@
Applied General Topology

c© Universidad Politécnica de Valencia

Volume 4, No. 1, 2003

pp. 157–192

Di-uniform texture spaces

Selma Özçağ and Lawrence M. Brown
∗

Abstract. Textures were introduced by the second author as a
point-based setting for the study of fuzzy sets, and have since proved
to be an appropriate framework for the development of complement-free
mathematical concepts. In this paper the authors lay the foundation
for a theory of uniformities in a textural context. Analogues are given
for both the diagonal and covering approaches to the classical theory of
uniform structures, the notion of uniform topology is generalized and
an analogue given for the well known result that a topological space is
uniformizable if and only if it is completely regular. Finally a textural
analogue of the classical interplay between uniformities and families of
pseudo-metrics is presented.

2000 AMS Classification: Primary: 54E15, 54A05. Secondary: 06D10,
03E20, 54A40, 54D10, 54D15, 54E35.

Keywords: Uniformity, texturing, direlation, dicover, difunction, di-uniform
texture space, uniform ditopology, uniform bicontinuity, initial di-uniformity,
separation, dimetric, complement-free, point-free, fuzzy set.

1. Introduction

Textures were introduced by the second author as a point-based setting for
the study of fuzzy sets, and have since proved to be an appropriate setting for
the development of complement-free mathematical concepts. In this paper the
authors lay the foundation for a theory of uniformities imposed on textures.
Analogues are given for both the diagonal and covering approaches to the clas-
sical theory of uniform structures, the notion of uniform topology is generalized
and an analogue given for the well known result that a topological space is uni-
formizable if and only if it is completely regular. Finally the notion of pseudo
dimetric is given and a pseudo metrization theorem for di-uniformities and for
ditopologies is presented.

∗Dedicated to the memory of Professor Doğan Çoker.



158 S. Özçağ and L. M. Brown

Let S be a non-empty set. We recall [2] that a texturing on S is a point
separating, complete, completely distributive lattice S of subsets of S with
respect to inclusion, which contains S, ∅, and for which arbitrary meet

∧
coincides with intersection

⋂
and finite joins ∨ coincide with unions ∪. The

pair (S, S) is then called a texture.
In general a texturing of S need not be closed under set complementation.

The sets

Ps =
⋂
{A ∈ S | s ∈ A}, Qs =

∨
{Pu | u ∈ S, s /∈ Pu}, s ∈ S,

are important in the study of textures, and the following facts concerning these
so called p–sets and q–sets will be used extensively below.

Lemma 1.1. [1]
(1) s /∈ A =⇒ A ⊆ Qs =⇒ s /∈ A[ for all s ∈ S, A ∈ S.
(2) A[ = {s | A 6⊆ Qs} for all A ∈ S.
(3) For Ai ∈ S, i ∈ I we have (

∨
i∈I Ai)

[ =
⋃
i∈I A

[
i.

(4) A is the smallest element of S containing A[ for all A ∈ S.
(5) For A, B ∈ S, if A 6⊆ B then there exists s ∈ S with A 6⊆ Qs and

Ps 6⊆ B.
(6) A =

⋂
{Qs | Ps 6⊆ A} for all A ∈ S.

(7) A =
∨
{Ps | A 6⊆ Qs} for all A ∈ S.

Here A[ is defined by

A[ =
⋂{⋃

{Ai | i ∈ I} | {Ai | i ∈ I}⊆ S, A =
∨
{Ai | i ∈ I}

}
and known as the core of A ∈ S. The above lemma exposes an important formal
duality in (S, S), namely that between

⋂
and

∨
, Qs and Ps, and Ps 6⊆ A and

A 6⊆ Qs. Indeed, it is to emphasize this duality that we normally write Ps 6⊆ A
in preference to s /∈ A.

Lemma 1.1 (5) is particularly useful in establishing inclusion by reductio ad
absurdum, and will be used without comment in the sequel.

The simplest example of a texture is (X,P(X)), for which Px = {x} and
Qx = X \ {x}, x ∈ X. A natural texturing of the unit interval I = [0, 1] is
defined by

I = {[0,r) | r ∈ I}∪{[0,r] | r ∈ I}.
For the texture (I, I) we have Pr = [0,r] and Qr = [0,r), r ∈ I. This texture
will prove useful in the later sections. Both (X,P(X)) and (I, I) have the
property that join coincides with union (equivalently, that Ps 6⊆ Qs for all s),
but certainly this is not the case in general.

The definition of a diagonal uniformity on a set S involves binary relations
on S, but the standard theory of binary relations and functions is largely inap-
propriate for general textures (S, S) because of their lack of symmetry. With
this in mind, the second author has recently introduced notions of relation and
corelation [1] for textures, based on the duality mentioned above. It is shown



Di-uniform Texture Spaces 159

in [1] that, working in terms of direlations, which are pairs consisting of a rela-
tion and a corelation, a theory is obtained which resembles in many important
respects that of classical binary relations and functions. It will be appropriate,
therefore, to base our textural analogue of a diagonal uniformity on the con-
cept of direlation and for the convenience of the reader we recall some basic
definitions and results from [1]. The reader is referred to [1] for more details,
motivation and examples.

For textures (S, S), (T, T) we denote by S ⊗ T the product texturing of
S × T [4]. Thus, S ⊗ T consists of arbitrary intersections of sets of the form
(A × T) ∪ (S × B), A ∈ S, B ∈ T. For s ∈ S, Ps and Qs will always denote
the p-sets and q-sets for the texture (S, S), while for t ∈ T , Pt and Qt will
denote the p-sets and q-sets for (T, T). We reserve the notation P(s,t), Q(s,t),
s ∈ S, t ∈ T , for the p-sets, q-sets in (S ×T, S ⊗ T). On the other hand, P (s,t)
and Q(s,t) will denote the p-sets and q-sets for the texture (S ×T, P(S) ⊗ T).
Hence (see [1]) we have P (s,t) = {s}×Pt and Q(s,t) = [(S\{s})×T ]∪ [S×Qt].
Likewise, P (t,s) and Q(t,s) are the p-sets and q-sets for (T ×S, P(T)⊗S). It is
easy to verify that P (s,t) 6⊆ Q(s′,t′) ⇐⇒ s = s′ and Pt 6⊆ Qt′. Again, we will
use this fact, and its companion P (t,s) 6⊆ Q(t′,s′) ⇐⇒ t = t′ and Ps 6⊆ Qs′,
without comment in what follows. Now let us recall:

Definition 1.2. [1] Let (S, S), (T, T) be textures. Then

(1) r ∈P(S) ⊗ T is called a relation on (S, S) to (T, T) if it satisfies
R1 r 6⊆ Q(s,t),Ps′ 6⊆ Qs =⇒ r 6⊆ Q(s′,t).
R2 r 6⊆ Q(s,t) =⇒ ∃s′ ∈ S such that Ps 6⊆ Qs′ and r 6⊆ Q(s′,t).

(2) R ∈P(S) ⊗ T is called a co-relation on (S, S) to (T, T) if it satisfies
CR1 P (s,t) 6⊆ R,Ps 6⊆ Qs′ =⇒ P (s′,t) 6⊆ R.
CR2 P (s,t) 6⊆ R =⇒ ∃s′ ∈ S such that Ps′ 6⊆ Qs and P (s′,t) 6⊆ R.

(3) A pair (r,R), where r is a relation and R a co-relation on (S, S) to
(T, T) is called a direlation on (S, S) to (T, T).

Normally, relations will be denoted by lower case and co-relations by upper
case letters, as in the above definition.

For direlations (p,P), (q,Q) on (S, S) to (T, T) we write (p,P) v (q,Q) if
and only if p ⊆ q and Q ⊆ P.

For a general texture (S, S) we define

i = iS =
∨
{P (s,s) | s ∈ S} and I = IS =

⋂
{Q(s,s) | s ∈ S}.

If we note that i 6⊆ Q(s,t) ⇐⇒ Ps 6⊆ Qt and P (s,t) 6⊆ I ⇐⇒ Pt 6⊆ Qs then
it is trivial to verify that i is a relation and I a co-relation on (S, S) to (S, S).
We refer to (i,I) as the identity direlation on (S, S).

A direlation (r,R) on (S, S) (that is, on (S, S) to (S, S)) is reflexive if r and
R are reflexive, that is if (i,I) v (r,R).



160 S. Özçağ and L. M. Brown

If (r,R) is a direlation on (S, S) to (T, T), the inverse (r,R)← = (R←,r←)
of (r,R) is the direlation on (T, T) to (S, S) defined by

r← =
⋂
{Q(t,s) | r 6⊆ Q(s,t)},

R← =
∨
{P (t,s) | P (s,t) 6⊆ R}.

A direlation (r,R) on (S, S) is called symmetric if (r,R) = (r,R)←, that is
if and only if R = r←. This notion of symmetry is quite different from the
classical notion of symmetry for relations. However, as we will see, it will play
the same role in the theory of textural uniformities as does classical symmetry
in the theory of uniformities.

Definition 1.3. Let (S, S), (T, T) be textures, r a relation and R a co-relation
on (S, S) to (T, T).

(1) For A ⊆ S the A–section of r is the element r(A) of T defined by

r(A) =
⋂
{Qt | ∀s, r 6⊆ Q(s,t) =⇒ A ⊆ Qs}∈ T.

(2) For A ⊆ S the A–section of R is the element R(A) of T defined by

R(A) =
∨
{Pt | ∀s,P (s,t) 6⊆ R =⇒ Ps ⊆ A}∈ T.

(3) For B ⊆ T the B–presection of r (B–presection of R) is the B–section
r←(B) ∈ S of the co-relation r← (respectively, the B–section R←(B) ∈
S of the relation R←) on (T, T) to (S, S).

The following lemma gives formulae for directly calculating the presections.

Lemma 1.4. For a relation r, a co-relation R and B ⊆ T we have:
(1) r←(B) =

∨
{Ps | ∀t, r 6⊆ Q(s,t) =⇒ Pt ⊆ B}∈ S.

(2) R←(B) =
⋂
{Qs | ∀t, P (s,t) 6⊆ R =⇒ B ⊆ Qt}∈ S.

The following results from [1] will prove useful later on.

Lemma 1.5. For a direlation (r,R) on (S, S) to (T, T) we have
(1) r 6⊆ Q(s,t) ⇐⇒ P (t,s) 6⊆ r← and P (s,t) 6⊆ R ⇐⇒ R← 6⊆ Q(t,s).
(2) r 6⊆ Q(s,t) ⇐⇒ r(Ps) 6⊆ Qt and P (s,t) 6⊆ R ⇐⇒ Pt 6⊆ R(Qs).

Proposition 1.6. With the notation as in Definition 1.3:
(1) For relations r1, r2 with r1 ⊆ r2, co-relations R1, R2 with R1 ⊆ R2,

A1, A2 in S with A1 ⊆ A2 and B1, B2 in T with B1 ⊆ B2 we have
r1(A1) ⊆ r2(A2), R1(A1) ⊆ R2(A2), r←2 (B1) ⊆ r←1 (B2) and R←2 (B1) ⊆
R←1 (B2).

(2) For any relation r we have r(∅) = ∅, A ⊆ r←(r(A)) for A ∈ S and
r(r←(B)) ⊆ B for B ∈ T.

(3) For any co-relation R we have R(S) = T , R←(R(A)) ⊆ A for A ∈ S
and B ⊆ R(R←(B)) for B ∈ T.

(4) For the identity direlation (i,I) on (S, S) and A ∈ S we have i(A) =
I(A) = A and hence i←(A) = I←(A) = A.



Di-uniform Texture Spaces 161

(5) If a relation r (co-relation R) on (S, S) is reflexive then for all A ∈ S
we have A ⊆ r(A) (R(A) ⊆ A).

(6) For a relation r and co-relation R on (S, S) to (T, T) we have

r(
∨
j∈J

Aj) =
∨
j∈J

r(Aj) and R(
⋂
j∈J

Aj) =
⋂
j∈J

R(Aj)

for any Aj ∈ S, j ∈ J.
(7) For a relation r and co-relation R on (S, S) to (T, T) we have

r←(
⋂
j∈J

Bj) =
⋂
j∈J

r←(Bj) and R
←(
∨
j∈J

Bj) =
∨
j∈J

R←(Bj)

for any Bj ∈ T, j ∈ J.

Another important concept for direlations is that of composition. We recall
the following:

Definition 1.7. [1] Let (S, S), (T, T), (U, U) be textures.
(1) If p is a relation on (S, S) to (T, T) and q a relation on (T, T) to (U, U)

then their composition is the relation q ◦ p on (S, S) to (U, U) defined
by

q ◦p =
∨
{P (s,u) | ∃t ∈ T with p 6⊆ Q(s,t) and q 6⊆ Q(t,u)}.

(2) If P is a co-relation on (S, S) to (T, T) and Q a co-relation on (T, T)
to (U, U) then their composition is the co-relation Q ◦ P on (S, S) to
(U, U) defined by

Q◦P =
⋂
{Q(s,u) | ∃t ∈ T with P (s,t) 6⊆ P and P (t,u) 6⊆ Q}.

(3) With p, q; P, Q as above, the composition of the direlations (p,P),
(q,Q) is the direlation

(q,Q) ◦ (p,P) = (q ◦p,Q◦P).

It is shown in [1] that the operation of taking the composition of direlations
is associative, and that the identity direlations are identities for this operation.

If (r,R) is a direlation on (S, S) then (r,R) ◦ (r,R) = (r ◦r,R◦R) is also a
direlation on (S, S), which we denote by (r,R)2. We give the obvious meaning
to (r,R)n for any n = 3, 4, . . .. The direlation (r,R) on (S, S) is called transitive
if (r,R)2 v (r,R).

We will also have occasion to consider the greatest lower bound of direlations.
We give the definition for two direlations, but it may be extended in the obvious
way to any family of direlations.

Definition 1.8. [1] Let (p,P), (q,Q) be direlations on (S, S) to (T, T). Then

puq =
∨
{P (s,t) | ∃v ∈ S with Ps 6⊆ Qv and p,q 6⊆ Q(v,t)},

P tQ =
⋂
{Q(s,t) | ∃v ∈ S with Pv 6⊆ Qs and P (v,t) 6⊆ P,Q}, and

(p,P) u (q,Q) = (puq,P tQ).



162 S. Özçağ and L. M. Brown

Proposition 1.9. With the notation as in Definition 1.8,
(1) puq is a relation on (S, S) to (T, T). It is the greatest lower bound of p

and q in the set of all relations on (S, S) to (T, T), ordered by inclusion.
(2) P t Q is a co-relation on (S, S) to (T, T). It is the least upper bound

of P and Q in the set of all co-relations on (S, S) to (T, T), ordered by
inclusion.

(3) The direlation (p,P) u (q,Q) is the greatest lower bound of (p,P) and
(q,Q) on the set of all direlations on (S, S) to (T, T), ordered by the
relation v.

(4) (puq)← = p← tq← and (P tQ)← = P← uQ←.
(5) For A ∈ S, (puq)(A) ⊆ p(A) ∩q(A) and P(A) ∪Q(A) ⊆ (P tQ)(A).
(6) For B ∈ T, p←(B) ∪ q←(B) ⊆ (p u q)←(B) and (P t Q)←(B) ⊆

P←(B) ∩Q←(B).
(7) Let (p1,P1), (p2,P2) be direlations on (S, S) to (T, T) and (q1,Q1),

(q2,Q2) direlations on (T, T) to (U, U). Then ((q1,Q1) u (q2,Q2)) ◦
((p1,P1) u (q2,Q2)) v ((q1,Q1) ◦ (p1,P1)) u ((q2,Q2) ◦ (p2,P2)).

