Ditopological texture spaces and fuzzy topology—III: Separation axioms

Fuzzy Sets and Systems 157 (2006) 1886 – 1912 www.elsevier.com/locate/fss

Ditopological texture spaces and fuzzy topology—III: Separation axioms Lawrence M. Browna,∗ , Rıza Ertürka , Senol ¸ Dostb a Mathematics Department, Faculty of Science, Hacettepe University, Beytepe, Ankara, Turkey b Department of Secondary Science and Mathematics Education, Hacettepe University, Beytepe, Ankara, Turkey

Received 18 July 2005; received in revised form 2 February 2006; accepted 2 February 2006 Available online 28 February 2006

Abstract This is the third of three papers which develop various fundamental aspects of the theory of ditopological texture spaces and relate this work with the theory of fuzzy topological spaces. In this paper, the authors define and study basic separation axioms in general ditopological texture spaces, relating these to known separation axioms for bitopological and topological spaces, and to the point-free separation axioms of B. Hutton and I. Reilly for fuzzy topological spaces. © 2006 Elsevier B.V. All rights reserved. MSC: primary 54A05;54A40; 54D10; 54D15; secondary 06D10;03E20; 54E15; 54E55 Keywords: Texture; Ditopology; Bitopology; Separation axioms; Hutton spaces; L-topological spaces

1. Introduction This is the last of a series of three papers which develop various fundamental aspects of the theory of ditopological texture spaces in a categorical setting, and present important links with the theory of L-topological spaces. In the first paper in this series, subtitled Basic Concepts, the authors presented the notions of direlation and difunction between textures and introduced the category dfTex of textures and difunctions, the construct fTex and several related categories. In the second paper, subtitled Topological Considerations, the category dfDitop of ditopological texture spaces and bicontinuous difunctions was defined and the forgetful functor U : dfDitop → dfTex shown to be topological. Several important properties were discussed, including the existence of products and coproducts, and an equivalence with the category of classical Hutton spaces was established. Finally, a consideration of L-valued sets and topologies from the viewpoint of ditopological texture spaces was presented. Apart from a section dealing with dineighbourhood systems, originally scheduled for the second paper but omitted due to lack of space, the present paper deals solely with separation axioms for ditopological texture spaces. However, since ditopological texture spaces include topological spaces, bitopological spaces and Hutton spaces as special cases, we will motivate our definitions by comparing them with known separation properties in these spaces and to this end we recall the necessary functors between the corresponding categories. ∗ Corresponding author.

E-mail addresses: [email protected] (L.M. Brown), [email protected] (R. Ertürk), [email protected] (S. ¸ Dost). 0165-0114/$ - see front matter © 2006 Elsevier B.V. All rights reserved. doi:10.1016/j.fss.2006.02.001

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It is well known that the construct Top (concrete category over Set) of topological spaces and continuous functions may be embedded as a full subconstruct of Bitop, the construct of bitopological spaces in the sense of Kelly [18] and pairwise continuous (also called, bicontinuous) functions. We denote this embedding by B, so for a given topological space (X, T ) we have B(X, T ) = (X, T , T ), while for a morphism f in Top we have B(f ) = f . Now, regarding the functor T given in [11] as mapping from Bitop to dfDitop we have Top B / Bitop T / dfDitop. Here, for a bitopological space (X, u, v), we have T(X, u, v) = (X, P(X), u, v c ) and for f ∈ MorBitop we have T(f ) = (f, f  ). Hence, T ◦ B is essentially just the functor Tc of [11] regarded as mapping from Top to dfDitop. These functors will enable us to relate known separation properties for topological and bitopological spaces with our proposed axioms for ditopological texture spaces. Given a property Q of a bitopological space (X, u, v) we adopt Kopperman’s convention [19] of denoting by Q∗ the property obtained by interchanging u and v, and of defining pairwise Q to mean Q and Q∗. According to [19] (X, u, v) is: Weakly symmetric (ws) if x ∈ / y u ⇒ y ∈ / xv . u Pseudo Hausdorff (pH) if x ∈ / y ⇒ ∃ U ∈ u, V ∈ v with x ∈ U , y ∈ V and U ∩ V = ∅.19 v Regular if given x ∈ U ∈ u we have G ∈ u with x ∈ G ⊆ G ⊆ U . Completely regular if whenever x ∈ U ∈ u there is a pairwise continuous function f mapping (X, u, v) into the bitopological unit interval with f (x) = 1 and f (y) = 0 for all y ∈ / U. v (v) Normal if whenever U ∈ u, K is v-closed and K ⊆ U we have G ∈ u with K ⊆ G ⊆ G ⊆ U .

(i) (ii) (iii) (iv)

Thus regular, completely regular, normal and their pairwise forms are essentially as defined in [18], pairwise ws is the pairwise R0 of [20], and pairwise pH is pairwise R1 in the sense of [24] (called preseparated in [3]). On the other hand, in [19] the Tn axioms T0 , T1 = T0 + ws, T2 = T0 + pH, T3 = T0 + regular, T4 = T1 + normal are based on the T0 axiom x ∈ y u ∩ y v and y ∈ x u ∩ x v implies x = y. Hence, Kopperman’s T0 is the weak pairwise T0 axiom of [25]. For the purpose of this study, we will find it convenient to base the Tn axioms on the stronger T0 axiom x ∈ y v and y ∈ x u implies x = y, as in [15]. Where necessary we refer to Kopperman’s axioms as the weak Tn axioms, noting that pairwise weakly T1 , pairwise weakly T2 have also be called semi-pairwise T1 , semi-pairwise T2 , respectively, by the first author in [4]. Clearly, the strong and weak forms of these axioms coincide when u = v, so there is no difference between the two forms of these axioms when restricted to a topological space. In just the same way the relation between Hutton algebras and Hutton textures detailed in [10, Example 1.1(2)], and exemplified in the equivalence between the category cdfDitop and the category H of classical (complemented) Hutton spaces [2] given in [11, Corollary 5.7], will be used to show that our ditopological separation axioms correspond exactly to the point-free axioms of Hutton and Reilly [17]. Indeed, it was precisely with the aim of improving the results of the second author in this direction that the notion of ditopological texture space was developed to replace the bitopological representation of a Hutton space given in [13,14]. We now mention some examples of ditopological texture spaces which will be useful in the sequel. Example 1.1. (1) A ditopology (
, ) on (S, S) is called discrete if
= S and codiscrete if  = S, [11]. (2) A ditopology (
, ) on (S, S) is called indiscrete if
= {S, ∅} and co-indiscrete if  = {S, ∅}, [11]. (3) Let L = (0, 1], L = {(0, r] | 0
r
1}, ((0, r]) = (0, 1 − r], 0
r
1. Then (L, L, ) is the Hutton texture of the Hutton algebra (I, ) [10]. Two natural complemented ditopologies on (L, L, ) are the discrete, codiscrete ditopology (L, L) and the indiscrete, co-indiscrete ditopology ({L, ∅}, {L, ∅}). (4) The unit interval texture (I, I, ,
I , I ) is defined by I = [0, 1], I = {[0, r), [0, r] | 0
r
1}, ([0, r)) = [0, 1 − r], ([0, r]) = [0, 1 − r), 0
r
1. The natural ditopology is given by
I = {[0, r) | 0
r
1} ∪ {I} and I = {[0, r] | 0
r
1} ∪ {∅}, [11]. (5) [11, Example 2.14]: The Hutton texture (MI , MI , I ) of the Hutton algebra (I, ) may be represented by MI = ([0, 1] × {0}) ∪ ((0, 1] × {1}),

MI = {Ar | r ∈ [0, 1]} ∪ {Br | r ∈ [0, 1]},

where Ar = ([0, r] × {0}) ∪ ((0, r] × {1}),

Br = ([0, r) × {0}) ∪ ((0, r] × {1}),

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and I (Ar ) = B1−r , I (Br ) = A1−r . It is easy to see that P(r,0) = Ar ,

Q(r,0) = Br

and

P(r,1) = Br ,

Q(r,1) = Br .

Finally, if we set
I = {Br | r ∈ [0, 1]} ∪ {MI }

and

I = {Ar | r ∈ [0, 1]} ∪ {∅}

we obtain a natural complemented ditopology on (MI , MI , I ). (6) Let S = {a, b, c} and define S = {∅, {a, b}, {b}, {c}, {b, c}, S}. It is easy to verify that S is a texturing of S. Clearly
= {∅, {b}, {a, b}, S},  = S, is a ditopology on (S, S).
We recall that for a ditopological texture space (S, S,
, ) the closure of A ∈ S is given by [A] = {K ∈  | A ⊆ K}  and the interior by ]A[ = {G ∈
| G ⊆ A}. We will make constant reference to the earlier papers [10,11] in this series. Otherwise, this paper is reasonably self-contained, although the reader may also wish to refer to [5–9,12,21–23] for additional motivation and results from the theory of textures and ditopological texture spaces. Our general reference for concepts and results from category theory is [1], while notions from lattice theory not defined here are as given in [16].

2. Dineighbourhood systems Definition 2.1. Let (
, ) be a ditopology on (S, S). (1) If s ∈ S
, a neighbourhood of s is a set N ∈ S for which there exists G ∈
satisfying Ps ⊆ G ⊆ N
Qs . (2) If s ∈ S, a coneighbourhood of s is a set M ∈ S for which there exists K ∈  satisfying Ps
M ⊆ K ⊆ Qs . Note that although conhds are defined for all s ∈ S it will be apparent later that we need only consider the conhds of points in S
. Again the formal duality between these two concepts is clear. We denote the set of nhds (conhds) of s by (s) ((s)), respectively. We will also refer to ((s), (s)), s ∈ S
, as the dinhd system of (
, ). Theorem 2.2. For a ditopology (
, ) on (S, S) let the families (s), s ∈ S
and (s), s ∈ S be defined as above. (1) For each s ∈ S
we have (s) = ∅ and these families satisfy the following conditions: (i) N ∈ (s) ⇒ N
Qs . (ii) N ∈ (s), N ⊆ N  ∈ S ⇒ N  ∈ (s). (iii) N1 , N2 ∈ (s), N1 ∩ N2
Qs ⇒ N1 ∩ N2 ∈ (s). (iv) (a) N ∈ (s) ⇒ ∃ N  ∈ S, Ps ⊆ N  ⊆ N , so that N 
Qt ⇒ N  ∈ (t), ∀ t ∈ S
. (b) For N ∈ S and N
Qs , if there exists N  ∈ S, Ps ⊆ N  ⊆ N , which satisfies N 
Qt ⇒ N  ∈ (t), ∀ t ∈ S
, then N ∈ (s). Moreover, the sets G in
are characterized by the condition that G ∈ (s) for all s with G
Qs . (2) For each s ∈ S we have (s) = ∅ and these families satisfy the following conditions: (i) M ∈ (s) ⇒ Ps
M. (ii) M ∈ (s), M ⊇ M  ∈ S ⇒ M  ∈ (s). (iii) M1 , M2 ∈ (s) ⇒ M1 ∪ M2 ∈ (s). (iv) (a) M ∈ (s) ⇒ ∃ M  ∈ S, M ⊆ M  ⊆ Qs , so that Pt
M  ⇒ M  ∈ (t), ∀ t ∈ S. (b) For M ∈ S and Ps
M, if there exists M  ∈ S, M ⊆ M  ⊆ Qs , which satisfies Pt
M  ⇒ M  ∈ (t), ∀ t ∈ S, then M ∈ (s). Moreover, the sets K in  are characterized by the condition that K ∈ (s) for all s with Ps
K.

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Conversely, if (s), s ∈ S
and (s), s ∈ S are non-empty families of sets in S which satisfy conditions (1) and (2) above, respectively, then there exists a ditopology (
, ) on (S, S) for which (s) ((s)) are the families of nhds (resp., conhds). Proof. (1) (i), (ii) and (iii) are trivial, so we concentrate on (iv). For (a), take N ∈ (s). Then we have G ∈
with Ps ⊆ G ⊆ N
Qs , and it is immediate that N ∗ = G has the required property. For (b), take N, s and N ∗ with the stated properties. Then for each t ∈ S with Ps
Qt we have N ∗
Qt , so N ∗ ∈ (t) and we have Gt ∈
with Pt ⊆ Gt ⊆ N ∗
Qt . Then, by [10, Theorem 1.2(7)], we have   {Pt | Ps
Qt } ⊆ {Gt | Ps
Qt } ⊆ N ∗ , Ps =  so G = {Gt | Ps
Qt } ∈
satisfies Ps ⊆ G ⊆ N
Qs since N ∗ ⊆ N and N
Qs . Hence N ∈ (s), as required. Now, we must verify that G ∈
has the characterization stated. It is immediate that if G ∈
then G ∈ (s) whenever G
Qs , so let us prove the converse. Then for G
Qs wehave G ∈ (s), so we have Gs ∈
with Ps ⊆ Gs ⊆ G
Qs and proceeding as in the proof of (iv)(b) we obtain G = {Gs | G
Qs } ∈
as required. (2) The proof is dual to that for (1). Finally, to establish the converse, suppose that (s) and (s) are non-empty families satisfying conditions (1), (2), respectively. Then it is a straightforward matter to verify that
= {G ∈ S | G
Qs ⇒ G ∈ (s)} and  = {K ∈ S | Ps
K ⇒ K ∈ (s)} define a ditopology (
, ) on (S, S) for which (s), s ∈ S
((s), s ∈ S) are the nhd (respectively, conhd) systems.  For future applications it will be convenient to consider a generalization of the notions of nhd and conhd system. First, let us note that the pair of sets (Ps , Qs ) associated with a point s ∈ S
of S has the properties Ps = ∅, Qs = S and Ps
A ⇒ A ⊆ Qs ∀A ∈ S. This leads to the following: Definition 2.3. The pair (L, M), L, M ∈ S is called a dipod if it satisfies L = ∅, M = S and L
A ⇒ A ⊆ M, ∀A ∈ S. A family (Lj , Mj ), j ∈ J , of dipods is called complete if   {Mj | Lj
A} A= {Lj | A
Mj } and A = for all A ∈ S. It is clear from [10, Theorem 1.2] that (Ps , Qs ), s ∈ S
, is a complete family of dipods, as is any family of dipods containing it. If (L, M) is a dipod we may define the nhds and conhds of (L, M) on analogy with those of a point s by replacing Ps by L and Qs by M. In particular, for a complete family (Lj , Mj ), j ∈ J , of dipods a theorem analogous to Theorem 2.2 may be given for the families (j ) = (Lj ) of nhds and (j ) = (Mj ) of conhds of the dipods (Lj , Mj ), j ∈ J . The relation between the dinhd systems ((s), (s)), s ∈ S
and ((j ), (j )), j ∈ J , for a complete family (Lj , Mj ) of dipods is given in the next lemma. Lemma 2.4. For the texture (S, S) let (Lj , Mj ), j ∈ J , be a complete family of dipods, ((j ), (j )), j ∈ J the dinhd system of (Lj , Mj ), j ∈ J and ((s), (s)), s ∈ S
the dinhd system for a ditopology (
, ) on (S, S). Then (i) (ii) (iii) (iv)

N ∈ (j ), Lj
Qs N ∈ (s), Ps
Mj M ∈ (j ), Ps
Mj M ∈ (s), Lj
Qs

⇒ N ∈ (s). ⇒ N ∈ (j ). ⇒ M ∈ (s). ⇒ M ∈ (j ).

