@
Appl. Gen. Topol. 23, no. 1 (2022), 45-54

doi:10.4995/agt.2022.16128

© AGT, UPV, 2022

Investigation of topological spaces using relators

Gergely Pataki

Department of Analysis, Budapest University of Technology and Economics, Hungary (pataki@math.bme.hu)

Department of Mathematics and Modelling, Hungarian University of Agriculture and Life Sciences,

Hungary

Communicated by H. Dutta

Abstract

In this paper, we define uniformities and topologies as relators and show
the equivalences of these definitions with the classical ones. For this,
we summarize the essential properties of relators, using their theory

from earlier works of Á. Száz. Moreover, we prove implications be-
tween important topological properties of relators and disprove others.
Finally, we show that our earlier analogous definition [G. Pataki, In-
vestigation of proximal spaces using relators, Axioms 10, no. 3 (2021):
143.] for uniformly and proximally filtered property is equivalent to the
topological one.
At the end of this paper, uniformities and topologies are defined in the
same way. This will give us new possibilities to compare these and
other topological structures.

2020 MSC: 54E15; 54A05; 54G15; 54G20; 54D10.

Keywords: (generalized) uniformities; (generalized) topologies; relators.

1. Introduction

At the beginning of the 20th century some mathematicians tried to define
abstract topological structures. The most relevant results due to Poincaré 1895,
Fréchet 1906, Hausdorff 1914, and Kuratowski 1922.

Uniform spaces in terms of relations were introduced by Weil in 1937 [12].
Uniform and topological spaces formulated the recently usable form by Bour-

baki in 1953 [1].

Received 26 August 2021 – Accepted 23 November 2021

http://dx.doi.org/10.4995/agt.2022.16128
https://orcid.org/0000-0002-4630-9949


G. Pataki

After the works of Davis, Pervin, and Nakano [2], [8], and [4], in 1987 Száz
[9] introduced the notion of relators and relator spaces in the following way.

Definition 1.1. A nonvoid family R of relations on a nonvoid set X is called
a relator on X, and the ordered pair (X,R) is called a relator space.

In the last decades, a few authors investigated the interpretation of well-
known topological properties in terms of relators.

For more details, see, for instance, [7] and [6], but for the readers’ conve-
nience, we summarize the necessary notions and notations.

Remark 1.2. With the usual notations, the statement R is a relator on X
means that

X 6= ∅, ∅ 6= R⊂ Exp(X2),
where Exp(X) is the power set of X, and X2 = X ×X.

If R is a relation on X, x ∈ X, and A ⊂ X, then the sets

R(x) = {y ∈ X : (x,y) ∈ R}, and R[A] =
⋃
x∈A

R(x)

are called the images of x and A under R, respectively.

2. Preliminary Concepts

Definition 2.1. If R and S are relations on X, then the composition of R and
S can be defined, such that (R◦S)(x) = R[S(x)] for all x ∈ X.

Moreover, let R−1 = {(y,x) : (x,y) ∈ R}, R0 = ∆X = {(x,x) : x ∈ X} and
Rn = R◦Rn−1, for all n = 1, 2, . . . . Finally, we say that R is

• reflexive if R0 ⊂ R,
• symmetric if R−1 ⊂ R,
• transitive if R2 ⊂ R.

Lemma 2.2. If R is a relation on X, and A,B ⊂ X, then

R[A] ⊂ B ⇐⇒ R−1[X \B] ⊂ X \A.

Definition 2.3. If R is a relator on X, then the relators

R∗ = {S ⊂ X2 : ∃R ∈R : R ⊂ S},

and

R∧ = {S ⊂ X2 : ∀x ∈ X : ∃R ∈R : R(x) ⊂ S(x)},

are called the uniform and the topological refinements of R, respectively. For
more details, see [7].

Moreover, for all n = −1, 0, 1, 2, . . . , we define

Rn = {Rn : R ∈R} .