The notion of difunction is derived from that of direlation as follows.

Definition 1.10. [1] Let (f,F) be a direlation on (S, S) to (T, T). Then (f,F)
is called a difunction on (S, S) to (T, T) if it satisfies the following two condi-
tions.

DF1 For s,s′ ∈ S, Ps 6⊆ Qs′ =⇒ ∃t ∈ T with f 6⊆ Q(s,t) and P (s′,t) 6⊆ F.
DF2 For t,t′ ∈ T and s ∈ S, f 6⊆ Q(s,t) and P (s,t′) 6⊆ F =⇒ Pt′ 6⊆ Qt.

Difunctions are preserved under composition. It is easy to see that the
identity direlation (iS,IS) on (S, S) is in fact a difunction on (S, S) to (S, S).
In this context we refer to (iS,IS) as the identity difunction on (S, S).

If (f,F) : (S, S) → (T, T) is a difunction, A ∈ S, then f(A) is called the
image and F(A) the co-image of A. Likewise, for B ∈ T, f←(B) is called the
inverse image and F←(B) the inverse co-image of B. It is shown in [1] that
f←(B) = F←(B) for all B ∈ T, that is the inverse image and inverse co-image
coincide.

Since a texturing is generally not closed under set complementation, when
discussing topological concepts we cannot insist that closed sets should be the
complement of open sets. This leads to the notion of a dichotomous topology,
or ditopology for short [2]. This is a pair (τ,κ) of subsets of S, where the set of
open sets τ satisfies

(1) S, ∅ ∈ τ,
(2) G1, G2 ∈ τ =⇒ G1 ∩G2 ∈ τ and
(3) Gi ∈ τ, i ∈ I =⇒

∨
i Gi ∈ τ,

and the set of closed sets κ satisfies
(1) S, ∅ ∈ κ,
(2) K1, K2 ∈ κ =⇒ K1 ∪K2 ∈ κ and
(3) Ki ∈ κ, i ∈ I =⇒

⋂
i Ki ∈ κ.



Di-uniform Texture Spaces 163

The reader is referred to [2, 3, 6, 7] for some results on ditopological texture
spaces and their relation with fuzzy topologies.

A subset β of τ is called a base of τ if every set in τ can be written as a join
of sets in β, while a subset β of κ is a base of κ if every set in κ can be written
as an intersection of sets in β.

For the unit interval texture (I, I) mentioned above, we may define a natural
ditopology (τI,κI) by

τI = {[0,s) | s ∈ I}∪{I}, κI = {[0,s] | s ∈ I}∪{∅}.

Continuity of difunctions is the subject of the following definition.

Definition 1.11. [6] Let (Sk, Sk,τk,κk), k = 1, 2, be ditopological texture
spaces and (f,F) a difunction on (S1, S1) to (S2, S2). Then

(1) (f,F) is continuous if G ∈ τ2 =⇒ F←(G) ∈ τ1.
(2) (f,F) is cocontinuous if K ∈ κ2 =⇒ f←(K) ∈ κ1.
(3) (f,F) is bicontinuous if it is continuous and cocontinuous.

The reader is referred to [9] for general terms related to lattice theory.
This paper comprises part of the first author’s research towards her PhD

thesis to be submitted to Hacettepe University.
We pause here to mention our motivation for introducing textures as a sub-

strate for topology.
Ditopological texture spaces were conceived as a point-set setting for the

study of fuzzy topology, and provide a unified setting for the study of topology,
bitopology and fuzzy topology. Some of the links with Hutton spaces, L–fuzzy
sets and topologies are expressed in a categorical setting in [6]. Here it is the
choice of bicontinuous difunctions for the morphisms on the textural side which
makes possible a correspondence with the point-free concept of Hutton space.

Despite the close links with fuzzy sets and topologies, the development of
the theory of ditopological texture spaces has proceeded largely independently,
and has concentrated on the development of concepts which help to compensate
for the possible lack of complementation. One such is that of direlation and
difunction, another that of dicover ([2,3], see §2 below). Both play a crucial
role in this paper. If one takes the view that a texturing S can provide a much
more economic computational model than P(S), it is important that we do not
lose power in other directions. For example I is certainly much simpler than
P(I), but if we consider only ordinary open covers (and closed cocovers) it is
trivial that for the usual ditopology, every closed subset is compact (and every
open set cocompact). However this is non-trivially equivalent [2] to the fact
that every open, coclosed dicover has a finite, cofinite subcover and this, via
a bitopological argument, can be shown to be equivalent to the compactness
of I under its usual topology. Hence this compactness property of (I, I) in its
dicovering form is as powerful as that of I, and we will see later that with an
appropriate di-uniformity (I, I) can again play the same role as I does in the
usual theory of uniformities.



164 S. Özçağ and L. M. Brown

Duality is an important element in defining such concepts. When applied
to ditopologies it often gives rise to pairs of properties, such as compact –
cocompact, regular – coregular. In the case of uniform ditopologies, as we will
see, it actually links the open and closed sets via symmetry, and this causes the
ditopology to be simultaneously completely regular and completely coregular.

A form of duality also plays a role in Giovanni Sambin’s basic picture for
formal topology [11]. There are clear parallels here which warrant further study.
Likewise, links with the theory of locales and with domain theory have yet to
be worked out. Finally, complement free textural concepts can be expected
to find applications in negation free logics, and indeed (ditopological) textures
themselves could well prove to be useful models for certain classes of such logics.

2. Direlations and Dicovers

As mentioned in the introduction, the entourages of a diagonal uniformity
in the classical sense [13] will be replaced by direlations in the textural setting.
A second important formulation of the theory of uniformities is that of the
covering uniformity [12], so we will require an appropriate notion of cover in
order to obtain an analogous description for textures. In this section we show
that the notion of dicover, used in [2] to characterize the important form of
compactness mentioned above and in [3] to describe various covering properties
of ditopological texture spaces, is associated in a natural way with direlations.
Hence this notion will form the basis for our description of covering uniformities
in the textural sense.

Let us recall [2,3] that by a dicover of the texture (S, S) we mean a family
C = {(Ai,Bi) | i ∈ I} of elements of S × S which satisfies

⋂
i∈I1 Bi ⊆

∨
i∈I2 Ai

for all partitions (I1,I2) of I, including the trivial partitions. An important
example is the family P = {(Ps,Qs) | s ∈ S[}, which is shown in [3] to be a
dicover for any texture (S, S). If D is a dicover we often write L D M in place
of (L,M) ∈ D. We recall the following notions for dicovers given in [3].

(1) C is a refinement of D if given i ∈ I we have L D M so that Ai ⊆ L
and M ⊆ Bi. In this case we write C ≺ D.

(2) The star and co-star of C ∈ S with respect to C are respectively the
sets

St(C,C) =
∨
{Ai | i ∈ I, C 6⊆ Bi}∈ S, and

CSt(C,C) =
⋂
{Bi | i ∈ I, Ai 6⊆ C}∈ S.

We say that C is a delta refinement of D, and write C ≺(∆) D, if
C∆ = {(St(C,Ps), CSt(C,Qs)) | s ∈ S[}≺ D.

We say that C is a star refinement of D, and write C ≺(?) D, if
C? = {(St(C,Ai), CSt(C,Bi)) | i ∈ I}≺ D.

Before describing the link between direlations and dicovers, it will be appro-
priate for us to define a particular class of dicovers that will arise naturally in
this connection.



Di-uniform Texture Spaces 165

Definition 2.1. A family C ⊆ S×S is called an anchored dicover if it satisfies:
(1) P ⊆ C, and
(2) Given A C B there exists s ∈ S satisfying

(a) A 6⊆ Qu =⇒ ∃A′ C B′ with A′ 6⊆ Qu and Ps 6⊆ B′, and
(b) Pv 6⊆ B =⇒ ∃A′′ C B′′ with Pv 6⊆ B′′ and A′′ 6⊆ Qs.

Since P is a dicover, we see by (1) that an anchored dicover is a dicover. It
is straightforward to verify that P itself is anchored. The notion of anchored
dicover enables us to improve ([3], Lemma 4.7 (3)). Since these results will be
useful later on we present the modified lemma in full.

Lemma 2.2. Let C, D, E be dicovers on (S, S).
(1) C ≺(?) D =⇒ C ≺ D.
(2) If P ≺ C then C ≺(?) D =⇒ C ≺(∆) D
(3) If C is anchored then

(i) C ≺(∆) D =⇒ C ≺ D.
(ii) C ≺(∆) D ≺(∆) E =⇒ C ≺(?) E.

Proof. (1) and (2) are proved in [3] so we concentrate on (3).
(i). Take A C B and s ∈ S as in Definition 2.1 (2). It will suffice to show

A ⊆ St(C,Ps) and CSt(C,Qs) ⊆ B. If A 6⊆ St(C,Ps) then we have u ∈ S with
A 6⊆ Qu and Pu 6⊆ St(C,Ps). By (2)(a) there exists A′ C B′ with A′ 6⊆ Qu and
Ps 6⊆ B′. But then A′ ⊆ St(C,Ps) so Pu 6⊆ A′, which gives the contradiction
A′ ⊆ Qu. The inclusion CSt(C,Qs) ⊆ B is proved likewise.

(ii). Take A C B and for s ∈ S satisfying Definition 2.1 (2), choose L D M so
that St(D,Ps) ⊆ L and M ⊆ CSt(D,Qs). It will suffice to show St(C,A) ⊆ L
and M ⊆ CSt(C,B). We prove the first inclusion, the second being dual.
Suppose St(C,A) 6⊆ L and take w ∈ S with St(C,A) 6⊆ Qw and Pw 6⊆ L. Now
we have A1 C B1 satisfying A1 6⊆ Qw and A 6⊆ B1. Let us choose u ∈ S with
A 6⊆ Qu and Pu 6⊆ B1. By condition (2)(a) we have A′ C B′ with A′ 6⊆ Qu
and Ps 6⊆ B′. Choose U D V with St(C,Pu) ⊆ U and V ⊆ CSt(C,Qu). Then
A1 ⊆ St(C,Pu) ⊆ U and V ⊆ CSt(C,Qu) ⊆ B′. Since Ps 6⊆ B′ we now have
Ps 6⊆ V , and so U ⊆ St(D,Ps) ⊆ L, whence A1 ⊆ L. A1 6⊆ Qw and Pw 6⊆ L
now give a contradiction, and the proof is complete. �

Let us now show that we may associate an anchored dicover with each re-
flexive direlation (d,D) on (S, S).

Proposition 2.3. Let (d,D) be a reflexive direlation on (S, S) and for s ∈ S
let d[s] = d(Ps) and D[s] = D(Qs). Then

γ(d,D) = {(d[s],D[s]) | s ∈ S[}
is an anchored dicover of (S, S).

Proof. Set C = γ(d,D). Since (d,D) is reflexive, Ps ⊆ d(Ps) = d[s] and
D[s] = D(Qs) ⊆ Qs by Proposition 1.6 (5)). Hence, P ≺ C.

Let us associate s with d[s] C D[s] and take d[s] 6⊆ Qu. Now d[s] = d(Ps) =
d(
∨
{Ps′ | Ps 6⊆ Qs′}) =

∨
{d(Ps′) | Ps 6⊆ Qs′} by Proposition 1.6 (6), so there



166 S. Özçağ and L. M. Brown

exists s′ ∈ S with Ps 6⊆ Qs′ and d[s′] 6⊆ Qu. Since D[s′] ⊆ Qs′ we also have
Ps 6⊆ D[s′], whence d[s′] C D[s′] satisfies the condition in Definition 2.1 (2a).
The proof of (2b) is dual to this. �

Let us denote by RDR the family of reflexive direlations and by ADC the
family of anchored dicovers on (S, S). The above proposition can now be seen
as giving us a mapping γ : RDR → ADC.

Proposition 2.4. For (d,D), (e,E) ∈ RDR we have (d,D)◦(d,D)← v (e,E)
=⇒ γ(d,D) ≺(∆) γ(e,E).

Proof. We establish St(γ(d,D),Ps) ⊆ e[s] for s ∈ S[. Suppose this is not so.
Then we have z ∈ S with d[z] 6⊆ e[s] and Ps 6⊆ D[z]. Take t ∈ S with d[z] 6⊆ Qt
and Pt 6⊆ e[s], and then t′ ∈ S satisfying d[z] 6⊆ Qt′ and Pt′ 6⊆ Qt. From
d[z] 6⊆ Qt′ we obtain d 6⊆ Q(z,t′), while Ps 6⊆ D[z] implies P (z,s) 6⊆ D, and
hence D← 6⊆ Q(s,z). Thus P (s,t′) ⊆ D← ◦d ⊆ e, so e 6⊆ Q(s,t) which gives the
contradiction Pt ⊆ e[s].

In just the same way E[s] ⊆ CSt(γ(d,D),Qs), and the proof is complete. �

Now let us show that a dicover gives rise to a reflexive, symmetric direlation
in a natural way.

Proposition 2.5. Let C = {(Aj,Bj) | j ∈ J} be a dicover on (S, S) and define
δ(C) = (d(C),D(C)) by

d(C) =
∨
{P (s,t) | ∃j ∈ J with Aj 6⊆ Qt and Ps 6⊆ Bj},

D(C) =
⋂
{Q(s,t) | ∃j ∈ J with Pt 6⊆ Bj and Aj 6⊆ Qs}.

Then δ(C) is a reflexive and symmetric direlation on (S, S).

Proof. Write d = d(C), D = D(C) for short. First we verify that d is a relation
on (S, S), leaving the proof that D is a co-relation to the reader. Take s,t ∈ S
with d 6⊆ Q(s,t). Then we have t′ ∈ S and j ∈ J satisfying P (s,t′) 6⊆ Q(s,t),
Aj 6⊆ Qt′ and Ps 6⊆ Bj. If Ps′ 6⊆ Qs then Ps ⊆ Ps′, whence Ps′ 6⊆ Bj and so
P (s′,t′) ⊆ d, which gives d 6⊆ Q(s′,t). This establishes R1. On the other hand,
since Ps 6⊆ Bj, we have s′ ∈ S satisfying Ps 6⊆ Qs′ and Ps′ 6⊆ Bj. As before,
d 6⊆ Q(s′,t), which verifies R2.

To show d is reflexive, suppose i 6⊆ d and take s,t ∈ S with i 6⊆ Q(s,t) and
P (s,t) 6⊆ d. Then Ps 6⊆ Qt and for all j ∈ J we have Aj ⊆ Qt or Ps ⊆ Bj.
Put J1 = {j ∈ J | Ps ⊆ Bj} and let J2 = J \J1. Then (J1,J2) is a partition
of J, so Ps ⊆

⋂
j∈J1 Bj ⊆

∨
j∈J2 Aj ⊆ Qt, since C is a dicover. This gives the

contradiction Ps ⊆ Qt, so d is reflexive. The proof that D is reflexive is dual
to this. Hence, (d,D) is reflexive.