Proof. (i) For N ∈ (j ) we have G ∈
with Lj ⊆ G ⊆ N
Mj . Now Lj
Qs gives G
Qs so Ps ⊆ G ⊆ N
Qs , whence N ∈ (s). (ii), (iii) and (iv) are proved similarly. 

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L.M. Brown et al. / Fuzzy Sets and Systems 157 (2006) 1886 – 1912

This lemma shows that ((j ), (j )), j ∈ J , merely represents a rearrangement of the dinhd system of (
, ). One important property of dipods is that if (L, M) is a dipod and  is a complementation on S then ((M), (L)) is also a dipod. Moreover, if (Lj , Mj ), j ∈ J is a complete family of dipods, so is ((Mj ), (Lj )), j ∈ J . On the other hand, if (
, ) is a complemented ditopology under  then (((j )), ((j ))), j ∈ J is the system of dinhds for ((Mj ), (Lj )). The proofs of these statements are straightforward and are omitted. These remarks, in conjunction with Lemma 2.4, may be used to establish the equivalence of certain local ditopological properties for complemented ditopologies. For an example, the reader is referred to [12] on local compactness in texture spaces. We end this section by showing that continuity and cocontinuity have natural characterizations in terms of nhds and conhds. The following lemma will be useful in this respect.


Lemma 2.5. Let (f, F ) is a difunction from (S1 , S1 ) to (S2 , S2 ), and s ∈ S1 . Then (f → Ps , F → Qs ) is a dipod in (S2 , S2 ). Moreover, if (f, F ) is surjective then (f → Ps , F → Qs )s∈S
is a complete system of dipods. 1

Proof. Since Ps = ∅ we have f → Ps = ∅ by [10, Proposition 2.28(1)], and by the same proposition F → Qs = S2 since
Qs = S1 for s ∈ S1 . Finally, suppose that for some A ∈ S2 we have f → Ps
A and A
F → Qs . Take t, t  ∈ S2 with f → Ps
Qt , Pt
A and A
Qt  , Pt 
F → Qt  . From [10, Lemma 2.6] we have f
Q(s,t) and P (s,t  )
F , so Pt 
Qt by
→ condition DF2. But now A
Qt  leads to A
Qt , which contradicts Pt
A. Hence (f → s ), s ∈ S1 , is a dipod. Ps , F Q ← ← ← Finally, let (f, F ) be surjective and take B ∈ S2 . Then F B ∈ S1 , so F B = {Ps | F B
Qs , s ∈ S
}. By [10, Corollary 2.33] it is easy to verify that B
F → Qs ⇐⇒ F ← B
Qs . Also, since (f, F ) is surjective, by [10, Corollary 2.33],    {Ps | B
F → Qs } = B = f → (F ← B) = f → {f → Ps | B
F → Qs , s ∈ S
} on using [10, Corollary 2.12(2)]. Dually, B =




{F → Qs | f → Ps
B}.



Now we may give: Theorem 2.6. Let (f, F ) : (S1 , S1 ,
1 , 1 ) → (S2 , S2 ,
2 , 2 ) be a difunction. (1) The following are equivalent: (i) (f, F ) is continuous.
(ii) Given s ∈ S1 , N ∈ (f → Ps )) ⇒ ∃ N ∗ ∈ (s) so that f → N ∗ ⊆ N .
(iii) Given s ∈ S1 , N ∈ (f → Ps ) ∩
2 ⇒ ∃ N ∗ ∈ (s) so that f → N ∗ ⊆ N . (2) The following are equivalent: (i) (f, F ) is cocontinuous.
(ii) Given s ∈ S1 , M ∈ (F → Qs ) ⇒ ∃ M ∗ ∈ (s) so that M ⊆ F → M ∗ .
(iii) Given s ∈ S1 , M ∈ (F → Qs ) ∩ 2 ⇒ ∃ M ∗ ∈ (s) so that M ⊆ F → M ∗ . Proof. We prove (1), leaving the essentially dual proof of (2) to the interested reader.
(i) ⇒ (ii) Take s ∈ S1 and N ∈ (f → Ps ). Then we have G ∈
2 satisfying f → Ps ⊆ G ⊆ N
F → Qs . From f → Ps ⊆ G we have Ps ⊆ f ← (f → Ps ) ⊆ f ← G by [10, Lemma 2.9(1)], whence Ps ⊆ F ← G ⊆ F ← N . Also, F ← N
Qs . For if we suppose F ← N ⊆ Qs then N ⊆ F → (F ← N ) ⊆ F → Qs by the same proposition, which is a contradiction. Since (f, F ) is continuous, F ← G ∈
1 . Hence, if we set N ∗ = F ← N then N ∗ ∈ (s) and f → N ∗ = f → (F ← N) = f → (f ← N ) ⊆ N , again by the same proposition. (ii) ⇒ (iii) Immediate. (iii) ⇒ (i) Suppose that the condition given in (iii) is satisfied and take G ∈
2 . We must show that F ← G ∈
1 . By Theorem 2.2(1) it will be sufficient to show that F ← (G)
Qs

⇒

F ← (G) ∈ (s).

Take s ∈ S1 satisfying F ← G
Qs . Using [10, Theorem 2.24(2b) and Lemma 2.9(2)] it is easy to deduce that f → Ps ⊆
G
F → Qs , whence G ∈ (f → Ps ). Clearly s ∈ S1 , so by hypothesis we have N ∗ ∈ (s) satisfying f → N ∗ ⊆ G. But

L.M. Brown et al. / Fuzzy Sets and Systems 157 (2006) 1886 – 1912

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now, by [10, Lemma 2.9(1)] we have N ∗ ⊆ f ← (f → N ∗ ) ⊆ f ← G = F ← G, whence F ← G ∈ (s) by Theorem 2.2. This verifies that (f, F ) is continuous.  Corollary 2.7.
(1) (f, F ) is continuous if and only if given s ∈ S1 and t ∈ S2 with Pt
F → Qs , N ∈ (t) ⇒ ∃ N ∗ ∈ (s) so that → ∗ f N ⊆ N. (2) (f, F ) is cocontinuous if and only if given s ∈ S1 and t ∈ S2 with f ← Ps
Qt , M ∈ (t) ⇒ ∃ M ∗ ∈ (s) so that M ⊆ F → M ∗ .


Proof. Again we consider only (1). If (f, F ) is continuous, s ∈ S1 , t ∈ S2 satisfy Pt
F → Qs and N ∈ (t), then it is
easy to see that N ∈ (f → Ps ), and so N ∗ ∈ (s) exists with f → N ∗ ⊆ N . Conversely, if for s ∈ S2 , G ∈
2 satisfies → → → f Ps ⊆ G
F Qs then there exists t ∈ S2 with G
Qt and Pt
F Qs . Then G ∈ (t) and by hypothesis we have N ∗ ∈ (s) with f → N ∗ ⊆ G. An examination of the proof of Theorem 2.6 shows that this is sufficient to ensure that (f, F ) is continuous.  3. Regularity properties The topological R0 , R1 and regular axioms, and their bitopological analogues mentioned in the Introduction, have provided the motivation for the definitions below. It will be noted that the basic duality involving the p-sets and q-sets has been exploited to give two dual forms of each axiom. Definition 3.1. A ditopology (
, ) on the texture (S, S) is said to be: (a) (b) (c) (d) (e) (f)

R0 if G ∈
, G
Qs ⇒ [Ps ] ⊆ G. Co-R0 if F ∈ , Ps
F ⇒ F ⊆ ]Qs [. R1 if G ∈
, G
Qs , Pt
G ⇒ ∃H ∈
with H
Qs , Pt
[H ]. Co-R1 if F ∈ , Ps
F , F
Qt ⇒ ∃K ∈  with Ps
K, ]K[
Qt . Regular if G ∈
, G
Qs ⇒ ∃H ∈
with H
Qs , [H ] ⊆ G. Coregular if F ∈ , Ps
F ⇒ ∃K ∈  with Ps
K, F ⊆]K[.

For each property P, if (
, ) is P and co-P we say (
, ) is bi-P. Hence, we may speak of bi-R0 , bi-R1 and biregular ditopological spaces. It is clear that regular ⇒ R1 ⇒ R0 , and coregular ⇒ co-R1 ⇒ co-R0 . Note 3.2. It is clear from the definitions that: (a) If (
, ) is R0 (R1 , regular) and ∗ is a cotopology with  ⊆ ∗ then (
, ∗ ) is also R0 (respectively, R1 , regular). (b) If (
, ) is co-R0 (co-R1 , coregular) and
∗ is a topology with
⊆
∗ then (
∗ , ) is also co-R0 (respectively, co-R1 , coregular). Example 3.3. (1) Every discrete ditopology is coregular, and hence co-R0 and co-R1 . (2) Every codiscrete ditopology is regular, and hence R1 and R0 . (3) The ditopological texture space (S, S,
, ) of Examples 1.1(6) is codiscrete and hence regular by (2). On the other hand (
, ) is not co-R0 , and hence neither co-R1 nor coregular. To see this we need only take F = {c} ∈  in Definition 3.1(b). Since Pa = {a, b} we have Pa
F . On the other hand Qa = {b, c}, so ]Qa [ = {b} and we see that F
]Qa [. (4) The unit interval (I, I,
I , I ) is biregular. Let us establish regularity, leaving the dual proof of coregularity to the reader. For G = I in Definition 3.1(e) we may take H = I, so suppose G ⊂ I. Hence G = [0, r), r
1. If G
Qs for some s ∈ I then s < r since Qs = [0, s), and we may choose h ∈ I with s < h < r. Setting H = [0, h) now gives H ∈
I , H
Qs and [H ] = [0, h] ⊆ G as required.

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L.M. Brown et al. / Fuzzy Sets and Systems 157 (2006) 1886 – 1912

We now consider the above properties in more detail. First, we make explicit the link between the R0 axioms and known topological and bitopological axioms. Let (X, u, v) be a bitopological space and (X, P(X), u, v c ) its image in dfDitop under the functor T mentioned in the Introduction. Clearly, (X, P(X), u, v c ) is R0 in the sense of Definition 3.1 if and only if x ∈ G ∈ u ⇒ x v ⊆ G, which is equivalent to (X, u, v) being ws. Likewise (X, P(X), u, v c ) is co-R0 if and only if (X, u, v) is ws∗. A topological (X, T ) space is called R0 if x ∈ G ∈ T ⇒ {x} ⊆ G. Hence, (X, T ) is R0 if and only if its image (X, T , T ) under B in Bitop is pairwise ws, and hence if and only if its image (X, P(X), T , T c ) under T ◦ B = Tc is bi-R0 . Lemma 3.4. Let (
, ) be a ditopology on the texture (S, S). (1) The following are equivalent: (i) (
, ) is R0 .  (ii) For G ∈
there are sets Fj ∈ , j ∈ J , with G = j ∈J Fj . (iii) Condition (ii) holds for all G in some subbase of
. (iv) Given G ∈
, s ∈ S with G
Qs there exists F ∈  with F ⊆ G and F
Qs . (2) The following are equivalent: (i) (
, ) is co-R0 .
(ii) For F ∈  there are sets Gj ∈
, j ∈ J , with F = j ∈J Gj . (iii) Condition (ii) holds for all F in some subbase of . (iv) Given F ∈ , s ∈ S with Ps
F there exists G ∈
with F ⊆ G and Ps
G. Proof. We prove (1), leaving the dual proof of (2) to the reader. (i) ⇒ (ii) Take G ∈
. Then by (i) and [10, Theorem 1.2(7)] we have   {[Ps ] | G
Qs } ⊆ G, G= {Ps | G
Qs } ⊆  whence G = G
Qs [Ps ]. Since [Ps ] ∈  for each s, this proves (ii). (ii) ⇒ (iii) Immediate. (iii) ⇒ (iv) Take G ∈
and s ∈ S with G
Qs . Let  be the subbase of
mentioned in (iii). Since G
can be written as a
join of finite meets of elements of , there
exists a finite index set K and Gk ∈ , k ∈ K, so that k∈K Gk ⊆ G and k∈K Gk
Qs . Take s  ∈ S satisfying k∈K Gk
Qs  and Ps 
Qs . By hypothesis we may write Gk = j ∈Jk Fjk ,

Fjk ∈ , and from Gk
Qs  there exists jk ∈ Jk with Fjkk
Qs  . Let F = k∈K Fjkk . Then F ∈ , F ⊆ k∈K Gk ⊆ G and Ps  ⊆ F whence F
Qs . This establishes (iv). (iv) ⇒ (i) Take G ∈
and s ∈ S with G
Qs . By (iv) we have F ∈  with F ⊆ G and F
Qs . Then, in particular, Ps ⊆ F ⊆ G, whence [Ps ] ⊆ G as required.  Corollary 3.5. In a complemented ditopology (
, ) on (S, S, ) the notions of R0 , co-R0 and bi-R0 are equivalent. Proof. Let (
, ) be R0 and take F ∈ . Then (F ) ∈
so by Lemma 3.4((1)(ii)) we may write (F ) = Fj ∈ . Then ⎞ ⎛    Fj ⎠ = (Fj ) = Gj , F = ((F )) =  ⎝ j ∈J

j ∈J



j ∈J

Fj ,

j ∈J

where Gj = (Fj ) ∈
. This proves that (
, ) is co-R0 by Lemma 3.4((2)(ii)). The proof of the converse is similar.  We note that Lemma 3.4((1)(ii)) gives a point-free characterization of the R0 axiom, and likewise ((2)(ii)) gives a point-free characterization for co-R0 . In particular, this enables us to associate these axioms with the R0 axiom of Hutton and Reilly [17].