© AGT, UPV, 2022 Appl. Gen. Topol. 23, no. 1 46



Investigation of topological spaces using relators

Remark 2.4. ∗ and ∧ are really refinements as we defined in [7], that is, they
are self-increasing in the sense that

R⊂S∗ ⇐⇒ R∗ ⊂S∗ and R⊂S∧ ⇐⇒ R∧ ⊂S∧,
or equivalently, they are expansive, increasing, and idempotent, in the sense
that

R⊂R∗, R⊂S =⇒ R∗ ⊂S∗, R∗∗ = R∗

and

R⊂R∧, R⊂S =⇒ R∧ ⊂S∧, R∧∧ = R∧,
for all R and S relators on X.

Moreover, ∧ is ∗-dominating, ∗-invariant, ∗-absorbing and ∗-compatible,
that is if R is a relator on X, then

R∗ ⊂R∧, R∧ = R∧∗, R∧ = R∗∧, R∧∗ = R∗∧.
For all n = −1, 0, 1, 2, . . . the mapping R 7→Rn of relators on X is increas-

ing.
Finally, ∗ is inversion-compatible, that is, for all R relators on X

R∗−1 = R−1∗.
And we have that for all R relators on X

R2∗ = R∗2∗.

The following examples show, that the analog assertions are not true for ∧.

Example 2.5. Let X = {1, 2, 3}, and
R = {∆X ∪{(1, 2)}, ∆X ∪{(3, 2)}

is an elementwise reflexive and transitive relator on X. Now, R∧−1 6⊂ R−1∧,
since ∆X ∈ R∧−1 however ∆X /∈ R−1∧. Moreover, if S = R−1, then ∆X ∈
S−1∧ \S∧−1.

Note, that R =
{
���
���
���

,
���
���
���

}
.

Example 2.6. Let X = {1, 2, 3, 4}, and
R = {∆X ∪{(1, 2), (4, 2), (2, 1), (2, 4)}, ∆X ∪{(1, 3), (4, 3), (3, 1), (3, 4)}}

is an elementwise reflexive and symmetric relator on X. Now, R∧2 6⊂ R2∧,
since R = X2 \{(1, 4), (4, 1)}∈R∧2 however R /∈R2∧.

Note, that R =

{
����
����
����
����

,
����
����
����
����

}
, and R =

����
����
����
����

.

Definition 2.7. If R is a relator on X, then for any x ∈ X and A ⊂ X, we
write:

x ∈ intR(A) if R(x) ⊂ A for some R ∈R,
and

A ∈TR if A ⊂ intR(A).

© AGT, UPV, 2022 Appl. Gen. Topol. 23, no. 1 47



G. Pataki

The relation intR is called the topological interior, and the elements of TR are
called the topologically open subsets induced by R on X.

Theorem 2.8.

int : Exp(Exp(X2)) \{∅}→ Exp(Exp(X) ×X)

is a ∧–increasing set-valued function for relators on X in the sense that

S ⊂R∧ ⇐⇒ intS ⊂ intR
for any two relators R and S on X.

Moreover, it follows that if R is a relator on X, then R∧ is the largest relator
on X such that

intR = intR∧ .

Theorem 2.9.

T : Exp(Exp(X2)) \{∅}→ Exp(Exp(X))

is an increasing set-valued function for relators on X in the sense that

S ⊂R =⇒ TS ⊂TR
for any two relators R and S on X.

Moreover, if R is a relator on X, then

TR = TR∧.

3. A new form of generalized topologies

Definition 3.1. Let R be a relator on X, and let � ∈{∗,∧} be a refinement
for relators on X. We define the followings.

• R is �-reflexive, if R⊂R0�;
• R is �-symmetric, if R⊂R−1�;
• R is �-transitive, if R⊂R2�;
• R is �-fine, if R = R�.

For instance, we say that R is uniformly symmetric or topologically transi-
tive instead of ∗-symmetric or ∧-transitive, respectively.

Following Weil, we say that the relator R on X is a generalized uniformity
on X, and the relator space (X,R) is a generalized uniform space if it is

• uniformly reflexive;
• uniformly symmetric;
• uniformly transitive;
• uniformly fine.

