of filters is easier to handle and preferred in the modern topology. It is because states or probabilities are mappings (functions) defined on elements but not on subsets of quantum structures. Note also, that connections between order convergence of filters and nets are not trivial. For instance, if a filter order converges to some point of a poset then the associated net need not order converge (see e.g, [12]).

On the other hand certain topological properties of studied structures characterize also their certain algebraic properties and conversely. For instance a known fact is that a Boolean algebra $B$ is atomic iff the interval topology $\tau_{i}$ on $B$ is Hausdorff (see [20, Corollary 3.4]). This is not more valid for lattice effect algebras (even MV-algebras). By Frink’s Theorem the interval topology $\tau_{i}$ on $B$ (more generally on any lattice $L$) is compact iff it is a complete lattice [5]. In [16] it was proved that if a lattice effect algebra $E$ (more generally any basic algebra) is compactly generated then $E$ is atomic.

We are going to prove that on an Archimedean atomic lattice effect algebra $E$ the interval topology $\tau_{i}$ is Hausdorff and $E$ is (o)-continuous if and only if $E$ is almost orthogonal. Moreover, if $E$ is complete then $\tau_{i}$ is compact and coincides with the order topology $\tau_{o}$ on $E$ and this compact topology $\tau_{i}=\tau_{o}$ is compatible with a uniformity on $E$ induced by a separating function family on $E$ corresponding to compact and cocompact elements of $E$.

As the main corollary of that we obtain that every Archimedean atomic block-finite lattice effect algebra $E$ has Hausdorff interval topology and hence both topologies $\tau_{i}$ and $\tau_{o}$ are Hausdorff and they coincide. In this case almost orthogonality of $E$ and s-compact generation by finite elements of $E$ are equivalent. As an application the state smearing theorem for these effect algebras is formulated. Moreover, the continuity of $\oplus$-operation in $\tau_{i}$ and $\tau_{o}$ on them is shown.

Definition 1.1**.**

A partial algebra $(E;\oplus,0,1)$ is called an effect algebra if [math], $1$ are two distinct elements and $\oplus$ is a partially defined binary operation on $E$ which satisfy the following conditions for any $a,b,c\in E$:

(Ei)ii

$b\oplus a=a\oplus b$* if $a\oplus b$ is defined,*

(Eii)i

$(a\oplus b)\oplus c=a\oplus(b\oplus c)$* if one side is defined,*

(Eiii)

for every $a\in E$ there exists a unique $b\in E$ such that $a\oplus b=1$ (we put $a^{\prime}=b$),

(Eiv)i

if $1\oplus a$ is defined then $a=0$.

We often denote the effect algebra $(E;\oplus,0,1)$ briefly by $E$. In every effect algebra $E$ we can define the partial order $\leq$ by putting

$a\leq b$ and $b\ominus a=c$ iff $a\oplus c$ is defined and $a\oplus c=b$, we set $c=b\ominus a$ .

If $E$ with the defined partial order is a lattice (a complete lattice) then $(E;\oplus,0,1)$ is called a lattice effect algebra (a complete lattice effect algebra).

Recall that a set $Q\subseteq E$ is called a sub-effect algebra of the effect algebra $E$ if

1. (i)

$1\in Q$ 2. (ii)

if out of elements $a,b,c\in E$ with $a\oplus b=c$ two are in $Q$, then $a,b,c\in Q$.

If $Q$ is simultaneously a sublattice of $E$ then $Q$ is called a sub-lattice effect algebra of $E$.

We say that a finite system $F=(a_{k})_{k=1}^{n}$ of not necessarily different elements of an effect algebra $(E;\oplus,0,1)$ is $\oplus$-orthogonal if $a_{1}\oplus a_{2}\oplus\dots\oplus a_{n}$ (written $\bigoplus\limits_{k=1}^{n}a_{k}$ or $\bigoplus F$) exists in $E$. Here we define $a_{1}\oplus a_{2}\oplus\dots\oplus a_{n}=(a_{1}\oplus a_{2}\oplus\dots\oplus a_{n-1})\oplus a_{n}$ supposing that $\bigoplus\limits_{k=1}^{n-1}a_{k}$ exists and $\bigoplus\limits_{k=1}^{n-1}a_{k}\leq a^{\prime}_{n}$. An arbitrary system $G=(a_{\kappa})_{\kappa\in H}$ of not necessarily different elements of $E$ is $\oplus$-orthogonal if $\bigoplus K$ exists for every finite $K\subseteq G$. We say that for a $\oplus$-orthogonal system $G=(a_{\kappa})_{\kappa\in H}$ the element $\bigoplus G$ exists iff $\bigvee\{\bigoplus K\mid K\subseteq G$, $K$ is finite$\}$ exists in $E$ and then we put $\bigoplus G=\bigvee\{\bigoplus K\mid K\subseteq G\}$ (we write $G_{1}\subseteq G$ iff there is $H_{1}\subseteq H$ such that $G_{1}=(a_{\kappa})_{\kappa\in H_{1}}$).

Recall that elements $x$ and $y$ of a lattice effect algebra are called compatible (written $x\leftrightarrow y$) if $x\vee y=x\oplus(y\ominus(x\wedge y))$ [13]. For $x\in E$ and $Y\subseteq E$ we write $x\leftrightarrow Y$ iff $x\leftrightarrow y$ for all $y\in Y$. If every two elements are compatible then $E$ is called an MV-effect algebra. In fact, every MV-effect algebra can be organized into an MV-algebra (see [2]) if we extend the partial $\oplus$ to a total operation by setting $x\oplus y=x\oplus(x^{\prime}\wedge y)$ for all $x,y\in E$ (also conversely, restricting a total $\oplus$ into partial $\oplus$ for only $x,y\in E$ with $x\leq y^{\prime}$ we obtain a MV-effect algebra).

Moreover, in [23] it was proved that every lattice effect algebra is a set-theoretical union of MV-effect algebras called blocks. Blocks are maximal subsets of pairwise compatible elements of $E$, under which every subset of pairwise compatible elements is by Zorn’s Lemma contained in a maximal one. Further, blocks are sub-lattices and sub-effect algebras of $E$ and hence maximal sub-MV-effect algebras of $E$. A lattice effect algebra is called block-finite if it has only finitely many blocks.

Finally note that lattice effect algebras generalize orthomodular lattices [10] (including Boolean algebras) if we assume existence of unsharp elements $x\in E$, meaning that $x\wedge x^{\prime}\neq 0$. On the other hand the set $S(E)=\{x\in E\mid x\wedge x^{\prime}=0\}$ of all sharp elements of a lattice effect algebra $E$ is an orthomodular lattice [8]. In this sense a lattice effect algebra is a “smeared” orthomodular lattice, while an MV-effect algebra is a “smeared” Boolean algebra. An orthomodular lattice $L$ can be organized into a lattice effect algebra by setting $a\oplus b=a\vee b$ for every pair $a,b\in L$ such that $a\leq b^{\perp}$.

For an element $x$ of an effect algebra $E$ we write ${\mathop{\rm ord}}(x)=\infty$ if $nx=x\oplus x\oplus\dots\oplus x$ ($n$-times) exists for every positive integer $n$ and we write ${\mathop{\rm ord}}(x)=n_{x}$ if $n_{x}$ is the greatest positive integer such that $n_{x}x$ exists in $E$. An effect algebra $E$ is Archimedean if ${\mathop{\rm ord}}(x)<\infty$ for all $x\in E$. We can show that every complete effect algebra is Archimedean (see [22]).

An element $a$ of an effect algebra $E$ is an atom if $0\leq b<a$ implies $b=0$ and $E$ is called atomic if for every nonzero element $x\in E$ there is an atom $a$ of $E$ with $a\leq x$. If $u\in E$ and either $u=0$ or $u=p_{1}\oplus p_{2}\oplus\dots\oplus p_{n}$ for some not necessarily different atoms $p_{1},p_{2},\dots,p_{n}\in E$ then $u\in E$ is called finite and $u^{\prime}\in E$ is called cofinite. If $E$ is a lattice effect algebra then for $x\in E$ and an atom $a$ of $E$ we have $a\leftrightarrow x$ iff $a\leq x$ or $a\leq x^{\prime}$. It follows that if $a$ is an atom of a block $M$ of $E$ then $a$ is also an atom of $E$. On the other hand if $E$ is atomic then, in general, every block in $E$ need not be atomic (even for orthomodular lattices [1]).