To show (d,D) is symmetric it will suffice to verify that d← = D. Suppose
that d← 6⊆ D and take u,v ∈ S satisfying d← 6⊆ Q(u,v) and P (u,v) 6⊆ D. We
have u′ ∈ S with Pu′ 6⊆ Qu and P (u′,v) 6⊆ D by CR2. There exists t ∈ S with
P (u′,v) 6⊆ Q(u′,t) and j ∈ J for which Pt 6⊆ Bj and Aj 6⊆ Qu′, whence P (t,u′) ⊆ d



Di-uniform Texture Spaces 167

and so d 6⊆ Q(t,u). This is easily seen to be equivalent to P (u,t) 6⊆ d← and so
d← ⊆ Q(u,t). Finally Qt ⊆ Qv, which gives the contradiction d← ⊆ Q(u,v).

Finally, suppose D 6⊆ d← and take u,v ∈ S with D 6⊆ Q(u,v) and P (u,v) 6⊆
d←. As above we have d 6⊆ Q(v,u), whence t ∈ S and j ∈ J so that P (v,t) 6⊆
Q(v,u), Aj 6⊆ Qt and Pv 6⊆ Bj. Since Qu ⊆ Qt we also have Aj 6⊆ Qu, whence
we have the contradiction D ⊆ Q(u,v) from the definition of D. �

If we denote by DC the set of dicovers and by SRDR the set of symmetric
reflexive direlations on (S, S), this proposition defines a mapping δ : DC →
SRDR.

Proposition 2.6. For C, D ∈ DC, C ≺(?) D =⇒ δ(C) ◦ δ(C) v δ(D).

Proof. Suppose d(C) ◦ d(C) 6⊆ d(D). Then we have s,u ∈ S so that P (s,u) 6⊆
d(D) and there exists t ∈ S satisfying d(C) 6⊆ Q(s,t) and d(C) 6⊆ Q(t,u). Now
we have t′ ∈ S so that Pt′ 6⊆ Qt and there exists A1CB1 for which A1 6⊆ Qt′,
Ps 6⊆ B1. Also we have u′ ∈ S so that Pu′ 6⊆ Qu and there exists A2CB2
for which A2 6⊆ Qu′, Pt 6⊆ B2. Since C ≺ (?) D we may choose CDE with
St(C,A1) ⊆ C and E ⊆ CSt(C,B1). Hence, since we clearly have A1 6⊆ B2,
A2 ⊆ St(C,A1) ⊆ C and E ⊆ CSt(C,B1) ⊆ B1. Thus C 6⊆ Qu′ and Ps 6⊆ E,
and we obtain the contradiction P (s,u) ⊆ P (s,u′) ⊆ d(D).

This establishes d(C) ◦d(C) ⊆ d(D), and the proof of D(D) ⊆ D(C) ◦D(C)
is dual to this. �

Let us now discuss the relation between the mappings γ and δ.

Theorem 2.7. Let (S, S) be a texture. With the notation above,
(1) δ(γ(d,D)) = (d,D) ◦ (d,D) for all (d,D) ∈ SRDR.
(2) γ(δ(C)) = C∆ for all C ∈ DC.

Proof. (1). Take (d,D) ∈ SRDR and suppose d(γ(d,D)) 6⊆ d◦d. Then we have
s,t ∈ S satisfying P (s,t) 6⊆ d◦d for which we have z ∈ S[ satisfying d[z] 6⊆ Qt
and Ps 6⊆ D[z]. However, d[z] 6⊆ Qt ⇐⇒ d 6⊆ Q(z,t) and Ps 6⊆ D[z] ⇐⇒
P (z,s) 6⊆ D = d← ⇐⇒ d 6⊆ Q(s,z) by Lemma 1.5, since (d,D) is symmetric,
and we obtain the contradiction P (s,t) ⊆ d◦d.

Conversely, suppose d◦d 6⊆ d(γ(d,D)). Then we have s,u ∈ S with P (s,u) 6⊆
d(γ(d,D)) for which we have t ∈ S satisfying d 6⊆ Q(s,t) and d 6⊆ Q(t,u). Firstly,
d 6⊆ Q(t,u) gives us d[t] 6⊆ Qu. Secondly, d 6⊆ Q(s,t) implies t ∈ S[ and also gives
P (t,s) 6⊆ d← = D since (d,D) is symmetric. Hence we obtain Ps 6⊆ D[t]. We
deduce that P (s,u) ⊆ d(γ(d,D)), which is a contradiction.

This completes the proof that d(γ(d,D)) = d◦d, and the proof of D(γ(d,D))
= D ◦D is dual to this, so δ(γ(d,D)) = (d,D) ◦ (d,D).

(2). Let C = {(Ai,Bi) | i ∈ I} and take s ∈ S[. Suppose that d(C)[s] 6⊆
St(C,Ps). Then we have t ∈ S with d(C)[s] 6⊆ Qt and Pt 6⊆ St(C,Ps). Now we
have w ∈ S with d(C) 6⊆ Q(w,t) and Ps 6⊆ Qw. Thus for some t′ ∈ S and i ∈ I,



168 S. Özçağ and L. M. Brown

P (w,t′) 6⊆ Q(w,t), Ai 6⊆ Qt′ and Pw 6⊆ Bi. We deduce that Ps 6⊆ Bi, and hence
Pt ⊆ Pt′ ⊆ Ai ⊆ St(C,Ps), which is a contradiction.

Conversely, suppose St(C,Ps) 6⊆ d(C)[s]. Then we have i ∈ I with Ai 6⊆
d(C)[s] and Ps 6⊆ Bi. Hence for some t ∈ S we have Ai 6⊆ Qt and d(C) 6⊆
Q(z,t) =⇒ Ps ⊆ Qz for all z ∈ S. Take s′, t′ ∈ S satisfying Ps 6⊆ Qs′,
Ps′ 6⊆ Bi and Ai 6⊆ Qt′, Pt′ 6⊆ Qt. Now Ai 6⊆ Qt′, Ps′ 6⊆ Bi gives P (s′,t′) ⊆
d(C), so d(C) 6⊆ Q(s′,t) and putting z = s′ in the above implication gives the
contradiction Ps ⊆ Qs′.

This completes the proof that d(C)[s] = St(C,Ps). likewise, D(C)[s] =
CSt(C,Qs), so γ(δ(C)) = C∆. �

Corollary 2.8. With the notation above:
(1) (d,D) v δ(γ(d,D)) for all (d,D) ∈ RSDR.
(2) C∆ is anchored for all dicovers C. If C is anchored then C ≺ γ(δ(C)).

Proof. (1). Clear by Theorem 2.7 (1) since (d,D) v (d,D)◦(d,D) when (d,D)
is reflexive.

(2). The first statement is clear from Theorem 2.7 (2) and Proposition 2.3.
For the second we need only note that C ≺ (∆) C∆ and apply Lemma 2.2 (1)
when C is anchored to give C ≺ C∆. Hence C ≺ γ(δ(C)) by Theorem 2.7 (2). �

Let us recall from [3] that the meet of two dicovers C and D is the dicover
C ∧ D = {(A∩C,B ∪D) | A C B, C D D}. As might be expected, this notion
is closely related to that of the greatest lower bound for direlations.

Proposition 2.9. Let (S, S) be a texture. With the notation above,
(1) For (d,D), (e,E) ∈ RDR we have γ((d,D)u(e,E)) ≺ γ(d,D)∧γ(e,E).
(2) For C, D ∈ DC we have δ(C ∧ D) v δ(C) uδ(D).

Proof. (1). Since γ((d,D) u (e,E)) = {((d u e)[s], (D t E)[s]) | s ∈ S[}, the
result follows trivially from Proposition 1.9 (5).

(2). If d(C ∧ D) 6⊆ d(C) u d(D) then we have s,t ∈ S satisfying P (s,t) 6⊆
d(C)ud(D) for which we have ACB, CDE satisfying A∩C 6⊆ Qt and Ps 6⊆ B∪E.
Take t′ ∈ S satisfying A∩C 6⊆ Qt′ and Pt′ 6⊆ Qt. Now P (s,t′) ⊆ d(C), d(D), so
d(C), d(D) 6⊆ P (s,t), which leads to the contradiction P (s,t) ⊆ d(C) ud(D).

This verifies d(C∧D) ⊆ d(C)ud(D), and the proof of D(C)tD(D) ⊆ D(C∧D)
is dual to this, so (2) is proved. �

3. Direlational and Dicover Uniformities

We now have the tools necessary to define direlational and dicover unifor-
mities on a texture, and to prove their equivalence.

Definition 3.1. Let (S, S) be a texture and U a family of direlations on (S, S).
If U satisfies the conditions

(1) (i,I) v (d,D) for all (d,D) ∈ U. That is, U ⊆ RDR.
(2) (d,D) ∈ U, (e,E) ∈ DR and (d,D) v (e,E) implies (e,E) ∈ U.



Di-uniform Texture Spaces 169

(3) (d,D), (e,E) ∈ U implies (d,D) u (e,E) ∈ U.
(4) Given (d,D) ∈ U there exists (e,E) ∈ U satisfying (e,E) ◦ (e,E) v

(d,D).
(5) Given (d,D) ∈ U there exists (c,C) ∈ U satisfying (c,C)← v (d,D).

then U is called a direlational uniformity on (S, S), and (S, S, U) is known as a
direlational uniform texture space.

It will be noted that this definition is formally the same as the usual def-
inition of a diagonal uniformity, and the notions of base and subbase may be
defined in the obvious way. Exactly as for diagonal uniformities we have the
following lemma.

Lemma 3.2. A direlational uniformity U on (S, S) has a base of symmetric
direlations.

Proof. Take (d,D) ∈ U. By condition (5) we have (e,E) ∈ U with (e,E)← v
(d,D), so (e,E) v (d,D)← and (d,D)← ∈ U by condition (2). But now
(f,F) = (d,D)u(d,D)← ∈ U by condition (3), and clearly (f,F) is symmetric
and satisfies (f,F) v (d,D). �

The following example of a direlational uniformity will prove important later
on.

Example 3.3. Let (I, I) be the unit interval texture and for � > 0 define d� =
{(r,s) | r,s ∈ I, s < r + �}, D� = {(r,s) | r,s ∈ I, s ≤ r − �}. Clearly (d�,D�)
is a reflexive, symmetric direlation on (I, I). Moreover, (d�,D�)2 v (d2�,D2�),
while for � ≤ δ, (d�,D�) v (dδ,Dδ) and so (d�,D�)u(dδ,Dδ) = (d�,D�). Hence

UI = {(d,D) | (d,D) ∈ DR and there exists � > 0 with (d�,D�) v (d,D)}

is a direlational uniformity on (I, I). We will call UI the usual direlational
uniformity on (I, I).

Definition 3.4. Let (S, S, U) be a direlational uniform texture space and C a
dicover of S. Then C is called uniform if γ(c,C) ≺ C for some (c,C) ∈ U.

Lemma 3.5. Let (S, S, U) be a direlational uniform texture space and υ the
family of uniform dicovers. Then υ has the following properties:

(1) Given C ∈ υ there exists D ∈ υ ∩ ADC with D ≺ C.
(2) C ∈ υ, D ∈ DC and C ≺ D implies D ∈ υ.
(3) C, D ∈ υ implies C ∧ D ∈ υ.
(4) Given C ∈ υ there exists D ∈ υ with D ≺(?) C.

Proof. (1). By hypothesis there exists (c,C) ∈ U with γ(c,C) ≺ C. But
D = γ(c,C) ∈ υ ∩ ADC and D ≺ C.

(2). Immediate.
(3). Take C, D ∈ υ and (c,C), (d,D) ∈ U with γ(c,C) ≺ C, γ(d,D) ≺ D.

Then (c,C) u (d,D) ∈ U by Definition 3.1 (3), and γ((c,C) u (d,D)) ≺ C ∧ D
by Proposition 2.9 (1), so C ∧ D ∈ υ.



170 S. Özçağ and L. M. Brown

(4). Take C ∈ υ and (c,C) ∈ U with γ(c,C) ≺ C. By Definition 3.1 (4) we
have (d,D) ∈ U with (d,D) ◦ (d,D) v (c,C), and then by Definition 3.1 (5)
we have (e,E) ∈ U with (e,E)← v (d,D). If we let (f,F) = (d,D) u (e,E)
then (f,F) ∈ U and (f,F) ◦ (f,F)← v (c,C), so γ(f,F) ≺ (∆) γ(c,C) ≺ C
by Proposition 2.4. In exactly the same way we may find (g,G) ∈ U with
γ(g,G) ≺ (∆) γ(f,F). If we let D = γ(g,G) then D ∈ υ is anchored by
Proposition 2.3, so by Lemma 2.2 (3 ii) we have D ≺(?) C. �

This leads to the following definition.

Definition 3.6. Let (S, S) be a texture. If υ is a family of dicovers of S
satisfying conditions (1)–(4) of Lemma 3.5 we say υ is a dicovering uniformity
on (S, S), and call (S, S,υ) a dicovering uniform texture space.

We can now see Lemma 3.5 as associating a dicovering uniformity with a
given direlational uniformity. The following theorem expresses the equivalence
of these two concepts.

Theorem 3.7. Let (S, S) be a texture.
(1) To each direlational uniformity U on (S, S) we may associate a dicover-

ing uniformity υ = Γ(U) = {C ∈ DC | ∃(c,C) ∈ U with γ(c,C) ≺ C}.
(2) To each dicovering uniformity υ on (S, S) we may associate a direla-

tional uniformity U = ∆(υ) = {(d,D) ∈ RDR | ∃C ∈ υ with δ(C) v
(d,D)}.

(3) ∆(Γ(U)) = U for every direlational uniformity U on (S, S).
(4) Γ(∆(υ)) = υ for every dicovering uniformity υ on (S, S).

Proof. (1). This is just Lemma 3.5.
(2). We need to establish the conditions (1)–(5) of Definition 3.1 for U =

∆(υ). Conditions (1) and (2) are an immediate consequence of the definition
of ∆(υ), and (3) follows trivially from Proposition 2.9 (2). Take (d,D) ∈ ∆(υ).
Then we have C ∈ υ satisfying δ(C) v (d,D). Now (c,C) = δ(C) ∈ ∆(υ), and
since (c,C) is symmetric by Proposition 2.5 we have (c,C)← = (c,C) v (d,D),
which proves (5). Finally we have E ∈ υ satisfying E ≺(?) C, and then (e,E) =
δ(E) ∈ ∆(υ) and (e,E)◦(e,E) v (d,D) by Proposition 2.6, so (4) is established
also.

(3). First take (d,D) ∈ ∆(Γ(U)). Then we have C ∈ Γ(U) with δ(C) v
(d,D), and then (c,C) ∈ U with γ(c,C) ≺ C. Without loss of generality we
may take (c,C) ∈ RSDR since the symmetric elements of U form a base, so by
Corollary 2.8,

(c,C) v δ(γ(c,C)) v δ(C) v (d,D),
which shows (d,D) ∈ U. Conversely, take (d,D) ∈ U and choose (e,E) ∈ U
with (e,E) symmetric so that (e,E)◦(e,E) v (d,D). Then δ(γ(e,E)) v (d,D)
by Theorem 2.7 (1), and we have established (d,D) ∈ ∆(Γ(U)).