Moreover, we say that the relator R on X, is a generalized topology on X,
and the relator space (X,R) is a generalized topological space if it is

• topologically reflexive;
• topologically transitive;
• topologically fine.

© AGT, UPV, 2022 Appl. Gen. Topol. 23, no. 1 48



Investigation of topological spaces using relators

Definition 3.2. If X is an arbitrary set, and T ⊂ Exp(X) satisfies the follow-
ing axioms, then we say that it is a generalized set-topology on X.

(1) T is closed under arbitrary union, that is
⋃
A∈T for any A⊂T ;

(2) X ∈T .

The following relations were investigated by Davis, Pervin, and Száz.

Definition 3.3. Let A be a subset of X. Then, the relation

DA = A
2 ∪ (X \A) ×X

is called the Davis–Pervin relation on X generated by A.

Some parts of the following theorem were proved in [10] and [11].

Proposition 3.4. If R is a relator on X, and

ψ : P(P(X)) →P(P(X2)), ψ(T ) = {DA : A ∈T},

then

(1) TR is a generalized set-topology for all R relators on X;
(2) ψ(T )∧ is a generalized topology for all ∅ 6= T ⊂P(X);
(3) If T is a generalized set-topology on X, then T = Tψ(T );
(4) If R is a generalized topology on X, then R = ψ(TR)∧.

Proof. (1) If A ⊂ TR and x ∈
⋃
A, then there exists an A ∈ A such that

x ∈ A. It follows that R(x) ⊂ A ⊂
⋃
A for some R ∈R, since A ∈TR.

Therefore,
⋃
A∈TR.

X ∈TR means only that the triviality R(x) ⊂ X for all x ∈ X and
R ∈R. Note that R 6= ∅.

(2) ψ(T )∧ is obviously reflexive and topologically fine. For proving topo-
logically transitivity of ψ(T )∧, that is ψ(T )∧ ⊂ ψ(T )∧2∧ or equiva-
lently ψ(T ) ⊂ ψ(T )∧2∧ let R ∈ ψ(T ) be arbitrary. By the definition
of the Davis–Pervin relations, we have that R2 = R and by using the
expansivity of ∧ the proof is complete.

(3) At first, we prove that T ⊂ Tψ(T ) for all T ⊂ P(X). If A ∈ T , then
DA ∈ ψ(T ). Since DA[A] = A, we have that A ∈Tψ(T ).

On the other hand, let A ∈ Tψ(T ). If A = X, then A ∈ T , since T
is a generalized set-topology.

If A 6= X, and x ∈ A, then there exists a Ux ∈ T such that x ∈ Ux
and Ux = DUx (x) ⊂ A.

A =
⋃
x∈A
{x}⊂

⋃
x∈A

Ux ⊂
⋃
x∈A

A = A

implies that A ∈T since T is closed under arbitrary union.
(4) Let R ∈ ψ(TR)∧ and x ∈ X. There exists an A ∈ TR such that DA ∈

ψ(TR) and DA(x) ⊂ R(x). If x ∈ A then S(x) ⊂ A for some S ∈ R
since A is topologically open, and DA(x) = A implies S(x) ⊂ R(x).
If x /∈ A, then DA(x) = X therefore obviously S(x) ⊂ X = R(x) for

© AGT, UPV, 2022 Appl. Gen. Topol. 23, no. 1 49



G. Pataki

some S ∈R. We proved that ψ(TR)∧ ⊂R∧ for an arbitrary R relator
on X, that is

ψ(TR)∧ ⊂R
if R is topologically fine.

For the converse inclusion let R ∈ R, x ∈ X and A = intR(R(x)).
By the definition of interior, for all y ∈ A there exists an S ∈ R such
that S(y) ⊂ R(x). Topological transitivity of R implies that there
exists a Q ∈ R such that Q[Q(y)] = Q2(y) ⊂ S(y) ⊂ R(x), that is
Q(y) ⊂ A.