The following theorem is well known.

Theorem 1.2**.**

[25, Theorem 3.3]* Let $(E;\oplus,0,1)$ be an Archimedean atomic lattice effect algebra. Then to every nonzero element $x\in E$ there are mutually distinct atoms $a_{\alpha}\in{E}$, $\alpha{\in\mathcal{E}}$ and positive integers $k_{\alpha}$ such that*

[TABLE]

under which $x\in S(E)$ iff $k_{\alpha}=n_{a_{\alpha}}={\mathop{\rm ord}}(a_{\alpha})$ for all $\alpha\in{\mathcal{E}}$.

Definition 1.3**.**

(1)* An element $a$ of a lattice $L$ is called compact iff, for any $D\subseteq L$ with $\bigvee D\in L$, if $a\leq\bigvee D$ then $a\leq\bigvee F$ for some finite $F\subseteq D$.*

(2)* A lattice $L$ is called compactly generated iff every element of $L$ is a join of compact elements.*

The notions of cocompact element and cocompactly generated lattice can be defined dually. Note that compact elements are important in computer science in the semantic approach called domain theory, where they are considered as a kind of primitive elements.

2 Characterizations of interval topologies on bounded lattices

The order convergence of nets ((o)-convergence), interval topology $\tau_{i}$ and order-topology $\tau_{o}$ ((o)-topology) can be defined on any poset. In our observations we will consider only bounded lattices and we will give a characterization of interval topologies on them.

Definition 2.1**.**

Let $L$ be a bounded lattice. Let $\mathcal{H}=\{[a,b]\subseteq L|a,b\in L$ with $a\leq b\}$ and let $\mathcal{G}=\{\bigcup^{n}_{k=1}[a_{k},b_{k}]|[a_{k},b_{k}]\in\mathcal{H},k=1,2,...,n\}$. The interval topology $\tau_{i}$ of $L$ is the topology of $L$ with $\mathcal{G}$ as a closed basis, hence with $\mathcal{H}$ as a closed subbasis.

From definition of $\tau_{i}$ we obtain that $U\in\tau_{i}$ iff for each $x\in U$ there is $F\in\mathcal{G}$ such that $x\in L\backslash F\subseteq U$.

Definition 2.2**.**

Let $L$ be a poset.

• (i)

A net $(x_{\alpha})_{\alpha\in\mathcal{E}}$ of elements of $L$ ** order converges* ($(o)$-converges, for short) to a point $x\in L$ if there exist nets $(u_{\alpha})_{\alpha\in\mathcal{E}}$ and $(v_{\alpha})_{\alpha\in\mathcal{E}}$ of elements of $L$ such that*

[TABLE]

where $x\uparrow u_{\alpha}$ means that $u_{\alpha_{1}}\leq u_{\alpha_{2}}$ for every ${\alpha_{1}}\leq{\alpha_{2}}$ and $x=\bigvee\{x_{\alpha}\mid\alpha\in\mathcal{E}\}$. The meaning of $v_{\alpha}\downarrow x$ is dual.

We write $x_{\alpha}\stackrel{{\scriptstyle(o)}}{{\rightarrow}}x,{\alpha\in\mathcal{E}}$ in $L$.

• (ii)

A topology $\tau_{0}$ on $L$ is called the order topology on $L$ iff

1. (a)

for any net $(x_{\alpha})_{\alpha\in\mathcal{E}}$ of elements of $L$ and $x\in L$: $x_{\alpha}\stackrel{{\scriptstyle(o)}}{{\rightarrow}}a$ in L $\Rightarrow$ $x_{\alpha}\stackrel{{\scriptstyle\tau_{0}}}{{\rightarrow}}x$, ${\alpha\in\mathcal{E}}$, where $x_{\alpha}\stackrel{{\scriptstyle\tau_{0}}}{{\rightarrow}}x$ denotes that $(x_{\alpha})_{\alpha\in\mathcal{E}}$ converges to $x$ in the topological space $(L,\tau_{0})$, 2. (b)

if $\tau$ is a topology on $L$ with property (a) then $\tau\subseteq\tau_{o}$.

Hence $\tau_{o}$ is the strongest (finest, biggest) topology on $L$ with property (a).

Recall that, for a directed set $(\mathcal{E},\leq)$, a subset $\mathcal{E}^{\prime}\subseteq\mathcal{E}$ is called cofinal in $\mathcal{E}$ iff for every $\alpha\in\mathcal{E}$ there is $\beta\in\mathcal{E}^{\prime}$ such that $\alpha\leq\beta$. A special kind of a subnet of a net $(x_{\alpha})_{\alpha\in\mathcal{E}}$ is net $(x_{\beta})_{\beta\in\mathcal{E}^{\prime}}$ where $\mathcal{E}^{\prime}$ is a cofinal subset of $\mathcal{E}$. This kind of subnets works in many cases of our considerations.

In what follows we often use the following useful characterization of topological convergence of nets:

Lemma 2.3**.**

For a net $(x_{\alpha})_{\alpha\in\mathcal{E}}$ of elements of a topological space $(X,\tau)$ and $x\in X$:

[TABLE]

Proof.

$\Rightarrow$: It is trivial.

$\Leftarrow$: Let for every $\mathcal{E}^{\prime}\subseteq\mathcal{E}$, where $\mathcal{E}^{\prime}$ is cofinal in $\mathcal{E}$ there exist $\mathcal{E}^{\prime\prime}\subseteq\mathcal{E}^{\prime}$, $\mathcal{E}^{\prime\prime}$ cofinal in $\mathcal{E}^{\prime}$ and $x_{\gamma}\stackrel{{\scriptstyle\tau}}{{\rightarrow}}x,\gamma\in\mathcal{E}^{\prime\prime}$, and let $x_{\alpha}\not\stackrel{{\scriptstyle\tau}}{{\rightarrow}}x,\alpha\in\mathcal{E}$. Then there exist $U(x)\in\tau$ such that for all $\alpha\in\mathcal{E}$ there exist $\beta_{\alpha}\in\mathcal{E}$ with $\beta_{\alpha}\geq\alpha$ and $x_{\beta_{\alpha}}\not\in U(x)$. Let $\mathcal{E}^{\prime}=\{\beta_{\alpha}\in\mathcal{E}|\alpha\in\mathcal{E},\beta_{\alpha}\geq\alpha,x_{\beta_{\alpha}}\not\in U(x)\}$ then $x_{\beta_{\alpha}}\not\stackrel{{\scriptstyle\tau}}{{\rightarrow}}x,{\beta_{\alpha}}\in\mathcal{E}^{\prime}$ and for all cofinal $\mathcal{E}^{\prime\prime}\subseteq\mathcal{E}^{\prime}$: $x_{\gamma}\not\stackrel{{\scriptstyle\tau}}{{\rightarrow}}x,\gamma\in\mathcal{E}^{\prime\prime}$. Hence there exists $\mathcal{E}^{\prime}\subseteq\mathcal{E}$ cofinal in $\mathcal{E}$ and for all $\mathcal{E}^{\prime\prime}\subseteq\mathcal{E}^{\prime}$, $\mathcal{E}^{\prime\prime}$ cofinal in $\mathcal{E}^{\prime}$: $x_{\gamma}\not\stackrel{{\scriptstyle\tau}}{{\rightarrow}}x,\gamma\in\mathcal{E}^{\prime\prime}$ a contradiction. ∎

Further, let us recall the following well known facts:

Lemma 2.4**.**

Let $L$ be a bounded lattice. Then

1. (i)

$F\subseteq L$* is $\tau_{0}$-closed iff for every net $(x_{\alpha})_{\alpha\in\mathcal{E}}$ of elements of $L$ and $x\in L$:*

$(x_{\alpha}\in F,x_{\alpha}\stackrel{{\scriptstyle(o)}}{{\rightarrow}}x,\alpha\in\mathcal{E})\Rightarrow x\in F$. 2. (ii)

For every $a,b\in L$ with $a\leq b$ the interval $[a,b]$ is $\tau_{0}$-closed. 3. (iii)

$\tau_{i}\subseteq\tau_{o}$. 4. (iv)

For any net $(x_{\alpha})_{\alpha\in\mathcal{E}}$ of elements of a $L$ and $x\in L$:

[TABLE] 5. (v)

If $\tau_{i}$ is Hausdorff then $\tau_{0}=\tau_{i}$ (see **[4]**). 6. (vi)

The interval topology $\tau_{i}$ of a lattice $L$ is compact iff $L$ is a complete lattice (see **[5]**).