(4). First take C ∈ Γ(∆(υ)). Then we have (c,C) ∈ ∆(υ) with γ(c,C) ≺ C,
and then D ∈ υ with δ(D) v (c,C). Without loss of generality we may take



Di-uniform Texture Spaces 171

D ∈ ADC since the anchored elements of υ form a base, so by Corollary 2.8 (2).
D ≺ γ(δ(D)) ≺ γ(c,C) ≺ C,

whence C ∈ υ. Conversely, take C ∈ υ and choose E ∈ υ with E ≺ (?) C.
Without loss of generality we may assume E is anchored, so by Lemma 2.2 (2)
we have E ≺ (∆) C, whence E∆ ≺ C. Now Theorem 2.7 (2) gives γ(δ(E)) ≺ C,
so C ∈ Γ(∆(υ)), as required. �

We will use the term di-uniformity to refer to direlational and dicovering
uniformities in general.

Example 3.8. Consider the texture (I, I). The dicovering uniformity υI corre-
sponding to the direlational uniformity UI of Example 3.3 has a base consisting
of the dicovers D�, � > 0, where

D� = {([0,r + �), [0,r − �]) | r ∈ I},
and [0,r + �) is understood to be [0, 1] when r + � > 1 and [0,r − �] is ∅ if
r − � < 0.

4. The Uniform Ditopology

We begin by associating a ditopology with a direlational uniformity.

Proposition 4.1. Let (S, S, U) be a direlational uniform texture space. Then
the family (ηU(s),µU(s)), s ∈ S[, defined by

ηU(s) = {N ∈ S | N 6⊆ Qs, Ps 6⊆ Qt =⇒ ∃(d,D) ∈ U, d[t] ⊆ N},
µU(s) = {M ∈ S | Ps 6⊆ M, Pt 6⊆ Qs =⇒ ∃(d,D) ∈ U, M ⊆ D[t]},

is the dineighbourhood system for a ditopology on (S, S).

Proof. We must verify that the family ηU(s), s ∈ S[, satisfies the following
conditions [6]:

(1) N ∈ ηU(s) =⇒ N 6⊆ Qs.
(2) N ∈ ηU(s), N ⊆ N′ ∈ S =⇒ N′ ∈ ηU(s).
(3) N1,N2 ∈ ηU(s), N1 ∩N2 6⊆ Qs =⇒ N1 ∩N2 ∈ ηU(s).
(4) (a) N ∈ ηU(s) =⇒ ∃N? ∈ S, Ps ⊆ N? ⊆ N so that N? 6⊆ Qt =⇒

N? ∈ ηU(t), t ∈ S[.
(b) For N ∈ S and N 6⊆ Qs, if there exists N? ∈ S,Ps ⊆ N? ⊆ N

which satisfies N? 6⊆ Qt =⇒ N? ∈ ηU(t), t ∈ S[, then N ∈ ηU(s).
Conditions (1) and (2) are immediate from the definitions, and (3) follows at
once from the inclusion (due)(Pt) ⊆ d(Pt) ∩e(Pt) (Proposition 1.9 (5)).

(4) (a). Take N ∈ ηU(s) and define

N? =
∨
{Pz | Pz 6⊆ Qt =⇒ ∃(d,D) ∈ U with d[t] ⊆ N}.

Clearly Ps ⊆ N? ⊆ N and if N? 6⊆ Qt it is easy to show that N ∈ ηU(t).
(4) (b). Take N ∈ S with N 6⊆ Qs and N? ∈ S with Ps ⊆ N? ⊆ N and

satisfying N? 6⊆ Qt =⇒ N? ∈ ηU(t). To show N ∈ ηU(s) take t ∈ S with



172 S. Özçağ and L. M. Brown

Ps 6⊆ Qt. Since Ps ⊆ N? we have N? 6⊆ Qt. Choose t′ ∈ S with N? 6⊆ Qt′ and
Pt′ 6⊆ Qt. By hypothesis N ∈ ηU(t′) so Pt′ 6⊆ Qt now gives (d,D) ∈ U with
d[t] ⊆ N. Since N 6⊆ Qs this shows that N ∈ ηU(s), as required.

In just the same way the sets µU(s) satisfy the dual of conditions (1)–(4)
above, and this completes the proof (cf. [6]). �

Definition 4.2. Let (S, S, U) be a direlational uniform texture space and ηU(s),
µU(s) defined as above. The ditopology with dineighbourhood system {(ηU(s),
µU(s)) | s ∈ S[} is called the uniform ditopology of U and denoted by (τU,κU).
Lemma 4.3. Let (S, S, U) be a direlational uniform texture space with uniform
ditopology (τU,κU).

(i) G ∈ τU ⇐⇒ (G 6⊆ Qs =⇒ ∃(d,D) ∈ U with d[s] ⊆ G).
(ii) K ∈ κU ⇐⇒ (Ps 6⊆ K =⇒ ∃(d,D) ∈ U with K ⊆ D[s]).

Proof. We prove (i), leaving (ii) to the reader.
It is shown in [6] that the open sets are characterized by the property that

G 6⊆ Qs =⇒ G ∈ ηU(s). Take G ∈ τU and s ∈ S with G 6⊆ Qs. Now we
have s′ ∈ S with G 6⊆ Qs′ and Ps′ 6⊆ Qs. By the above G ∈ ηU(s′) and now
Ps′ 6⊆ Qs implies there exists (d,D) ∈ U with d[s] ⊆ G.

Conversely suppose G has the property stated in (i). Then if G 6⊆ Qs we
have (d,D) ∈ U with d[s] ⊆ G. Now if Ps 6⊆ Qt we have Pt ⊆ Ps and so
d[t] ⊆ d[s] ⊆ G, which shows that G ∈ ηU(s). Thus G ∈ τU. �

Proposition 4.4. Let υ be a dicovering uniformity on (S, S). Denote by (τ,κ)
the uniform ditopology of the direlational uniformity ∆(υ). Then:

(i) G ∈ τ ⇐⇒ (G 6⊆ Qs =⇒ ∃C ∈ υ with St(C,Ps) ⊆ G).
(ii) K ∈ κ ⇐⇒ (Ps 6⊆ K =⇒ ∃C ∈ υ with K ⊆ CSt(C,Qs)).

Proof. (i). Take G ∈ τ and G 6⊆ Qs. Then by Lemma 4.3 we have (d,D) ∈ ∆(υ)
with d[s] ⊆ G. We may take (e,E) ∈ ∆(υ) with (e,E) ◦ (e,E)← v (d,D) and
as in the proof of Proposition 2.4 we have St(γ(e,E),Ps) ⊆ d[s]. There exists
C ∈ υ with δ(C) v (e,E) and without loss of generality we may assume C is
anchored. Hence by Corollary 2.8 (2), C ≺ γ(δ(C)) ≺ γ(e,E), so St(C,Ps) ⊆
d[s] ⊆ G.

Conversely, suppose that given G 6⊆ Qs there exists C ∈ υ with St(C,Ps) ⊆
G. If we set (d,D) = δ(C) then (d,D) ∈ ∆(υ) and by Theorem 2.7 (2) we have
d[s] = St(C,Ps) ⊆ G. Hence G ∈ τ.

(ii). The proof is dual to (i), and is omitted. �

This justifies the following definition.

Definition 4.5. Let υ be a dicovering uniformity on (S, S). Then the ditopol-
ogy (τυ,κυ) defined by

τυ = {G ∈ S | G 6⊆ Qs =⇒ ∃C ∈ υ, St(C,Ps) ⊆ G},
κυ = {K ∈ S | Ps 6⊆ K =⇒ ∃C ∈ υ, K ⊆ CSt(C,Qs)},

is called the uniform ditopology of υ.



Di-uniform Texture Spaces 173

In just the same way the dineighbourhood system (ηυ(s),µυ(s)), s ∈ S[, for
(τυ,κυ) is given by

ηυ(s) = {N ∈ S | N 6⊆ Qs, Ps 6⊆ Qt =⇒ ∃C ∈ υ, St(C,Pt) ⊆ N},
µυ(s) = {M ∈ S | Ps 6⊆ M, Pt 6⊆ Qs =⇒ ∃C ∈ υ, M ⊆ CSt(C,Qt}.

We omit the details.
The following lemma enables us to generate open sets and closed sets for the

uniform ditopology of a dicovering uniformity.

Lemma 4.6. Let υ be a dicovering uniformity on (S, S) and take L ∈ S.
(1) The set

G = G(L) =
∨
{Pu | ∃D ∈ υ, St(D,Pu) ⊆ L}

is open for the uniform ditopology.
(2) The set

K = K(L) =
⋂
{Qu | ∃D ∈ υ, L ⊆ CSt(D,Qu)}

is closed for the uniform ditopology.

Proof. We establish (1), leaving the dual proof of (2) to the reader.
Take G 6⊆ Qs. Then we have u ∈ S and D ∈ υ satisfying Pu 6⊆ Qs and

St(D,Pu) ⊆ L. Take E ∈ υ with E ≺(?) D. By Definition 4.5 it will be sufficient
to show that St(E,Ps) ⊆ G. If this is not so then we have A0 E B0 with Ps 6⊆ B0
and A0 6⊆ G so we may take v ∈ S with A0 6⊆ Qv and Pv 6⊆ G. If we can show
that St(E,Pv) ⊆ St(D,Pu) we will obtain an immediate contradiction to the
definition of G, so take A1 E B1 with Pv 6⊆ B1, and choose A′0 D B′0 satisfying
St(E,A0) ⊆ A′0 and B′0 ⊆ CSt(E,B0). Since CSt(E,B0) ⊆ B0, Ps 6⊆ B0 and
Pu 6⊆ Qs we see that Pu 6⊆ B′0, whence A1 ⊆ St(E,A0) ⊆ A′0 ⊆ St(D,Pu),
using the evident fact that A0 6⊆ B1. This establishes the required inclusion
and completes the proof. �

Corresponding results for direlational uniformities may easily be formulated
and the details are left to the interested reader.

It is well known that a classical uniformity has a base of open members
and a base of closed members. We now establish an analogous result for di-
uniformities. We confine our attention to the dicovering case since there is a
well established meaning to the notions of openness and closedness for dicovers
[3]. Namely, a dicover C of the ditopological texture space (S, S,τ,κ) is open
(respectively, closed , co-open, coclosed ) if A C B =⇒ A ∈ τ (A ∈ κ, B ∈ τ,
B ∈ κ). First we require the following lemma.

Lemma 4.7. Let υ be a dicovering uniformity on (S, S), C ∈ υ and L ∈ S.
Consider the uniform ditopology on (S, S). Then:

(1) L ⊆ ] St(C,L)[ and [CSt(C,L)] ⊆ L.
(2) [L] ⊆ St(C,L) and CSt(C,L) ⊆ ]L[.



174 S. Özçağ and L. M. Brown

Proof. 1. If H = H(St(C,L)) is the open set defined in Lemma 4.6 (1) it is
trivial to verify that L ⊆ H ⊆ St(C,L), whence L ⊆ ] St(C,L)[. The second
inclusion follows in the same way from Lemma 4.6 (2).

2. If K = K(L) is the closed set defined in Lemma 4.6 (2) it is trivial to
verify that L ⊆ K ⊆ St(C,L), whence [L] ⊆ St(C,L). The second inclusion
follows in the same way from Lemma 4.6 (1). �

Proposition 4.8. A dicovering uniformity has a base of open, coclosed dicovers
and a base of closed, co-open dicovers.

Proof. Trivial from Lemma 4.7. �

Definition 4.9. A ditopological texture space (S, S,τ,κ) is called di-uniformiz-
able if there exists a di-uniformity on (S, S) whose uniform ditopology coincides
with (τ,κ).

We recall the following regularity axioms for ditopological texture spaces.

Definition 4.10. [3] Let (τ,κ) be a ditopology on (S, S). Then (τ,κ) is called
(1) Regular if G ∈ τ, G 6⊆ Qs =⇒ ∃H ∈ τ with H 6⊆ Qs, [H] ⊆ G.
(2) Coregular if F ∈ κ, Ps 6⊆ F =⇒ ∃K ∈ κ with Ps 6⊆ K, F ⊆ ]K[.
(3) Biregular if it is regular and coregular.

Using Proposition 4.8 it is straightforward to verify that a di-uniformizable
ditopology is biregular. However we will shortly prove a more powerful result,
and so omit the details.

Definition 4.11. [7] Let (τ,κ) be a ditopology on (S, S). Then (τ,κ) is called
(1) Completely regular if given G ∈ τ, G 6⊆ Qs, there exists a bicontin-

uous difunction (f,F) : (S, S) → (I, I) satisfying Ps ⊆ f←(P0) and
F←(Q1) ⊆ G.

(2) Completely coregular if given K ∈ τ, Ps 6⊆ K, there exists a bicon-
tinuous difunction (f,F) : (S, S) → (I, I) satisfying K ⊆ f←(P0) and
F←(Q1) ⊆ Qs.

(3) Completely biregular if it is completely regular and completely coregu-
lar.

We end this section by showing that a di-uniformizable ditopology is com-
pletely biregular. Since it is easy to see that the complete regularity condi-
tions imply the corresponding regularity conditions it will follow that a di-
uniformizable ditopology is biregular. We choose to work with direlational
uniformities. First we require the following lemma, which is the textural ana-
logue of the Metrization Lemma ([10], Page 185).

Lemma 4.12. Let (S, S) be a texture and rn, n ∈ N, a sequence of reflexive
relations satisfying r3n+1 ⊆ rn, n ∈ N. Define the function ϕ : S×S → [0, 1] by

ϕ(u,v) =




0 if rn 6⊆ Q(u,v) ∀n ∈ N,
1 if rn ⊆ Q(u,v) ∀n ∈ N,

2−n if ∃n ∈ N, rn 6⊆ Q(u,v), rn+1 ⊆ Q(u,v).



Di-uniform Texture Spaces 175

Then there exists a function q : S ×S → [0,∞) satisfying
(1) 1

2
ϕ(u,v) ≤ q(u,v) ≤ ϕ(u,v), ∀u,v ∈ S.

(2) Pu 6⊆ Qv =⇒ q(u,v) = 0 ∀u,v ∈ S.
(3) q(u,v) ≤ q(u,w) + q(w,v) ∀u,v,w ∈ S.

Proof. We consider chains u0,u1, . . . ,un of elements of S and write

s(u0,u1, . . . ,un) =
n−1∑
i=1

ϕ(ui,ui+1), n > 0, s(u0,u0) = ϕ(u0,u0) = 0.

Consider the function q : S ×S → [0,∞) defined by
q(u,v) = inf{s(u0, . . . ,un) | u = u0 and v = un, n ∈ N}.

(1) It is clearly sufficient to prove that ϕ(u,v) ≤ 2s(u0, . . . ,un) for any chain
u0,u1, . . . ,un with u = u0 and v = un. The proof is by induction on n ∈ N
and follows essentially the same steps as the proof of the Metrizaton Lemma.
We therefore omit the details.

(2) For each n ∈ N we have iS ⊆ rn so Pu 6⊆ Qv implies iS 6⊆ Q(u,v), and
hence rn 6⊆ Q(u,v). By (1) we now have 0 ≤ q(u,v) ≤ ϕ(u,v) = 0, whence
q(u,v) = 0.