It follows that A ∈ TR, and then DA ∈ ψ(TR). Since x ∈ A and
R is reflexive we have that DA(x) = A ⊂ R(x). We proved that
R ∈ ψ(TR)∧ for an arbitrary R ∈R.

�

Several papers including [10] used {R ◦ S : R,S ⊂ R} instead of our R2.
Because of the definition of uniformities, we need our one, but later in Theorem
4.8 we will see that in some cases, these definitions are equivalent.

Theorem 3.5. If R is a relator on X, then R 7→ TR is a bijection of the set
of generalized topologies on X onto the set of generalized set-topologies on X.

Proof. The injectivity and surjectivity are followed by Proposition 3.4 (4) and
(3), respectively. �

4. A new form of topologies

Definition 4.1. Let A be a family of sets, or equivalently A ⊂ Exp(X) for
some set X. We call

Φ(A) =
{⋂
B : ∅ 6= B ⊂A, and B is finite

}
the filtered family of sets generated by A.

Moreover, we say that A is filtered if Φ(A) = A.

Remark 4.2. Since Φ is a refinement for relators on X, we write RΦ instead of
Φ(R), if R is a relator on X. Note also that R is filtered iff RΦ ⊂R.

Moreover, note that Φ is an inversion compatible refinement for relators on
X, that is, if R is a relator on X, then R−1Φ = RΦ−1.

Finally, if R is finite, then RΦ∗ = {
⋂
R}∗.

Remark 4.3. By [6] we know that if R is a relator on X, then R∗Φ = RΦ∗,
R2Φ ⊂RΦ2∗. Now, we can state a similar assertion for topological refinement.

Lemma 4.4. If R is a relator on X, then R∧Φ ⊂RΦ∧.

Proof. It is easy to see that

∃∅ 6= S ⊂R∧ finite :
⋂
S = R =⇒

=⇒ ∀x ∈ X : ∃∅ 6= Q⊂R finite :
⋂
Q(x) ⊂ R(x).

�

© AGT, UPV, 2022 Appl. Gen. Topol. 23, no. 1 50



Investigation of topological spaces using relators

Proposition 4.5. If R is a relator on X and � ∈ {∗,∧}, then the following
assertions are equivalent.

(1) RΦ ⊂R�;
(2) there exists an S relator on X, such that SΦ ⊂S� = R�.

Proof. We need only to prove the (2) =⇒ (1) implication. For this, we note
that if (2) is true, then

RΦ ⊂R�Φ = S�Φ ⊂SΦ� ⊂S�� = S� = R�.

�

Because of the above Proposition, the following definition seems unduly
overcomplicated, but from [6] we know, that this is reasonable.

Definition 4.6. If � is a refinement for relators on X, then we say that
the R relator on X is �-filtered if there exists an S relator on X such that
SΦ ⊂S� = R�.

We use the uniformly filtered and topologically filtered notions instead of
∗-filtered and ∧-filtered.

Note that by Proposition 4.5 we have that R is a uniformly (topologically)
filtered relator on X iff RΦ ⊂ R∗ (RΦ ⊂ R∧), which can be found in the
definition of uniformities by Weil.

Definition 4.7. If the generalized uniformity/generalized topology R on X is
also uniformly/topologically filtered, then we say that R is a uniformity/topo-
logy on X, and (X,R) is a uniform space/topological space.

Moreover, if T is a generalized set-topology on X, such that T is filtered,
then we say that T is a set-topology on X.

Several papers including [7] investigated the following properties.

Theorem 4.8. If R is topologically fine and topologically reflexive (or topolog-
ically filtered), then the following assertions are equivalent.

(1) R is topologically transitive;
(2) ∀x ∈ X,R ∈R : ∃P,Q ∈R : P [Q(x)] ⊂ R(x);
(3) ∀x ∈ X,R ∈R : x ∈ intR(intR(R(x)));
(4) ∀A ⊂ X : intR(A) ∈TR;
(5) ∀x ∈ X,R ∈R : intR(R(x)) ∈TR.