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Corollary 3.6. A topology T on the Hutton algebra (L, ) is R0 in the sense of [17] if and only if the corresponding complemented ditopological simple texture is R0 (equivalently, co-R0 , bi-R0 ). Proof. Immediate from [17, Definition (R0 , T1 )] and Lemma 3.4((1)(ii)).



Now, we turn our attention to the R1 axioms, first looking at the relation with known topological and bitopological properties. Let (X, u, v) be a bitopological space and (X, P(X), u, v c ) its image under T. Then (X, P(X), u, v c ) is R1 if and v only if for x ∈ U ∈ u, y ∈ / U we have U1 ∈ u with x ∈ U1 , y ∈ / U1 . This is easily seen to be equivalent to pH, so c (X, P(X), u, v ) is R1 if and only if (X, u, v) is pH, and likewise it is co-R1 if and only if (X, u, v) is pH∗. Again a topological space (X, T ) is R1 if and only if the bitopological space (X, T , T ) is pairwise pH and therefore if and only if the ditopological texture space (X, P(X), T , T c ) is bi-R1 . Lemma 3.7. Let (
, ) be a ditopology on the texture (S, S). (1) The following are equivalent: (i) (
, ) is R1 . (ii) Given G ∈
, s, t ∈ S with G
Qs and Pt
G we have H ∈
with Ps ⊆ H ⊆ [H ] ⊆ Qt . (iii) For G ∈
we have Hji ∈
, j ∈ Ji , i ∈ I , with G=



Hji =

i∈I j ∈Ji



[Hji ].

i∈I j ∈Ji

(iv) Condition (iii) holds for all G in some subbase of
. (2) The following are equivalent: (i) (
, ) is co-R1 . (ii) Given F ∈ , s, t ∈ S with Ps
F and F
Qt we have K ∈  with Pt ⊆ ]K[ ⊆ K ⊆ Qs . (iii) For F ∈  we have Kji ∈ , j ∈ Ji , i ∈ I with F =



Kji =

i∈I j ∈Ji



]Kji [.

i∈I j ∈Ji

(iv) Condition (iii) holds for all F in some subbase of . Proof. We prove (1), leaving (2) to the interested reader. (i) ⇒ (ii) Trivial since H
Qs implies Ps ⊆ H and Pt
[H ] implies [H ] ⊆ Qt . (ii) ⇒ (iii) For any s, t ∈ S with G
Qs and Pt
G we may chose Hts ∈
with Ps ⊆ Hts ⊆ [Hts ] ⊆ Qt . Clearly 

Ps ⊆

Hts ⊆

Pt
G



[Hts ] ⊆ G,

Pt
G

whence 

G⊆

Ps ⊆

G
Qs





Hts ⊆

G
Qs Pt
G





[Hts ] ⊆ G.

G
Qs Pt
G

(iii) ⇒ (iv) Immediate. (iv) ⇒ (i) Take G ∈
and s, t ∈ S with G
Q
s , Pt
G. If  is
the subbase mentioned in (iv) we have, as in the proof of Lemma 3.4, Gk ∈ , k ∈ K satisfying k∈K Gk ⊆ G and k∈K Gk
Qs . Now, from Pt
G we have k ∈ K with Pt
Gk . By hypothesis, we may write Gk in the form Gk =

 i∈I j ∈Ji

Hji =

 i∈I j ∈Ji

[Hji ],

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L.M. Brown et al. / Fuzzy Sets and Systems 157 (2006) 1886 – 1912


where Hji ∈
for all j ∈ Ji , i ∈ I . From Gk
Qs , we obtain i ∈ I with j ∈Ji Hji
Qs . On the other hand Pt
Gk
gives Pt
j ∈Ji [Hji ], so there exists j ∈ Ji for this i with Pt
[Hji ]. Since for this i and j we also have Hji
Qs , we have established that (
, ) is R1 .  Corollary 3.8. For a complemented ditopology on (S, S, ) the notions of R1 , co-R1 and bi-R1 are equivalent. Proof. Immediate from the point-free characterization in Lemma 3.7((1)(iii)). Corollary 3.9. A topology T on the Hutton algebra (L, ) is R1 in the sense of [17] if and only if the corresponding complemented ditopological simple texture is R1 (equivalently, co-R1 , bi–R1 ). Proof. Immediate from [17, Definition (R1 , T2 )] and Lemma 3.7((1)(iii)).



Finally, we turn to the regularity axioms, first looking at their relation with topological and bitopological regularity. It is immediate from the definitions that (X, P(X), u, v c ) is regular (coregular) if and only if (X, u, v) is regular (respectively, regular∗). Likewise, (X, T ) is regular if and only if (X, T , T ) is pairwise regular and hence if and only if (X, P(X), T , T c ) is biregular. Lemma 3.10. Let (
, ) be a ditopology on the texture (S, S). (1) The following are equivalent: (i) (
, ) is regular. (ii) Given G ∈
, s ∈ S with G
Qs there exists H ∈
with Ps ⊆ H ⊆ [H ] ⊆ G. (iii) For G ∈
we have Hj ∈
, j ∈ J , with   G= Hj = [Hj ]. j ∈J

j ∈J

(iv) Condition (iii) holds for all G in some subbase of
. (2) The following are equivalent: (i) (
, ) is coregular. (ii) Given F ∈ , s ∈ S with Ps
F there exists K ∈  with F ⊆ ]K[ ⊆ K ⊆ Qs . (iii) For F ∈  we have Kj ∈ , j ∈ J , with   F = Kj = ]Kj [. j ∈J

j ∈J

(iv) Condition (iii) holds for all F is some subbase of . Proof. As for the previous lemmas. The details are left to the interested reader.



Corollary 3.11. For a complemented ditopology on (S, S, ) the notions of regular, coregular and biregular are equivalent. Proof. Straightforward.



Corollary 3.12. A topology T on the Hutton algebra (L, ) is regular in the sense of [17] if and only if the corresponding complemented ditopological simple texture is regular (equivalently, coregular, biregular). Proof. Immediate from [17, Definition (R, T3 )] and Lemma 3.7((1)(iii)).



Let us now verify that the above properties are both additive and multiplicative in the class of ditopological texture spaces.

L.M. Brown et al. / Fuzzy Sets and Systems 157 (2006) 1886 – 1912

1895

Theorem 3.13. Let (Sj , Sj ,
j , j ), j ∈ J , be non-empty disjoint ditopological texture spaces and (S, S,
, ) their disjoint sum. Then (1) (S, S,
, ) is R0 (co-R0 ) if and only if (Sj , Sj ,
j , j ) is R0 (respectively, co-R0 ) for all j ∈ J . (2) (S, S,
, ) is R1 (co-R1 ) if and only if (Sj , Sj ,
j , j ) is R1 (respectively, co-R1 ) for all j ∈ J . (3) (S, S,
, ) is regular (coregular) if and only if (Sj , Sj ,
j , j ) is regular (respectively, coregular) for all j ∈ J . Proof. We consider (1), leaving (2) and (3) to the interested reader. First, suppose  that (S, S,
, ) is R0 and take k ∈ J . Choose G ∈ Sk and s ∈ Sk satisfying G
Qks . Then G ∈
and G
Qs = Qks ∪ j ∈J \{k} Sj since G ⊆ Sk . Hence [Ps ] ⊆ G. But [Ps ] = [Psk ]k since s ∈ Sk , so [Psk ]k ⊆ G and we see that (Sk , Sk ,
k , k ) is R0 . Here, and below, [ ]k denotes closure in (Sk , Sk ,
k , k ). Conversely, suppose that (Sj , Sj ,
j , j ) is R0 for each j ∈ J and take G ∈
, s ∈ S with G
Qs . Let k be the unique element of J for which s ∈ Sk . By the definition of
we have G ∩ Sk ∈
k , and clearly G ∩ Sk
Qks . Since (Sk , Sk ,
k , k ) is R0 we have [Ps ] = [Psk ]k ⊆ G ∩ Sk ⊆ G, whence (S, S,
, ) is R0 as required. The proof for co-R0 is dual to the above.  Before looking at the preservation of the regularity axioms by products, we will establish the more general result that they are preserved by initial ditopologies. Here, by the initial ditopology on (S, S) generated by the family (fj , Fj ) : (S, S) → (Sj , Sj ,
j , j ), j ∈ J , we mean the ditopology (
, ) on (S, S) with subbase {Fj← G | G ∈
j , j ∈ J } and cosubbase {fj← K | K ∈ j , j ∈ J }. Theorem 3.14. Let (Sj , Sj ,
j , j ), j ∈ J , be ditopological texture spaces, (S, S) a texture and for each j ∈ J let (fj , Fj ) : (S, S) → (Sj , Sj ,
j , j ) be a difunction. We denote by (
, ) the initial ditopology on (S, S) generated by the difunctions (fj , Fj ) and the spaces (Sj , Sj ,
j , j ). Then if for each j ∈ J the spaces (Sj , Sj ,
j , j ) are R0 (co-R0 , R1 , co-R1 , regular, coregular), the initial ditopology (
, ) is R0 (respectively, co-R0 , R1 , co-R1 , regular, coregular). Proof. We establish the results for R0 , R1 and regular, leaving the dual proofs for co-R0 , co-R1 and coregular to the reader. Suppose first that the given spaces are R0 . In view of Lemma 3.4((1)(iii)) it will be sufficient to show that Fj← G, G ∈
j , canbe written as a join of elements of . However, (S ) is R0 so by Lemma 3.4((1)(ii)) we may j , Sj ,
j , j  write G = i∈I Ki for Ki ∈ j , i ∈ I . Hence Fj← G = Fj← ( i∈I Ki ) = i∈I Fj← Ki by [10, Corollary 2.26], and Fj← Ki = fj← Ki ∈  by the definition of , so we deduce (
, ) is R0 . 

Suppose now that the spaces are R1 . Then we may write G = i∈I k∈Ki Hki = i∈I k∈Ki [Hki ] with Hki ∈
j for all i, k. Hence     Fj← G = Fj← Hki = Fj← [Hki ], i∈I k∈Ki

i∈I k∈Ki



again using [10, Corollary 2.26]. Now, for each i, k we have Fj← Hki ⊆ Fj← [Hki ] and so k∈Ki Fj← Hki ⊆ k∈Ki




← i Fj← [Hki ]. Since the set on the right belongs to  as above, we obtain k∈Ki Fj← Hki ⊆ k∈Ki Fj Hk ⊆ k∈Ki Fj← [Hki ]. Taking the join over i ∈ I and using the above equalities for Fj← G now leads to ⎡ ⎤     ⎣ Fj← Hki = Fj← Hki ⎦ . Fj← G = i∈I k∈Ki

i∈I

k∈Ki

Since Fj← Hki ∈
for all i, k we deduce (
, ) is R1 by Lemma 3.7((1)(iv)).   Finally, let the given spaces be regular, whence we may write G = k∈K Gk = k∈K [Gk ] for Gk∈
j . Arguing  as above, we obtain from Fj← G = k∈K Fj← Gk = k∈K Fj← [Gk ] that Fj← G = k∈K Fj← Gk = k∈K [Fj← Gk ], whence (
, ) is regular by Lemma 3.10((1)(iv)).  In categorical terms, the above result may be expressed in terms of initial closedness [1, Definition 21.29].

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Corollary 3.15. Denote by R any of the full concrete subcategories of dfDitop whose objects are (co-, bi-)R0 , (co-, bi-)R1 or (co-, bi-) regular ditopological texture spaces, and let E : R → dfDitop be the inclusion functor. Then (R, U ◦ E) is initially closed in (dfDitop, U). Theorem 3.16. With the notation as in Corollary 3.15, each of the categories R is topological. Proof. By [11, Theorem 3.6] the concrete category (dfDitop, U) is topological, so the result follows from Corollary 3.15 and [1, Proposition 21.30].  In view of the Topological Duality Theorem [1, 21.9], the existence of unique U ◦ E-initial lifts of U ◦ E-structured sources implies the existence of unique U ◦ E-final lifts of U ◦ E-structured sinks. Theorem 3.17. With the notation as in Corollary 3.15, each of (R, U ◦ E) is concretely reflective in (dfDitop, U). Proof. Immediate from Corollary 3.15 and [1, Proposition 21.31].



