Construction of Boolean contact algebras

Ivo Duntsch & Michael Winter?

Department of Computer Science,Brock University,

St. Catharines, Ontario, Canada, L2S 3A1{duentsch,mwinter}@brocku.ca

Abstract. We consider Boolean algebras endowed with a contact re-lation which are abstractions of Boolean algebras of regular closed setstogether with Whitehead’s connection relation [12], in which two non-empty regular closed sets are connected if they have a non-empty in-tersection. These are standard examples for structures used in qualita-tive reasoning, mereotopology, and proximity theory. We exhibit variousmethods how such algebras can be constructed and give several non-standard examples, the most striking one being a countable model of theRegion Connection Calculus in which every proper region has infinitelymany holes.

1 Introduction

The origins of Boolean contact algebras go back to the works of Lesniewski [5] onmereology and Leonard and Goodman [4] on the calculus of individuals on theone hand, and, on the other hand, the efforts of e.g. de Laguna [2], Tarski [11] andWhitehead [12] to use regions instead of points as the basic entity of geometry. Acentral role played the notion of “connection” (or “contact”) of regions, which,in its simplest form, is a reflexive and symmetric relation C among non-emptyregions, satisfying an additional extensionality axiom [? ]. In order to formal-ize mereological structures (which were, basically, complete Boolean algebras Bfrom which the smallest element has been removed) together with Whitehead’sconnection relation C, Clarke [1] proposed additional axioms, among them acompatibility axiom a ≤ b⇐⇒ {u : aCu} ⊆ {u : bCu} and a summation axiomaC(u + v) ⇐⇒ aCu or aCv. A subsequent development, the Region Connec-tion Calculus (RCC) [8], also supposed that each proper non-zero region wasconnected to its complement. Boolean algebras of regular closed sets of regularT1 spaces, together with Whitehead’s connection aCb ⇐⇒ a ∩ b 6= ∅ served asstandard models for these ”connection algebras”. Recently, Stell [9] gave an ax-iomatization of the RCC as a Boolean algebra enhanced with a contact relationwhich satisfied the axioms given in [8]. In a separate development, proximities

? Both authors gratefully acknowledge support from the Natural Sciences and Engi-neering Research Council of Canada.

on power set algebras were investigated which exhibited many similarities to themereo-topological contact relations [3, 7, 10]. Two questions arose:

1. Does every RCC algebra have representation as a dense subalgebra of thealgebra of all regular closed sets of a topological space with Whitehead’sconnection relation?

2. What are some of the algebraic constructions of Boolean contact algebraswhen one disregards the topological aspect?

Problem 1. has been answered in the affirmative by ? ], with an earlier resulton a special class of RCC algebras having been obtained by ? ]. In the presentpaper, we address the remaining question. We exhibit various methods how suchalgebras can be constructed and give several non-standard examples, the moststriking one being a countable model of the Region Connection Calculus in whichevery proper region has infinitely many holes.

2 Boolean contact algebras

If T is a family of sets, then T+ denotes T \ {∅}. For M ⊆ U we set 1′M ={〈x, x〉 : x ∈ M}. If M = U we just write 1′. ω is the ordered set of naturalnumbers.

In the sequel, 〈B,+, ·, ∗, 0, 1〉 will denote a Boolean algebra (BA); here, ∗ is theBoolean complent; we will usually identify algebraic structures with their baseset. B+ is the set of all non-zero elements of B. Furthermore, if a ∈ B andT ⊆ B, we let a ≤ T if and only if a ≤ b for all b ∈ T . To avoid trivialities, weassume that all Boolean algebras under discussion have at least four elements.

If M ⊆ B+, we call M dense in B, if for all b ∈ B+ there is some m ∈ M suchthat m ≤ b.

A Boolean contact algebra (BCA) is a structure 〈B,C〉 such that B is a Booleanalgebra, and C is a binary relation on B such that

C0. (∀x)0(−C)xC1. (∀x)[x 6= 0⇒ xCx]C2. (∀x)(∀y)[xCy ⇒ yCx]C3. (∀x)(∀y)(∀z)[xCy ∧ y ≤ z ⇒ xCz.C4. (∀x)(∀y)(∀z)[xC(y + z)⇒ (xCy ∨ xCz)]C5. (∀x 6= 1)(∃y)[y 6= 0 ∧ ¬xCy].

Here and below, −C is the set complement of C in the power set of B × B.Observe that the class of BCAs is finitely axiomatizable by ∀∃ formulas. We alsoconsider the properties

Table 1. Some relations derived from C

P = −(C ◦ −C), part of(2.1)

PP = P ∩ −1′. proper part of(2.2)

O = P ˘ ◦ P overlap(2.3)

EC = C ∩ −O external contact(2.4)

ECD = −(P ◦ P ˘) ∩ −(P ˘ ◦ P ) complement(2.5)

ECN = EC ∩ −ECD non-exhaustive external contact(2.6)

TPP = PP ∩ (EC ◦ EC) tangential proper part(2.7)

NTPP = PP ∩ −TPP non–tangential proper part(2.8)

DC = −C disconnected(2.9)

C6. (∀x)(∀y)[x(−C)y =⇒ (∃z)(x(−C)z ∧ y(−C)z∗)] (The interpolation axiomof [7]). (The connection axiom).

It is well known that for BCAs, C6 is equivalent to

C6’. (∀x)(∀y)[xNTPPy ⇒ (∃z)(xNTPPzNTPPy)].

Figure 1 illustrates the interpolation axiom which has been used mainly in prox-imity structures. Table 1 shows some relations which are definable from C with

Fig. 1. The interpolation axiom

relational operators.

The following Lemma lists some easy properties of C and its derivatives, theproof of which is left to the reader:

Lemma 1. 1. a(−C)b⇒ a · b = 0.2. aCb or aCc⇒ aC(b + c).

3. a ≤ b⇒ C(a) ⊆ C(b).4. If a 6= 0, then aPb⇐⇒ a ≤ b.5. aOb⇐⇒ a · b 6= 0.6. xNTPPy ⇐⇒ x(−C)y∗.