Finally, let us note that compact Hausdorff topological space is always normal. Thus separation axiom $T_{2},T_{3}$ and $T_{4}$ are trivially equivalent for the interval topology of a complete lattice $L$.

Theorem 2.5**.**

Let $L$ be a complete lattice with interval topology $\tau_{i}$. If $F\subseteq L$ is a complete sub-lattice of $L$ then

1. (a)

$\tau^{F}_{i}=\tau_{i}\cap F$* is the interval topology of $F$,* 2. (b)

for any net $(x_{\alpha})_{\alpha\in\mathcal{E}}$ of elements of $F$ and $x\in F$:

[TABLE]

Proof.

(a): Let $\mathcal{H}$ and $\mathcal{H}_{F}$ be a closed subbasis of $\tau_{i}$ and $\tau^{F}_{i}$ respectively. Then evidently $\mathcal{H}\cap F=\{[a,b]\cap F|[a,b]\in\mathcal{H}\}$ is a closed subbasis of $\tau_{i}\cap F$. Further for $[c,d]_{F}\in\mathcal{H}_{F}$ we have $[c,d]_{F}=\{x\in F|c\leq x\leq d\}=[c,d]\cap F\in\mathcal{H}\cap F$. Conversely, since $F$ is a complete sub-lattice of $L$, if $[a,b]\in\mathcal{H}$ then $[a,b]\cap F=\{x\in F|a\leq x\leq b\}$ and either $[a,b]\cap F=\emptyset$ or there exist $c=\wedge\{x\in F|a\leq x\leq b\}$ and $d=\vee\{x\in F|a\leq x\leq d\}$ and $[a,b]\cap F=[c,d]_{F}\in\mathcal{H}_{F}$. This proves that $\tau^{F}_{i}=\tau_{i}\cap F$.

(b): This is an easy consequence of (a). ∎

3 Hausdorff interval topology of almost orthogonal

Archimedean atomic lattice effect algebras and their order continuity

The atomicity of Boolean algebra $B$ is equivalent with Hausdorffness of interval topology on $B$ (see [11], [29] and [20, Corollary 3.4]). This is not more valid for lattice effect algebras, even also for MV-algebras.

Example 3.1**.**

Let $M=[0,1]\subseteq{\mathbb{R}}$ be a standard MV-effect algebra, i.e., we define $a\oplus b=a+b$ iff $a+b\leq 1$, $a,b\in M$. Then $M$ is a complete (o)-continuous lattice with $\tau_{i}=\tau_{o}$ being Hausdorff and with (o)-convergence of nets coinciding with $\tau_{o}$-convergence. Nevertheless, $M$ is not atomic. **

We have proved in [16] that a complete lattice effect algebra is atomic and (o)-continuous lattice iff $E$ is compactly generated. Nevertheless, in such a case, the interval topology on $E$ need not be Hausdorff.

Example 3.2**.**

Let $E$ be a horizontal sum of infinitely many finite chains $(P_{i},\bigoplus_{i},0_{i},1_{i})$ with at least 3 elements, $i=1,2,...,n,\dots,$ (i. e., for $i=1,2,...,n,\dots,$, we identify all $0_{i}$ and all $1_{i}$ as well, $\bigoplus_{i}$ on $P_{i}$ are preserved and any $a\in P_{i}\backslash\{0_{i},1_{i}\}$, $b\in P_{j}\backslash\{0_{j},1_{j}\}$ for $i\not=j$ are noncomparable). Then $E$ is an atomic complete lattice effect algebra, $E$ is not block-finite and the interval topology $\tau_{i}$ on $E$ is compact. Nevertheless, $\tau_{i}$ is not Hausdorff because e.g., for $a\in P_{i},b\in P_{j},i\not=j$, $a,b$ noncomparable, we have $[a,1]\cap[0,b]=\emptyset$ and there is no finite family $\mathcal{I}$ of closed intervals in $E$ separating $[a,1],[0,b]$ (i.e., the lattice $E$ can not be covered by a finite number of closed intervals from $\mathcal{I}$ each of which is disjoint with at least one of the intervals $[a,1]$ and $[0,b]$). This implies that $\tau_{i}$ is not Hausdorff by [20, Lemma 2.2]. Further $E$ is compactly generated by finite elements (hence (o)-continuous). It follows by [16] that the order topology $\tau_{o}$ on $E$ is a uniform topology and (o)-convergence of nets on $E$ coincides with $\tau_{o}$-convergence.

In what follows we shall need an extension of [26, Lemma 2.1 (iii)].

Lemma 3.3**.**

Let $E$ be a lattice effect algebra, $x,y\in E$. Then $x\wedge y=0$ and $x\leq y^{\prime}$ iff $kx\wedge ly=0$ and $kx\leq(ly)^{\prime}$, whenever $kx$ and $ly$ exist in $E$.

Proof.

Let $x\leq y^{\prime}$, $x\wedge y=0$ and $2y$ exists in $E$. Then $x\oplus y=(x\vee y)\oplus(x\wedge y)=x\vee y\leq y^{\prime}$ and hence there exists $x\oplus 2y=(x\vee y)\oplus y=(x\oplus y)\vee 2y=x\vee y\vee 2y=x\vee 2y$, which gives that $x\leq(2y)^{\prime}$ and $x\wedge 2y=0$. By induction, if $ly$ exists then $x\oplus ly=x\vee ly$ and hence $x\leq(ly)^{\prime}$ and $x\wedge ly=0$.

Now, $x\leq(ly)^{\prime}$ iff $ly\leq x^{\prime}$ and because $x\wedge ly=0$, we obtain by the same argument as above that $ly\oplus kx=ly\vee kx$, hence $kx\leq(ly)^{\prime}$ and $ly\wedge kx=0$ whenever $kx$ exists in $E$.

Conversely, $kx\wedge ly=0$ implies that $x\wedge y=0$ and $kx\leq(ly)^{\prime}$ implies $x\leq kx\leq(ly)^{\prime}\leq y^{\prime}$. ∎

In next we will use the statement of Lemma 3.3 in the following form: For any $x,y\in E$ with $x\wedge y=0$, $x\not\leq y^{\prime}$ iff $kx\not\leq(ly)^{\prime}$, whenever $kx$ and $ly$ exist in $E$.

Definition 3.4**.**

Let $E$ be an atomic lattice effect algebra. $E$ is said to be almost orthogonal if the set $\{b\in E\mid b\not\leq a^{\prime},b\ \mbox{is an atom}\}$ is finite for every atom $a\in E$.

Note that our definition of almost orthogonality coincides with the usual definition for orthomodular lattices (see e.g. [17, 18]).

Theorem 3.5**.**

Let $E$ be an Archimedean atomic lattice effect algebra. Then $E$ is almost orthogonal if and only if for any atom $a\in E$ and any integer $l$, $1\leq l\leq n_{a}$, there are finitely many atoms $c_{1},\dots,c_{m}$ and integers $j_{1},\dots,j_{m}$, $1\leq j_{1}\leq n_{c_{1}},\dots,1\leq j_{m}\leq n_{c_{m}}$ such that $j_{k}{}c_{k}\not\leq(la)^{\prime}$ for all $k\in\{1,\dots,m\}$ and, for all $x\in E$, $x\not\leq(la)^{\prime}$ implies $j_{k_{0}}{}c_{k_{0}}\leq x$ for some ${k_{0}}\in\{1,\dots,m\}$.