More particularly, [1, Theorem 21.35(4)] guarantees for each R the existence of a concrete functor from (dfDitop, U) to (R, U ◦ E) that leaves each R-object fixed. We identify this functor in case R is the category of R0 ditopological texture spaces, leaving a discussion of the remaining cases to the interested reader. Proposition 3.18. Let R be the full subcategory of dfDitop whose objects are R0 spaces, take (S, S,
, ) ∈ Ob dfDitop and let
0 = {G ∈
| G
Qs ⇒ [Ps ] ⊆ G}. Then (i, I ) : (S, S,
, ) → (S, S,
0 , ) is a universal arrow which defines a concrete functor from (dfDitop, U) to (R, U ◦ E) leaving the objects of R fixed. Proof. It is straightforward to verify that
0 is the finest topology on (S, S) coarser than
for which (
0 , ) is R0 , and we omit the details. Since
0 ⊆
, we see that (i, I ) : (S, S,
, ) → (S, S,
0 , ) is bicontinuous. To establish the universal property take (S1 , S1 ,
1 , 1 ) ∈ Ob R and let (f, F ) : (S, S,
, ) → (S1 , S1 ,
1 , 1 ) be bicontinuous. It will clearly be sufficient to show that (f, F ) is bicontinuous regarded as a difunction from (S, S,
0 , ) to (S1 , S1 ,
1 , 1 ). (i,I ) / (S, S,
0 , ) (S, S,
, )U  UUUU UUUU  UUUU  (f,F ) U U U (f,F ) UUUU  U* (S1 , S1 ,
1 , 1 )

However, the initial ditopology (
∗ , ∗ ) on (S, S) generated by the difunction (f, F ) and the R0 space (S1 , S1 ,
1 , 1 ) is a R0 space coarser than (
, ). Hence (
∗ , ) is R0 by Note 3.2(a), and we deduce
∗ ⊆
0 , which gives the required result. Since this universal arrow is identity-carried it certainly gives rise to a concrete functor, while the evident fact that (
, ) is R0 if and only if
0 =
shows that it leaves fixed the objects of R.  Theorem 3.19. Let (Sj , Sj ,
j , j ), j ∈ J , be non-empty ditopological texture spaces and (S, S,
, ) their product. Then (1) (S, S,
, ) is R0 (co-R0 ) if and only if (Sj , Sj ,
j , j ) is R0 (respectively, co-R0 ) for all j ∈ J . (2) (S, S,
, ) is R1 (co-R1 ) if and only if (Sj , Sj ,
j , j ) is R1 (respectively, co-R1 ) for all j ∈ J . (3) (S, S,
, ) is regular (coregular) if and only if (Sj , Sj ,
j , j ) is regular (respectively, coregular) for all j ∈ J . Proof (Sufficiency). Clear from Theorem 3.14 since the product ditopology is the initial topology generated by the projection difunctions ( j , j ) and the spaces (Sj , Sj ,
j , j ).

L.M. Brown et al. / Fuzzy Sets and Systems 157 (2006) 1886 – 1912

1897

Necessity: We prove the result for R0 , leaving the remaining cases to the interested reader. Suppose that (S, S,
, )
is R0 . Take k ∈ J , G ∈
k and s ∈ Sk with G
Qs . For j ∈ J \ {k} choose aj ∈ Sj and define u = (uj ) ∈ S by  s if j = k, uj = aj otherwise.  Since Qu = j ∈J E(j, Quj ) by [10, Proposition 1.3(2)], and Quj = Qaj =Sj for j ∈ J \ {k} we have E(k, G)
Qu . Since (
, ) is R0 and E(k, G) ∈
we have [Pu ] ⊆ E(k, G). But Pu = j ∈J Puj by [10, Proposition 1.3(1)], so [Pu ] = j ∈J [Puj ] by [11, Proposition 3.16(1)]. Hence [Ps ] ⊆ G, and we have established that (
k , k ) is R0 .  Proposition 3.20. In each of the categories R defined in Corollary 3.15, the ditopological products and ditopological sums are, respectively, concrete products and concrete coproducts. Proof. Clear by [11, Propositions 3.12, 3.26], Theorems 3.19 and 3.13.



Finally, let us consider preservation of the regularity properties under dihomeomorphisms, that is under isomorphisms in the category dfDitop. As shown in [5], characteristic properties of a dihomeomorphism are that it is surjective, injective, bicontinuous, open (= co-open) and closed (= coclosed). Theorem 3.21. Let (Sk , Sk ,
k , k ), k = 1, 2 be ditopological texture spaces and (f, F ) : (S1 , S1 ) → (S2 , S2 ) a dihomeomorphism. Then (S1 , S1 ,
1 , 1 ) is R0 (co-R0 , R1 , co-R1 , regular, coregular) if and only if (S2 , S2 ,
2 , 2 ) is R0 (respectively, co-R0 , R1 , co-R1 , regular, coregular). Proof. We give the proof for the R0 axiom, leaving the remaining cases to the interested reader. It will clearly be sufficient to show that if (S1 , S1 ,
1 , 1 ) is R0 then (S2 , S2 ,
2 , 2 ) is R0 . To show this take ← G ∈
2 . By  continuity of (f, F ) we have F G ∈
1 , so by Lemma 3.4((1)(ii)) we have Fj ∈ 1 ,→j ∈←J with ← F G = j ∈J F j . Now, since (f, F ) is surjective we have by [10, Corollary 2.33(1)] that G = f (F G) = f → ( j ∈J Fj ) = j ∈J f → Fj by [10, Corollary 2.12(2)]. Since (f, F ) is closed, f → Fj ∈ 2 for all j ∈ J , whence (S2 , S2 ,
2 , 2 ) is R0 by Lemma 3.4((1)(ii)).  An examination of the above proof shows that the following more general result holds, and other results of a similar nature may be obtained for the other axioms. Proposition 3.22. Let (Sk , Sk ,
k , k ), k = 1, 2 be ditopological texture spaces and (f, F ) : (S1 , S1 ) → (S2 , S2 ) a surjective difunction. (1) If (
1 , 1 ) is R0 and (f, F ) is continuous and closed then (
2 , 2 ) is R0 . (2) If (
1 , 1 ) is co-R0 and (f, F ) is cocontinuous and coclosed then (
2 , 2 ) is co-R0 . Note 3.23. Since it is easy to show that a fDitop isomorphism gives rise to a dfDitop isomorphism, the regularity axioms are also preserved under fDitop isomorphisms (see [10,11]). 4. Point separation properties If (S, S) is a texture then for s, t ∈ S, s = t is equivalent to Ps = Pt . In the course of our study of point separation properties, however, we will often have cause to consider the condition Qs = Qt . It is necessary, therefore, for us to consider the relation between these two conditions. If Ps ⊆ Pt then it is immediate from the definition of q-set that Qs ⊆ Qt . The following example shows that the converse is not true in general. Example 4.1. Consider the texture (MI , MI , I ) of Examples 1.1(5). Here, for (r, 0), (r, 1) ∈ MI = ([0, 1] × {0}) ∪ ((0, 1] × {1}) we have Q(r,0) = Q(r,1) but P(r,0)
P(r,1) .

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Definition 4.2. The texture (S, S) is called coseparated if Qs ⊆ Qt ⇒ Ps ⊆ Pt for all s, t ∈ S. Since the texture of Example 4.1 is simple, we see that not every simple texture is coseparated. On the other hand: Lemma 4.3. Every plain texture is coseparated. The converse is false in general. Proof. Let (S, S) be plain and take s, t ∈ S with Qs ⊆ Qt . Suppose that Ps
Pt and take u ∈ S with Ps
Qu and Pu
Pt . From Ps
Qu we have Qu ⊆ Qs ⊆ Qt , so from Pu
Pt we have Pt ⊆ Qu ⊆ Qt , which contradicts the fact that (S, S) is plain. To see that a coseparated texture need not be plain, consider the texture (L, L, ) of Examples 1.1(3). Since for this texture we clearly have Pr = Qr = (0, r] for all r ∈ L = (0, 1] we certainly have Pr ⊆ Pt ⇐⇒ Qr ⊆ Qt for all r, t ∈ L. On the other hand, this texture is not plain.  We note in passing that the unit interval texture (I, I, ), being plain, is coseparated by Lemma 4.3. However, this texture is shown in [11, Example 2.14] to be isomorphic in dfTex to (MI , MI , I ), so by Example 4.1 dfTex isomorphisms do not preserve the property of being coseparated. Before presenting our definition of the T0 property we give two general theorems which will be found useful. Theorem 4.4. Let A ⊆ S contain S, ∅ and be closed under arbitrary joins. Then the following are equivalent:
(1) ∀A ∈ S ∃Aj ∈ A, j ∈ J , with A = j ∈J Aj . (2) For s, t ∈ S, Ps
Pt ⇒ ∃A ∈ A with Pt ⊆ A and Ps
A. (3) For s, t ∈ S, Ps
Pt ⇒ ∃ A ∈ A with Pt ⊆ A ⊆ Qs . (4) There exists a complete family of dipods (Lk , Mk )k∈K satisfying Lk 
Lk (5) (6) (7) (8)

⇒

∃A ∈ A

with Lk ⊆ A ⊆ Mk  .


 j j ∀ A ∈ S there exists Ai ∈ A, j ∈ J , i ∈ Ij , with A = j ∈J i∈Ij Ai . For s, t ∈ S, Qs
Qt ⇒ ∃ A ∈ A with Ps
A
Qt . For s, t ∈ S, Qs
Qt ⇒ ∃ A ∈ A with Pt ⊆ A ⊆ Qs . Qt ∈ A ∀t ∈ S.

If (S, S) is coseparated then the following is also equivalent to the above: (9) For s, t ∈ S, Ps
Pt ⇒ ∃ A ∈ A with Ps
A
Qt .
Proof. (1) ⇒ (2) Applying (1) with A = Pt gives Ps
Pt = j ∈J Aj for some Aj ∈ A, j ∈ J . Hence Ps
Aj for some j ∈ J , and clearly Pt ⊆ Aj . (2) ⇒ (3) Immediate. (3) ⇒ (4) By (3), (Ps , Qs )s∈S
is a complete family of dipods satisfying the given condition. (4) ⇒ (5) Let (Lk , Mk )k∈K be a complete family of dipods satisfying the stated condition and take A
∈ S. Let J = {j ∈ K | A
Mj } and for each j ∈ J let Ij = {i ∈ K | Li
A}. Then by definition A = j ∈J Lj = i∈Ij Mi . j

j

For j ∈ J , i ∈ Ij we have Li
Lj so by hypothesis there exists Ai ∈ A with Lj ⊆ Ai ⊆ Mi . Hence Lj ⊆

 i∈Ij

j

Ai ⊆



Mi = A,

i∈Ij


j which gives A = j ∈J i∈Ij Ai , as required.  (5) ⇒ (6) Take Qs
Qt . By definition, Qs = {Pu | Ps
Pu } so we have u ∈ S with Ps
Pu and Pu
Qt . Now,

j j j j we may write Pu = j ∈J i∈Ij Ai , Ai ∈ A, so we have j ∈ J with i∈Ij Ai
Qt , whence Ps
Ai
Qt for some i ∈ Ij . (6) ⇒ (7) Trivial.

L.M. Brown et al. / Fuzzy Sets and Systems 157 (2006) 1886 – 1912

1899

 (7) ⇒ (8) Qs = {Pt | Qs
Qt } and for Qs
Qt we may choose At ∈ A with Pt ⊆ At ⊆ Qs . Hence Qs = {At | Qs
Qt } ∈ A.
(8) ⇒ (1) For A ∈ S we have A = {Qs | s ∈ / A}, and Qs ∈ A. If (S, S) is coseparated then (9) is clearly equivalent to (6).  Theorem 4.5. Let B ⊆ S contain S, ∅ and be closed under arbitrary intersections. Then the following are equivalent:  (1) ∀ B ∈ S ∃ Bj ∈ B, j ∈ J , with B = j ∈J Bj . (2) For s, t ∈ S, Qs
Qt ⇒ ∃ B ∈ B with Ps
B
Qt . (3) For s, t ∈ S, Qs
Qt ⇒ ∃ B ∈ B with Pt ⊆ B ⊆ Qs . (4) There exists a complete family of dipods (Lk , Mk )k∈K satisfying Mk 
Mk

⇒

∃B ∈ B

with Lk ⊆ B ⊆ Mk  .
 j j (5) ∀ B ∈ S there exists Bi ∈ B, j ∈ J , i ∈ Ij , with B = j ∈J i∈Ij Bi . If (S, S) is coseparated, each of the following is also equivalent to the above: (6) For s, t ∈ S, Ps
Pt ⇒ ∃ B ∈ B with Ps
B
Qt . (7) For s, t ∈ S, Ps
Pt ⇒ ∃ B ∈ B with Pt ⊆ B ⊆ Qs . (8) Ps ∈ B ∀s ∈ S. Proof. Essentially dual to the proof of Theorem 4.4.