(3) Immediate from the definition of q. �

Lemma 4.13. If we consider a sequence of corelations Rn, n ∈ N, satisfying
Rn ⊆ R3n+1 and define

ϕ∗(u,v) =




0 if P (u,v) 6⊆ Rn ∀n ∈ N,
1 if P (u,v) ⊆ Rn ∀n ∈ N,

2−n if ∃n ∈ N, P (u,v) 6⊆ Rn, P (u,v) ⊆ Rn+1,
we obtain q∗ : S ×S → [0,∞) satisfying

(1) 1
2
ϕ∗(u,v) ≤ q∗(u,v) ≤ ϕ∗(u,v), ∀u,v ∈ S.

(2) Pv 6⊆ Qu =⇒ q∗(u,v) = 0 ∀u,v ∈ S.
(3) q∗(u,v) ≤ q∗(u,w) + q∗(w,v) ∀u,v,w ∈ S.

In case Rn = r←n then we clearly have ϕ
∗(u,v) = ϕ(v,u) and q∗(u,v) = q(v,u)

for all u,v ∈ S.

Now we may give:

Theorem 4.14. A diuniformizable ditopological texture space is completely
biregular.

Proof. Let (S, S,τ,κ) be a ditopological texture space and U a compatible dire-
lational uniformity.

To show that (τ,κ) is completely regular take G ∈ τ and a ∈ S with G 6⊆
Qa. Then there exists (r,R) ∈ U with r(Pa) ⊆ G. Let (r0,R0) = (r,R).
By Definition 3.1 there exists (r1,R1) ∈ U such that (r1,R1)3 v (r0,R0),
(r2,R2) ∈ U with (r2,R2)3 v (r1,R1), and so on. Hence we obtain a sequence
(rn,Rn) of reflexive direlations satisfying (rn+1,Rn+1)3 v (rn,Rn), and by



176 S. Özçağ and L. M. Brown

Lemma 3.2 there is no loss of generality in assuming that the (rn,Rn) are
symmetric, i.e. Rn = r←n for each n ∈ N. Let ϕ and q be the functions given
in Lemma 4.12 for the sequence rn, n ∈ N of reflexive relations and define
θ : S → [0, 1] by

θ(s) = 2q(a,s) ∧ 1.
We take the texture I on I = [0, 1] and verify that the point function θ satisfies
the condition Pu 6⊆ Qv =⇒ Pθ(u) 6⊆ Qθ(v) of ([1], Theorem 3.14). However
if Pu 6⊆ Qv then q(a,v) ≤ q(a,u) + q(u,v) = q(a,u) by Lemma 4.12 (2),(3), so
θ(v) ≤ θ(u), which is equivalent to Pθ(u) 6⊆ Qθ(v) in (I, I). It follows that

f =
∨
{P (s,t) | ∃v ∈ S with Ps 6⊆ Qv and t ≤ θ(v)},

F =
⋂
{Q(s,t) | ∃v ∈ S with Pv 6⊆ Qs and θ(v) ≤ t},

define a difunction (f,F) : (S, S) → (I, I). If we take the usual ditopology
(τI,κI) on (I, I) then (f,F) is bicontinuous. We prove continuity, leaving the
dual proof of cocontinuity to the reader.

For s ∈ S suppose F←([0,r)) 6⊆ Qs. Then we have t ∈ I with P (s,t) 6⊆
F and t < r. From the definition of F we have v ∈ S and t′ ∈ I with
P (s,t) 6⊆ Q(s,t′), Pv 6⊆ Qs and θ(v) ≤ t′. From Pt 6⊆ Qt′ we have t′ ≤ t and
so θ(v) < r. Clearly θ(v) < 1 and so θ(v) = 2q(a,v) < r, whence there exists
n with 2(q(a,v) + 2−n) < r. We verify that rn(Ps) ⊆ F←([0,r)). Suppose
the contrary and take w ∈ S with rn(Ps) 6⊆ Qw and Pw 6⊆ F−1([0,r)). Now
we have z ∈ S with rn 6⊆ Q(z,w) and Pv 6⊆ Qz, whence rn 6⊆ Q(v,w) and so
q(v,w) ≤ ϕ(v,w) ≤ 2−n by Lemma 4.12. Hence we have

θ(w) ≤ 2q(a,w) ≤ 2(q(a,v) + q(v,w)) ≤ 2(q(a,v) + 2−n) < r.
On the other hand, from Pw 6⊆ F←([0,r)) we have w′ ∈ S with Pw 6⊆ Qw′ for
which

(4.1) P (w′,u) 6⊆ F =⇒ u ≤ r.
Choose r′ ∈ I satisfying θ(w) < r′ < r. Then Pr′ 6⊆ Qθ(w) and Pw 6⊆ Qw′, so by
the definition of F we have F ⊆ Q(w′,r′), which is equivalent to P (w′,r′) 6⊆ F.
Applying implication (4.1) with u = r′ now gives the contradiction r ≤ r′,
and we have proved F←([0,r)) 6⊆ Qs =⇒ rn(Ps) ⊆ F←([0,r)). Hence
F←([0,r)) ∈ τ since (rn,Rn) ∈ U and τ = τU. This proves continuity since
F←(I) = S ∈ τ.

It remains to show that Pa ⊆ f←(P0) and F←(Q1) ⊆ G. Suppose first that
Pa 6⊆ f←(P0). Then we have b ∈ I with f 6⊆ Q(a,b) and Pb 6⊆ P0, that is b > 0.
By the definition of f we have b′ ∈ I with P (a,b′) 6⊆ Q(a,b) and v ∈ S with
Pa 6⊆ Qv satisfying b′ ≤ θ(v). Hence 0 < b ≤ b′ ≤ θ(v) whence q(a,v) > 0.
However Pa 6⊆ Qv implies q(a,v) = 0 by Lemma 4.12, which is a contradiction.

If now we suppose F←([0, 1)) 6⊆ G then we have s ∈ S satisfying F←([0, 1)) 6⊆
Qs and Ps 6⊆ G. Hence we have t ∈ I with P (s,t) 6⊆ F and [0, 1) 6⊆ Qt, that
is t < 1. From the definition of F we now have t′ ∈ I with P (s,t) 6⊆ Q(s,t′)



Di-uniform Texture Spaces 177

and v ∈ S with Pv 6⊆ Qs and θ(v) ≤ t′. Hence θ(v) ≤ t′ ≤ t < 1, whence
2q(a,v) < 1 and so ϕ(a,v) < 1 by Lemma 4.12. Hence there exists n ∈ N
with rn 6⊆ Q(a,v) and so r = r0 6⊆ Q(a,v). This leads to r(Pa) 6⊆ Qa and so
G 6⊆ Qv. On the other hand Pv 6⊆ Qs and Ps 6⊆ G give Pv 6⊆ G, and we have
the contradiction G ⊆ Qv.

This completes the proof of complete regularity, and complete coregularity
can be proved in a similar way using the conjugate functions ϕ∗ and q∗ of
Lemma 4.13. �

The converse of the above proposition is also true, but we postpone the proof
until we have discussed initial di-uniformities in the next section.

Definition 4.15. A direlational uniformity U satisfying
d

U = (i,I) is called
separated.

We recall from [7] the following characteristic property of T0 ditopological
spaces:

(τ,κ) is T0 if and only if Qs 6⊆ Qt =⇒ ∃B ∈ τ ∪κ with Ps 6⊆ B 6⊆ Qt.

Theorem 4.16. Let U be a direlational uniformity. Then the uniform ditopol-
ogy (τU,κU) is T0 if and only if U is separated.

Proof. =⇒. We know that i ⊆
d
{d | (d,D) ∈ U}, so suppose

d
{d | (d,D) ∈

U} 6⊆ i. Then we have s,t ∈ S with P (s,t) 6⊆ i for which we have s′ ∈ S
satisfying d 6⊆ Q(s′,t) for all (d,D) ∈ U. Now P (s,t) 6⊆ i implies Pt 6⊆ Ps, which
with Ps 6⊆ Qs′ gives Qt 6⊆ Qs. Since (τU,κU) is T0 we have B ∈ τU ∪ κU
satisfying Pt 6⊆ B 6⊆ Qs. There are two cases to consider:

(a) B ∈ τU. Now B 6⊆ Qs′ implies d[s′] ⊆ B for some (d,D) ∈ U. It follows
that Pt 6⊆ d[s′] = d(Ps′). But now d(Ps′) ⊆ Qt, so by Lemma 1.5(2)
we have d ⊆ Q(s′,t), which is a contradiction.

(b) B ∈ κU. Noting that U has a base of symmetric direlations, a dual
argument again leads to a contradiction.

This completes the proof of
d
{d | (d,D) ∈ U} = i, and

⊔
{D | (d,D) ∈ U} = I

is dual.

⇐=. Take s,t ∈ S with Qs 6⊆ Qt. By the definition of Qs there exists u ∈ S
with Ps 6⊆ Pu and Pu 6⊆ Qt. Take s′, t′ ∈ S satisfying Ps 6⊆ Qs′, Ps′ 6⊆ Pu and
Pu 6⊆ Qt′, Pt′ 6⊆ Qt. Then P (u,s′) 6⊆ i =

d
{d | (d,D) ∈ U} since Ps′ 6⊆ Pu, so

as Pu 6⊆ Qt′ there exists (e,E) ∈ U with e ⊆ Q(t′,s′), whence e(Pt′) ⊆ Qs′.

Now let G =
∨
{Pz | z ∈ S, ∃(d,D) ∈ U with d[z] ⊆ Qs′}. It may be shown

that G ∈ τU (compare Lemma 4.6). Clearly Pt′ ⊆ G, and so G 6⊆ Qt. On the
other hand if d[z] ⊆ Qs′ then Pz ⊆ d[z] ⊆ Qs′ so G ⊆ Qs′ and hence Ps 6⊆ G.
This verifies that (τU,κU) is T0. �



178 S. Özçağ and L. M. Brown

5. Uniform Bicontinuity and Initial Di-uniformities

In order to define uniform bicontinuity it will be necessary to say what we
mean by the inverse of a direlation and of a dicover under a difunction. We
begin with the following:

Definition 5.1. Let (S, S), (T, T) be textures, (r,R) a direlation on (T, T) and
(f,F) a difunction on (S, S) to (T, T). Then

(f,F)−1(r) =
∨
{P (s1,s2) | ∃Ps1 6⊆ Qs′1 so that P (s′1,t1) 6⊆ F, f 6⊆ Q(s2,t2)

=⇒ P (t1,t2) ⊆ r},

(f,F)−1(R) =
⋂
{Q(s1,s2) | ∃Ps′1 6⊆ Qs1 so that f 6⊆ Q(s′1,t1),P (s2,t2) 6⊆ F,

=⇒ R ⊆ Q(t1,t2)},
(f,F)−1(r,R) = ((f,F)−1(r), (f,F)−1(R)).

Remark 5.2. In Definition 5.1, P (t1,t2) ⊆ r may be replaced by r 6⊆ Q(t1,t2)
and R ⊆ Q(t1,t2) by P (t1,t2) 6⊆ R. Indeed, if s

′
1, s2 satisfy the conditions in the

definition of (f,F)−1(r), and P (s′1,t1) 6⊆ F, f 6⊆ Q(s2,t2), then we may choose
t′2 with f 6⊆ Q(s2,t′2), P (s2,t′2) 6⊆ Q(s2,t2). This gives us P (t1,t′2) ⊆ r, and so
r 6⊆ Q(t1,t2) since Pt′2 6⊆ Qt2 . The opposite direction is trivial, and the second
property is dual.

It is trivial to verify that (f,F)−1(r,R) is indeed a direlation on (S, S), and
we omit the proof. Let us examine the properties of this inverse mapping.

Proposition 5.3. Let (f,F) be a difunction on (S, S) to (T, T). Then

(f,F)−1(iT ,IT ) = (iS,IS),

where (iS,IS), (iT ,IT ) are the identity direlations on (S, S), (T, T) respectively.

Proof. To establish (f,F)−1(iT ) = iS we first suppose that iS 6⊆ (f,F)−1(iT ).
Then iS 6⊆ Q(s,s′) and P (s,s′) 6⊆ (f,F)−1(iT ) for some s,s′ ∈ S. We have
Ps 6⊆ Qs′ since iS 6⊆ Q(s,s′). By Definition 5.1 there exists w1,w2 ∈ T satisfying
P (s′,w1) 6⊆ F, f 6⊆ Q(s′,w2) and P (w1,w2) 6⊆ iT . On the other hand DF2 implies
Pw1 6⊆ Qw2 . Hence iT 6⊆ Q(w1,w2) which contradicts P (w1,w2) 6⊆ iT .

The proof of the reverse inclusion is similar and the proof of the dual equality
(f,F)−1(IT ) = IS is left to the reader. �

Corollary 5.4. Let (f,F) be a difunction on (S, S) to (T, T). If (r,R) is
a reflexive direlation on (T, T) then (f,F)−1(r,R) is a reflexive direlation on
(S, S).

Proof. Let (r,R) be reflexive. Then

(iT ,IT ) v (r,R) =⇒ (iS,IS) = (f,F)−1(iT ,IT ) v (f,F)−1(r,R)
by the proposition so (f,F)−1(r,R) is reflexive. �



Di-uniform Texture Spaces 179

Proposition 5.5. Let (f,F) be a difunction on (S, S) to (T, T) and (r,R) a
direlation on (T, T). Then

((f,F)−1(r,R))← = (f,F)−1((r,R)←).

Proof. It is clearly sufficient to establish ((f,F)−1(r))← = (f,F)−1(r←), since
the dual equality ((f,F)−1(R))← = (f,F)−1(R←) then follows by replacing r
by R← and taking the inverse of both sides.

First suppose that ((f,F)−1(r))← 6⊆ (f,F)−1(r←). Then ((f,F)−1(r))← 6⊆
Q(s,s′) and P (s,s′) 6⊆ (f,F)−1(r←) for some s,s′ ∈ S. Since r← is a corelation,
by Remark 5.2 we have u,v ∈ S with P (s,s′) 6⊆ Q(s,v), Pu 6⊆ Qs so that

(5.2) f 6⊆ Q(u,t1), P (v,t2) 6⊆ F =⇒ P (t1,t2) 6⊆ r
←

for all t1, t2 ∈ T . On the other hand ((f,F)−1(r))← 6⊆ Q(s,s′) gives (f,F)−1(r) ⊆
Q(s′,s) and so P (s′,u) 6⊆ (f,F)−1(r). Since Ps′ 6⊆ Qv we have w1,w2 ∈ T for
which P (v,w1) 6⊆ F, f 6⊆ Q(u,w2) and r ⊆ Q(w1,w2) by Remark 5.2. Putting
t1 = w2, t2 = w1 in the implication (5.2) now gives P (w2,w1) 6⊆ r

←, so giving
the contradiction r 6⊆ Q(w1,w2). Hence ((f,F)

−1(r))← ⊆ (f,F)−1(r←).
The proof of (f,F)−1(r←) ⊆ ((f,F)−1(r))← is similar, and is omitted. �

Corollary 5.6. Let (f,F) be a difunction on (S, S) to (T, T). If (r,R) is a
symmetric direlation on (T, T) then (f,F)−1(r,R) is a symmetric direlation on
(S, S).