To illustrate the properties of a contact relation, we show how the axioms trans-late into the properties of the overlap relation O in Table 2

Table 2. Contact vs Overlap

C0. aCb⇒ a, b 6= 0. a · b 6= 0⇒ a, b 6= 0.C1. a 6= 0⇒ aCa. a 6= 0⇒ a · a 6= 0.C2. C is symmetric. a · b 6= 0⇒ b · a 6= 0.C3. aCb and b ≤ c⇒ aCc. a · b 6= 0 and b ≤ c⇒ a · c 6= 0.C4. aC(b + c)⇒ aCb or aCc. a · (b + c) 6= 0⇒ a · b 6= 0 or a · c 6= 0.C5. C(a) ⊆ C(b)⇒ a ≤ b. (∀x)(a · x 6= 0⇒ b · x 6= 0)⇒ a ≤ b.C6. a(−C)b⇒ (∃c)[a(−C)c and c∗(−C)b].

a · b = 0⇒ (∃c)[a · c = 0 and c∗ · b = 0].C7. a 6= 0, 1⇒ aCa∗. But: a · a∗ = 0.

Proposition 1. 1. O is the smallest contact relation on B.2. If B is a finite–cofinite algebra, then O is the only contact relation on B.

Proof. 1. Let C be a contact relation on B, and a · b 6= 0. Then, (a · b)C(a · b)by C1, and, since a · b ≤ b, we have aCb by C3. Observe that O can be definedby the Boolean operators alone and does not depend on C.

2. Suppose that B is a finite–cofinite algebra. Let C be a contact relation andC 6= O. Then, there are a, b 6= 0 such that aCb and a·b = 0. Since C is summativeand one of a and b is a finite sum of atoms, we can assume w.l.o.g. that b is anatom, and since C is expansive by C3, we can assume that a = b∗. Now, a isan antiatom connected to its complement, and therefore it is connected to everynon-zero element, contradicting extensionality.

A BCA 〈B,C〉 which satisfies ?? is called an RCC–algebra. These are the modelsof the Region Connection Calculus (RCC) of Randell et al. [8] which has receivedsome prominence in the field of qualitative spatial reasoning.

If b ∈ B+, we let Ib = {a ∈ B : a(−C)b}. The next lemma describes someproperties of the relation −C:

Lemma 2. Ib is an ideal below b∗ and sup Ib = b∗.

Proof. 1. If b(−C)a, then a · b = 0, i.e. a ≤ b∗. If a ∈ Ib, and c ≤ a, then c ∈ Ibby C3, and, if a, c ∈ Ib, then a + c ∈ Ib by C4.

Clearly, a ≤ b∗ for all a ∈ Ib. Conversely, suppose that a ≤ c for all a ∈ Ib,and assume that b∗ · c∗ 6= 0, i.e. b + c 6= 1. Then, there is some d 6= 0 such thatd(−C)(b + c). It follows that d ∈ Ib and d · c = 0, contradicting d 6= 0 ≤ c.

3 A one dimensional RCC algebra

Contact relations were originally defined on n–dimensional spaces where n = 2, 3.However, since BCAs are first order structures, they have countable models bythe Lowenheim–Skolem Theorem. In this Section, we describe a simple count-able model which may be regarded as one-dimensional. Other constructions ofcountable BCAs can be found in [? ].

Let L be a dense linear order with smallest element m. Suppose that ∞ is asymbol not in L, and set L+ = L ∪ {∞} with x �∞ for all x ∈ L. An intervalof L is a set of the form [s, t) = {u ∈ L : s ≤ u � t}, where s, t ∈ L. IntAlg(L)is the collection of all finite unions of intervals

[x00, x

10) ∪ [x0

1, x11) ∪ . . . ∪ [x0

t(x), x1t(x)),(3.1)

together with the empty set. It is well known that IntAlg(L) is a Boolean algebra[see ? , p.10], called the interval algebra of L. Each nonzero x ∈ IntAlg(L) canbe written in the form (3.1) in such a way that xi

j ∈ L+, x0j < x1

j < x0j+1, and

the intervals [x0j , x

1j ) are pairwise disjoint. The representation of x in this form is

unique, and we call it the standard representation. In the sequel, we shall assumethat all elements of IntAlg(L) are in standard representation.

For each x ∈ IntAlg(L)+, we let

rel(x) = {x0j : j ≤ t(x)} ∪ {x1

j : j ≤ t(x)}rel+(x) = rel(x) \ {m,∞},

be the set of relevant points of x, with and without extreme elements, and

Int(x) = {[x0j , x

1j ) : j ≤ t(x)}

be the set of relevant intervals of x.

If s ∈ rel(x) we call s left–relevant (right–relevant), if [s, t) ∈ Int(x) ([t, s) ∈Int(x)) for some t ∈ rel(x), and the set of all left–relevant (right–relevant) ele-ments of x is denoted by rell(x) (relr(x)). Observe that s ∈ rell(x) iff s = min(x)or sup{t ∈ x : t � s} 6= s. Similarly, s ∈ relr(x) iff s = sup(x) or inf{t ∈ x : s �t} 6= s.

Lemma 3. Suppose that x, y ∈ IntAlg(L), x, y 6= 0, 1.

1. If {m,∞} ∩ rel(x) ∩ rel(y) 6= ∅, then x · y 6= 0.2. rel(x+y) = (rell(x)∩rell(y))∪(relr(x)∩relr(y))∪{min{rel(x), rel(y)},max{rel(x), rel(y)}}∪(rel(x)⊕ (y ∪ rel(y)))∪ (rel(y)⊕ (x∪ rel(x))). Here u⊕ v is the symmetricdifference of u and v.

3. If x 6= y, then,

(x · y = 0 and rel(x) ∩ rel(y) = ∅)⇒(rel(x) ∪ rel(y) ⊆ rel(x + y)).

4. If x ≤ y ≤ z and rel(x) ∩ rel(z) 6= ∅, then rel(x) ∩ rel(y) 6= ∅.5. rel+(x) = rel+(x∗).

Proof. 1 Let m ∈ relx∩ rel y. Then, there are s, t such that [m, s) ∈ Int(x) and[m, t) ∈ Int(y), and it follows that x ∩ y 6= ∅. The other case is similar.

2 “⊆”: Suppose that s 6= min(x+y), sup(x+y), s 6∈ (rell(x)∩rell(y))∪(relr(x)∩relr(y)) ∪ (rel(x)⊕ (y ∪ rel(y))) and s ∈ rel(x + y). Since x + y = x ∪ y, we haves ∈ rel(x) ∪ rel(y). Assume that s ∈ rel(y) ∩ (x ∪ rel(x)). First, suppose that sis left–relevant for y, w.l.o.g [s, t) ∈ Int(y). Then, s is not left–relevant for x byour hypothesis. If s ∈ x, then the relevant interval of y containing s will be asubset of a relevant interval of x + y, and since s is not left–relevant for y, itwill be an inner point of some relevant interval of y. Hence, s 6∈ rel(x + y), acontradiction. If s ∈ rel(x), then s is right–relevant for x, and thus, there is somer such that [r, s) ∈ Int(x). Hence, the combined interval [r, s)∪ [s, t) = [r, t) willbe contained in a relevant interval of x + y, and s 6∈ rel(x + y). The case that sis right–relevant for y is analogous.