Let (
∪ )∨ denote the set of arbitrary joins of sets in
∪ , and (
∪ )∩ the set of arbitrary intersections of sets in
∪ . If we suppose that Theorem 4.4 holds for A = (
∪ )∨ , then (1) says that any element of S can be written
 
j j j in the form j ∈J ( i∈Ij Ci ) with Ci ∈
∪ . By complete distributivity this set is equal to ∈ Ij ( j ∈J C(j ) ), so Theorem 4.5 holds with B = (
∪ )∩ . The converse follows in the same way, so the conditions in these two theorems are equivalent for this choice of A and B. This leads to the following definition. Definition 4.6. A ditopological space which satisfies the equivalent conditions obtained by setting A = (
∪ )∨ in Theorem 4.4, and B = (
∪ )∩ in Theorem 4.5 is said to be T0 . The following theorem gives some characteristic properties of T0 ditopological texture spaces. Several others may be read off easily from Theorems 4.4 and 4.5. Theorem 4.7. The following are characteristic properties of T0 ditopological texture spaces:  (1) Ps
Pt ⇒ ∃ Cj ∈
∪ , j ∈ J , with Pt ⊆ j ∈J Cj ⊆ Qs .
(2) Qs
Qt ⇒ ∃ Cj ∈
∪ , j ∈ J , with Pt ⊆ j ∈J Cj ⊆ Qs .
 j j (3) For A ∈ S there exist Ci ∈
∪ , j ∈ J , i ∈ Ij , with A = j ∈J i∈Ij Ci . (4) Qs
Qt ⇒ ∃ C ∈
∪  with Ps
C
Qt . (5) [Ps ] ⊆ [Pt ] and ]Qs [ ⊆ ]Qt [⇒ Qs ⊆ Qt . (6) For all s ∈ S we have Qs = j ∈J Cj for Cj ∈
∪ . If (S, S) is coseparated the following condition also characterizes the T0 property:
(7) For all s ∈ S we have Ps = j ∈J Cj for Cj ∈
∪ . Proof. We need only verify (5) since the other conditions follow from Theorem 4.4 with A = (
∪ )∨ , or from Theorem 4.5 with B = (
∪ )∩ . Now, if Qs
Qt we have C ∈
∪  with Ps
C
Qt by (4). If C ∈
then Ps
C ⇒ C ⊆ Qs ⇒ C ⊆ ]Qs [, and we have ]Qs [
]Qt [. If C ∈  then C
Qt ⇒ Pt ⊆ C ⇒ [Pt ] ⊆ C, and so [Ps ]
[Pt ]. Thus (4) ⇒ (5). Conversely, if (5) is satisfied and Qs
Qt then [Ps ]
[Pt ] or ]Qs [
]Qt [. In the first case, Pt ⊆ [Pt ] ⊆ Qt and in the second, Pt ⊆ ]Qs [ ⊆ Qt . Hence (5) ⇒ (2). 

1900

L.M. Brown et al. / Fuzzy Sets and Systems 157 (2006) 1886 – 1912

Corollary 4.8. A topology T on the Hutton algebra (L, ) is T0 in the sense of [17] if and only if the corresponding complemented ditopological simple texture is T0 . Proof. Clear from [17, Definition (T0 )] and Theorem 4.7(3).



Notes 4.9. (1) It will be observed that T0 is essentially a self-dual condition, so we have no separate co-T0 condition. (2) Let the image (X, P(X), u, v c ) of the bitopological space (X, u, v) under T be T0 , and take x = y in X. Then Qx
Qy in (X, P(X)), so by Theorem 4.7(4) we have C ∈ u ∪ v c with x ∈ / C, y ∈ C. If C ∈ u then y ∈ / x u , while if v c C ∈ v we have x ∈ / y , so (X, u, v) is T0 in the strong sense used here (see the Introduction). Conversely, if (X, u, v) is T0 then it is easy to see that (X, P(X), u, v c ) is T0 . In particular, (X, T ) is T0 if and only if (X, P(X), u, T , T c ) is T0 . (3) For s, t ∈ S let Qs
Qt . Then if (
, ) is T0 , Condition (4) gives either an (open) nhd of t which is not a nhd of s, or a (closed) conhd of s which is not a conhd of t. Conversely, the existence of such a nhd or conhd for any such s, t establishes Condition (2), showing that (
, ) is T0 . Thus, (
, ) is T0 if and only Qs
Qt ⇒ (t)
(s) or (s)
(t). In particular, for a coseparated space, (
, ) is T0 if and only if given s, t ∈ S, s = t, we have (s) = (t) or (s) = (t). (4) The space (MI , MI , I ,
I , I ) is T0 because, for example, Qs ∈
I ⊆
I ∪ I for each s ∈ MI . For this space, we also have Ps ∈
I ∪ I for all s ∈ MI , whence Theorem 4.7(7) holds, even though this texture is not coseparated. The authors do not know of an example of a T0 ditopological texture space not satisfying this condition. (5) It is shown in [5] that the category dfDitop0 of T0 ditopological texture spaces and bicontinuous difunctions is a full reflexive subcategory of dfDitop, a reflection being obtained by taking a quotient ditopological texture space with respect to an appropriate equivalence direlation. (6) The T0 property is preserved under dihomeomorphisms, the proof being easily obtained from the characterization in Theorem 4.7(3) and the properties of a dihomeomorphism. Hence, T0 is also preserved under fDitop isomorphisms. Now, we turn our attention to the T1 axiom. Definition 4.10. A ditopological texture space is called: (1) T1 if it is T0 and R0 . (2) Co-T1 if it is T0 and co-R0 . (3) Bi-T1 if it is T0 and bi-R0 . Theorem 4.11. Let (
, ) be a ditopology on (S, S). (1) (
, ) is T1 if and only if it satisfies the conditions of Theorem 4.5 with B = . In particular, the following are characteristic of a T1 ditopological space.  (i) For any A ∈ S we have Fi ∈ , i ∈ I , with A = i∈I Fi . (ii) For s, t ∈ S, Qs
Qt ⇒ ∃ F ∈  with Ps
F
Qt . (iii) If (S, S) is coseparated then Ps ∈  for each s ∈ S. (2) (
, ) is co-T1 if and only if it satisfies the conditions of Theorem 4.4 with A =
. In particular, the following are characteristic of a co-T1 ditopological space.
(i) For any A ∈ S we have Gi ∈
, i ∈ I , with A = i∈I Gi . (ii) For s, t ∈ S, Qs
Qt ⇒ ∃ G ∈
with Ps
G
Qt . (iii) Qs ∈
for all s ∈ S. Proof. We prove (2), leaving the proof of (1) to the reader. Let (
, ) be T0 and co-R0 . Take s, t ∈ S with Qs
Qt . By Theorem 4.7(4), we have B ∈
∪  with Ps
B
Qt . If B ∈
then G = B ∈
satisfies Pt ⊆ G ⊆ Qs . On the other hand, if B ∈  then Ps
B implies B ⊆ ]Qs [ by co-R0 , whence for G = ]Qs [ we again have Pt ⊆ G ⊆ Qs . This establishes Theorem 4.4(7) for A =
. Conditions (i), (ii) and (iii) are now, respectively, just conditions (1), (6) and (8) of Theorem 4.4. Conversely, let (
, ) satisfy Theorem 4.4 with A =
. Then, in particular, it satisfies this theorem with A = (
∪ )∨ , To show it is co-T1 it remains to show it is co-R0 . However, by Theorem 4.4(1) applied to A = F ∈  so (
, ) is T0 .
we have F = j ∈J Gj , Gj ∈
, whence the result follows by Lemma 3.4((2)(ii)). 

L.M. Brown et al. / Fuzzy Sets and Systems 157 (2006) 1886 – 1912

1901

Corollary 4.12. For a complemented ditopological texture space the notions of T1 , co-T1 and bi-T1 coincide. Proof.
This is clear by Corollary 3.5. For a direct proof we need  only note that under a complementation  the condition A = j ∈J Gj , Gj ∈
, transforms to the condition (A) = j ∈J (Gj ), (Gj ) ∈ , and conversely.  Corollary 4.13. A topology T on the Hutton algebra (L, ) is T1 in the sense of [17] if and only if the corresponding complemented ditopological simple texture is T1 (equivalently, co-T1 , bi-T1 ). Proof. Clear from [17, Definition (R0 , T1 )] and Definition 4.10.



The following example shows that even for a complemented bi-T1 ditopology on a texture which is not coseparated, the sets Ps need not be closed. Example 4.14. Consider the complemented ditopological space (MI , MI , I ,
I , I ). For all r, 0
r
1, we have Q(r,0) = Br ∈
I and for 0 < r
1 we have Q(r,1) = Br ∈
I , so (
I , I ) is co-T1 by Theorem 4.11((2)(iii)). /  for Hence (
I , I ) is bi-T1 by Corollary 4.13. However, although P(r,0) = Ar ∈ , 0
r
1 we have P(r,1) = Br ∈ 0 < r
1. Now, let us consider the T2 (Hausdorff) axiom. In parallel with Definition 4.10 we have the following. Definition 4.15. A ditopological texture space is called: (1) T2 if it is T0 and R1 . (2) Co-T2 if it is T0 and co-R1 . (3) Bi-T2 if it is T0 and bi-R1 . It is immediate from the definitions that T2 ⇒ T1 ⇒ T0

co-T2 ⇒ co-T1 ⇒ T0 ,

and

whence bi-T2 ⇒ bi-T1 ⇒ T0 . Proposition 4.16. For a complemented ditopology on (S, S, ) the T2 , co-T2 and bi-T2 properties are equivalent. Proof. Clear from Lemma 3.8.



Theorem 4.17. The following are equivalent for a ditopology (
, ) on (S, S). (1) (
, ) is bi-T2 . (2) For s, t ∈ S, Qs
Qt ⇒ ∃ H ∈
, K ∈  with H ⊆ K, Ps
K and H
Qt . 
(3) For A ∈ S there exist Hji ∈
, Kji ∈ , i ∈ I , j ∈ Ji , with Hji ⊆ Kji for all i, j and A = i∈I j ∈Ji Hji = 
i i∈I j ∈Ji Kj . Proof. (1) ⇒ (2) Let Qs
Qt . Since (
, ) is T0 we have B ∈
∪  with Ps
B
Qt by Theorem 4.7(4). If B ∈
, we have H ∈
with Ps
[H ], H
Qt by the R1 axiom, whence (2) is satisfied for K = [H ]. On the other hand, if B ∈  then we have K ∈  with Ps
K, ]K[
Q whence (2) is satisfied with H =]K[.  t by the co-R1 axiom,
(2) ⇒ (3) For A ∈ S we may write A = {Pt | A
Qt } = {Qs | Ps
A}. For s, t satisfying A
Qt and Ps
A we have Qs
Qt and so there exist Hst ∈
, Kst ∈  with Hst ⊆ Kst and Ps
Kst , Hst
Qt . It is trivial to verify that     A= Hst = Kst . A
Qt Ps
A

A
Qt Ps
A

(3) ⇒ (1) Left to the reader.



Corollary 4.18. A coregular T1 space is bi-T2 and a regular co-T1 space is bi-T2 .

1902

L.M. Brown et al. / Fuzzy Sets and Systems 157 (2006) 1886 – 1912

Proof. Let (S, S,
, ) be coregular and T1 and take s, t ∈ S with Qs
Qt . By Theorem 4.11((1)(ii)) we have F ∈  with Ps
F
Qt , whence by the definition of coregularity we have K ∈  with Ps
K and F ⊆]K[. Setting H =]K[ gives H ∈
, H ⊆ K and H
Qt , so (S, S,
, ) is bi-T2 by Theorem 4.17. The proof of the second result is dual, and is omitted.  Corollary 4.19. A topology T on the Hutton algebra (L, ) is T2 in the sense of [17] if and only if the corresponding complemented ditopological simple texture is T2 (equivalently, co-T1 , bi-T1 ). Proof. Clear from [17, Definition (R1 , T2 )] and Definition 4.15.



Notes 4.20. (1) Since the formation of the T1 , T2 axioms and their variants are consistent with the formation of the corresponding bitopological and topological axioms, the correspondence between these axioms is guaranteed by our earlier comments. (2) By our earlier results, the T1 and T2 axioms and their variants are preserved under dihomeomorphisms, and hence under fDitop isomorphisms. Now, let us consider sums and products of ditopological texture spaces in relation to the axioms considered in this section. In view of Theorems 3.13 and 3.19, it will be sufficient to consider the T0 axiom only. Theorem 4.21. Let (Sj , Sj ,
j , j ), j ∈ J , be non-empty disjoint ditopological texture spaces and (S, S,
, ) their disjoint sum. Then (S, S,
, ) is T0 if and only if (Sj , Sj ,
j , j ) is T0 for all j ∈ J . Proof. Let (
j , j ) be T0 for all j ∈ J and take s, t ∈ S with Qs
Qt . Let j, k ∈ J be the unique indices satisfying s ∈ Sj , t ∈ Sk . If j = k then Qkt = Sk , so B = Sk satisfies B ∈
∪  and Ps
B
Qt . On the other hand, if j = k then Qks
Qkt and we have Bk ∈
k ∪ k with Psk
Bk
Qkt . In this case, B = Bk satisfies B ∈
∪  and Ps
B
Qt . Hence (
, ) is T0 by Theorem 4.7(4). The converse is easily shown using the same characterization, and the details are omitted.  Corollary 4.22. Let (Sj , Sj ,
j , j ), j ∈ J , be non-empty disjoint ditopological texture spaces and (S, S,
, ) their disjoint sum. Then (1) (S, S,
, ) is T1 (co-T1 , bi-T1 ) if and only if (Sj , Sj ,
j , j ) is T1 (co-T1 , bi-T1 ) for all j ∈ J . (2) (S, S,
, ) is T2 (co-T2 , bi-T2 ) if and only if (Sj , Sj ,
j , j ) is T2 (co-T2 , bi-T2 ) for all j ∈ J . Proof. Immediate from Theorems 4.21 and 3.13.