Proof. (1) =⇒ (2) =⇒ (3) =⇒ (4) =⇒ (5) are quite obvious (without extra
conditions). Note only for proving (3) =⇒ (4), that if x ∈ intR(A), that is there
exists an R ∈ R such that R(x) ⊂ A, then by (3) x ∈ intR(intR(R(x))) ⊂
intR(intR(A)) and it follows that there exists an S ∈ R such that S(x) ⊂
intR(A).

Therefore we prove only the (5) =⇒ (1) implication.
For this, let R ∈ R and x ∈ X be fixed and use (5). We have x ∈

intR(R(x)) ∈ TR, that is there exists a Q ∈ R such that Q(x) ⊂ intR(R(x)).

© AGT, UPV, 2022 Appl. Gen. Topol. 23, no. 1 51



G. Pataki

For such a Q, and for all y ∈ Q(x) there exists a Py ∈R such that Py(y) ⊂ R(x).
Now define the S relation on X by the following.

S(y) =



Q(y), if y = x,

Py(y), if y ∈ Q(x) \{x},
X, else.

Note that because of R is topologically fine, we have that S ∈R∧ = R.
If R is topologically reflexive, then intR(R(x)) ⊂ R(x), therefore

S2(x) = S[Q(x)] =
⋃

y∈Q(x)\{x}

Py(y) ∪Q(x) ⊂ R(x).

If R is topologically filtered, then write Q ∩ R in place of Q. Note that
Q∩R ∈RΦ ⊂R∧ = R. In this case

S2(x) = S[Q(x)] ⊂
⋃

y∈Q(x)\{x}

Py(y) ∪Q(x) ⊂ R(x).

�

Proposition 4.9. Let R be a relator on X.
(1) If R is topologically filtered, then TR is filtered.
(2) If R is topologically reflexive and topologically transitive, moreover TR

is filtered, then R is topologically filtered.

Proof. (1) If A ∈ Φ(TR), then there exists a nonvoid finite subset B of TR
such that A =

⋂
B. Let x ∈ A be an arbitrary fixed point. For all

B ∈ B we have that x ∈ B ∈ TR, therefore there exists SB ∈ R such
that SB(x) ⊂ B. Now, with R =

⋂
B∈B SB ∈ R

Φ ⊂ R∧ we can see
that R(x) ⊂

⋂
B = A, that is A ∈TR∧ = TR.

(2) If R ∈RΦ, then let S ⊂R nonvoid and finite such that R =
⋂
S. Let

x ∈ X be an arbitrary fixed point. We need to show that there exists
a P ∈R such that P(x) ⊂ R(x).

Topologically transitivity of R, the filtered property of TR and The-
orem 4.8 give that U =

⋂
S∈S intR(S(x)) ∈ TR. It is easy to see, that

x ∈ intR(S(x)) for all S ∈ R, that is x ∈ U, therefore there exists a
P ∈R such that P(x) ⊂ U.

On the other hand, since R is topologically reflexive, we have that
intR(S(x)) ⊂ S(x), and it follows that U ⊂ (

⋂
S) (x) = R(x).

�

It gives the following.

Theorem 4.10. If R is a relator on X, then R 7→TR is a bijection of the set
of topologies on X onto the set of set-topologies on X.

Proof. Proposition 4.9 yields that the range of the bijection in Theorem 3.5
restricted to the set of topologies on X is the set of set-topologies on X. �

© AGT, UPV, 2022 Appl. Gen. Topol. 23, no. 1 52



Investigation of topological spaces using relators

5. A new form of S0-topologies

Following notations of [3], (X,R) is called quasi-uniformities, iff R is a
uniformly reflexive, uniformly transitive, uniformly filtered, and uniformly fine
relator on X.

By Definition 4.7, we have that (X,R) is a topology, iff R is a topologically
reflexive, topologically transitive, topologically filtered and topologically fine
relator on X.

Uniformities have the symmetric property. Let us see topologies with this.