“⊇”: Clearly, min{rel(x), rel(y)} ∈ rel(x+y) and max{rel(x), rel(y)} ∈ rel(x+y).Suppose that s ∈ rell(x) ∩ rell(y), and that s 6= min(x + y). Then, sup{t � s :t ∈ x ∪ y} � s, so that s ∈ rel(x + y). Similarly, relr(x) ∩ relr(y) ⊆ rel(x + y). Ifs ∈ rel(x) ⊕ (y ∪ rel(y)) and s ∈ rell(x), then sup{t ∈ y : t � s} 6= s, and thus,s ∈ rel(x + y). The other cases are analogous.

3. Let x ∩ y = rel(x) ∩ rel(y) = ∅. If s ∈ rel(x), then, by our hypothesis,s 6∈ (y ∪ rel(y), and thus, s ∈ rel(x + y) by 2. If s ∈ rel(y), then s 6∈ (x ∪ rel(x).

4. Let a ∈ rel(x) ∩ rel(z), and assume a 6∈ rel(y). Then, there is some [y0j , y1j )

such that y0j � a � y1j . Since y ≤ z, [y0j , y1j ) is totally contained in an interval of

z, contradicting a ∈ rel(z).

5. We only show ⊆. If x ∈ {0, 1}, then rel(x) = rel(x∗) = {m,∞} and there isnothing more to show. If [s, t) ∈ Int(x) and s 6= m, then there is some [v, w) ∈Int(x∗) such that w = s; if t 6=∞, then there is some [v, w) ∈ Int(x∗) such thatt = v. It follows that {s, t} \ {0,∞} ⊆ rel(x∗).

Next, we define a binary relation C on IntAlg(L) by

xCy ⇐⇒ x · y > 0 or rel(x) ∩ rel(y) 6= ∅.(3.2)

Proposition 2. 〈B,C〉 satisfies C0 – ??.

Proof. Clearly, C satisfies C0 – C3. For C4, Let x, y, z ∈ B0 and xC(y + z). Ifx ·(y+z) 6= 0, we are done. Otherwise, rel(x)∩rel(y+z) 6= ∅. By Lemma 3(1), wehave rel(y+z) ⊆ rel(y)∪rel(z), and thus, rel(x)∩rel(y) 6= ∅ or rel(x)∩rel(z) 6= ∅.

Suppose that x 6= 0, and let [s, t) ∈ Int(x∗). If s � u � v � t and y = [u, v),then y 6= 0 and x(−C)y, which shows C5.

Since C is a contact relation, we show C6’. Thus, let x 6= 0 and xNTPPy. By(3.4) below (whose proof only requires that C is a contact relation), we havex ≤ y and rel+(x) ∩ rel+(y) = ∅. If m 6∈ rel(x), set Im = ∅. Otherwise, x ≤ yimplies that there are s, t such that s ≤ t and [m, s) ∈ Int(x), [m, t) ∈ Int(y).Now, rel+(x) ∩ rel+(y) = ∅ implies that s � t, and by the density of L we canchoose some s � r � t; now, set Im = [m, r). If [s, t) is the i–th interval of x withs 6= m, t 6=∞, there is some [u, v) ∈ Int(y) such that u � s � t � v. Again usingthe density of L we find p, q such that u � p � s � t � q � v; set Ii = [p, q).Suppose that in this way we define I1, I2, . . . , Ik. If ∞ 6∈ rel(x), set I∞ = ∅.Otherwise, there are s, t such that s � t and [s,∞) ∈ Int(y), [t,∞) ∈ Int(x).Choose some s � r � t and set I∞ = [r,∞). Now, let z = Im∪ I1∪ . . .∪ Ik ∪ I∞.By the construction, x ≤ z ≤ y and rel+(x)∩ rel+(z) = rel+(y)∩ rel+(z) = ∅, sothat xNTPPzNTPPy by (3.4).

Finally, if x ∈ B, x 6= 0, 1, then rel(x) ∩ rel(x∗) 6= ∅ by Lemma 3(5), and hence,xCx∗.

Lemma 4. Let x, y ∈ B+. Then,

xECy ⇐⇒ x · y = 0 and rel+(x) ∩ rel+(y) 6= ∅.(3.3)

xNTPPy ⇐⇒ x ≤ y and rel+(x) ∩ rel+(y) = ∅.(3.4)

Proof. (3.3) follows immediately from the definition and the fact that x · y = 0and rel(x) ∩ rel(y) 6= ∅ imply rel+(x) ∩ rel+(y) 6= ∅.

For (3.4)., suppose that xNTPPy. Then, x(−C)y∗ by Lemma 1(6), and thedefinition of C implies x · y∗ = 0, i.e. x ≤ y, and rel(x) ∩ rel(y∗) = ∅. Then, inparticular, rel+(x)∩rel+(y∗) = ∅, and by Lemma 3.5 we have rel+(x)∩rel+(y) =∅.

Conversely, suppose that x ≤ y, rel+(x) ∩ rel+(y) = ∅ and assume that xCy∗.Since x ·y∗ = 0, we have rel(x)∩rel(y∗) 6= ∅, and thus, rel(x)∩rel(y∗) ⊆ {m,∞}.In both cases, x · y∗ 6= 0, a contradiction.

4 Extensions and restrictions

In this Section, we suppose that C is a contact relation on the Boolean algebraB, and that A is a subalgebra of B. Our first two results deal with the casethat A is dense in B, each having a contact relation, which can be extended,respectively, restricted to the other.

Proposition 3. Let A be a dense subalgebra of B, and D be the restriction ofC to A×A. Then, D is a contact relation. If C satisfies ??, then so does D.

Proof. Since C0 – C4 and ?? are universal, we only need to show C5: Let a ∈A, a 6= 1; then, there is some b ∈ B such that b 6= 0 and a(−C)b. Since A isdense in B, there is some p ∈ A, p 6= 0, such that p ≤ b; it follows that a(−C)p,and hence, a(−D)p, since a, p ∈ A. The final part follows from D ⊆ C.

Problem 1. Is C6 inherited as well?

The next result shows that the other direction works as well:

Proposition 4. Let 〈A,D〉 be a BCA, and A be a dense subalgebra of B. ThenC defined by

aCb⇐⇒ (∀s, t ∈ A)[a ≤ s and b ≤ t⇒ sDt]

is a contact relation on B such that (A×A) ∩C = D, and if D satisfies ??, sodoes C. Furthermore, if E is a contact relation on B such that E∩ (A×A) = D,then E ⊆ C.