Theorem 4.23. Let (Sj , Sj ,
j , j ), j ∈ J , be non-empty ditopological texture spaces and (S, S,
, ) their product. Then (S, S,
, ) is T0 if and only if (Sj , Sj ,
j , j ) is T0 for all j ∈ J . Proof. First, suppose that (
j , j ) is T0 for all j ∈ J and take s = (sj ), t = (tj ) ∈ S  with Qs
Qt . By Theorem 4.7(4), it will be sufficient to find B ∈
∪  satisfying Ps
B
Qt . However, since Qs = j ∈J E(j, Qsj )
Qt there exists j ∈ J with E(j, Qsj )
Qt and so, in particular, E(j, Qsj )
E(j, Qtj ), whence Qsj
Qtj and by Theorem 4.7(4) we have Bj ∈
j ∪ j satisfying Psj
Bj
Qtj . Clearly B = E(j, Bj ) now has the required properties, so (
, ) is T0 . Conversely, let (
, ) be T0 , take any j ∈ J and for sj , tj ∈ Sj suppose that Qsj
Qtj . For k ∈ J \ {j } choose


uk ∈ Sk , which is possible since Sk = ∅. Let s = (si ), t = (ti ) ∈ S be defined by  si =

sj if i = j, ui if i = j,

 ti =

tj if i = j, ui if i = j.

It is easy to verify that Qs
Qt , so by Theorem 4.7(4) there exists B ∈
∪  satisfying Ps
B
Qt . Suppose first that n B ∈
. Then by the definition n of product topology we have j1 , j2 , . . . , jn ∈ J and Bjk ∈
jk , 1
k
n, so that E(j , B ) ⊆ B and k j k k=1 k=1 E(jk , Bjk )
Qs . We now have k, 1
k
n, for which Ps
E(jk , Bjk )
Qt , so

L.M. Brown et al. / Fuzzy Sets and Systems 157 (2006) 1886 – 1912

1903

Psjk
Bjk
Qsjk . This leads to an immediate contradiction if jk = j since then sjk = tjk . Hence jk = j , and we have found Bj ∈
j satisfying Psj
Bj
Qtj . In case B∈, a dual proof gives Bj ∈j with Psj
Bj
Qtj , so by Theorem 4.7(4), (
j , j ) is T0 for all j ∈I .  Corollary 4.24. Let (Sj , Sj ,
j , j ), j ∈ J , be non-empty ditopological texture spaces and (S, S,
, ) their product. Then (1) (S, S,
, ) is T1 (co-T1 , bi-T1 ) if and only if (Sj , Sj ,
j , j ) is T1 (co-T1 , bi-T1 ) for all j ∈ J . (2) (S, S,
, ) is T2 (co-T2 , bi-T2 ) if and only if (Sj , Sj ,
j , j ) is T2 (co-T2 , bi-T2 ) for all j ∈ J . Proof. Immediate from Theorems 4.23 and 3.19.



Proposition 4.25. Denote by T any of the full concrete subcategories of dfDitop whose objects are T0 , (co-, bi-) T1 or (co-, bi-) T2 ditopological texture spaces. Then in each of the categories T the ditopological products and ditopological sums are concrete products and concrete coproducts, respectively. Proof. Clear by [11, Propositions 3.12, 3.26], Theorem 4.23, Corollary 4.24 and Theorem 4.21, Corollary 4.22.



5. Further regularity properties and normality Definition 5.1. A ditopological texture space is called T3 (co-T3 , bi-T3 ) if it is T0 and regular (respectively, coregular, biregular). It is immediate from the definitions that T3 ⇒ T2

and

co-T3 ⇒ co-T2 ,

so bi-T3 ⇒ bi-T3 . By Corollary 3.11, we have at once: Proposition 5.2. For a complemented ditopological texture space the notions of T3 , co-T3 and bi-T3 are equivalent. Definition 5.3. Let (
, ) be a ditopology on (S, S). Then (
, ) is called (1) Completely regular if given G ∈
, G
Qs , there exists a bicontinuous difunction (f, F ) : (S, S,
, ) → (I, I,
I , I ) satisfying Ps ⊆ f ← P0 and F ← Q1 ⊆ G. (2) Completely coregular if given K ∈
, Ps
K, there exists a bicontinuous difunction (f, F ) : (S, S,
, ) → (I, I,
I , I ) satisfying K ⊆ f ← P0 and F ← Q1 ⊆ Qs . (3) Completely biregular if it is completely regular and completely coregular.  (4) A T0 completely regular (completely coregular, completely biregular) space is called T3 1 2  bi-T3 1 .

respectively, co-T3 1 , 2

2

Notes 5.4. (1) It is clear from the definitions that: (a) If (
, ) is completely regular and ∗ is a cotopology satisfying  ⊆ ∗ , then (
, ∗ ) is also completely regular. (b) If (
, ) is completely coregular and
∗ is a topology satisfying
⊆
∗ , then (
∗ , ) is also completely coregular. (2) Suppose that the image (X, P(X), u, v c ) of (X, u, v) under T is completely regular. To show that (X, u, v) is completely regular take x ∈ U ∈ u. Then U
Qx so we have a bicontinuous difunction (w, W ) : (X, P(X), u, v c ) → (I, I,
I , I ) with Px ⊆ w ← P0 and W ← Q1 ⊆ U . Since (X, P(X)) is plain, by [10, Proposition 3.7, Lemmas 3.4 and 3.9], there exists a function : X → I with w← B = −1 [B] = W ← B for all B ∈ I. The bicontinuity of (w, W ) now easily implies the pairwise continuity of : (X, u, v) → (I,
I , cI ). Finally x ∈ Px ⊆ w ← P0 = −1 [{0}] gives

1904

L.M. Brown et al. / Fuzzy Sets and Systems 157 (2006) 1886 – 1912

(x) = 0, while y ∈ / U ⇒ y ∈ / W ← Q1 = −1 [[0, 1)] ⇒ (y) = 1. Since Kopperman takes the bitopological c interval to be (I, I ,
I ), setting f = 1 − shows that (X, u, v) is completely regular. Conversely, if (X, u, v) is completely regular then (X, P(X), u, v c ) is completely regular, and likewise (X, u, v) is completely regular∗ if and only if (X, P(X), u, v c ) is completely coregular. The details are left to the interested reader. (3) It is shown in [21] that a ditopological texture space is di-uniformizable if and only if it is completely biregular. Hence, a necessary and sufficient condition for a ditopological texture space (S, S,
, ) to be completely biregular is that there should exist a di-uniformity on (S, S) whose uniform ditopology coincides with (
, ). (4) Exactly as for the classical case the unit interval I in Definition 5.3 may be replaced by an arbitrary bounded interval Ia,b = [a, b], where a, b ∈ R and a < b. Specifically, we define a texturing Ia,b on Ia,b by Ia,b = {[a, r) | a
r
b} ∪ {[a, r] | a
r
b} and the complementation a,b : Ia,b → Ia,b by a,b ([a, r)) = [a, a + b − r], a,b ([a, r]) = [a, a + b − r), a
r
b. Clearly the mapping

a,b : I → Ia,b , s  → (b − a)s + a is a cfTex isomorphism from (I, I, ) to (Ia,b , Ia,b , a,b ), and hence the corresponding difunction (fa,b , Fa,b ) = (f a,b , F a,b ) is a cdfTex isomorphism from (I, I, ) to (Ia,b , Ia,b , a,b ). Finally, setting
a,b = {[a, r) | a
r
b} ∪ {Ia,b },

a,b = {[a, r] | a
r
b} ∪ {∅}

turns (Ia,b , Ia,b , a,b ,
a,b , a,b ) into a complemented ditopological texture space isomorphic to (I, I, ,
I , I ) in both cfDitop and cdfDitop. If (S, S,
, ) is completely regular and G
Qs for G ∈
, s ∈ S, there exists a bicontinuous difunction (u, U ) : (S, S,
, ) → (I, I,
I , I ) satisfying Ps ⊆ u← P0 and U ← Q1 ⊆ G. It follows easily that (v, V ) = (fa,b , Fa,b ) ◦ (u, U ) : (S, S,
, ) → (Ia,b , Ia,b ,
a,b , a,b ) is a bicontinuous difunction satisfying Ps ⊆ v ← Pa and V ← Qb ⊆ G. Conversely, the existence of such a (v, V ) gives the bicontinuous difunction (u, U ) = (fa,b , Fa,b )← ◦ (v, V ) : (S, S,
, ) → (I, I,
I , I ) satisfying Ps ⊆ u← P0 and U ← Q1 ⊆ G. A similar result holds for complete coregularity, and we have established the following: (i) (S, S,
, ) is completely regular iff given G ∈
, G
Qs , there exists a bicontinuous difunction (f, F ) : (S, S,
, ) → (Ia,b , Ia,b ,
a,b , a,b ) satisfying Ps ⊆ f ← Pa and F ← Qb ⊆ G. (ii) (S, S,
, ) is completely coregular iff given K ∈
, Ps
K, there exists a bicontinuous difunction (f, F ) : (S, S,
, ) → (Ia,b , Ia,b ,
a,b , a,b ) satisfying K ⊆ f ← Pa and F ← Qb ⊆ Qs . Example 5.5. The usual direlational uniformity [21, Example 3.3] on (I, I) is easily shown to be compatible with the ditopology (
I , I ), so (I, I,
I , I ) is completely biregular by Notes 5.4(2). A direct proof of complete regularity would proceed as follows. Take G = [0, t) ∈
I and s ∈ I with G
Qs . Then s < t and we may consider the difunction (us,t , Us,t ) on (I, I) corresponding to the point function s,t : I → I given by ⎧ if r < s, ⎨0

s,t (r) = (r − s)/(t − s) if s
r
t, ⎩ 1 if r > t. ← ← Clearly u← s,t [0, b] = [0, b(t − s) + s], Us,t [0, b) = [0, b(t − s) + s) so (us,t , Us,t ) is bicontinuous, and also us,t P0 = ← [0, s] = Ps , Us,t Q1 = [0, t) = G, so (us,t , Us,t ) has the required properties. Finally, (us,1 , Us,1 ) has the required properties when G = I, so (
I , I ) is completely regular. A direct proof of complete coregularity is similar, and is omitted.

Proposition 5.6. A completely regular space is regular, and a completely coregular space coregular.

L.M. Brown et al. / Fuzzy Sets and Systems 157 (2006) 1886 – 1912

1905

Proof. Let (S, S,
, ) be completely regular, and take G ∈
and s ∈ S with G
Qs . Let (f, F ) : (S, S,
, ) → (I, I,
I , I ) be a bicontinuous difunction satisfying Ps ⊆ f ← P0 and F ← Q1 ⊆ G. Let H = F ← Q1/2 . Then H ∈
since Q1/2 ∈
I and (f, F ) is continuous. Clearly Ps ⊆ f ← P0 ⊆ f ← Q1/2 ⊆ f ← P1/2 ⊆ f ← Q1 ⊆ G since f ← Q1 = F ← Q1 . Finally, f ← P1/2 ∈  since (f, F ) is cocontinuous and P1/2 ∈ I , while f ← Q1/2 = F ← Q 1 = H , so we obtain Ps ⊆ H ⊆ [H ] ⊆ G. This shows that (
, ) is regular by Lemma 3.10(1). The proof of the second result is dual to the above, and is omitted. 

2

It is now immediate that: Corollary 5.7. T

1 32

⇒ T3 and co-T

1 32

⇒ co-T3 .

Proposition 5.8. For a complemented ditopology (
, ) on (S, S, ), the notions of complete regularity, complete coregularity and complete biregularity coincide. Proof. Suppose that the complemented ditopology (
, ) is completely regular and take K ∈ , s ∈ S, with Ps
K. Then (K)
(Ps ) and we may choose t ∈ S with (K)
Qt and Pt
(Ps ). Since (
, ) is complemented we have (K) ∈
, whence there exists a bicontinuous difunction (f, F ) : (S, S, ,
, ) → (I, I, ,
I , I ) satisfying Pt ⊆ f ← P0 and F ← Q1 ⊆ (K). Now, we have (f ← P0 ) ⊆ (Pt ), while Pt
(Ps ) gives Ps
(Pt ), so Ps
(f ← P0 ) and therefore (f ← P0 ) ⊆ Qs . On the other hand K ⊆ (F ← Q1 ), and it remains only to note that by [10, Lemma 2.20(2)] we have (f ← P0 ) = (f ← (Q1 )) = (f  )← Q1 , (F ← Q1 ) = (F ← (P0 )) = (F  )← P0 to see that the difunction (f, F ) = (F  , f  ) : (S, S, ,
, ) → (I, I, ,
I , I ) satisfies (f  )← Q1 ⊆ Qs and K ⊆ (F  )← P0 . Finally, again by [10, Lemma 2.20(2)] it is trivial to show that (f, F ) is bicontinuous, so (
, ) is completely coregular. The proof of the opposite implication is dual to the above, and is omitted.  Theorem 5.9. Let (Sj , Sj ,
j , j ), j ∈ J , be ditopological texture spaces, (S, S,
, ) their disjoint sum. Then (S, S,
, ) is completely regular (completely coregular) if and only if (Sj , Sj ,
j , j ) is completely regular (respectively, completely coregular) for each j ∈ J . Proof. There will be no loss of generality in assuming the spaces (Sj , Sj ,
j , j ) are non-empty. We give the proof for complete regularity, the complete coregularity case being dual. Necessity: Suppose  (S, S,
, ) is completely regular and take k ∈ J , Gk ∈
k and sk ∈ Sk with Gk
Qsk . Let G = (Gk × {k}) ∪ j ∈J \{k} (Sj × {j }). Then G ∈
and for s = (sk , k) we have G
Qs by [10, Proposition 1.5(2)], so there exists a bicontinuous difunction (u, U ) : (S, S,
, ) → (I, I,
I , I ) satisfying Ps ⊆ u← P0 and U ← Q1 ⊆ G. Denoting by (ek , Ek ) the bicontinuous embedding of (Sk , Sk ,
k , k ) in (S, S,
, ) (see [11]), we obtain the bicontinuous difunction (u, U ) ◦ (ek , Ek ) : (Sk , Sk ,
k , k ) → (I, I,
I , I ). Clearly Psk = ek← Ps ⊆ ek← u← P0 = (u ◦ ek )← P0 and (U ◦ Ek )← Q1 = Ek← U ← Q1 ⊆ Ek← G = Gk so (
k , k ) is completely regular.  Sufficiency: Let (
j , j ) be completely regular for each j ∈ J and take G ∈
, s ∈ S with G
Qs . Then G = j ∈J (Gj × {j }), Gj ∈
j , and s = (sk , k) for some k ∈ J . Let J = {j ∈ J | Gj = ∅}. Clearly, Gk
Qsk so k ∈ J  and for j ∈ J  \ {k} we choose sj ∈ Sj with Gj
Qsj . By hypothesis, we have a bicontinuous difunction ← (uj , Uj ) : (Sj , Sj ,
j , j ) → (I, I,
I , I ) for all j ∈ J  satisfying Psj ⊆ u← j P0 and Uj Q1 ⊆ Gj . Consider the function  : I → S defined by  ← / B, j ∈J  (uj B) × {j } if 1 ∈ (B) = S otherwise,