Proof. Immediate. �

Proposition 5.7. Let (f,F) be a difunction on (S, S) to (T, T), (p,P) and
(q,Q) direlations on (T, T). Then

(f,F)−1(p,P) ◦ (f,F)−1(q,Q) v (f,F)−1((p,P) ◦ (q,Q)).

Proof. Suppose that (f,F)−1(p)◦(f,F)−1(q) 6⊆ (f,F)−1(p◦q). By the defini-
tion of composition of relations we have s,u,z ∈ S with P (s,u) 6⊆ (f,F)−1(p◦q),
(f,F)−1(q) 6⊆ Q(s,z) and (f,F)−1(p) 6⊆ Q(z,u). By R2 there exists s′ ∈ S with
Ps 6⊆ Qs′ and (f,F)−1(q) 6⊆ Q(s′,z). By Definition 5.1, P (s,u) 6⊆ (f,F)−1(p◦q)
gives w1,w2 ∈ S satisfying P (s′,w1) 6⊆ F, f 6⊆ Q(u,w2) and P (w1,w2) 6⊆ p◦q.

On the other hand from (f,F)−1(q) 6⊆ Q(s′,z) we have z′,s′′ ∈ S with
P (s′,z′) 6⊆ Q(s′,z), Ps′ 6⊆ Qs′′ for which

(5.3) P (s′′,t1) 6⊆ F, f 6⊆ Q(z′,t2) =⇒ q 6⊆ Q(t1,t2)
for all t1, t2 ∈ T by Remark 5.2. Likewise, (f,F)−1(p) 6⊆ Q(z,u) gives u′,z′′ ∈ S
with P (z,u′) 6⊆ Q(z,u), Pz 6⊆ Qz′′ for which

(5.4) P (z′′,v1) 6⊆ F, f 6⊆ Q(u′,v2) =⇒ p 6⊆ Q(v1,v2)
for all v1,v2 ∈ T . Finally Pz′ 6⊆ Qz′′ so by DF1 we have v ∈ T for which
f 6⊆ Q(z′,v) and P (z′′,v) 6⊆ F.



180 S. Özçağ and L. M. Brown

By CR1, P (s′,w1) 6⊆ F and Ps′ 6⊆ Qs′′ gives P (s′′,w1) 6⊆ F so implication (5.3)
may be applied with t1 = w1, t2 = v to give q 6⊆ Q(w1,v). Likewise (5.4) may be
applied with v1 = v, v2 = w2 to give p 6⊆ Q(v,w2). This gives the contradiction
P (w1,w2) ⊆ p◦q and we have shown (f,F)

−1(p)◦(f,F)−1(q) ⊆ (f,F)−1(p◦q).
The proof of (f,F)−1(P ◦ Q) ⊆ (f,F)−1(P) ◦ (f,F)−1(Q) is dual to the

above, and is omitted. �

Corollary 5.8. Let (f,F) be a difunction on (S, S) to (T, T). If (r,R) is a
transitive direlation on (T, T) then (f,F)−1(r,R) is a transitive direlation on
(S, S).

Proof. Straightforward. �

Now let us make the following definition.

Definition 5.9. Let U be a direlational uniformity on (S, S), V a direlational
uniformity on (T, T) and (f,F) a difunction from (S, S) to (T, T). If (d,D) ∈
V =⇒ (f,F)−1(d,D) ∈ U the difunction (f,F) is said to be U–V uniformly
bicontinuous.

Example 5.10. Let U be a direlational uniformity on (S, S). Then the identity
difunction (i,I) on (S, S) is U–U–uniformly bicontinuous. To see this it is clearly
sufficient to note that

(i,I)−1(r,R) = (r,R)
for all direlations (r,R) on (S, S). The proof of this equality is straightforward
and is left to the interested reader.

Now let us consider the composition of uniformly bicontinuous difunctions.
The following lemma will be useful.

Lemma 5.11. Let (S, S), (T, T) and (W, W) be textures, (f,F) a difunction on
(S, S) to (T, T), (g,G) a difunction on (T, T) to (W, W) and (r,R) a direlation
on (W, W). Then

(f,F)−1((g,G)−1(r,R)) = ((g,G) ◦ (f,F))−1(r,R).

Proof. First suppose that (f,F)−1((g,G)−1(r)) 6⊆ (g◦f,G◦F)−1(r). Then we
have s,s′,u ∈ S with P (s,u) 6⊆ (g ◦f,G◦F)−1(r), Ps 6⊆ Qs′, satisfying

(5.5) P (s′,t1) 6⊆ F, f 6⊆ Q(u,t2) =⇒ (g,G)
−1(r) 6⊆ Q(t1,t2)

for all t1, t2 ∈ T by Remark 5.2. Now P (s,u) 6⊆ (g ◦f,G◦F)−1(r), Ps 6⊆ Qs′,
gives w1,w2 ∈ W for which P (s′,w1) 6⊆ G◦F, g◦f 6⊆ Q(u,w2) and r ⊆ Q(w1,w2).

Now we obtain w′1 ∈ W , v1 ∈ T with P (s′,v1) 6⊆ F, P (v1,w′1) 6⊆ G, and
w′2 ∈ W , v2 ∈ T with f 6⊆ Q(u,v2) and g 6⊆ Q(v2,w′2). Hence we may apply (5.5)
with t1 = v1, t2 = v2 to give (g,G)−1(r) 6⊆ Q(v1,v2). Hence we have v

′
1,v
′
2 ∈ T

satisfying P (v1,v′2) 6⊆ Q(v1,v2), Pv1 6⊆ Qv′1 , for which

(5.6) P (v′1,z1) 6⊆ G, g 6⊆ Q(v′2,z2) =⇒ r 6⊆ Q(z1,z2)



Di-uniform Texture Spaces 181

for all z1,z2 ∈ W by Remark 5.2. Now P (v1,w′1) 6⊆ G, Pv1 6⊆ Qw′1 and Pv1 6⊆ Qv′1
gives P (v′1,w1) 6⊆ G. Also, g 6⊆ Q(v2,w′2), Pv′2 6⊆ Qv2 and Pw′2 6⊆ Qw2 gives f 6⊆
Q(v′2,w2)

. Hence we may apply (5.6) with z1 = w1, z2 = w2 to give r 6⊆ Q(w1,w2),
which is a contradiction. Hence (f,F)−1((g,G)−1(r)) ⊆ (g ◦ f,G ◦ F)−1(r),
and the proof of the reverse inclusion is similar and is omitted.

The proof of the dual equality (f,F)−1((g,G)−1(R)) = (g ◦f,G◦F)−1(R)
is left to the interested reader. �

The following is now immediate from the definitions:

Proposition 5.12. Uniform bicontinuity is preserved under composition of
difunctions.

As expected we also have:

Proposition 5.13. Let (τk,κk), k = 1, 2, be the uniform ditopology of the dire-
lational uniformity Uk on (Sk, Sk), and let (f,F) be a difunction on (S1, S1) to
(S2, S2). Then if (f,F) is U1–U2 uniformly bicontinuous it is (τ1,κ1)–(τ2,κ2)
bicontinuous.

Proof. Take G ∈ τ2 and s ∈ S1 with F←(G) 6⊆ Qs. Then we have t ∈ S2
with P (s,t) 6⊆ F and G 6⊆ Qt, whence by Lemma 4.3 there exists (d,D) ∈ U2
satisfying d[t] ⊆ G. Let (e,E) = (f,F)−1(d,D) ∈ U1. It may be shown that
e[s] ⊆ F←(G), whence F←(G) ∈ τ1, again by Lemma 4.3. Hence (f,F) is
τ1–τ2 continuous, and the proof of κ1–κ2 cocontinuity is dual to this. �

Now let us turn our attention to the notion of initial di-uniformity.

Theorem 5.14. Let (S, S) be a texture, Vi, i ∈ I, direlational uniformities on
the textures (Ti, Ti) and (fi,Fi) difunctions on (S, S) to (Ti, Ti), i ∈ I. Then
the family

(fi,Fi)
−1(di,Di), (di,Di) ∈ Vi, i ∈ I

is a subbase for a direlational di-uniformity U on (S, S).

Proof. Let U = {(d,D) ∈ DR | ∃i1, . . . , in ∈ I, (dik,Dik) ∈ Vik, 1 ≤ k ≤
n with

dn
k=1(fik,Fik)

−1(dik,Dik) v (d,D)}. We must verify conditions (1)–
(5) of Definition 3.1. Clearly (1) is immediate from Proposition 5.3 and (2), (3)
are trivial from the definition of U. If (d,D) ∈ U then we have (dik,Dik) ∈ Vik,
1 ≤ k ≤ n, with

dn
k=1(fik,Fik)

−1(dik,Dik) v (d,D). For each k we may
choose (eik,Eik) ∈ Vik satisfying (eik,Eik) ◦ (eik,Eik) v (dik,Dik). If we set



182 S. Özçağ and L. M. Brown

(e,E) =
dn
k=1(fik,Fik)

−1(eik,Eik) then (e,E) ∈ U and

(e,E) ◦ (e,E) v
nl

k=1

((fik,Fik)
−1(eik,Eik) ◦ (fik,Fik)

−1(eik,Eik))

v
nl

k=1

(fik,Fik)
−1((eik,Eik) ◦ (eik,Eik))

v
nl

k=1

(fik,Fik)
−1(dik,Dik) v (d,D),

by Proposition 1.9 (7) and Proposition 5.7. Hence (4) is satisfied. Finally, (5)
may be verified in a similar way using Proposition 1.9 (4) and Proposition 5.5.

�

Definition 5.15. The di-uniformity U on (S, S) defined in Theorem 5.14 is
called the initial direlational di-uniformity on (S, S) defined by the spaces
(Ti, Ti, Vi) and the difunctions (fi,Fi), i ∈ I.

Clearly the initial di-uniformity is the coarsest di-uniformity on (S, S) for
which the difunctions (fi,Fi) are U–Vi uniformly bicontinuous for all i ∈ I.

We are now in a position to prove the converse of Theorem 4.14, and so
complete our characterization of di-uniformizable ditopological texture spaces.

Theorem 5.16. (S, S,τ,κ) is di-uniformizable if and only if it is completely
biregular.

Proof. It remains to show that if (τ,κ) is completely biregular then there ex-
ists a compatible di-uniformity. Let U denote the initial direlational unifor-
mity generated by the family of all bicontinuous difunctions from (S, S,τ,κ) to
(I, I,τI,κI). We show that (τ,κ) = (τU,κU).

First take G ∈ τU and G 6⊆ Qs. Then there exist z,s′,s′′,s′′′,w ∈ S so
that G 6⊆ Qz, Pz 6⊆ Qs′, Ps′ 6⊆ Qs′′, Ps′′ 6⊆ Qs′′′, Ps′′′ 6⊆ Qw and Pw 6⊆ Qs.
Choose (d,D) ∈ U with d[z] ⊆ G. Now there exist (τ,κ)– (τI,κI) bicontinuous
difunctions (f1,F1), . . . , (fn,Fn) and � > 0 for which

e = (f1,F1)
−1(d�) u . . .u (fn,Fn)−1(d�) ⊆ d.

Since Ps′′′ 6⊆ Qw, by DF1 there exists ri ∈ I for each i = 1, . . . ,n, so that
fi 6⊆ Q(s′′′,ri) and P (w,ri) 6⊆ Fi. We show that

(a)
⋂n
i=1 F

←
i ([0,ri + �)) ⊆ e[z] ⊆ d[z] ⊆ G, and

(b)
⋂n
i=1 F

←
i ([0,ri + �)) 6⊆ Qs,

from which it follows at once that G ∈ τ.
Suppose that (a) is false. Take u,u′,u′′ ∈ S with

⋂n
i=1 F

←
i ([0,ri+�)) 6⊆ Qu′′,

Pu′′ 6⊆ Qu, Pu 6⊆ Qu′ and Pu′ 6⊆ e[z]. Now for each i = 1, . . . ,n we have ti ∈ I
with P (u′′,ti) 6⊆ Fi and [0,ri + �) 6⊆ Qti, that is ti < ri + �. Take any v1,v2 ∈ I
with

P (s′′′,v1) 6⊆ Fi and fi 6⊆ Q(u′′,v2).



Di-uniform Texture Spaces 183

By DF1 we have Pv1 6⊆ Qri and Pti 6⊆ Qv2 , whence ri ≤ v1 and v2 ≤ ti. Thus
v2 ≤ ti < ri + � ≤ v1 + �, so P (v1,v2) ⊆ d�. Since Ps′′ 6⊆ Qs′′′ and Pu′′ 6⊆ Qu we
deduce that

(fi,Fi)
−1(d�) 6⊆ Q(s′′,u) ∀i = 1, . . . ,n,

whence P (s′,u) ⊆ (f1,F1)−1(d�) u . . .u (fn,Fn)−1(d�). Since Pu 6⊆ Qu′ we now
have e 6⊆ Q(s′,u′). On the other hand Pu′ 6⊆ e[z] gives v ∈ S with Pu′ 6⊆ Qv for
which e 6⊆ Q(x,v) =⇒ Pz ⊆ Qx∀x ∈ S. From the above we have e 6⊆ Q(s′,v)
so setting x = s′ in the above implication leads to the contradiction Pz ⊆ Qs′.

To prove (b) it will suffice to show Pw ⊆
⋂n
i=1 F

←
i ([0,ri + �)), so assume the

contrary. Now we have j, 1 ≤ j ≤ n, and Pw 6⊆ Qw′ for which P (w′,y) 6⊆ Fj =⇒
[0,rj + �) ⊆ Qy. However P (w,rj) 6⊆ Fj and Pw 6⊆ Qw′ gives P (w′,rj) 6⊆ Fj, and
we obtain the contradiction [0,rj + �) ⊆ [0,rj) by taking y = rj in the above
implication.

Conversely, take G ∈ τ and s ∈ S with G 6⊆ Qs. Since (τ,κ) is completely
regular there exists a bicontinuous difunction (f,F) on (S, S) to (I, I) for which
f←(P0) 6⊆ Qs and F←(Q1) ⊆ G. Take � > 0 and define e = (f,F)−1(d�). Then
e ∈ U and we will show that e[s] ⊆ G, whence G ∈ τU.

It will be sufficient to show e[s] ⊆ F←(Q1), so assume the contrary and take
v ∈ S with e[s] 6⊆ Qv and Pv 6⊆ F←(Q1). The latter gives us v′ ∈ S with

(5.7) P (v′,w) 6⊆ F =⇒ Q1 ⊆ Qw =⇒ w = 1,

and the former gives z ∈ S with e 6⊆ Qv and Ps 6⊆ Qz. Now we have v′′ ∈ S
with P (z,v′′) 6⊆ Q(z,v) and z′ ∈ S with Pz 6⊆ Qz′ so that

(5.8) P (z′,t1) 6⊆ F, f 6⊆ Q(v′,t2) =⇒ P (t1,t2) ⊆ d� =⇒ t2 < t1 + �.

Clearly Pv′′ 6⊆ Qv′ so by DF1 there exists t′ ∈ I with f 6⊆ Q(v′′,t′) and P (v′,t′) 6⊆
F. Now (7) with w = t′ gives t′ = 1 and so f 6⊆ Q(v′′,1).

On the other hand, from f←(P0) 6⊆ Qs we have s′ ∈ S with f←(P0) 6⊆ Qs′,
Ps′ 6⊆ Qs. Hence we have u ∈ S with Pu 6⊆ Qs′ so that

(5.9) f 6⊆ Q(u,w) =⇒ Pw ⊆ P0 =⇒ w = 0.