Proof. C0 – C3 and C∩(A×A) = D are obvious. For C4 suppose that aC(b+c),and assume that a(−C)b and a(−C)c. Then, there are s0, s1, t0, t1 ∈ A such that

a ≤ s0, b ≤ t0, s0(−C)t0, a ≤ s1, c ≤ t1, s1(−D)t1.

It follows that a ≤ s0 · s1, b+ c ≤ t0 + t1, and thus, by our hypothesis, we have(s0 · s1)D(t0 + t1). Since D satisfies C3, we have w.l.o.g. (s0 · s1)Dt0. But then,s0Dt0 by C2 and C3 for D, a contradiction.

For C5, suppose that C(a) ⊆ C(b); we show that b∗ ≤ a∗. Since A is densein B, it suffices to show that u ≤ b∗ implies u ≤ a∗ for all u ∈ A. Thus, letu ∈ A, u ≤ b∗. By Lemma 2, we know that u = sup{t ∈ A : t(−D)u∗}; thereforeit is enough to show that t(−D)u∗ implies t ≤ a∗. Thus, let t ∈ A, t(−D)u∗.Since b ≤ u∗, we have b(−C)t, and hence, C(a) ⊆ C(b) implies a(−C)t. It followsthat a · t = 0, i.e. t ≤ a∗.

Suppose that D satisfies ??, i.e. aDa∗ for all a ∈ A, a 6= 0, 1. If b ∈ B, b 6= 0, 1,then, by definition of C, there are s, t ∈ A such that b ≤ s, b∗ ≤ t, and s(−D)t.Since s+t = 1, it follows that s∗ ≤ t, and C3 implies that s(−D)s∗, contradictingthat D satisfies ??.

Next, let E be a contact relation on B whose restriction to A is D, and supposethat aEb. If a ≤ s ∈ A, b ≤ t ∈ A, then sEt by (C3), and E ∩ (A × A) = Dimplies that sDt.

Problem 2. Does D satisfy C6?

There may be different contact relations on B whose restriction to A gives D:

Example 1. Let B be the interval algebra of the non-negative reals R+, and Athe interval algebra of the non-negative rationals Q+. Then, A is dense in B. LetM be the set of all intervals of R+ whose set of critical points includes

√2, i.e.

all intervals of the form [x,√

2) or [√

2, y) where 0 ≤ x �√

2 � y. Let O be theoverlap relation on B, and set C = O ∪ (M ×M). Our aim is to show that C isa contact relation on B. Since

√2 6∈ Q, we see that C ∩ (A×A) = O ∩ (A×A).

Clearly, C satisfies C0 – C2. To show C3, let aCb and b ≤ c. If a ∩ b 6= ∅, thereis nothing to show. If a ∩ b = ∅, then let w.l.o.g. a = [x,

√2), b = [

√2, y). Since

b ≤ c, we have either√

2 ∈ c, in which case a ∩ c 6= ∅, or√

2 is the left endpointof c, and c ∈ M . Next, let aC(b ∪ c), and a ∩ (b ∪ c) = ∅. Then, a, b ∪ c ∈ M ,which implies

√2 ∈ rel(a) and

√2 ∈ rel(b) or

√2 ∈ rel(c). It follows that aCb or

aCc. Finally, C5 follows from the fact that for each b ∈ B there is some c ∈ Bdisjoint from b, whose set of critical points does not include

√2. �

The next result shows an extension of a contact relation in case C6 is not fulfilled.

Proposition 5. Let B be atomless, and set C+ = C∪{〈a, b〉 : (∀c)[aCc or c∗Cb]}.Then, C+ is a contact relation on B.

Proof. Clearly, C+ satisfies C0 – C2.

C3: Suppose that aC+b and b ≤ d. If aCb, then bCd, since C is a contactrelation. Otherwise, aCc or c∗Cb for all c ∈ B. Since b ≤ d, c∗Cb implies c∗Cd,and thus, aCd.

C4: Let aC+(b+c), and assume that a(−C+)b and a(−C+)c. Then, in particular,a(−C)b and a(−C)c, and there are d, e such that

a(−C)d, d∗(−C)b, a(−C)e, e∗(−C)c.

It follows from C4 that a(−C)(d + e) and from C3 that (d∗ · e∗)(−C)b, (d∗ ·e∗)(−C)c. Together with C4, the latter imply (d∗·e∗)(−C)(b+c). Now, aC+(b+c)implies that aCm or m∗C(b + c) for all m ∈ B. Setting m = d + e leads to acontradiction.

C5: Suppose that C+(a) ⊆ C+(b), and assume that a 6≤ b, i.e. a · b∗ 6= 0. SinceC + (a · b∗) ⊆ C+(a), we can suppose w.l.o.g that a · b = 0. Assume that b(−C)cfor some 0 6= c ≤ a. Since aCx for every 0 6= x ≤ a and C(a) ⊆ C+(b), we have

b(−C)d⇒ xCd∗

for all 0 6= x ≤ c ≤ a and all d ∈ B. In particular, xCc∗ for all 0 6= x ≤ c,and thus, dCc∗ for all d 6= 0, contradicting c 6= 0. It follows that bCc for all0 6= c ≤ a. This is not possible.

Note that in case C satisfies C6, we have C = C+. We do not know, whetherC+ satisfies C6, regardless of C.

Now we show that, given a contact relation C on an atomless Boolean algebra,it is possible to extend C in such a way, that for two given x, y ∈ B+, there is acontact relation R containing C in which xRy. If x · y 6= 0, then xCy already, sothe interesting case is when x and y are disjoint. In this case, there are ultrafiltersF and G with x ∈ F and y ∈ G.

Proposition 6. Let B be atomless, F,G be distinct ultrafilters of B, and R =C ∪ (F ×G) ∪ (G× F ). Then, R is a contact relation on B. Furthermore, if Csatisfies ??, so does R.