Proof.

$\Longrightarrow$: Assume that $E$ is almost orthogonal. Let $a\in E$ be an atom, $1\leq l\leq n_{a}$. We shall denote $A_{a}=\{b\in E\mid b\ \mbox{is}\ \mbox{an}\ \mbox{atom},b\not\leq a^{\prime}\}$. Clearly, $A_{a}$ is finite i.e. $A_{a}=\{b_{1},\dots,b_{n}\}$ for suitable atoms $b_{1},\dots,b_{n}$ from $E$.

Let $b\in E$ be an atom, $1\leq k\leq n_{b}$ and $kb\not\leq(la)^{\prime}$. Either $b=a$ or $b\not={}a$ and in this case we have by Lemma 3.3 (iv) that $b\not\leq a^{\prime}$. Hence either $b=a$ or $b\in A_{a}$. Let us put \{c_{1},\dots,c_{m}\}=\left\{\begin{array}[]{l l}A_{a}&\mbox{if}\ a\in S(E)\\ A_{a}\cup\{a\}&\mbox{otherwise}\end{array}\right.. In both cases we have that $a\in\{c_{1},\dots,c_{m}\}$.

Now, let $x\in E$ and $x\not\leq(la)^{\prime}$. By Theorem 1.2 there is an atom $c\in E$ and an integer $1\leq j\leq n_{c}$ such that $jc\leq x$ and $jc\not\leq(la)^{\prime}$. Either $c=a$ or $c\not\leq a$. In the first case we have that $j\geq(n_{a}-l+1)$ i.e. $x\geq(n_{a}-l+1)a$. In the second case we get that $c\not\leq a^{\prime}$ i.e. $c\in A_{a}$ and $x\geq b_{i}$ for suitable $i\in\{1,\dots,n\}$. Hence it is enough to put $j_{k}=1$ if $c_{k}\in A_{a}$ and $j_{k}=(n_{a}-l+1)$ if $c_{k}=a$.

$\Longleftarrow$: Conversely, let $a\in E$ be an atom. Then there are finitely many atoms $c_{1},\dots,c_{m}$ and integers $j_{1},\dots,j_{m}$, $1\leq j_{1}\leq n_{c_{1}},\dots,1\leq j_{m}\leq n_{c_{m}}$ such that $j_{k}{}c_{k}\not\leq a^{\prime}$ for all $k\in\{1,\dots,m\}$ and, for all $x\in E$, $x\not\leq a^{\prime}$ implies $j_{k_{0}}{}c_{k_{0}}\leq x$ for some ${k_{0}}\in\{1,\dots,m\}$. Let us check that $A_{a}\subseteq\{c_{1},\dots,c_{m}\}$. Let $b\in A_{a}$. Then $b\geq j_{k_{0}}{}c_{k_{0}}\geq c_{k_{0}}$ for some ${k_{0}}\in\{1,\dots,m\}$. Hence $b=c_{k_{0}}$. This yields $A_{a}$ is finite. ∎

Lemma 3.6**.**

Let $E$ be an almost orthogonal Archimedean atomic lattice effect algebra. Then, for any atom $a\in E$ and any integer $l$, $1\leq l\leq n_{a}$ there are finitely many atoms $b_{1},\dots,b_{n}$ and integers $j_{1},\dots,j_{n}$, $1\leq j_{1}\leq n_{b_{1}},\dots,1\leq j_{n}\leq n_{b_{n}}$ such that

[TABLE]

Hence $[0,(la)^{\prime}]$ is a clopen subset in the interval topology.

Proof.

Let $a\in E$ be an atom, $1\leq l\leq n_{a}$. By Definition 3.5, let $\{j_{1}{}b_{1},\dots,j_{n}{}b_{n}\}$ be the finite set of non-orthogonal finite elements to $la$ of the form $j_{k}b_{k}$, $1\leq j_{k}\leq n_{b_{k}}$ minimal such that $b_{1},\dots,b_{n}$ are atoms different from $a$. We put $D=[0,(la)^{\prime}]\cup\left(\bigcup_{k=1}^{n}[j_{k}b_{k},1]\cup[(n_{a}+1-l)a,1]\right)$. Let us check that $D=E$. Clearly, $D\subseteq E$. Now, let $z\in E$. Then by Theorem 1.2 there are mutually distinct atoms $c_{\gamma}\in{E}$, $\gamma{\in\mathcal{E}}$ and integers $t_{\gamma}$ such that

[TABLE]

Either $t_{\gamma}c_{\gamma}\leq(la)^{\prime}$ for all $\gamma\in{\mathcal{E}}$ and hence $z\in[0,(la)^{\prime}]$ or there exists $\gamma_{0}\in{\mathcal{E}}$ such that $t_{\gamma_{0}}c_{\gamma_{0}}\not\leq(la)^{\prime}$. Hence, by almost orthogonality, either $j_{k_{0}}b_{k_{0}}\leq t_{\gamma_{0}}c_{\gamma_{0}}\leq z$ for some $k_{0}\in\{1,\dots,n\}$ or $(n_{a}+1-l)a\leq t_{\gamma_{0}}c_{\gamma_{0}}\leq z$. In both cases we get that $z\in D$.

Now, assume that $y\in{}[0,(la)^{\prime}]\cap\left(\bigcup_{k=1}^{n}[j_{k}{}b_{k},1]\cup[(n_{a}+1-l)a,1]\right)$. Then $(n_{a}+1-l)a\leq y\leq(la)^{\prime}$ or $j_{k}{}b_{k}\leq y\leq(la)^{\prime}$ for some $k\in\{1,\dots,n\}$. In any case we have a contradiction. ∎

Proposition 3.7**.**

Let $E$ be an almost orthogonal Archimedean atomic lattice effect algebra. Then, for any not necessarily different atoms $a,b\in E$ and any integers $l,k$; $1\leq l\leq n_{a}$, $1\leq k\leq n_{b}$, the interval $[kb,(la)^{\prime}]$ is clopen in the interval topology.

Proof.

From Lemma 3.6 we have that $[0,(la)^{\prime}]$ is a clopen subset. Since a dual of an almost orthogonal Archimedean atomic lattice effect algebra is an almost orthogonal Archimedean atomic lattice effect algebra as well, we have that $[kb,1]$ is again clopen in the interval topology. Hence also $[kb,(la)^{\prime}]$ is clopen in the interval topology. ∎

Theorem 3.8**.**

Let $E$ be an almost orthogonal Archimedean atomic lattice effect algebra. Then the interval topology $\tau_{i}$ on $E$ is Hausdorff.

Proof.

Let $x,y\in E$ and $x\not=y$. Then (without loss of generality) we may assume that $x\not\leq y$. Then by [25, Theorem 3.3] there exists an atom $b$ from $E$ and an integer $k$, $1\leq k\leq n_{b}$ such that $kb\leq x$ and $kb\not\leq y$. Applying the dual of [25, Theorem 3.3] there exists an atom $a$ from $E$ and an integer $l$, $1\leq l\leq n_{a}$ such that $y\leq(la)^{\prime}$ and $kb\not\leq(la)^{\prime}$. Clearly, $x\in[kb,1]$, $y\in[0,(la)^{\prime}]$.

Assume that there is an element $z\in E$ such that $z\in[kb,1]\cap[0,(la)^{\prime}]$. Then $kb\leq z\leq(la)^{\prime}$, a contradiction. Hence by Proposition 3.7, $[kb,1]$ and $[0,(la)^{\prime}]$ are disjoint open subsets separating $x$ and $y$. ∎

Theorem 3.9**.**

Let $E$ be an almost orthogonal Archimedean atomic lattice effect algebra. Then $E$ is compactly generated and therefore (o)-continuous.

Proof.

It is enough to check that, for any atom $a\in E$ and any integer $l$, $1\leq l\leq n_{a}$ the element $la$ is compact in $E$ since any element of $E$ is a join of such elements (see Theorem 1.2 resp. [25, Theorem 3.3]).