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L.M. Brown et al. / Fuzzy Sets and Systems 157 (2006) 1886 – 1912

for B ∈ I. We claim that  preserves arbitrary joins and arbitrary intersections. Take A. Note that  B ∈ I,  ∈  1 ∈ ∈A B ⇐⇒ ∃  ∈ A with 1 ∈ B since (I, I) is plain, and that in that case ( ∈A B ) = S = ∈A (B ). Hence suppose that / B A. Then  for all  ∈    1∈        ← ←  uj B = B × {j } = uj B × {j } ∈A

j ∈J 

=

 

∈A

j ∈J 

(u← j B ) × {j }

=

∈A j ∈J 



∈A

(B )

∈A



by [10, Corollary 2.26], so  preserves joins. Now, let us verify that ( ∈A B ) = ∈A (B ). Clearly, we may assume without loss of generality that 1 ∈ / B for  all  ∈ A. Hence       ← ← (B ) = (uj B ) × {j } = uj B × {j } ∈A

∈A j ∈J 

=





u← j

j ∈J 



j ∈J 





B × {j } = 

∈A

∈A



 B

∈A

by [10, Corollary 2.26], as required. Now, let (u, U ) : (S, S) → (I, I) be the difunction (f  , F  ) defined in [11, Proposition 4.1]. Then for B ∈ I we have u← B = U ← B = (B), so the bicontinuity of the difunctions (uj , Uj ), j ∈ J  , immediately implies that of (u, U ). Finally,  ← (u← Ps = Psk × {k} ⊆ (u← k P0 ) × {k} ⊆ j P0 ) × {j } ⊆ u P0 j ∈J 

since 1 ∈ / P0 , and U ← Q1 =



(Uj← Q1 ) × {j } ⊆

j ∈J 



Gj × {j } = G

j ∈J 

since 1 ∈ / Q1 , so (
, ) is completely regular.



Before considering the preservation of complete regularity and complete coregularity under initial structures we prove the following useful result. Lemma 5.10. Let (S, S) be a texture, (T , T ) a texture for which (T , ⊆) is a chain and (gk , Gk ) : (S, S)
→ (T , T ) difunctions k = 1, 2, . . . , n. Then there exists a difunction (g, G) : (S, S) → (T , T ) for which g ← B = nk=1 gk← B,
for n ← G B = k=1 G← k B for all B ∈ T .
Proof. Define the mapping  : T → S by (B) = nk=1 gk← B for B ∈ T . We establish that  preserves arbitrary intersections and arbitrary joins. By the preservation of intersections by inverse images [10, Corollary 2.26], it is immediate that  also preserves  intersections. On the other hand, inverse images preserve inclusion, whence it is clear that j ∈J (Bi ) ⊆ ( j ∈J Bj ) for Bj ∈ T , j ∈ J . It remains, therefore, to prove the reverse inclusion. Now, ⎞ ⎛ n n  n       Bj ⎠ = gk← Bj = gk← Bj = gk← B(k) ⎝ j ∈J

k=1

j ∈J

k=1 j ∈J

∈J n k=1

by the preservation of joins by inverse images [10, Corollary 2.26] and the fact that (S, ⊆) is completely distributive. Take  ∈ J n . Then, since (T , ⊆) is a chain, the finite set {B(k) | k = 1, 2, . . . , n} contains a set Bj with B(k) ⊆ Bj for all k, 1
k
n. Hence n  k=1

gk← B(k) ⊆

n  k=1

gk← Bj = (Bj ),

  
 so ( j ∈J Bj ) = ∈J n nk=1 gk← B(k) ⊆ ∈J n (Bj ) ⊆ j ∈J (Bj ), as required.

L.M. Brown et al. / Fuzzy Sets and Systems 157 (2006) 1886 – 1912

1907

The proof is now completed by noting that the difunction (g, G) = (f  , F  ) : (S, S) → (T , T ) defined in [11, Proposition 4.1] has the property g ← B = (B) for all B ∈ T , the remaining equality being an immediate consequence of the equality of the inverse image and inverse co-image for difunctions.  Theorem 5.11. Let (Sj , Sj ,
j , j ), j ∈ J , be ditopological texture spaces, (S, S) a texture and (fj , Fj ) : (S, S) → (Sj , Sj ,
j , j ) difunctions. Then if the ditopologies (
j , j ) are completely regular (completely coregular) for all j ∈ J , the initial ditopology (
, ) on (S, S) generated by the given spaces and difunctions is completely regular (respectively, completely coregular). Proof. We prove complete regularity, leaving the dual proof of complete coregularity to the interested reader.
Take H ∈
and s ∈ S with H
Qs . Now we have Hk ∈ Sk , k = 1, 2, . . . , n, with nk=1 Fk← Hj ⊆ H and
n ← ← k=1 Fk Hj
Qs , so Fk Hj
Qs for k = 1, 2, . . . , n and we have sk ∈ Sk with P (s,sk )
Fk and Hk
Qsk . Since (
k , k ) is completely regular we have a bicontinuous difunction (uk , Uk ) : (Sk , Sk ,
k , k ) → (I, I,
I , I ) satisfying ← Psk ⊆ u← k P0 and Uk Q1 ⊆ Hk . Let (gk , Gk ) = (uk , Uk ) ◦ (fk , Fk ). Then for each k, (gk , Gk ) : (S, S)
→ (I, I), and (I, ⊆) is a
chain, so by Lemma 5.10 there exists a difunction (g, G) : (S, S) → (I, I) with g ← B = nk=1 gk← B and G← B = nk=1 G← k B for all B ∈ I. Clearly, the bicontinuity of (fk , Fk ) and (uk , Uk ) for each k imply that of (gk , Gk ), and hence (g, G) : (S, S,
, ) → (I, I,
I , I ) is bicontinuous. It remains to show that Ps ⊆ g ← P0 and G← Q1 ⊆ H . Suppose that the first inclusion is false. Then for some k, 1
k
n, we have Ps
gk← P0 = fk← u← k P0 by [10, P . But by the condition DF2 for (fk , Fk ) Lemma 2.16(2)]. Hence there exists sk ∈ Sk with fk
Q(s,sk ) and Psk
u← 0 k ← we obtain Psk
Qsk , whence Psk ⊆ Psk and we have the contradiction Psk
uk P0 .

n
n ← ← ← Finally, G← Q1 = nk=1 G← k=1 Fk Uk Q1 ⊆ k=1 Fk Hk ⊆ H , which completes the proof that k Q1 = (S, S,
, ) is completely regular.  In categorical terms the above result may be expressed as follows. Corollary 5.12. Denote by CR any of the full concrete subcategories of dfDitop whose objects are completely regular, completely coregular or completely biregular ditopological texture spaces, and let E : CR → dfDitop be the inclusion functor. Then (CR, U ◦ E) is initially closed in (dfDitop, U). Theorem 5.13. With the notation as in Corollary 5.12, each of the categories CR is topological. Proof. By [11, Theorem 3.6] the concrete category (dfDitop, U) is topological, so the result follows from Corollary 5.12 and [1, Proposition 21.30].  In view of the topological duality theorem [1, 21.9], the existence of unique U ◦ E-initial lifts of U ◦ E-structured sources implies the existence of unique U ◦ E-final lifts of U ◦ E-structured sinks. Theorem 5.14. With the notation as in Corollary 5.12, each of (CR, U ◦ E) is concretely reflective in (dfDitop, U). Proof. Immediate from Corollary 5.12 and [1, Proposition 21.31].



More particularly, [1, Theorem 21.35(4)] guarantees for each CR the existence of a concrete functor from (dfDitop, U) to (CR, U ◦ E) that leaves each CR-object fixed. We identify this functor in case CR is the category of completely regular ditopological texture spaces, leaving a discussion of the remaining cases to the interested reader. Proposition 5.15. Let CR be the full subcategory of dfDitop whose objects are completely regular spaces, take (S, S,
, ) ∈ ObdfDitop and let (
r , r ) be the initial ditopology on (S, S) generated by the bicontinuous difunctions into (I, I,
I , I ). Then (i, I ) : (S, S,
, ) → (S, S,
r , ) is a universal arrow which defines a concrete functor from (dfDitop, U) to (CR, U ◦ E) leaving the objects of CR fixed. Proof. By Example 5.5 the unit interval (I, I,
I , I ) is completely biregular, so by Theorem 5.11 the ditopology (
r , r ) is also completely biregular. Since clearly r ⊆ , we see by Notes 5.4((1)(a)) that (
r , ) is completely regular. Moreover
r ⊆
, so the identity difunction (i, I ) : (S, S,
, ) → (S, S,
r , ) is bicontinuous.

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L.M. Brown et al. / Fuzzy Sets and Systems 157 (2006) 1886 – 1912

To establish the universal property, let (T , T , , ) be an arbitrary completely regular ditopological space and (h, H ) : (S, S,
, ) → (T , T , , ) bicontinuous. We must show the existence of a unique morphism (g, G) making the following diagram commutative. (i,I )

/ (S, S,
r , ) (S, S,
, )U  UUUU UUUU  UUUU  (g,G) (h,H ) UUUUU  UU*  (T , T , , ) It is clear that the only difunction satisfying the equality (g, G) ◦ (i, I ) = (h, H ) is (g, G) = (h, H ), so it will be sufficient to show that (h, H ) is (
r , )–(, ) bicontinuous. However, cocontinuity is given, so we only need to establish continuity. Take G ∈  and s ∈ S with H ← G
Qs , and choose s  ∈ S with H ← G
Qs  and Ps 
Qs . Now, we have t ∈ T with P (s  ,t)
H , G
Qt , whence we may take t  ∈ T satisfying G
Qt  and Pt 
Qt . Since (, ) is completely regular we have a bicontinuous difunction (u, U ) : (T , T , , ) → (I, I,
I , I ) with Pt  ⊆ u← P0 and U ← Q1 ⊆ G. Noting that (u, U ) ◦ (h, H ) : (S, S,
, ) → (I, I,
I , I ) is a continuous difunction and Q1 ∈
I , we have (U ◦ H )← Q1 ∈
r by the definition of
r . Hence to establish that H ← G ∈
r it will be sufficient to show that (U ◦ H )← Q1 ⊆ H ← G

and (U ◦ H )← Q1
Qs .

Clearly, (U ◦ H )← Q1 = H ← U ← Q1 ⊆ H ← G since U ← Q1 ⊆ G. The second result will follow if we can prove that Ps  ⊆ (U ◦ H )← Q1 , so we assume the contrary. This now gives us s  ∈ S with Ps 
Qs  for which P (s  ,w)
H ⇒ U ← Q1 ⊆ Qw for all w ∈ T . Since P (s  ,t)
H and Ps 
Qs  , we obtain P (s  ,t)
H by CR1, whence U ← Q1 ⊆ Qt by the above implication with w = t. But now Pt  ⊆ u← P0 = U ← P0 ⊆ U ← Q1 ⊆ Qt , which contradicts Pt 
Qt and completes the proof that (i, I ) : (S, S,
, ) → (S, S,
r , ) is a universal arrow. Since this universal arrow is identity-carried it gives rise to a concrete functor. Moreover, (
, ) is clearly completely regular if and only if
r =
, so the objects of CR remain fixed.  Theorem 5.16. Let (Sj , Sj ,
j , j ) be completely regular (completely coregular) ditopological texture spaces for all j ∈ J , and (S, S,
, ) their product. Then (S, S,
, ) is completely regular (respectively, completely coregular). Proof. Immediate from Theorem 5.11 since the product ditopology is the initial ditopology under the projection difunctions.  The authors do not know if the converse of this theorem is true for non-empty ditopological spaces. Proposition 5.17. In each of the full concrete subcategories of dfDitop whose objects   are T3 (co-T3 , bi-T3 ), completely regular (completely coregular, completely biregular) or T 1 co-T 1 , bi-T 1 ditopological texture spaces, 32

32

32

the ditopological products and ditopological sums are concrete products and concrete coproducts, respectively. Proof. In view of [11, Propositions 3.12, 3.26], for the T3 axioms the results follow from Theorems 3.13, 3.19 and 4.21, 4.23, while for the complete regularity (T 1 ) axioms the results follow from Theorems 5.9, 5.16 (plus, Theorems 4.21, 4.23). 

32

We end with a discussion of normality for ditopological texture spaces. The reader may refer to [7] for some higher separation properties (dicover normality, full normality, etc.). Definition 5.18. A ditopology (
, ) is called normal if given F ∈ , G ∈
with F ⊆ G we have H ∈
, K ∈  with F ⊆ H ⊆ K ⊆ G. A T1 normal space is called T4 and a co-T1 normal space is called co-T4 .