Proof. Clearly, R satisfies C0 – C2. Suppose aRb and b ≤ c, and a(−C)b. Ifw.l.o.g. 〈a, b〉 ∈ F ×G, then c ∈ G, and it follows that 〈a, c〉 ∈ F ×G ⊆ R. Next,suppose that aR(b+c), and w.l.o.g. 〈a, b+c〉 ∈ F×G. Since G is an ultrafilter, wehave b ∈ G or c ∈ G, and therefore, aRb or aRc. Finally, let a 6= 0, 1. By C5 thereis some b 6= 0 such that a(−C)b. If a 6∈ F ∪G, then 〈a, b〉 6∈ (F ×G) ∪ (G× F ),and it follows that a(−R)b. Otherwise, let w.l.o.g. a ∈ F . Since B is atomlessand b 6= 0, there are p, q 6= 0 such that p + q = b, p · q = 0. Since a(−C)b, wehave a(−C)p and a(−C)q. Now, p · q = 0 and the fact that G is a proper filterimply that we cannot have both p, q ∈ G, so let w.l.o.g. p 6∈ G. Observing thata · b = 0 and p ≤ b, we have p 6∈ F , and thus, 〈a, p〉 6∈ C ∪ (F ×G)∪ (G×F ) = R.The final claim follows from the fact that C ⊆ R.

Let A be dense in B. Since every ultrafilter on A can be extended to B invarious ways, this is another example that distinct contact relations on B canhave identical restrictions to A. We can also show

Proposition 7. The class of Boolean contact algebras is not closed under unionsof chains.

Proof. Suppose that B is countable, and enumerate B \ {0, 1} by {xi : i ∈ ω}.For each i ∈ ω choose ultrafilters Fi, Gi such that xi ∈ Fi, x

∗i ∈ Gi. Set C0 = C,

and Ci+1 = Ci ∪ (Fi × Gi) ∪ (Gi × Fi). By Proposition 6, each Ci is a contactrelation on B, but

⋃i∈ω Ci = B+ ×B+, violating C5.

Example 2. The following is an example where 〈B,C〉 satisfies C6, but an ultra-filter extension as in Proposition 6 does not: Suppose that L = [0, 1) is the realleft-closed right open unit interval, B = IntAlg(L) and C the contact relationdefined by (3.2). Recall that for each a ∈ L, a � 1, Fa = {x ∈ B : a ∈ x} isan ultrafilter of B. Choose some a ∈ B, a 6= 0, and set F = F0, G = Fa. ByProposition 6, the relation D = C ∪ (F × G) ∪ (G × F ) is a contact relationon B. To show that 〈B,D〉 does not satisfy C6, choose some 0 � b � c � a,and let x = [0, b), y = [c, a). Then, x(−D)y, since x(−C)y and y 6∈ F ∪ G.Since a ∈ rel(y), and a ∈ t for all t ∈ G, we have yCt for all t ∈ G. Assumethat C6 holds. Then x(−D)y implies that there is some z 6= ∅ with x(−D)z andy(−D)z∗. Since x ∈ F , we cannot have z ∈ G, and therefore, z∗ ∈ G. But thiscontradicts that yCt for all t ∈ G. �

The direct product of two BCAs with the product relation is not necessarilya BCA, since e.g. C4 may be violated. Nevertheless, using a somewhat moreinvolved construction, we show that a direct product of Boolean algebras whichare BCAs can be made into a BCA. Suppose that 〈A,CA〉 and 〈B,CB〉 are BCAs,and let D = A×B be the direct product of A and B. Let F be an ultrafilter onA, G be an ultrafilter on B, and define a relation CD on D by 〈a0, b0〉CD〈a1, b1〉if one of the following conditions is true:

a0 = a1 = 1 and (b0CBb1 or b0 = b1 = 0),(4.1)

(a0 = a1 = 0 or a0CAa1) and b0 = b1 = 1,(4.2)

{a0, a1} 6= {1} and {b0, b1} 6= {1}(4.3)

and (a0CAa1 or b0CBb1),

a0 ∈ F and b0 = 0 and a1 = 0 and b1 ∈ G(4.4)

a0 = 0 and b0 ∈ G and a1 ∈ F and b1 = 0).(4.5)

Proposition 8. 〈D,CD〉 is a BCA, and both 〈A,CA〉 and 〈B,CB〉 are isomor-phic to substructures of 〈D,CD〉. If one of CA and CB satisfy ??, so does CD.

Proof. Let eA : A→ D be defined by

eA(x) =

{〈x, 1〉, if x ∈ F,

〈x, 0〉, otherwise.(4.6)

It is easy to see (and well known) that eA is an embedding of Boolean algebras,and eA preserves CA by (4.3). Conversely, suppose that eA(x)CDeA(y). If x, y ∈F , then eA(x) = 〈x, 1〉, eA(y) = 〈y, 1〉, and hence, xCAy by (4.2). If x 6∈ F andy ∈ F , then x 6= 1, y 6= 0, and eA(x) = 〈x, 0〉, eA(y) = 〈y, 1〉. The only possiblecase for eA(x)CDeA(y) is (4.3), and thus, xCAy, since 0(−CB)1. The case x 6∈F, y ∈ F is shown analogously. If x, y 6∈ F , then eA(x) = 〈x, 0〉, eA(y) = 〈y, 0〉,and xCAy follows from (4.3) and 0(−CB)0. Similarly, 〈B,CB〉 is isomorphic toa substructure of 〈D,CD〉.

Next, we show that 〈D,CD〉 is a BCA:

(C0): This follows immediately from the definition of CD.

(C1): Suppose that 〈x, y〉 6= 〈0, 0〉, and let w.l.o.g. x 6= 0. If x 6= 1, then〈x, y〉CD〈x, y〉 by (4.3). If x = 1 and y = 0, then 〈x, y〉CD〈x, y〉 by (4.1), and ify 6= 0, then 〈x, y〉CD〈x, y〉 by yCBy and (4.1).

(C2): This follows from the fact that (4.1) – (4.5) are invariant against substi-tuting a0 for a1 and b0 for b1.

(C3): Let 〈x, y〉 6∈ {〈0, 0〉, 〈1, 1〉}, and suppose w.l.o.g. that x 6= 0. If x 6= 1,then xCAx

∗, and thus, 〈x, y〉CD〈x∗, y∗〉 by (4.3). If x = 1, then y 6= 1. If y = 0,then 〈x, y〉CD〈x∗, y∗〉 by (4.4), and if y 6= 0, then yCBy

∗, and 〈x, y〉CD〈x∗, y∗〉by (4.3).