Let $x=\bigvee_{\alpha\in\mathcal{E}}x_{\alpha}$ for some net $(x_{\alpha})_{\alpha\in\mathcal{E}}$ in $E$, $la\leq x$, i.e., $(la)^{\prime}\geq x^{\prime}\downarrow x^{{}^{\prime}}_{\alpha}$.

By Lemma 3.6 we have $E=[0,(la)^{\prime}]\cup\left(\bigcup_{k=1}^{n}[j_{k}{}b_{k},1]\cup[(n_{a}+1-l)a,1]\right)$, $[0,(la)^{\prime}]\cap\left(\bigcup_{k=1}^{n}[b_{k},1]\cup[(n_{a}+1-l)a,1]\right)=\emptyset$, $b_{1},\dots,b_{n}$ are atoms of $E$, $1\leq j_{k}\leq n_{b_{k}}$, $1\leq k\leq n$.

Since ${\mathcal{E}}$ is directed upwards, there exists a cofinal subset $\mathcal{E}^{\prime}\subseteq\mathcal{E}$ such that $x^{{}^{\prime}}_{\beta}\in[0,(la)^{\prime}]$ for all $\beta\in\mathcal{E}^{\prime}$ or there exists $k_{0}\in\{1,2,...,n\}$ such that $x_{\beta}\in[j_{k_{0}}b_{k_{0}},1]$ for all $\beta\in\mathcal{E}^{\prime}$ or $x^{{}^{\prime}}_{\beta}\in[(n_{a}+1-l)a,1]$ for all $\beta\in\mathcal{E}^{\prime}$. If $x^{{}^{\prime}}_{\beta}\in[0,(la)^{\prime}]$ for all $\beta\in\mathcal{E}^{\prime}$ then clearly $la\leq x_{\beta}$ for all $\beta\in\mathcal{E}^{\prime}$. If there exists $k_{0}\in\{1,2,...,n\}$ such that $x^{{}^{\prime}}_{\beta}\in[j_{k_{0}}b_{k_{0}},1]$ for all $\beta\in\mathcal{E}^{\prime}$ or $x^{{}^{\prime}}_{\beta}\in[(n_{a}+1-l)a,1]$ for all $\beta\in\mathcal{E}^{\prime}$ we obtain that $x^{\prime}\in[j_{k_{0}}b_{k_{0}},1]$ or $x^{{}^{\prime}}\in[(n_{a}+1-l)a,1]$ which is a contradiction with $x^{{}^{\prime}}\in[0,(la)^{\prime}]$. ∎

Let $E$ be an Archimedean atomic lattice effect algebra. We put ${\mathcal{U}}=\{x\in E\mid x=\bigvee_{i=1}^{n}l_{i}a_{i},a_{1},\dots,a_{n}\ \mbox{are atoms of }\ E,1\leq l_{i}\leq n_{a_{i}},1\leq i\leq n,n\ \mbox{natural}$ $\mbox{number}\}$ and ${\mathcal{V}}=\{x\in E\mid x^{\prime}\in{\mathcal{U}}\}$. Then by [25, Theorem 3.3], for every $x\in L$, we have that

[TABLE]

Consider the function family $\Phi=\{f_{u}\mid u\ \in{\mathcal{U}}\}\cup\{g_{v}\mid v\in{\mathcal{V}}\}$, where $f_{u},g_{v}:L\to\{0,1\}$, $u\in{\mathcal{U}},v\in{\mathcal{V}}$ are defined by putting f_{u}(x)=\left\{\begin{array}[]{r c l}1&\hbox{iff}&u\leq x\\ 0&\hbox{iff}&u\not\leq x\end{array}\right.

and g_{v}(y)=\left\{\begin{array}[]{r c l}1&\hbox{iff}&x\leq v\\ 0&\hbox{iff}&x\not\leq v\end{array}\right.  for all $x,y\in L$.

Further, consider the family of pseudometrics on $L$: $\Sigma_{\Phi}=\{\rho_{u}\mid u\in{\mathcal{U}}\}\cup\{\pi_{v}\mid v\in{\mathcal{V}}\}$, where $\rho_{u}(a,b)=|f_{u}(a)-f_{u}(b)|$ and $\pi_{v}(a,b)=|g_{v}(a)-g_{v}(b)|$ for all $a,b\in L$.

Let us denote by $\mathcal{U}_{\Phi}$ the uniformity on $L$ induced by the family of pseudometrics $\Sigma_{\Phi}$ (see e.g. [3]). Further denote by $\tau_{\Phi}$ the topology compatible with the uniformity $\mathcal{U}_{\Phi}$.

Then for every net $(x_{\alpha})_{\alpha\in\mathcal{E}}$ of elements of $L$

[TABLE]

This implies, since $f_{u}$, $u\in{\mathcal{U}}$, and $g_{v}$, $v\in{\mathcal{V}}$, is a separating function family on $L$, that the topology $\tau_{\Phi}$ is Hausdorff. Moreover, the intervals $[u,v]=[u,1]\cap[0,v]=f_{u}^{-1}(\{1\})\cap g_{v}^{-1}(\{1\})$ are clopen sets in $\tau_{\Phi}$.

Definition 3.10**.**

Let $E$ be an Archimedean atomic lattice effect algebra. Let $\Phi$ be a separating function family on $E$ defined above. We will denote by $\tau_{\Phi}$ the uniform topology on $E$ defined by this function family, that means for every net $(x_{\alpha})_{\alpha\in\mathcal{E}}$ of elements of $L$

[TABLE]

Theorem 3.11**.**

Let $E$ be an almost orthogonal Archimedean atomic lattice effect algebra. Then $\tau_{i}=\tau_{o}=\tau_{\Phi}$.

Proof.

Since by Theorem 3.8, $\tau_{i}$ is Hausdorff we obtain by [4] that $\tau_{i}=\tau_{o}$. Further if $O\in\tau_{o}$ and $x\in O$ then by Theorem 1.2 we have $x=\bigvee\{u\in{\mathcal{U}}\mid u\leq x\}=\bigwedge\{v\in{\mathcal{V}}\mid x\leq v\}$, which by [12] implies that there exist finite sets $F\subseteq{\mathcal{U}}$, $G\subseteq{\mathcal{V}}$ such that $x\in[\bigvee F,\bigwedge G]\subseteq O$. Hence $\tau_{o}\subseteq\tau_{\Phi}$. By Theorem 3.9 and [16, Theorem 1] we obtain $\tau_{o}=\tau_{\Phi}$. ∎

Theorem 3.12**.**

Let $E$ be an Archimedean atomic block-finite lattice effect algebra. Then $\tau_{i}=\tau_{o}$ is a Hausdorff topology.

Proof.

As in [18], it suffices to show that for every $x,y\in E$, $x\not\leq y$ there are finitely many intervals, none of which contains both $x$ and $y$ and the union of which covers $E$.