Clearly Pu 6⊆ Qz so by DF1 we have t ∈ I so that f 6⊆ Q(u,t), P (z′,t) 6⊆ F. Now
(9) with w = t gives t = 0, so P (z′,0) 6⊆ F and (8) with t1 = 0, t2 = 1 gives the
contradiction 1 < �.

We have now established τ = τU, and a dual proof gives κ = κU, so the
proof is complete. �

Before leaving the topics of uniform bicontinuity and initial di-uniformity
we must see how these should be defined for dicovering di-uniformities. The
following gives a fairly obvious notion of inverse image of a dicover under a
difunction.



184 S. Özçağ and L. M. Brown

Definition 5.17. Let (S, S), (T, T) be textures, (f,F) a difunction on (S, S)
to (T, T) and C a dicover of (T, T). Then

(f,F)−1(C) = {(F←(A),f←(B)) | A C B}.

It is a straightforward matter to verify that (f,F)−1(C) is a dicover of (S, S),
but the authors do not know if this inverse operation preserves the property
of being anchored, or even of being refined by the dicover P. This will cause
some technical difficulties, but will not prevent us using this operation to char-
acterize uniform bicontinuity and initial diuniformities in terms of dicovers, as
we will see. We begin by relating this inverse image with that given earlier for
direlations.

Proposition 5.18. Let (f,F) : (S, S) → (T, T) be a difunction and (d,D) a
reflexive direlation on (T, T). Then

γ((f,F)−1(d,D)) ≺ ((f,F)−1(γ(d,D)))∆.

Proof. Let D = γ(d,D) and C = (f,F)−1(D). If we set c = (f,F)−1(d)
and C = (f,F)−1(D) we must verify c[s] ⊆ St(C,Ps) and CSt(C,Qs) ⊆ C[s].
We prove the first inclusion, the second being dual. Recall that c[s] = c(Ps)
and suppose that c(Ps) 6⊆ St(C,Ps). Now we have u ∈ T with c(Ps) 6⊆ Qu,
Pu 6⊆ St(C,Ps) and hence s′ ∈ S with c 6⊆ Q(s′,u), Ps 6⊆ Qs′. By Remark 5.2
we have P (s′,u′) 6⊆ Q(s′,u) and Ps′ 6⊆ Qs′′ for which

(5.10) P (s′′,t1) 6⊆ F, f 6⊆ Q(u′,t2) =⇒ d 6⊆ Q(t1,t2)
for all t1, t2 ∈ T . On the other hand, Pu 6⊆ St(C,Ps) =

∨
{f←(d[z]) | z ∈

T, Ps 6⊆ F←(D[z])} = f←(
∨
{d[z] | z ∈ T, Ps 6⊆ F←(D[z])}) by Proposi-

tion 1.6 (7), and so we have w ∈ T satisfying f 6⊆ Q(u,w) and Pw 6⊆
∨
{d[z] |

z ∈ T, Ps 6⊆ F←(D[z])}.
Since Pu′ 6⊆ Qu we have f 6⊆ Q(u′,w). On the other hand applying DF1

to Ps′ 6⊆ Qs′′ gives v ∈ T satisfying f 6⊆ Q(s′,v) and P (s′′,v) 6⊆ F. We may
now apply implication (5.10) with t1 = v, t2 = w to give d 6⊆ Q(v,w). This is
equivalent to d[v] 6⊆ Qw, and so Pw ⊆ d[v]. To obtain a contradiction it will
therefore suffice to show that F←(D[v]) ⊆ Qs′, for then Ps 6⊆ F←(D[v]) and
so Pw 6⊆ d[v].

Suppose, therefore, that F←(D[v]) 6⊆ Qs′. Then we have t ∈ T with P (s′,t) 6⊆
F and D[v] 6⊆ Qt. Using DF2 now gives Pt 6⊆ Qv, whence D[v] ⊆ Qv ⊆ Qt
since D is reflexive. This contradiction completes the proof. �

Proposition 5.19. Let (f,F) : (S, S) → (T, T) be a difunction and (c,C),
(d,D) reflexive direlations on (T, T). Then

(d,D) ◦ (d,D)← v (c,C) =⇒ δ((f,F)−1(γ(d,D))) v (f,F)−1(c,C).

Proof. Assume that (d,D)◦(d,D)← v (c,C), i.e. d◦D← ⊆ c and C ⊆ D◦d←.
Let D = γ(d,D), E = (f,F)−1(D), and assume that d(E) 6⊆ (f,F)−1(c).

Now we have s,s′ ∈ S with P (s,s′) 6⊆ (f,F)−1(c) and t ∈ T with Ps 6⊆



Di-uniform Texture Spaces 185

f←(D(Qt)), F←(d(Pt)) 6⊆ Qs′. Hence we have v ∈ T with f 6⊆ Q(s,v),
Pv 6⊆ D(Qt), and v′ ∈ T with P (s′,v′) 6⊆ F, d(Pt) 6⊆ Qv′. Also, by R2 for
the relation f, we have u ∈ S with Ps 6⊆ Qu and f 6⊆ Q(u,v). Hence, since
P (s,s′) 6⊆ (f,F)−1(c), there exists t1, t2 ∈ T such that P (u,t1) 6⊆ F, f 6⊆ Q(s′,t2)
and P (t1,t2) 6⊆ c.

On the other hand, from d(Pt) 6⊆ Qv′ we have d 6⊆ Q(t,v′) and from Pv 6⊆
D(Qt) we have P (t,v) 6⊆ D, that is D← 6⊆ Q(v,t). Since (f,F) is a difunction,
P (s′,v′) 6⊆ F and f 6⊆ Q(s′,t2) imply Pv′ 6⊆ Qt2 by DF2, so d 6⊆ Q(t,t2). Likewise,
Pt1 6⊆ Qv and so D← 6⊆ Q(t1,t) by R1 for the relation D

←. We now obtain
P (t1,t2) ⊆ d◦D

← ⊆ c, which is a contradiction.
The proof of D(E) ⊆ (f,F)−1(C) is dual to the above, and is omitted. �

With the notation of Theorem 3.7 we now have:

Proposition 5.20. Let (f,F) : (S, S) → (T, T) be a difunction. If U, V are
direlational di-uniformities on (S, S), (T, T), respectively, then (f,F) is U–V
uniformly bicontinuous if and only if C ∈ Γ(V) =⇒ (f,F)−1(C)∆ ∈ Γ(U).

Proof. Suppose (f,F) is U–V uniformly bicontinuous and take C ∈ Γ(V). Now
we have (c,C) ∈ V with γ(c,C) ≺ C, so (d,D) = (f,F)−1(c,C) ∈ U and
γ(d,D) ∈ Γ(U). However γ(d,D) ≺ (f,F)−1(C)∆ by Proposition 5.18, so
(f,F)−1(C)∆ ∈ Γ(U).

Conversely suppose C ∈ Γ(V) =⇒ (f,F)−1(C)∆ ∈ Γ(U) and take (e,E) ∈
V. Choose a symmetric (c,C) ∈ V with (c,C) ◦ (c,C) v (e,E) and (d,D) ∈
V with (d,D) ◦ (d,D)← v (c,C). By Corollary 5.6, (f,F)−1(c,C) is also
symmetric, and it is reflexive by Corollary 5.4. Hence by Theorem 2.7 (1) and
Proposition 5.7,

δ(γ((f,F)−1(c,C))) = (f,F)−1(c,C) ◦ (f,F)−1(c,C)
v (f,F)−1((c,C) ◦ (c,C))
v (f,F)−1(e,E).

By Proposition 5.19, δ((f,F)−1(γ(d,D)) v (f,F)−1(c,C) and so

δ((f,F)−1(γ(d,D))∆) = δ(γ(δ((f,F)−1(γ(d,D))))

v δ(γ((f,F)−1(c,C))
v (f,F)−1(e,E)

by Theorem 2.7 (2). Since (d,D) ∈ V, C = γ(d,D) ∈ Γ(V) and so, by hypothe-
sis, (f,F)−1(γ(d,D))∆ = (f,F)−1(C)∆ ∈ Γ(U). Hence δ((f,F)−1(γ(d,D))∆) ∈
∆(Γ(U)) = U by Theorem 3.7 (3). It follows from the above inclusion that
(f,F)−1 (e,E) ∈ U, and we have shown that (f,F) is U–V uniformly bicontin-
uous. �

This justifies the following definition.



186 S. Özçağ and L. M. Brown

Definition 5.21. Let υ, ν be dicovering uniformities on (S, S), (T, T) respec-
tively and (f,F) : (S, S) → (T, T) a difunction. Then (f,F) is called υ–ν
uniformly bicontinuous if C ∈ ν =⇒ (f,F)−1(C)∆ ∈ υ.

Finally we have the following:

Proposition 5.22. Let (S, S) be a texture and for i ∈ I let (Ti, Ti, Vi) be a
direlational di-uniform texture space and (fi,Fi) : (S, S) → (Ti, Ti) a difunc-
tion. If U is the initial direlational uniformity on (S, S) for the given system,
the family( n∧

k=1

(fik,Fik)
−1(Cik)

∆

)∆
, n ∈ N+, ik ∈ I, Cik ∈ Γ(Vik), 1 ≤ k ≤ n,

is a base for the dicovering di-uniformity Γ(U).

Proof. Take C ∈ Γ(U). Then there exists (e,E) ∈ U satisfying γ(e,E) ≺ C, and
hence ik ∈ I, (eik,Eik) ∈ Vik for 1 ≤ k ≤ n with

dn
k=1(fik,Fik)

−1(eik,Eik) v
(e,E). If we choose a symmetric (cik,Cik) ∈ Vik with (cik,Cik) ◦ (cik,Cik) v
(eik,Eik), and then (dik,Dik) ∈ Vik with (dik,Dik) ◦ (dik,Dik)

← v (cik,Cik),
we have δ((fik,Fik)

−1(γ(dik,Dik))
∆ v (fik,Fik)

−1(eik,Eik), exactly as in the
proof of Proposition 5.20. In view of Proposition 2.9 (2) we deduce

δ

( n∧
k=1

(fik,Fik)
−1(γ(dik,Dik))

∆

)
v

nl

k=1

δ
(
(fik,Fik)

−1(γ(dik,Dik))
∆
)

v
nl

k=1

(
(fik,Fik)

−1(eik,Eik)
)
v (e,E).

Applying γ to both sides and using Theorem 2.7 (2) now gives( n∧
k=1

(fik,Fik)
−1(Cik)

∆

)∆
≺ C,

where we have set Cik = γ(dik,Dik) ∈ Γ(Vik). It remains to show that the
dicover on the left belongs to Γ(U). Now D = γ(

dn
k=1(fik,Fik)

−1(dik,Dik)) ∈
Γ(U) is anchored by Proposition 2.3, and D ≺

∧n
k=1 γ((fik,Fik)

−1(dik,Dik) by
Proposition 2.8 (1), so by Lemma 2.2 (i) we have

D ≺
( n∧
k=1

γ((fik,Fik)
−1(dik,Dik))

)∆
≺
( n∧
k=1

(fik,Fik)
−1(Cik)

∆

)∆
by Proposition 5.18, and this gives the required result. �

In view of the above proposition, the following definition is compatible with
Definition 5.15.

Definition 5.23. Let (S, S) be a texture and for each i ∈ I let (Ti, Ti,νi) be a
dicovering di-uniform texture space and (fi,Fi) : (S, S) → (Ti, Ti) a difunction.



Di-uniform Texture Spaces 187

Then the covering di-uniformity υ on (S, S) with base{ n∧
k=1

(
(fi,Fi)

−1(Cik)
∆

)∆
| ik ∈ I, Cik ∈ νik, 1 ≤ k ≤ n, n ∈ N

+

}
is called the initial covering di-uniformity on (S, S) for the spaces (Ti, Ti,νi)
and difunctions (fi,Fi), i ∈ I.

It is not known if the above results and definitions can be simplified in
general.

6. Dimetrics and Diuniformities

Definition 6.1. Let (S, S) be a texture, ρ, ρ : S × S → [0,∞) two point
functions. Then ρ = (ρ,ρ) is called a pseudo dimetric on (S, S) if

M1 ρ(s,t) ≤ ρ(s,u) + ρ(u,t) ∀s,u,t ∈ S,
M2 Ps 6⊆ Qt =⇒ ρ(s,t) = 0 ∀s,t ∈ S,
DM ρ(s,t) = ρ(t,s) ∀s,t ∈ S,

CM1 ρ(s,t) ≤ ρ(s,u) + ρ(u,t) ∀s,u,t ∈ S,
CM2 Pt 6⊆ Qs =⇒ ρ(s,t) = 0 ∀s,t ∈ S.

In this case ρ is called the pseudo metric, ρ the pseudo cometric of ρ.

If ρ is a pseudo dimetric which satisfies the conditions

M3 Ps 6⊆ Qu, ρ(u,v) = 0, Pv 6⊆ Qt =⇒ Ps 6⊆ Qt ∀s,t,u,v ∈ S,
CM3 Pu 6⊆ Qs, ρ(u,v) = 0, Pt 6⊆ Qv =⇒ Pt 6⊆ Qs ∀s,t,u,v ∈ S,

it is called a dimetric.

When giving examples it will clearly suffice to give ρ satisfying the metric
conditions, since DM may then be used to define ρ, which will automatically
satisfy the cometric conditions. Note that for a pseudo dimetric to be a dimetric
it is sufficient that ρ(s,t) = 0 =⇒ Ps 6⊆ Qt, but example (4) below shows this
condition is not necessary in general.

Example 6.2. (1) Let (S, S) be any texture and define

ρ(s,t) =

{
0 if Ps 6⊆ Qt,
1 otherwise.

Clearly ρ defines a dimetric ρ, which we will call the discrete dimetric on (S, S).

(2) If d is a (pseudo) metric on X in the usual sense, ρ = (d,d) is a (pseudo)
dimetric on (X,P(X)).

(3) Consider the texture (I, I) and set ρ(s,t) = (t−s) ∨ 0. Then ρ defines a
dimetric ρ on (I, I), which we will call the usual dimetric on (I, I).

(4) Let (L, L) be the texture L = (0, 1], L = {(0,r] | 0 ≤ r ≤ 1}. Again
ρ(s,t) = (t−s) ∨ 0 defines a dimetric, called the usual dimetric on (L, L).



188 S. Özçağ and L. M. Brown

Note that (2) and (4) may be combined to give a rich supply of pseudo
dimetrics on the product of (X,P(X)) and (L, L). Since this is the texture
corresponding to the lattice of classic fuzzy sets on X [4,6] the connection with
fuzzy topology is clear, although we will not pursue this line of enquiry here.

As expected, a (pseudo) dimetric ρ gives rise to a ditopology, which we will
refer to as the (pseudo) metric ditopology of ρ.

Proposition 6.3. Let ρ be a pseudo dimetric on (S, S) and for s ∈ S[, � > 0
define

Nρ� (s) =
∨
{Pt | ∃u ∈ S with Ps 6⊆ Qu, ρ(u,t) < �},

Mρ� (s) =
⋂
{Qt | ∃u ∈ S with Pu 6⊆ Qs, ρ(u,t) < �}.