(C4): Suppose that 〈x, y〉CD(〈u, v〉+ 〈s, t〉), and that w.l.o.g. x 6= 0.

x 6= 1 and y 6= 1: In this case, (4.1) and (4.2) are not relevant. If y 6= 0, then wehave (4.3), and w.l.o.g. let xCAu+s; by (C4) for CA we can suppose w.l.o.g thatxCAu, and thus, 〈x, y〉CD〈u, v〉 by (4.3). If y = 0, then 〈x, 0〉CD〈u + s, v + t〉.If u + s = 0, then u = s = 0 and v + t ∈ G, and we can suppose w.l.o.g. thatv ∈ G; then, 〈x, 0〉CD〈0, v〉 = 〈u, v〉. If u+ s 6= 0, then we must have xCA(u+ s)by (4.3). It follows that xCAu or xCAs, and we can continue as above.

x = 1, 0 � y � 1: Let u + s 6= 0. If u + s = 1, then w.l.o.g we have yCAv by(4.1), and 〈x, y〉CD〈u, v〉 by (4.3). If 0 � u + s � 1, then w.l.o.g. we have xCAu,and 〈x, y〉CD〈u, v〉 by (4.3). If u + s = 0, then u = s = 0, and w.l.o.g. we haveyCBv by (4.3). Again by (4.3), this implies 〈x, y〉CD〈u, v〉.

x = 1, y = 0: In this case, v + t ∈ G. If w.l.o.g. v ∈ G, then 〈1, 0〉CD〈0, v〉 by(4.4).

x = 1, y = 1: If u + s = 1, then v + t = 6= 0 by (4.1), and thus w.l.o.g. wehave yCBv. It follows from (4.1) that 〈1, 1〉CD1, v. Similarly, we treat the casev + t = 1. If both u + s, v + t 6= 0, then, w.l.o.g. u + s 6= 0, and xCAu. Hence,〈1, 1〉CDu, v by (4.3).

(C5): Let 〈x, y〉 6∈ {〈0, 0〉, 〈1, 1〉}, and suppose w.l.o.g. that x 6= 0. If x 6= 1, with(C5) choose some z ∈ A such that x(−CA)z; then, 〈x, y〉(−CD)〈z, 0〉. If x = 1,then y 6= 1. If y = 0, choose some z ∈ B \G; then 〈x, y〉 6= 〈0, z〉. If y 6= 0, choosesome z ∈ B with y(−CB)z; then, 〈x, y〉(−CD)〈0, z〉.

(??): Let 〈x, y〉 6∈ {〈0, 0〉, 〈1, 1〉}. Then, {x, y} 6= {1} and {x∗, y∗} 6= {1}, and,since xCAx

∗ or yCBy∗, we have 〈x, y〉CD〈x∗, y∗〉 by (4.3).

5 An RCC algebra full of holes

Mormann [6] has introduced the “hole relation” H which is defined as

xHy ⇐⇒ xECy, x + y 6= 1, ECN(x) ⊆ O(y).

xHy is read as “x is a hole of y”.

In Figure 2, x is a hole of y; Figure 3 shows a hole in space. A region y is calledsolid, if it does not have any holes, i.e. if we cannot have xHy for any x ∈ B+.In other words, y is solid, if and only if for all x ∈ B+ with xECNy there issome z ∈ B+ such that xECNz and z · y = 0.

Mormann asked, whether every RCC algebra contained solid proper regions. Theaim of this Section is to construct an RCC algebra in which every proper regionhas a hole (and thus, infinitely many).

For later use we show

Lemma 5. Suppose that C satisfies ??, and let x, y 6= 0. Then,

xHy ⇐⇒ x · y = 0, x + y 6= 1, and xNTPP (x + y).

Fig. 2. A hole in the plane

.

Fig. 3. A hole in space

.

Proof. “⇒”: The first two assertions follow from xECNy. Assume that x(−NTPP )(x+y), i.e. xC(x+ y)∗. Then, xC(x∗ · y∗) which shows that xEC(x+ y)∗. If x+ (x+y)∗ = 1, then y ≤ x, contradicting y 6= 0, x · y = 0. Hence, xECN(x + y)∗, andthe definition of H implies that yO(x + y)∗, a contradiction.

“⇐”: Assume that x(−C)y, i.e. xNTPPy∗. By our hypothesis, we also havexNTPP (x + y). It follows that xNTPP (y∗ · (x + y)), i.e. xNTPPy∗ · x. Sincex · y = 0, we obtain xNTPPx, i.e. x(−C)x∗; this contradicts ??. Next, letxECNz; then, xCz and z ≤ x∗. Assume that z · y = 0, i.e. z ≤ y∗; then,z ≤ x∗ ·y∗ = (x+y)∗, and we obtain xC(x+y)∗. This contradicts xNTPP (x+y).

Our example of an RCC model which does not have any proper solid regions willinvolve the construction of two Boolean algebras, of which the first algebra servesas an index for a family of interval algebras, from which we will construct ourdesired model. As a preparation, we characterize the solid regions in an intervalalgebra. The result shows that not all intervals are solid, which is somewhat

surprising, and one wonders whether the terminology of “hole” and “solid” isadequate in this situation; an illustration is given in Figure 4.

Fig. 4. Some holes and solids in IntAlg([0, 1))

.

Lemma 6. Suppose that B = IntAlg(L), and m = minL.

1. x ∈ B+ is solid if and only if x has the form [m, s) or [t,∞) or [m, s)∪[t,∞).2. If yHx, then there is some z ∈ B+ such that zHx and z is an interval.

Proof. 1. “⇒”: Suppose that x ∈ B+ is solid, and assume that x is not aninterval, i.e. x = [x0

0, x10) ∪ . . . ∪ [x0

k, x1k) for some k 6= 0. Let y = [x1

0, x01). Then,

xECy, and yNTPP (x+y). Since x is solid, we must have y = x∗, and it followsthat x = [m,x1

0)∪[x01,∞). If x = [s, t) and m 6= s and t 6=∞, then y = [m, s) 6= 0,

yECNx and yNTPP (y + x), a contradiction.

“⇐′′: Let x = [m, s). If yECNx, then sup(y) 6=∞. Choose some sup(y) � t �∞; then,

[sup(y), t)ECNy and [sup(y), t)(−O)x

shows that y cannot be a hole of x. Similarly, one shows that each [t,∞) is solid.Finally, let x = [m, s) ∪ [t,∞). If yECNx, then sup(y) � t or min(y) s, orthere are [u, v), [p, q) ∈ Int(y) such that v � p; in any case we can use the sameargument as in the previous case.

2. Suppose that yHx, and that z = [s, t) is the leftmost interval of y. First,suppose that s = m, and assume that t 6∈ rel(x). Then, it follows from y · x = 0that there is some q such that t � q and q � min{r ∈ x : t � r}. Since L is densewe can also assume that q � min{r ∈ y : t � r}. Now set z = [t, q); then, zECyand z(−O)x, contradicting that y is a hole of x. If s 6= m, we can do a similarconstruction.