By [15], $E$ is a union of finitely many atomic blocks $M_{i}$, $i=1,2,\dots,n$. Choose $i\in\{1,2,\dots,n\}$. If $x,y\in M_{i}$ then there is an atom $a_{i}\in M_{i}$ and an integer $l_{i}$, $1\leq l_{i}\leq n_{a_{i}}$ such that $l_{i}a_{i}\leq x$, $l_{i}a_{i}\not\leq y$. Let us put $k_{i}=n-l_{i}+1$. Since $M_{i}$ is almost orthogonal (the only possible non-orthogonal $kb$ to $la$ for an atom $a$, $1\leq l\leq n_{a}$ is that $a=b$) we have by Lemma 3.6 that $M_{i}=([0,(k_{i}a_{i})^{\prime}]\cap M_{i})\cup([(n_{a_{i}}+1-k_{i})a_{i},1]\cap M_{i})$. Hence $M_{i}\subseteq[0,(k_{i}a_{i})^{\prime}]\cup[(n_{a_{i}}+1-k_{i})a_{i},1]$. Let us check that $[0,(k_{i}a_{i})^{\prime}]\cap[(n_{a_{i}}+1-k_{i})a_{i},1]=\emptyset$. Assume that $(n_{a_{i}}+1-k_{i})a_{i}\leq z\leq(k_{i}a_{i})^{\prime}$. Then $(n_{a_{i}}+1-k_{i})a_{i}\leq(k_{i}a_{i})^{\prime}$, a contradiction. Put $J_{i}=[0,(k_{i}a_{i})^{\prime}]$, $K_{i}=[(n_{a_{i}}+1-k_{i})a_{i},1]$. This yields $x\in K_{i}$, $y\in J_{i}$, $M_{i}\subseteq J_{i}\cup K_{i}$ and $J_{i}\cap K_{i}=\emptyset$. Let $x\not\in M_{i}$. Then there exists an atom $a_{i}\in M_{i}$ that is not compatible with $x$. Let us check that $x\not\in[0,(a_{i})^{\prime}]\cup[n_{a_{i}}a_{i},1]$. Assume that $x\in[0,(a_{i})^{\prime}]$ or $x\in[n_{a_{i}}a_{i},1]$. Then $x\leq(a_{i})^{\prime}$ or $a_{i}\leq n_{a_{i}}a_{i}\leq x$, i.e., in both cases we get that $x\leftrightarrow a_{i}$, a contradiction. Let us put $J_{i}=[0,(a_{i})^{\prime}]$, $K_{i}=[n_{a_{i}}a_{i},1]$. As above, $M_{i}\subseteq J_{i}\cup K_{i}$, $J_{i}\cap K_{i}=\emptyset$ and moreover $x\notin J_{i}\cup K_{i}$. The remaining case $y\not\in M_{i}$ can be checked by similar considerations. We obtain $E=\bigcup_{i=1}^{n}M_{i}\subseteq\bigcup_{i=1}^{n}(J_{i}\cup K_{i})\subseteq E$ and none of the intervals $J_{i},K_{i}$, $i=1,2,\dots,n$ contains both $x$ and $y$. ∎

4 Order and interval topologies

of complete atomic block-finite lattice effect algebras

We are going to show that on every complete atomic block-finite lattice effect algebra $E$ the interval topology is Hausdorff. Hence both topologies $\tau_{i}$ and $\tau_{o}$ are in this case compact Hausdorff and they coincide. Moreover, a necessary and sufficient condition for a complete atomic lattice algebra $E$ to be almost orthogonal is given.

For the proof of Theorems 4.2 and 4.3 we will use the following statement, firstly proved in the equivalent setting of D-posets in [19].

Theorem 4.1**.**

[19, Theorem 1.7]* Suppose that $(E;\oplus,0,1)$ is a complete lattice effect algebra. Let $\emptyset\not=D\subseteq E$ be a sub-lattice effect algebra. The following conditions are equivalent:*

1. (i)

For all nets $(x_{\alpha})_{\alpha\in\mathcal{E}}$ such that $x_{\alpha}\in D$ for all ${\alpha\in\mathcal{E}}$

[TABLE] 2. (ii)

For every $M\subseteq D$ with $\bigvee M=x$ in $E$ it holds $x\in D$. 3. (iii)

For every $Q\subseteq D$ with $\bigwedge Q=y$ in $E$ it holds $y\in D$. 4. (iv)

$D$* is a complete sub-lattice of $E$.* 5. (v)

$D$* is a closed set in order topology $\tau_{o}$ on $E$.*

Each of these conditions implies that $\tau_{o}^{D}=\tau_{o}^{E}\cap D$, where $\tau_{o}^{D}$ is an order topology on $D$.

Important sub-lattice effect algebras are blocks, $S(E)$, $B(E)=\bigcap\{M\subseteq E\mid M\ \mbox{block\/ of}\ E\}$ and $C(E)=B(E)\cap S(E)$ (see [6, 7, 13, 21, 23]).

Theorem 4.2**.**

Let $E$ be a complete lattice effect algebra. Then for every $D\in\{S(E),C(E),B(E)\}$ or $D=M$, where $M$ is a block of $E$, we have:

1. (1)

[TABLE] 2. (2)

If $\tau_{i}$ is Hausdorff then

[TABLE]

Proof.

The first part of the statement follows by Theorem 2.5 and the fact that if $E$ is a complete lattice effect algebra then $M$, $S(E)$, $C(E)$ and $B(E)$ are complete sub-lattices of $E$ (see [9, 24]). The second part follows by [4] since $\tau_{i}$ is Hausdorff implies $\tau_{i}=\tau_{o}$ and by Theorem 4.1. ∎

Theorem 4.3**.**

(i)* The interval topology $\tau_{i}$ on every Archimedean atomic MV-effect algebra $M$ is Hausdorff and $\tau_{i}=\tau_{o}=\tau_{\Phi}$.*

(ii)* For every complete atomic MV-effect algebra $M$ and for any net $(x_{\alpha})$ of $M$ and any $x\in M$,*

$x_{\alpha}\stackrel{{\scriptstyle\tau_{o}}}{{\longrightarrow}}x$* if and only if $x_{\alpha}\stackrel{{\scriptstyle(o)}}{{\longrightarrow}}x$ (briefly $\tau_{o}\equiv(o)$).*

Moreover, $\tau_{o}$ is a uniform compact Hausdorff topology on $M$.

(iii)* For every atomic block-finite lattice effect algebra $E$, $E$ is a complete lattice iff $\tau_{i}=\tau_{o}$ is a compact Hausdorff topology.*

Proof.

(i), (ii): This follows from the fact that every pair of elements of $M$ is compatible, hence every pair of atoms is orthogonal. Thus for (i) we can apply Theorem 3.11 and for (ii) we can use (i) and [16, Theorem 2] since $M$ is compactly generated by finite elements and $\tau_{i}$ is compact.

(iii) From Theorem 3.12 we know that $\tau_{i}=\tau_{o}$ is a Hausdorff topology. By Lemma 2.4 (vi) the interval topology $\tau_{i}$ on $E$ is compact iff $E$ is a complete lattice. ∎

In what follows we will need Corollary 4.5 of Lemma 4.4.

Lemma 4.4**.**

Let $E$ be an Archimedean atomic lattice effect algebra. Then

1. (i)

If $c,d\in E$ are compact elements with $c\leq d^{\prime}$ then $c\oplus d$ is compact. 2. (ii)

If $u=\bigoplus G$, where $G$ is a $\oplus$-orthogonal system of atoms of $E$, and $u$ is compact then $G$ is finite.

Proof.

(i) Let $c\oplus d\leq\bigvee D$. Let ${\cal E}=\{F\subseteq D:F\ \mbox{is finite}\}$ be directed by set inclusion and let for every $F\in{\cal E}$ be $x_{F}=\bigvee F$. Then $x_{F}\uparrow x=\bigvee D$. Since $c\leq\bigvee D$ and $d\leq\bigvee D$ there is a finite subset $F_{1}\subseteq D$ such that $c\vee d\leq\bigvee F_{1}$. Therefore, for $F\geq F_{1}$, $x_{F}\ominus c\uparrow x\ominus c,d\leq x\ominus c$. Then there is a finite subset $F_{2}\subseteq D$, $F_{1}\leq F_{2}$ such that $d\leq x_{F_{2}}\ominus c$. Hence $c\oplus d\leq x_{F_{2}}$.

(ii) Let $u\in E$, $u=\bigoplus G=\bigvee\{\bigoplus K\mid K\subseteq G$ is finite$\}$ where $G=(a_{\kappa})_{\kappa\in H}$ is a $\oplus$-orthogonal system of atoms. Clearly if $K_{1},K_{2}\subseteq G$ are finite such that $K_{1}\subseteq K_{2}$ then $\bigoplus K_{1}\leq\bigoplus K_{2}$.