L.M. Brown et al. / Fuzzy Sets and Systems 157 (2006) 1886 – 1912

1909

Notes 5.19. (1) Normality is self-dual, so there is no separate conormality axiom. Clearly, equivalent formulations of the normality condition for (
, ) are that given F ∈ , G ∈
with F ⊆ G there should exist H ∈
with F ⊆ H ⊆ [H ] ⊆ G, or that there should exist K ∈  with F ⊆ ]K[ ⊆ K ⊆ G. (2) Clearly, the bitopological space (X, u, v) is normal if and only if the ditopological texture space (X, P(X), u, v c ) is normal, and the topological space (X, T ) is normal if and only if (X, P(X), T , T c ) is normal. Likewise, the T4 axioms correspond. (3) It is clear from the definitions that a topology T on the Hutton algebra (L, ) is normal (T4 ) in the sense of [17] if and only if the corresponding complemented ditopological simple texture is normal (respectively, T4 ; equivalently co-T4 , bi-T4 ). Example 5.20. (1) Consider the ditopological texture space (S, S,
, ). If
⊆  or  ⊆
then (
, ) is normal. In particular, (
, S) and (S, ) are always normal. (2) The unit interval (I, I,
I , I ) is clearly normal. Lemma 5.21. A R0 normal space is regular and a co-R0 normal space is coregular. Proof. We prove the first result, the second being dual. Take G ∈
and s ∈ S with G
Qs . Then [Ps ] ⊆ G, and applying normality with F = [Ps ] ∈  gives H ∈
with Ps ⊆ [Ps ] ⊆ H ⊆ [H ] ⊆ G. Hence (
, ) is regular by Lemma 3.9((1)(ii)).  This immediately gives: Corollary 5.22. T4 ⇒ T3 and co-T4 ⇒ co-T3 . We now present a counterpart of Uryshon’s Lemma for ditopological texture spaces which will enable us to strengthen this last result. Theorem 5.23. The ditopological texture space (S, S,
, ) is normal if and only if given F ∈ , G ∈
with F ⊆ G there exists a bicontinuous difunction (w, W ) : (S, S,
, ) → (I, I,
I , I ) satisfying F ⊆ w ← P0 and W ← Q1 ⊆ G. Proof (Sufficiency). If (w, W ) satisfies the above conditions then H = W ← Q1/2 ∈
, K = w← P1/2 ∈  trivially satisfy F ⊆ H ⊆ K ⊆ G. Necessity: Suppose that (S, S,
, ) is normal and take F ∈ , G ∈
with F ⊆ G. Just as in the proof of the classical Uryshon’s Lemma we may inductively choose for each binary number  = k/2n , k = 2, 3, . . . , 2n − 1, n ∈ N+ , a set H () ∈
so that F ⊆ H () ⊆ [H ( )] ⊆ G for all binary numbers ,  with 0 <  <  < 1. Now, we define a function : S → I by setting (s) = inf{ | H ()
Qs }, where the infimum of an empty set of binary numbers is taken to be 1. If Ps
Qs  then Ps  ⊆ Ps , whence Qs  ⊆ Qs and we deduce that (s  )
(s). Hence P (s)
Q (s  ) , and we have shown that satisfies condition (a) of [10, Lemma 3.4]. We deduce from this lemma that the equalities  w = {P (s,t) | ∃ v ∈ S with Ps
Qv , P (v)
Qt }  = {P (s, (v)) | v ∈ S, Ps
Qv }, (1) W= =





{Q(s,t) | ∃ v ∈ S with Pv
Qs , Pt
Q (v) } {Q(s, (v)) | v ∈ S, Pv
Qs }

(2)

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L.M. Brown et al. / Fuzzy Sets and Systems 157 (2006) 1886 – 1912

define a difunction (w, W ) from (S, S) to (I, I). For r ∈ I we now have:
{[H ()] | r < }, w ← Pr =  W ← Qr = {H () |  < r}.

(3) (4)


We establish (3), leaving the proof of the dual result (4) to the interested reader. Suppose first that w ← Pr
{[H ()] | r < } and take a binary number  with r <  and w← Pr
[H ()]. Now, we have s ∈ S with w ← Pr
Qs and Ps
[H ()]. From w ← Pr
Qs , we have s  ∈ S with Ps 
Qs for which w
Q(s  ,t) ⇒ Pt ⊆ Pr for all t ∈ I, whence we obtain (s)
r since w
Q(s  , (s)) by (1). Hence (s) <  and we may choose  with  <  and H ( )
Qs by the definition of . However this gives Ps ⊆ H ( ) ⊆ [H

()], which is a contradiction. Now, suppose that {[H ()] | r < }
w ← Pr . Then {[H ()] | r < }
Qs and Ps
w ← Pr for some s ∈ S. This gives t ∈ I with w
Q(s,t) and Pt
Pr , whence r < t. Now, from (1) we have v ∈ S with Ps
Qv and P (s, (v))
Q(s,t) , from which we obtain t
(v). Also, (v)
(s) by condition (a). Thus r < (s) and we may choose binary numbers ,  satisfying r <  <  < (s). Finally, r <  gives [H ( )]
Qs , and we deduce from [H ( )] ⊆ H () that H ()
Qs , whence the definition of gives the contradiction (s)
. This completes the proof of (3). Since F ⊆ H () ⊆ [H ()] for all , equality (3) implies F ⊆ w← P0 . Also, the elements of I \ {∅} all have the form Pr , r ∈ I, and we see from (3) that w ← Pr ∈ . Hence (w, W ) is cocontinuous. Likewise (4) gives W ← Q1 ⊆ G since H () ⊆ G for all , while all the elements of
I \ {I} have the form Qr , r ∈ I, and we see from (4) that W ← Qr ∈
. Hence (w, W ) is also continuous, and the proof of the theorem is complete.  Corollary 5.24. A R0 normal space is completely regular and a co-R0 normal space completely coregular. Proof. Let (S, S,
, ) be R0 and normal, and take G ∈
, s ∈ S with G
Qs . Then [Ps ] ⊆ G, and applying Theorem 5.23 with F = [Ps ] gives a bicontinuous difunction (w, W ) : (S, S,
, ) → (I, I,
I , I ) with Ps ⊆ w← P0 and W ← Q1 ⊆ G. Hence (S, S,
, ) is completely regular. The proof of the second result is dual and is omitted.  Corollary 5.25. T4 ⇒ T3 1 ⇒ T3 and co-T4 ⇒ co-T3 1 ⇒ co-T3 . 2

2

Proof. A T4 space is T1 and normal, hence T0 , R0 and normal, so by Corollary 5.24 it is T0 and completely regular, i.e., T3 1 . Thus T4 ⇒ T3 1 , and T3 1 ⇒ T3 follows from Corollary 5.5. The remaining implications may be proved 2 2 2 likewise.  Example 5.26. By Examples 3.3 and 5.20, a codiscrete space is R0 and normal, hence completely regular by Corollary 5.24. Likewise, a discrete space is completely coregular.  Proposition 5.27. The T3 (co-T3 , bi-T3 ), completely regular (completely coregular, completely biregular), T3 1 co2  T3 1 , bi-T3 1 , normal and T4 (co-T4 , bi-T4 ) properties are preserved under dihomeomorphisms and hence under fDitop 2

2

isomorphisms. Proof. We sketch the proof for complete regularity, complete coregularity being dual and the remaining cases straightforward. Let (S1 , S1 ,
1 , 1 ) be completely regular and (f, F ) : (S1 , S1 ,
1 , 1 ) a dihomeomorphism. For G ∈
2 , s ∈ S2 with G
Qs we have F ← G ∈
1 and F ← G
F ← Qs since (f, F ) is bijective. Take s  ∈ S1 with F ← G
Qs  , Ps 
F ← Qs . Now, we have a bicontinuous difunction (w, W ) : (S1 , S1 ,
1 , 1 ) → (I, I,
I , I ) with Ps  ⊆ w← P0 and W ← Q1 ⊆ F ← G. It is not difficult to verify that the bicontinuous difunction (u, U ) = (w, W ) ◦ (f, F )← : (S2 , S2 ,
2 , 2 ) → (I, I,
I , I ) satisfies Ps ⊆ u← P0 and U ← Q1 ⊆ G, whence (S2 , S2 ,
2 , 2 ) is completely regular.  6. Conclusion The main difference between the textural and molecular lattice approach to the study of topologies on lattices is that with a texture the points are built into the structure from the beginning, whereas with molecular lattices they are added

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1911

later. This makes the textural approach somewhat closer to the classical philosophy of building a mathematical structure on a base set, but the difference is not merely philosophical. In the case where there is a point in S for each molecule in S, that is in the case where (S, S) is simple, the two approaches are certainly equivalent, but, in general, a texture can be built on a smaller set of points. This gives us the opportunity of constructing highly economic structures to represent certain traditional topological spaces within the theory. One striking example is the ditopological unit interval texture space (I, I, ,
I , I ) that has been shown in this paper to play an important role in the theory of ditopological texture spaces. This is certainly an intuitively satisfying, highly economic representation of the classical unit interval in general topology. It has the very desirable property of being plain, and therefore coseparated. As we have seen this space satisfies all the ditopological separation properties defined in this paper, and it is also known to be dicompact. Naturally, these separation properties are also satisfied by the Hutton texture (MI , MI , I ,
I , I ) of (I, ) since these spaces are isomorphic in dfDitop, but the extra points in the latter space not only make it less economic and less intuitive, they also lead to the loss of the plainness property and, perhaps more importantly, the property of being coseparated. In view of the fact that textures may be built on a set smaller than the set of molecules, one very important fact which emerges from this study is that the points are nonetheless sufficient to give pointed representations of the latticetheoretic or “point free” separation axioms of Hutton and Reilly. Naturally, these representations may be translated into the language of topological molecular lattices or of L-topological spaces. We do not go into details, preferring to leave the interested reader to carry out such translations as the need arises. However, the authors certainly hope that the material presented here will be of interest to those working outside the context of ditopological texture spaces, as well as to those wishing to continue the development of this theory. To conclude we will, however, mention one link with L-topologies which is particularly relevant. Let us recall from [11, Theorem 5.9] the functor F = F(L,,
0 ,0 ) , which we may regard as mapping from Top to dfDitop. Here, (L, ) is a fixed Hutton algebra, (
0 , 0 ) a fixed complemented ditopology on the Hutton texture (ML , ML , L ) of (L, ), and F maps a topological space (X, T ) to the ditopological product of (X, P(X), X , T , T c ) and (ML , ML , L ,
0 , 0 ). As we have shown, if (X, T ) satisfies one of the separation axioms considered here, (X, P(X), X , T , T c ) satisfies the ditopological form of the same axiom. Hence, if (
0 , 0 ) satisfies one of the ditopological separation axioms other than normality and its derivatives, we have seen that the product will also satisfy this axiom. In particular, this will hold if (
0 , 0 ) is discrete (and hence, codiscrete). Since, in particular, for L = I (in which case (ML , ML , L ) = (L, L, )) we know from [11] that the classical Lowen functor corresponds to F(L,,
0 ,0 ) for (
0 , 0 ) discrete and codiscrete, so this functor also preserves the I-topological counterparts of such axioms. References [1] J. Adámek, H. Herrlich, G.E. Strecker, Abstract and Concrete Categories, Wiley, New York, 1990. [2] D. Adnadjevic, A.P. Sostak, On inductive dimensions for fuzzy topological spaces, Fuzzy Topology, Fuzzy Sets and Systems 73 (1) (1995) 5–12. [3] L.M. Brown, Dual covering theory, confluence structures and the lattice of bicontinuous functions, Ph.D. Thesis, Glasgow University, 1981. [4] L.M. Brown, On stable local compactness, Proc. Third National Mathematics Symp., Van, Turkey, 1990, Centenial University Publications, vol. 6, 1991, pp. 39–43. [5] L.M. Brown, Quotients of textures and of ditopological texture spaces, Topology Proc. 29 (2) (2005) 337–368. [6] L.M. Brown, M. Diker, Ditopological texture spaces and intuitionistic sets, Fuzzy Sets and Systems 98 (1998) 217–224. [7] L.M. Brown, M. Diker, Paracompactness and full normality in ditopological texture spaces, J. Math. Anal. Appl. 227 (1998) 144–165. [8] L.M. Brown, R. Ertürk, Fuzzy sets as texture spaces, I. Representation theorems, Fuzzy Sets and Systems 110 (2) (2000) 227–236. [9] L.M. Brown, R. Ertürk, Fuzzy sets as texture spaces, II. Subtextures and quotient textures, Fuzzy Sets and Systems 110 (2) (2000) 237–245. [10] L.M. Brown, R. Ertürk, S. ¸ Dost, Ditopological texture spaces and fuzzy topology, I. Basic Concepts, Fuzzy Sets and Systems 147 (2) (2004) 171–199. [11] L.M. Brown, R. Ertürk, S. ¸ Dost, Ditopological texture spaces and fuzzy topology, II. Topological Considerations, Fuzzy Sets and Systems 147 (2) (2004) 201–231. [12] M. Diker, One-point compactifications of ditopological texture spaces, Fuzzy Sets and Systems 147 (2) (2004) 233–248. [13] R. Ertürk, Fuzzy topology and bitopological spaces, Ph.D. Thesis, Hacettepe University, 1992 (in Turkish). [14] R. Ertürk, Separation axioms in fuzzy topology characterized by bitopologies, Fuzzy Sets and Systems 58 (1993) 206–209. [15] P. Fletcher, H.B. Hoyle, C.W. Patty, The comparison of topologies, Duke Math. J. 36 (1969) 325–331. [16] G. Gierz, K.H. Hofmann, K. Keimel, J.D. Lawson, M. Mislove, D.S. Scott, A Compendium of Continuous Lattices, Springer, Berlin, 1980. [17] B. Hutton, I. Reilly, Separation axioms in fuzzy topological spaces, Fuzzy Sets and Systems 3 (1980) 93–104. [18] J.C. Kelly, Bitopological spaces, Proc. London Math. Soc. 13 (1963) 71–89.

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