Then βρ = {Nρ� (s) | s ∈ S[, � > 0} is a base and γρ = {Mρ� (s) | s ∈ S[, � > 0}
a cobase for a ditopology (τρ,κρ) on (S, S).

Proof. By M2 it is clear that Ps ⊆ Nρ� (s) for all s ∈ S[ and so
∨
βρ = S.

Now take s1,s2,s ∈ S[, �1,�2 > 0 with Nρ�1 (s1) ∩N
ρ
�2

(s2) 6⊆ Qs. Choose t ∈ S
with Nρ�1 (s1) ∩N

ρ
�2

(s2) 6⊆ Qt, Pt 6⊆ Qs and then for k = 1, 2 take tk ∈ S with
Ptk 6⊆ Qt so that for some Psk 6⊆ Qs′k, s

′
k ∈ S, we have ρ(s

′
k, tk) < �k. Since

ρ(tk, t) = 0 by M2 we deduce ρ(s′k, t) < �k for k = 1, 2 by M1, so we may
choose � ∈ R satisfying 0 < � < min(�1 −ρ(s′1, t), �2 −ρ(s′2, t)). However it is
now straightforward to verify that

Nρ� (t) ⊆ N
ρ
�1

(s1) ∩Nρ�2 (s2), N
ρ
� (t) 6⊆ Qs

whence by ([6], Theorem 4.3), βρ is a base for some topology τρ on (S, S). The
proof that γρ is a base for some cotopology κρ on (S, S) is dual to this, and is
omitted. �

Clearly the discrete dimetric on (S, S) gives rise to the discrete, codiscrete
ditopology. Likewise, the metric ditopology of the usual dimetric on (I, I) is the
usual ditopology on (I, I), while the same dimetric on (L, L) gives the discrete,
codiscrete ditopology.

Now let us verify that a pseudo dimetric also defines a di-uniformity.

Theorem 6.4. Let ρ be a pseudo dimetric on (S, S).
i) For � > 0 let

r� = r
ρ
� =

∨
{P (s,t) | ∃u ∈ S, Ps 6⊆ Qu and ρ(u,t) < �},

R� = R
ρ
� =

⋂
{Q(s,t) | ∃u ∈ S, Pu 6⊆ Qs and ρ(u,t) < �}.

Then the family {(rρ� ,Rρ� ) | � > 0} is a base for a direlational uniformity
Uρ on (S, S).

ii) The di-uniformity Uρ is separated if and only if ρ is a dimetric.
iii) The uniform ditopology of Uρ coincides with the pseudo metric ditop-

ology of ρ.



Di-uniform Texture Spaces 189

Proof. (i) It is trivial to verify that (r�,R�) is a direlation for all � > 0.

We must verify the conditions of Definition 3.1 for the family

Uρ = {(r,R) | ∃� > 0, (r�,R�) v (r,R)}.

Condition (1) is trivial from M1, CM1; and (2) follows from the definition.
Condition (3) is a consequence of (r�,R�) v (r�1,R�1 ) u (r�2,R�2 ), where � =
min(�1,�2), which is trivial since clearly � ≤ δ =⇒ (r�,R�) v (rδ,Rδ). To
prove (4) we need only show that (r�,R�)2 v (r2�,R2�). If r� ◦ r� 6⊆ r2� there
exists s,t ∈ S with P (s,t) 6⊆ r2� so that for some u,v ∈ S we have Ps 6⊆ Qu,
r� 6⊆ Q(u,v) and r� 6⊆ Q(v,t). By M1, M2 and the definition of r� we obtain
ρ(u,v) < �, ρ(v,t) < �, whence ρ(u,t) ≤ ρ(u,v) + ρ(v,t) < 2�. This gives the
contradiction P (s,t) ⊆ r2� so r� ◦ r� ⊆ r2�, and the dual result R2� ⊆ R� ◦ R�
is proved likewise. Finally (5) follows from (r�,R�)← = (r�,R�). To prove this
we need only show that r←� = R� for any � > 0. Suppose that R� 6⊆ r←� . Then
we have s,t ∈ S with R� 6⊆ Q(s,t) and P (s,t) 6⊆ r←� . Since r←� is a corelation,
P (s,t) 6⊆ r←� is equivalent to r� 6⊆ Q(t,s) and so we have s′ ∈ S satisfying
P (t,s′) 6⊆ Q(t,s) for which we have t′ ∈ S with Pt 6⊆ Qt′ and ρ(t′,s′) < �.
By M1 we have ρ(t,s′) < �, whence ρ(s′, t) < � by DM. Since Ps′ 6⊆ Qs we
obtain R� ⊆ Q(s,t), which is a contradiction. This establishes R� ⊆ r←� , and
the reverse inclusion is proved in the same way. This completes the proof that
Uρ is a direlational uniformity on (S, S).

(ii) It is sufficient to show that M3 is equivalent to
d

�>0
r� ⊆ i. Suppose

that M3 holds but
d

�>0
r� 6⊆ i. Now we have s,t ∈ S with

d

�>0
r� 6⊆ Q(s,t) and

P (s,t) 6⊆ i. Hence we have t′ ∈ S with P (s,t′) 6⊆ Q(s,t) so that for some s′ ∈ S
with Ps 6⊆ Qs′ we have r� 6⊆ Q(s′,t′) ∀� > 0. We deduce ρ(s′, t′) = 0 and so
Ps 6⊆ Qt by M3. However now i 6⊆ Q(s,t), which contradicts P (s,t) 6⊆ i. The
proof that

d

�>0
r� ⊆ i implies M3 is left to the interested reader.

(iii) By Lemma 4.3 the set G ∈ S is open for the uniform ditopology if and
only if G 6⊆ Qs =⇒ ∃� > 0 with r�[s] ⊆ G. Since Ps ⊆ r�(Ps) = r�[s], if we
can show that r�[s] is uniformly open it will follow by ([6], Theorem 4.2) that
the family r�[s], s ∈ S[, � > 0, is a base for τUρ. However, if we take r�[s] 6⊆ Qt,
we then have t′ ∈ S with P (s,t′) 6⊆ Q(s,t), so that ρ(s′, t′) < � for some s′ ∈ S
with Ps 6⊆ Qs′. Since Pt′ 6⊆ Qt we have ρ(t′, t) = 0 and so ρ(s′, t) < �, whence
we may choose δ > 0 with ρ(s′, t) + δ < � and it is now easy to show that
rδ[t] ⊆ r�[s]. This establishes that r�[s] is uniformly open, as required. Finally
it is straightforward to verify that

r�[s] = N
ρ
� (s),

so the family Nρ� (s), s ∈ S[, � > 0, is a base for both τUρ and τρ, whence these
topologies coincide. Likewise, the cotopologies coincide. �



190 S. Özçağ and L. M. Brown

Corollary 6.5. A pseudo metric ditopology is completely biregular. It is T0,
and hence bi–T3 12 [7] and in particular bi–T2 [7], if and only if ρ is a dimetric.

Proof. Immediate from Theorem 6.4, Theorem 4.14 and Theorem 4.16. �

For the dimetric of Example 6.2 (3) we obtain the discrete direlational uni-
formity with base {(i,I)}. The metric di-uniformity of the usual dimetric on
(I, I) is the usual di-uniformity, while the dicovering uniformity corresponding
to the usual dimetric on (L, L) has base {((0,s + �], (0,s− �]) | 0 < s < 1} (cf.
Example 3.8).

Definition 6.6. A direlational uniformity U on (S, S) is called (pseudo) metriz-
able if there exists a (pseudo) dimetric ρ with U = Uρ.

Theorem 6.7. A direlational uniformity U is pseudo metrizable if and only if
it has a countable base. It is metrizable if and only if it is also separated.

Proof. If U is pseudo metrizable there is a pseudo dimetric ρ with U = Uρ. But
now, for example, (rρ

1/n
,R

ρ
1/n

), n ≥ 1, is a countable base of Uρ, and hence of
U.

Conversely, let U have the countable base (bn,Bn), n ≥ 1. Take (d1,D1) ∈ U
symmetric with (d1,D1) v (b1,B1), and by induction for n > 1 choose a
symmetric (dn,Dn) ∈ U so that (dn,Dn)3 v (dn−1,Dn−1) u (bn,Bn). Then
(dn+1,Dn+1)3 v (dn,Dn) for all n = 1, 2, . . . and (dn,Dn) v (dn,Dn)3 v
(bn,Bn), so {(dn,Dn) | n = 1, 2, . . .} is also a base of U. Now let q, q∗ be
as defined in Lemma 4.12 and Lemma 4.13 for the sequence (dn,Dn). Clearly
ρ = (q,q∗) is a pseudo dimetric, so we may consider the direlational uniformity
Uρ. However, Lemma 4.12 (1) and Remark 4.13 (1) immediately give

(dn+1,Dn+1) v (r
ρ
2−n

,R
ρ
2−n

) v (dn,Dn),
whence U = Uρ.

The final statement is immediate from Theorem 4.16 and Corollary 6.5. �

Definition 6.8. A ditopology on (S, S) is called (pseudo) metrizable if it is the
metric ditopology of some (pseudo) dimetric on (S, S).

Theorem 6.9. The ditopology (τ,κ) on (S, S) is pseudo metrizable if and only
if there exists a family Cn, n = 1, 2, . . . of anchored dicovers of (S, S) satisfying
the conditions

(1) Cn+1 ≺(?) Cn for all n ≥ 1.
(2) G ∈ τ ⇐⇒ (G 6⊆ Qs =⇒ ∃n, St(Cn,Ps) ⊆ G).
(3) F ∈ κ ⇐⇒ (Ps 6⊆ F =⇒ ∃n, F ⊆ CSt(Cn,Qs)).

Proof. Suppose first that (τ,κ) = (τρ,κρ) for some pseudo dimetric ρ, and
consider the direlational uniformity Uρ on (S, S). Note that (r

ρ
4−n

,R
ρ
4−n

), n ≥ 1,
is a base of Uρ, whence Cn = γ(r

ρ
4−n

,R
ρ
4−n

), n ≥ 1, is a base of anchored
dicovers for υρ = Γ(Uρ). By the hypothesis and Theorem 6.4 (iii), (τ,κ) is the
uniform ditopology of Uρ, while Uρ = ∆(υρ) by Theorem 3.7 (3). Clearly (2)
and (3) now follow from Proposition 4.4, so it remains to show (1). However



Di-uniform Texture Spaces 191

noting that (rρ� ,R
ρ
� ) is symmetric we may apply Proposition 2.4 to (r

ρ
� ,R

ρ
� )

2 v
(rρ2�,R

ρ
2�) for � = 4

−(n+1) and � = 2 × 4−(n+1) to give
Cn+1 ≺(∆) γ(r

ρ
2×4−(n+1), R

ρ
2×4−(n+1) ) ≺(∆) Cn,

whence Cn+1 ≺(?) Cn by Lemma 2.2 (3 ii), since the dicovers are anchored.
Conversely, suppose that there exists a sequence of anchored dicovers Cn sat-

isfying (1)–(3). Then these form a base for a dicovering uniformity υ. Moreover,
by Proposition 4.4, conditions (2) and (3) imply that the uniform topology of
υ, and hence of U = ∆(υ), is (τ,κ). Clearly δ(Cn), n = 1, 2, . . . is a countable
base of U, so by Theorem 6.7 there is a pseudo dimetric ρ for which the uniform
ditopology of Uρ = U is the metric ditopology of ρ. Hence (τ,κ) = (τρ,κρ), so
(τ,κ) is pseudo metrizable. �

Clearly conditions (2) and (3) may also be given in terms of the dineigh-
bourhood system and Theorem 6.9 is then seen as a ditopological analogue of
the Alexandroff-Urysohn metrization theorem [12].

We end by showing that arbitrary di-uniformities may be defined using
pseudo dimetrics.

Definition 6.10. Let U be a direlational uniformity on (S, S). Then a pseudo
dimetric ρ on (S, S) is called uniform for U if (rρ� ,R

ρ
� ) ∈ U ∀� > 0.

For pseudo metrics ρ1, ρ2 on (S, S), ρ1 ∨ρ2 = (ρ1 ∨ρ2, ρ1 ∨ρ2) is a pseudo
metric on (S, S), and clearly (rρ1� ,R

ρ1
� ) u (rρ2� ,Rρ2� ) = (rρ1∨ρ2� ,Rρ1∨ρ2� ). Hence

the family G of pseudo dimetrics on (S, S) uniform for U has the property
ρ1,ρ2 ∈ G =⇒ ρ1 ∨ρ2 ∈ G. This leads to the following:

Theorem 6.11. Let G be a non-empty family of pseudo dimetrics on (S, S)
which is closed under finite suprema. Then

UG = {(r,R) | ∃ρ ∈ G, � > 0 with (rρ� ,R
ρ
� ) v (r,R)}

is a direlational uniformity on (S, S). Moreover, if U is a direlational unifor-
mity on (S, S) and G the set of pseudo dimetrics uniform for U then UG = U.

Proof. Straightforward. �

Acknowledgements. The authors would like to thank the referees for their
helpful comments

References

[1] L. M. Brown, Relations and functions on textures, preprint.

[2] L. M. Brown and M. Diker, Ditopological texture spaces and intuitionistic sets, Fuzzy
Sets and Systems 98 (1998), 217–224.



192 S. Özçağ and L. M. Brown

[3] L. M. Brown and M. Diker, Paracompactness and full normality in ditopological texture

spaces, J. Math. Anal. Appl. 227 (1998), 144–165.
[4] L. M. Brown and R. Ertürk, Fuzzy sets as texture spaces, I. Representation theorems,

Fuzzy Sets and Systems 110 (2) (2000), 227–236.
[5] L. M. Brown and R. Ertürk, Fuzzy sets as texture spaces, II. Subtextures and quotient

textures, Fuzzy Sets and Systems 110 (2) (2000), 237–245.
[6] L. M. Brown, R. Ertürk and Ş. Dost, Ditopological texture spaces and fuzzy topology, I.

General principles, submitted.
[7] L. M. Brown, R. Ertürk and Ş. Dost, Ditopological texture spaces and fuzzy topology, II.

Separation axioms, preprint, 2002.

[8] M. Diker, Connectedness in ditopological texture spaces, Fuzzy Sets and Systems 108

(1999), 223–230.
[9] G. Gierz, K. H. Hofmann, K. Keimel, J. D. Lawson, M. Mislove and D. S. Scott, A

compendium of continuous lattices (Springer–Verlag, Berlin, 1980).
[10] J. L. Kelley, General Topology (D. Van Nostrand, Princeton, 1955).

[11] G. Sambin, Some points in formal topology, (Proc. Dagstuhl Seminar Topology in Com-

puter Science, June 2000), Theoretical Computer Science, to appear.
[12] J. W. Tukey, Convergence and uniformity in topology, (Ann. of Math. Studies 2, 1940).

[13] A. Weil, Sur les espaces à structure uniforme et sur la topologie générale (Actualités
Sci. Ind. 551, Paris, 1937).

Received July 2002 Revised December 2002

S. Özçağ and L. M. Brown

Hacettepe University,
Faculty of Science,
Department of Mathematics,
06532 Beytepe,
Ankara,
Turkey.

E-mail address : sozcag@hacettepe.edu.tr
brown@hacettepe.edu.tr


	Di-uniform texture spaces. By S. Özçag and L. M. Brown