Assume that $u$ is compact. Hence there are finite $K_{1},K_{2},$ $\dots,K_{n}\subseteq G$ such that $u\leq\bigvee\{\bigoplus K_{i}\mid i=1,2,\dots,n\}$. Let $K_{0}=\bigcup\{K_{i}\mid i=1,2,\dots,n\}$. Then $K_{0}\subseteq G$, $K_{0}$ is finite and $\bigoplus K_{i}\leq\bigoplus K_{0}$, $i=1,2\dots,n$, which gives that $\bigvee\{\bigoplus K_{i}\mid i=1,2,\dots,n\}\leq\bigoplus K_{0}$. It follows that $u\leq\bigoplus K_{0}\leq u=\bigvee\{\bigoplus K\mid K\subseteq G$ is finite$\}$. Hence $u=\bigoplus K_{0}$, $K_{0}\subseteq G$ is finite. Further, for every finite $K\subseteq G\setminus K_{0}$ we have $\bigoplus K_{0}\subseteq\bigoplus(K_{0}\cup K)=\bigoplus K_{0}\oplus\bigoplus K\leq u=\bigoplus K_{0}$ , which gives that $\bigoplus K=0$. Hence $K=\emptyset$ and thus $G\setminus K_{0}=\emptyset$ which gives that $K_{0}=G$.∎

Corollary 4.5**.**

Let $E$ be an o-continuous Archimedean atomic lattice effect algebra. Then every finite element of $E$ is compact.

Proof.

Clearly, by [16, Theorem 7] we know that $E$ is compactly generated. Therefore, any atom of $E$ is compact. The compactness of every finite element follows by an easy induction. ∎

Theorem 4.6**.**

Let $E$ be an Archimedean atomic lattice effect algebra. Then the following conditions are equivalent:

1. (i)

$\tau_{i}=\tau_{o}=\tau_{\Phi}$. 2. (ii)

$E$* is o-continuous and $\tau_{i}$ is Hausdorff.* 3. (iii)

$E$* is almost orthogonal.*

Proof.

(i)$\implies$(ii): Since $\tau_{o}=\tau_{\Phi}$ we have by [16, Theorem 1] that $E$ is compactly generated and hence o-continuous. The condition $\tau_{i}=\tau_{\Phi}$ implies that $\tau_{i}$ is Hausdorff because $\tau_{\Phi}$ is Hausdorff.

(ii)$\implies$(i), (iii): Since $\tau_{i}$ is Hausdorff we obtain $\tau_{i}=\tau_{o}$ by [4]. Moreover, from [16, Theorem 7] and Corollary 4.5 the (o)-continuity of $E$ implies that $E$ is compactly generated by the elements from ${\mathcal{U}}$. This gives $\tau_{o}=\tau_{\Phi}$ from [16, Theorem 1].

Let $a\in E$ be an atom, $1\leq l\leq n_{a}$. Then the interval $[0,(la)^{\prime}]$ is a clopen set in the order topology $\tau_{o}=\tau_{\Phi}=\tau_{i}$. Hence there is a finite set of intervals in $E$ such that $0\in E\setminus\bigcup_{i=1}^{n}[u_{i},v_{i}]\subseteq[0,(la)^{\prime}]$. Thus $E\subseteq[0,(la)^{\prime}]\cup\bigcup_{i=1}^{n}[u_{i},v_{i}]\subseteq[0,(la)^{\prime}]\cup\bigcup_{i=1}^{n}[k_{i}{}b_{i},1]$, where $b_{i}\in E$ are atoms such that $k_{i}{}b_{i}\leq u_{i}$, $1\leq k_{i}\leq n_{b_{i}}$, $i=1,\dots,n$. This yields that $E$ is almost orthogonal.

(iii)$\implies$(ii): From Theorems 3.8 and 3.9 we have that $\tau_{i}$ is Hausdorff and $E$ is compactly generated, hence (o)-continuous. ∎

Corollary 4.7**.**

Let $E$ be a complete atomic lattice effect algebra. Then the following conditions are equivalent:

1. (i)

$E$* is almost orthogonal.* 2. (ii)

$\tau_{i}=\tau_{o}=\tau_{\Phi}\equiv(o)$. 3. (iii)

$E$* is (o)-continuous and $\tau_{i}$ is Hausdorff.*

Proof.

It follows from Theorems 4.6 and the fact that by (o)-continuity of $E$ [27, Theorem 8] we have $\tau_{o}\equiv(o)$. ∎

The next example shows that a complete block-finite atomic lattice effect algebra need not be (o)-continuous and almost orthogonal in spite of that $\tau_{i}=\tau_{o}$ is a compact Hausdorff topology.

Example 4.8**.**

Let $E$ be a horizontal sum of finitely many infinite complete atomic Boolean algebras $(B_{i},\bigoplus_{i},0_{i},1_{i})$, $i=1,2,...,n$. Then $E$ is an atomic complete lattice effect algebra, $E$ is not almost orthogonal, $E$ is not compactly generated by finite elements (hence $\tau_{o}\not=\tau_{\Phi}$), $E$ is block-finite, $\tau_{i}=\tau_{o}$ is Hausdorff by Theorem 3.12, and the interval topology $\tau_{i}$ on $E$ is compact.

5 Applications

Theorem 5.1**.**

Let $E$ be a block-finite complete atomic lattice effect algebra. Then the following conditions are equivalent:

1. (i)

$E$* is almost orthogonal.* 2. (ii)

$E$* is compactly generated.* 3. (iii)

$E$* is (o)-continuous.* 4. (iv)

$\tau_{i}=\tau_{o}=\tau_{\Phi}\equiv(o)$.

Proof.

By Theorem 3.12, $\tau_{i}=\tau_{o}$ is a Hausdorff topology. This by [16, Theorem 7] gives that (ii) $\Longleftrightarrow$ (iii) and by Corollary 4.7 we obtain that (i) $\Longleftrightarrow$ (iii) $\Longleftrightarrow$ (iv). ∎

In Theorem 5.1, the assumption that $E$ is atomic can not be omitted. For instance, every non-atomic complete Boolean algebra is (o)-continuous but it is not compactly generated, because in such a case $E$ must be atomic by [16, Theorem 6].

Remark 5.2**.**

If a $\oplus$-operation on a lattice effect algebra $E$ is continuous with respect to its interval topology $\tau_{i}$ meaning that $x_{\alpha}\leq y_{\alpha}^{{}^{\prime}}$, $x_{\alpha}\stackrel{{\scriptstyle\tau_{i}}}{{\rightarrow}}x$, $y_{\alpha}\stackrel{{\scriptstyle\tau_{i}}}{{\rightarrow}}y$, $\alpha\in\mathcal{E}$ implies $x_{\alpha}\oplus y_{\alpha}\stackrel{{\scriptstyle\tau_{i}}}{{\rightarrow}}x\oplus y$, then $\tau_{i}$ is Hausdorff (see [14]). Hence $\oplus$-operation on complete (o)-continuous atomic lattice effect algebras which are not almost orthogonal cannot be $\tau_{i}$-continuous, by [14] and Corollary 4.7. **

Theorem 5.3**.**

Let $E$ be a block-finite complete atomic lattice effect algebra. Let $(x_{\alpha})_{\alpha\in\mathcal{E}}$ and $(y_{\alpha})_{\alpha\in\mathcal{E}}$ be nets of elements of $E$ such that $x_{\alpha}\leq y_{\alpha}^{{}^{\prime}}$ for all ${\alpha\in\mathcal{E}}$.

If $x_{\alpha}\stackrel{{\scriptstyle\tau_{i}}}{{\rightarrow}}x$, $y_{\alpha}\stackrel{{\scriptstyle\tau_{i}}}{{\rightarrow}}y$, $\alpha\in\mathcal{E}$ then $x\leq y^{{}^{\prime}}$ and $x_{\alpha}\oplus y_{\alpha}\stackrel{{\scriptstyle\tau_{i}}}{{\rightarrow}}x\oplus y$, $\alpha\in\mathcal{E}$. Moreover, $\tau_{i}=\tau_{o}$.

Proof.