Our stragegy is to letQn be the rational interval [n, n+1) and Ln = IntAlg(Qn)).If x is a solid of L0, then

1. Replicate x in all intervals Q2n by setting

t2n(x) =

[2n, 2n + s), if x = [0, s),

[2n + s, 2n + 1), if x = [s, 1),

[2n, 2n + s) ∪ [2n + t, 2n + 1),

if x = [0, s) ∪ [t, 1).

to obtain x.2. Fill the space between t4n(x) and t4n+2(x) to obtain a hole of x.

see Figure 5.

Fig. 5. Destroying a solid

For 2 ≤ n, 0 ≤ k � n, let Mn,k = {q ∈ ω : q ≡ k mod n} be the kth–residueclass of n. B+ is the collection of all sets of the form Mn,k0 ∪ . . . ∪Mn,kt . Here,as below, we suppose that 0 ≤ k0 � k1 � . . . � kt � n. Furthermore, setB = B+ ∪ ∅.

Proposition 9. B is an atomless Boolean subalgebra of 2ω.

Proof. Consider a = Mn,k, and suppose that m is a multiple of n, say, m = r ·n.Then, a = Mm,k∪Mm,n+k∪Mm,2n+k∪. . .∪Mm,(r−1)·n+k. It follows that there issome 0 � b � a. Furthermore, if a = Mn,k1∪. . .Mn,kt and b = Mm,j1∪. . .Mm,js ,then we can write both a and b in terms of lcm(a, b); this shows that B is closedunder union. Since the sets Mn,k, k � n partition ω, B is also closed undercomplementation, and thus, it is a Boolean algebra.

For each n ∈ ω let tn : Q→ Q be the translation q 7→ q+n. Furthermore, we letLn be the interval algebra of the rational interval [n, n + 1) (with n + 1 takingthe place of ∞) endowed with the contact relation defined in (3.2), and definefn : L0 → Ln by fn(x) = {tn(q) : q ∈ x}. Clearly, fn is a BCA isomorphism,and n 6= m implies fn(x)∩fm(y) = ∅. Next, define g : B×L0 → 2Q by g(T, x) =⋃{fn(x) : n ∈ T}. Note that g(T, x) = ∅ if T = ∅ or x = ∅. Furthermore,

Lemma 7. 1. g(T, x) ∪ g(T, y) = g(T, x + y).2. g(T, x) ∪ g(S, x) = g(S ∪ T, x).3. −g(T, x) = g(−T, [0, 1)) ∪ g(T, x∗).4. g(T, x) ∩ g(S, y) = g(T ∩ S, x · y).

Proof. 1. – 3. are immediate, and we just show 4.:

“⊆”: Let a ∈ g(T, x) ∩ g(S, y).Then, there are n ∈ T, m ∈ S such that a ∈fn(x) ∩ fm(y). Since Ln and Lm are disjoint if n 6= m, we have n = m. Thus,a ∈ fn(x) ∩ fn(y) = fn(x · y), since fn is a Boolean homomorphism. Hence,a ∈ g(T ∩ S, x · y).

“⊇”: Let a ∈ g(T ∩ S, x · y). Then, there is some n ∈ T ∩ S such that a ∈fn(x · y) = fn(x) ∩ fn(y). Now, fn(x) ⊆ g(T, x) and fn(y) ⊆ g(S, y) give thedesired result.

Let A be the collection of all finite unions of sets of the form g(T, x), whereT ⊆ ω and x ∈ L0. To avoid confusion with the rational numbers 0 and 1, wewill denote the smallest element of A by 0, and its largest element by 1.

Proposition 10. A is a Boolean subalgebra of 2Q.

Proof. We need to show that each finite union of elementary products of the formg(T0, x0)∩ · · · ∩ g(Tn, xn)∩−g(S0, y0)∩ · · · ∩ −g(Sm, ym) is in A. By Lemma 7,

g(T0, x0) ∩ · · · ∩ g(Tn, xn) ∩ −g(S0, y0)

∩ · · · ∩ −g(Sm, ym) = g(T0 ∩ · · · ∩ Tk, x0 · . . . · xn)

∩ [g(−S0, 1) ∪ g(S0, y∗0) ∪ · · · ∪ (g(−Sm, 1) ∪ g(Sm, y∗m)],

whence the claim follows by distributivity .

If a ∈ A+, a = g(T0, x0)∪ . . .∪g(Tk, xk), and each Ti is a union of residue classesof ni, by taking lcm(n0, . . . , nk) we may assume that all the ni are equal, and,using Lemma 7, that the Ti are all different. Using Lemma 7 again shows thatwe can assume that the Ti are all pairwise disjoint.

Observe that each nonempty a ∈ A is a union of intervals of Q. Thus, the notionsrel(a) and Int(a) still make sense. As with IntAlg(L), define C on A+ by

aCb⇐⇒ (a ∩ b) ∪ (rel(a) ∩ rel(b)) 6= ∅.

We also will use the relations of Table 1. The proof of the following result issimilar to that of Lemma 4 and is left to the reader.

Lemma 8. Let x, y ∈ A+. Then,

xECy ⇐⇒ x · y = 0 and rel(x) ∩ rel(y) 6= ∅.(5.1)

xNTPPy ⇐⇒ x ≤ y and rel+(x) ∩ rel+(y) = ∅.(5.2)

Proposition 11. 〈A,C〉 is an RCC algebra which has no proper solid regions.

Proof. To show that 〈A,C〉 is an RCC model is analogous to Proposition 2 andis left to the reader. Suppose that 0 6= a 6= 1, a = g(T0, x0) ∪ g(T1, x1) ∪ . . . ∪g(Tr, xr). As in the proof of Proposition 9, we may assume that all Ti are unionsof residue classes to the same base n, and using Lemma 7, we can suppose thata has the form

a = g(Mn,k0 , x0) ∪ g(Mn,k1 , x1) ∪ . . . ∪ g(Mn,kt , xt),(5.3)

where 0 ≤ k0 � k1 � . . . � kt � n. For example, if, for simplicity, xi = [pi, qi),then

a = [k0 + p0, k0 + q0) ∪ . . . ∪ [kt + pt, kt + qt)

∪[n + k0 + p0, n + k0 + qi) ∪ . . .

∪[n + kt + pt, n + kt + qi)

∪[2n + k0 + p0, 2n + k0 + qi) ∪ . . .

∪[2n + kt + pt, 2n + kt + qi) ∪ . . .

In each of the following cases we define an element b which is easily seen to be ahole of a, the verification of which is left to the reader. The key observation hereis that we “fill” gaps in the even multiples of n and leave them untouched in theodd multiples. This assures that a·b = 0 and a+b 6= 1, while rel+(a)∩rel+(b) = ∅.