Definition 5.1. If T is a (generalized) set-topology on X, such that for all
x,y ∈ X

(∃U ∈T : x ∈ U ⊂ X \{y}) =⇒ (∃U ∈T : y ∈ U ⊂ X \{x}) ,

then we say that T is a (generalized) S0-set-topology on X.

Lemma 5.2. If R is a relator on X, then
⋂
R⊂

⋂
R∧.

Proof. On the contrary, assume that there exist an (x,y) ∈
⋂
R and an R ∈R∧

such that (x,y) /∈ R. In this case, y /∈ R(x) that is R(x) ⊂ X \{y}. It follows
that there exists an S ∈R such that S(x) ⊂ R(x), and this is a contradiction
because y /∈ S(x) means y /∈ (

⋂
R) (x). �

Note that by the above Lemma, we have that
⋂
R =

⋂
R∧ for all R relators

on X.

Proposition 5.3. If R is a generalized topology on X, then the following
assertions are equivalent.

(1)
⋂
R⊂

⋂
R−1;

(2)
⋂
R⊃

⋂
R−1;

(3)
⋂
R =

⋂
R−1;

(4) R is topologically symmetric;
(5) TR is a generalized S0-set-topology on X.

Proof. (1) ⇐⇒ (2) ⇐⇒ (3) is quite obvious since
⋂
R−1 = (

⋂
R)−1.

(4) =⇒ (2): By Lemma 5.2 and (4), we have that
⋂
R−1 ⊂

⋂
R−1∧ ⊂⋂

R.
(1) =⇒ (4): By (1)

⋂
R ⊂

⋂
R−1 ⊂ R−1 for all R ∈ R, that is R−1 ⊂

{
⋂
R}∗ = R∧−1∧−1 since [5] Definition 3.1. (3) and 4.1., Remark 4.2.

and Theorem 5.3. It follows that R⊂R∧−1∧ = R−1∧ because of R is
topologically fine.

(3) =⇒ (5): Let x,y ∈ X be fixed, and assume that there exists a U ∈TR
such that x ∈ U ⊂ X \ {y}. Since U ∈ TR hence y /∈ R(x) for
some R ∈ R, and hence y /∈ (

⋂
R) (x) =

(⋂
R−1

)
(x). In this case,

x /∈ (
⋂
R) (y), that is there exists an R ∈ R such that x /∈ R(y). By

Theorem 4.8 (5), we have that intR(R(y)) ∈TR. It is easy to see that
y ∈ intR(R(y)). Because of the topologically reflexivity of R it is also
easy to see that x /∈ intR(R(y)) that is intR(R(y)) ⊂ X \{x}.

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G. Pataki

(5) =⇒ (1): Let (x,y) ∈
⋂
R be arbitrary. If the U topologically open

subset contains x, then there exists an R ∈R such that y ∈ (
⋂
R)(x) ⊂

R(x) ⊂ U. Now, by using (5), we have that for all U ∈TR y ∈ U =⇒
x ∈ U.

If R ∈ R, then y ∈ intR(R(y)) ∈ TR and hence x ∈ intR(R(y)) ⊂
R(y) since R is topologically reflexive. It follows that (y,x) ∈

⋂
R that

is (x,y) ∈
⋂
R−1. It holds for an arbitrary (x,y) ∈

⋂
R, therefore (1)

is true.

�

The previous Proposition shows the appropriateness of the following.

Definition 5.4. If R is a topologically symmetric (generalized) topology on
X, then we say that R is a (generalized) S0-topology on X, and the ordered
pair (X,R) is called a (generalized) S0-topological space.

The previous Proposition gives the following.

Theorem 5.5. If R is a relator on X, then R 7→TR is a bijection of the set of
(generalized) S0-topologies on X onto the set of (generalized) S0-set-topologies
on X.

Proof. Proposition 5.3 yields that the range of the bijection in Theorem 3.5
restricted to the set of generalized S0-topologies on X is the set of generalized
S0-set-topologies on X.

On the other hand, Proposition 5.3 yields that the range of the bijection in
Theorem 4.10 restricted to the set of S0-topologies on X is the set of S0-set-
topologies on X. �

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