Since, by Theorem 3.12, $\tau_{i}$ is Hausdorff, we obtain that $\tau_{i}=\tau_{o}$ by [4]. Let $\{M_{1},\dots,M_{n}\}$ be the set of all blocks of $E$. Further, for every ${\alpha\in\mathcal{E}}$, elements of the set $\{x_{\alpha},y_{\alpha},x_{\alpha}\oplus y_{\alpha}\}$ are pairwise compatible. It follows that for every ${\alpha\in\mathcal{E}}$ there exists a block $M_{k_{\alpha}}$ of $E$, ${k_{\alpha}}\in\{1,\dots,n\}$ such that $\{x_{\alpha},y_{\alpha},x_{\alpha}\oplus y_{\alpha}\}\subseteq M_{k_{\alpha}}$. Let ${\mathcal{E}}^{\prime}$ be any cofinal subset of ${\mathcal{E}}$. Since ${\mathcal{E}}^{\prime}$ is directed upwards, there exists a block $M_{k_{0}}$ of $E$ and a cofinal subset ${\mathcal{E}}^{\prime\prime}$ of ${\mathcal{E}}^{\prime}$ such that $\{x_{\beta},y_{\alpha},x_{\beta}\oplus y_{\beta}\}\subseteq M_{k_{0}}$ for all $\beta\in{\mathcal{E}}^{\prime\prime}$. Otherwise we obtain a contradiction with the finiteness of the set $\{M_{1},\dots,M_{n}\}$. Further, by Theorem 2.5, we obtain that $\tau^{M_{k_{0}}}_{i}=\tau_{i}\cap M_{k_{0}}$, as $M_{k_{0}}$ is a complete sublattice of $E$ (see Theorem 4.2). It follows that the interval topology $\tau^{M_{k_{0}}}_{i}$ on the complete MV-effect algebra $M_{k_{0}}$ is Hausdorff. The last by [14, Theorem 3.6] gives that $x_{\beta}\oplus y_{\beta}\stackrel{{\scriptstyle\tau^{M_{k_{0}}}_{i}}}{{\rightarrow}}x\oplus y$, $\beta\in\mathcal{E}^{\prime\prime}$ and hence $x_{\beta}\oplus y_{\beta}\stackrel{{\scriptstyle\tau_{i}}}{{\rightarrow}}x\oplus y$, $\beta\in\mathcal{E}^{\prime\prime}$, as $\tau^{M_{k_{0}}}_{i}=\tau_{i}\cap M_{k_{0}}$. It follows that $x_{\alpha}\oplus y_{\alpha}\stackrel{{\scriptstyle\tau_{i}}}{{\rightarrow}}x\oplus y$, $\alpha\in\mathcal{E}$ by Lemma 2.3. ∎

In [22, Theorem 4.5] it was proved that a block-finite lattice effect algebra $(E;\oplus,0,1)$ has a MacNeille completion which is a complete effect algebra $(MC(E);\oplus,0,1)$ containing $E$ as a (join-dense and meet-dense) sub-lattice effect algebra iff $E$ is Archimedean. In what follows we put $\widehat{E}=MC(E)$.

Corollary 5.4**.**

Let $E$ be a block-finite Archimedean atomic lattice effect algebra. Then for any nets $(x_{\alpha})_{\alpha\in\mathcal{E}}$ and $(y_{\alpha})_{\alpha\in\mathcal{E}}$ of elements of $E$ with $x_{\alpha}\leq y_{\alpha}^{{}^{\prime}}$, ${\alpha\in\mathcal{E}}$: $x_{\alpha}\stackrel{{\scriptstyle\tau_{i}}}{{\rightarrow}}x$, $y_{\alpha}\stackrel{{\scriptstyle\tau_{i}}}{{\rightarrow}}y$, $\alpha\in\mathcal{E}$ implies $x_{\alpha}\oplus y_{\alpha}\stackrel{{\scriptstyle\tau_{i}}}{{\rightarrow}}x\oplus y$, $\alpha\in\mathcal{E}$.

Proof.

By [20, Lemma 1.1], for interval topologies $\widehat{\tau}_{i}$ on $\widehat{E}$ and ${\tau_{i}}$ on $E$, we have $\widehat{\tau_{i}}\cap E={\tau_{i}}$. Thus for $x_{\alpha},y_{\alpha},x,y\in E$ we obtain $x_{\alpha}\oplus y_{\alpha}\stackrel{{\scriptstyle\widehat{\tau}_{i}}}{{\rightarrow}}x\oplus y$, $\alpha\in\mathcal{E}$ which gives $x_{\alpha}\oplus y_{\alpha}\stackrel{{\scriptstyle{\tau}_{i}}}{{\rightarrow}}x\oplus y$, $\alpha\in\mathcal{E}$ by the fact that $\widehat{\tau_{i}}\cap E={\tau_{i}}$. ∎

Definition 5.5**.**

Let $E$ be a lattice.Then

1. (i)

An element $u$ of $E$ is called strongly compact (briefly s-compact) iff, for any $D\subseteq E$: $u\leq c\in E$ for all $c\geq D$ implies $u\leq\bigvee F$ for some finite $F\subseteq D$. 2. (ii)

$E$* is called s-compactly generated iff every element of $E$ is a join of s-compact elements.*

Theorem 5.6**.**

Let $E$ be a block-finite Archimedean atomic lattice effect algebra. Then the following conditions are equivalent:

1. (i)

$E$* is almost orthogonal.* 2. (ii)

$\widehat{E}=MC(E)$* is almost orthogonal.* 3. (iii)

$\widehat{E}=MC(E)$* is compactly generated.* 4. (iv)

${E}$* is s-compactly generated.*

Proof.

By J. Schmidt [30] a MacNeille completion $\widehat{E}$ of $E$ is (up to isomorphism) a complete lattice such that for every element $x\in\widehat{E}$ there exist $P,Q\subseteq E$ such that $x=\bigvee_{\widehat{E}}P=\bigwedge_{\widehat{E}}Q$ (taken in ${}_{\widehat{E}}$). Here we identify $E$ with $\varphi(E)$, where $\varphi:E\to{\widehat{E}}$ is the embedding (meaning that $E$ and $\varphi(E)$ are isomorphic lattice effect algebras). It follows that $E$ and ${\widehat{E}}$ have the same set of all atoms and coatoms and hence also the same set of all finite and cofinite elements, which implies that (i) $\Longleftrightarrow$ (ii).

Moreover, for any $A\subseteq E$ and $u\in E$, we have ($d\in E$, $A\leq d$ implies $u\leq d$) iff $u\leq\bigvee_{\widehat{E}}A$. Then $u$ is s-compact in $E$ iff $u$ is compact in $\widehat{E}$, which gives (iii) $\Longleftrightarrow$ (iv).

Finally (ii) $\Longleftrightarrow$ (iii) by Theorem 5.1. ∎

Definition 5.7**.**

Let $E$ be an effect algebra. A map $\omega:E\to[0,1]$ is called a state on $E$ if $\omega(0)=0$, $\omega(1)=1$ and $\omega(x\oplus y)=\omega(x)+\omega(y)$ whenever $x\oplus y$ exists in $E$.

Theorem 5.8**.**

(State smearing theorem for almost orthogonal block-finite Archimedean atomic lattice effect algebras)* Let $(E;\oplus,0,1)$ be a block-finite Archimedean atomic lattice effect algebra. If $E$ is almost orthogonal then:*

1. (i)

$E_{1}=\{x\in E\mid x\ \mbox{or}\ x^{\prime}\ \mbox{is finite}\}$* is a sub-lattice effect algebra of $E$.* 2. (ii)

If there exists an $(o)$-continuous state $\omega$ on $E_{1}$ (or on $S(E_{1})=S(E)\cap E_{1}$, or on $S(E)$) then there exists an $(o)$-continuous state $\widetilde{\omega}$ on $E$ extending $\omega$ and an $(o)$-continuous state $\widehat{\omega}$ on $\widehat{E}=MC(E)=MC(E_{1})$ extending $\widetilde{\omega}$.

Proof.

(i) By Theorem 5.6, ${E}$ is s-compactly generated and thus by [28, Theorem 2.7] $E_{1}$ is a sub-lattice effect algebra of $E$.

(ii) Since ${E}$ is s-compactly generated, we obtain the existence of $(o)$-continuous extensions $\widetilde{\omega}$ on $E$ and $\widehat{\omega}$ on $\widehat{E}$ by [28, Theorem 4.2]. ∎