Case 1. xi = [0, 1) for all i ≤ t: Since a 6= 1, we have {0, . . . , n − 1} 6= {k0, . . . , kt},and

a = [k0, k0 + 1) ∪ . . . ∪ [kt, kt + 1)∪[n + k0, n + k0 + 1) ∪ . . . ∪ [n + kt, n + kt + 1) ∪ . . .

(a) Suppose that kj+1 6= kj + 1 for some j � t. In this case, we set

b =g(M2n,kj+1, [0, 1)) ∪ g(M2n,kj+2, [0, 1))

∪ . . . ∪ g(M2n,kj+1−1, [0, 1))

=[kj + 1, kj + 2) ∪ . . . ∪ [kj+1 − 1, kj+1)

∪ [2n + kj + 1, 2n + kj + 2) ∪ . . .

=[kj + 1, kj+1) ∪ [2n + kj + 1, 2n + kj+1)

∪ . . .

(b) Suppose that ki + 1 = ki+1 for all i � t; then,

a =[k0, k0 + 1) ∪ [k0 + 1, k0 + 2) ∪ . . .

∪ [kt, kt + 1) ∪ [n + k0, n + k0 + 1)

∪ [n + k0 + 1, k0 + 2) ∪ . . .

∪ [n + kt, n + kt + 1)

∪ [2n + k0, 2n + k0 + 1)

∪ [2n + k0 + 1, 2k0 + 2)

∪ . . . ∪ [2n + kt, 2n + kt + 1) ∪ . . .

and thus,

a =[k0, kt + 1) ∪ [n + k0, n + kt + 1)

∪ [2n + k0, 2n + kt + 1) ∪ . . .

Since a 6= 1 we have k0 6= 0 or kt 6= n− 1.

i. kt = n− 1: Then,

a =[k0, n) ∪ [n + k0, 2n)

∪ [2n + k0, 3n) ∪ . . .

and k0 6= 0. Now, we set

b =g(M2n,0, [0, 1)) ∪ g(M2n,1, [0, 1)) ∪ . . .

∪ g(M2n,k0−1, [0, 1))

=[0, k0 − 1) ∪ [2n, 2n + k0 − 1)

∪ [4n, 4n + k0 − 1) ∪ . . .

ii. kt 6= n− 1: In this case, kt + 1 6= n, and we set

b =g(M2n,kt+1, [0, 1)) ∪ . . .

∪ g(M2n,n, [0, 1)) ∪ . . .

∪ g(M2n,n+k0, [0, 1))

=[kt + 1, kt + 2) ∪ . . .

∪ [n− 1, n) ∪ [n− 1, n)

∪ [n, n + 1) ∪ . . .

∪ [n, n + k0) ∪ . . .

=[kt + 1, n + k0)∪[2n + kt + 1, 2n + n + K0) ∪ . . .

Case 2. There is some j ≤ t such that xt 6= [0, 1):

(a) xj is not an interval: Then, there are s ∈ relr(xj), t ∈ rell(xj) with s � tand [s, t) ∩ xj = ∅. In this case,

b = g(M2n,kj, [s, t)) = [kj + s, kj + t)

∪[2n + kj + s, 2n + kj + t) ∪ . . .

is a hole of a.

(b) xj = [p, q): Then, p 6= 0 or q 6= 1.

i. p 6= 0:

A. j = 0: If t = 0, then [0, p) is a hole of a. Otherwise, let u =minx1, v = supxt and set

c =

g(M2n,k0+1, [0, 1)) ∪ . . .

∪g(M2n,k1−1, [0, 1), if k0 + 1 6= k1,

∅, otherwise,

and define

b =g(M2n,0, [0, 1)) ∪ . . .

∪ g(M2n,k0−1, [0, 1))

∪ g(M2n,k0, [0, p))

∪ g(M2n,k0, [q, 1)) ∪ c

∪ g(M2n,k1, [0, u))

∪ g(M2n,kt, [v, 1)).

B. j 6= 0: Let u = supxj−1 and set

b = g(M2n,kj−1, [u, 1)) ∪ g(M2n,kj

, [0, p)).

ii. p = 0, q 6= 1: If t = 0, then [q, 1) is a hole of a. Thus, let 0 � t, andu = minx0.

A. j = t: Set

b =g(M2n,kt, [q, 1)) ∪ g(M2n,kt+1, [0, 1))

∪ . . . ∪ g(M2n,n+k0−1, [0, 1))

∪ g(M2n,n+k0, [0, u)).

B. j 6= t: Let u = minxj+1, and

c =

g(M2n,kj+1, [0, 1)) ∪ . . .

∪g(M2n,kj+1−1, [0, 1),

if kj+1 6= kj + 1,

∅, otherwise.

Now set

b = g(M2n,kj, [q, 1)) ∪ c

∪g(M2n,kj+1, [0, u)).

This completes the proof.

This construction has interesting topological consequences. We know from [?], that 〈A,C〉 is isomorphic to a dense substructure of the Boolean algebraRegCl(X) of regular closed sets of some connected T1 space X. Now supposethat 〈A,C〉 itself is such a substructure of some such space X. Since A is densein RegCl(X), its non-zero elements are a basis for the closed sets. Since foreach x ∈ A+, its complement x∗ in RegCl(X) is topologically disconnected, weobserve that X has an open basis of disconnected regular open sets.

6 Conclusion and outlook

We have exhibited various ways in which new contact structures can be obtainedfrom a given one. Since the overlap relation O is a contact relation on everyBoolean algebra, the most interesting remaining question is

– Can one define on every atomless Boolean algebra an RCC contact relation?

Another open problem is the following:

– Can the completion A of the algebra constructed in Section 5 be equippedwith an RCC contact relation D whose restriction to A is C, and 〈A,D〉contains no proper solids?

If the answer is affirmative, then there would be a connected T1 space in whichevery proper non-empty regular open set would be topologically disconnected,and thus, X would be locally disconnected at each of its points. Furthermore, wewould have a standard complete RCC model without proper connected regions.

Acknowledgment

The results reported in this paper are part of an on-going project concernedwith the construction and topological and algebraic representation of contactand proximity algebras,and, vice versa, the algebraic properties of proximityalgebras induced by topological spaces, carried out jointly by the current authorsand G. Dimov and D. Vakarelov of Sofia University, Bulgaria, and we thank ourcolleagues for many constructive discussions.

We also would like to thank the referees for their constructive and thoughtfulcomments.

Bibliography

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