@ Appl. Gen. Topol. 22, no. 1 (2021), 169-181doi:10.4995/agt.2021.14446
© AGT, UPV, 2021

Metric spaces related to Abelian groups

Amir Veisi
a
and Ali Delbaznasab

b

a
Faculty of Petroleum and Gas, Yasouj University, Gachsaran, Iran (aveisi@yu.ac.ir)

b
Department of Mathematics, Shahid Bahonar University, Kerman, Iran (delbaznasab@gmail.com)

Communicated by F. Lin

Abstract

When working with a metric space, we are dealing with the additive
group (R, +). Replacing (R, +) with an Abelian group (G, ∗), offers
a new structure of a metric space. We call it a G-metric space and
the induced topology is called the G-metric topology. In this paper,
we are studying G-metric spaces based on L-groups (i.e., partially or-
dered groups which are lattices). Some results in G-metric spaces are
obtained. The G-metric topology is defined which is further studied
for its topological properties. We prove that if G is a densely or-
dered group or an infinite cyclic group, then every G-metric space
is Hausdorff. It is shown that if G is a Dedekind-complete densely
ordered group, (X, d) a G-metric space, A ⊆ X and d is bounded,
then f : X → G with f(x) = d(x, A) := inf{d(x, a) : a ∈ A} is
continuous and further x ∈ clXA if and only if f(x) = e (the iden-
tity element in G). Moreover, we show that if G is a densely or-
dered group and further a closed subset of R, K(X) is the family of
nonempty compact subsets of X, e < g ∈ G and d is bounded, then
d
′(A, B) < g if and only if A ⊆ Nd(B, g) and B ⊆ Nd(A, g), where

Nd(A, g) = {x ∈ X : d(x, A) < g}, dB(A) = sup{d(a, B) : a ∈ A} and
d
′(A, B) = sup{dA(B), dB(A)}.

2010 MSC: 54C40; 06F20; 16H20.

Keywords: G-metric space; L-group; Dedekind-complete group; densely or-
dered group; continuity.

Received 08 October 2020 – Accepted 29 January 2021

http://dx.doi.org/10.4995/agt.2021.14446


A. Veisi and A. Delbaznasab

1. Introduction

In this article, a group (G, ∗) (briefly, G) is an Abelian group and for read-
ability, we use g1g2 instead of g1 ∗ g2. Let X be a set and ≤ relation on X,
we recall that the pair (X, ≤) is a partially ordered set (in brief, a poset) if the
following conditions hold: x ≤ x, if x ≤ y and y ≤ x, then x = y; if x ≤ y and
y ≤ z, then x ≤ z. In a poset, the symbol a∨b denotes sup{a, b}, i.e., the small-
est element c, if one exists, such that c ≥ a and c ≥ b. Likewise, a∧b stands for
inf{a, b}. When both a∨b and a∧b exist, for all a, b ∈ A, then A is called a lat-
tice. A subset S is a sublattice of A provided that, for all x, y ∈ S, the elements
x ∨ y and x ∧ y of A belong to S. (Thus, it is not enough that x and y have a
supremum and infimum in S.) For instance, C(X), the ring of real-valued con-
tinuous functions on the topological space X is a lattice. If f, g ∈ C(X), then
f ∨ g = f+g+|f−g|

2
∈ C(X) (note, f ∧ g = −(−f ∨ −g) = f+g−|f−g|

2
∈ C(X)).

In fact, C(X) is a sublattice of RX, the ring of real-valued functions on the set
X (note, the partial ordering on RX is: f ≤ g if and only if f(x) ≤ g(x) for all
x in X). A poset in which every nonempty subset has both a supremum and an
infimum is said to be a lattice-complete. For example, P(X), the set of all sub-
sets of X with inclusion is lattice-complete. Union (resp. intersection) of sets is
the supremum (resp. the infimum) of them. A totally (or linearly) ordered set
is a poset in which every pair of elements is comparable, i.e., x ≤ y or y ≤ x for
all x and y in X. We use “ordered sets” instead of “totally ordered sets”. An
ordered set is often referred to as a chain. A lattice need not be an ordered set,
necessarily, but the converse is always true. We notice that C(X) and Cc(X),
its subalgebra consisting of elements with countable image, are lattices, while
they are not ordered sets, also, they are not lattice-complete necessarily. An or-
dered set is said to be Dedekind-complete provided that every nonempty subset
with an upper bound has a supremum, or equivalently, every nonempty subset
with a lower bound has an infimum. (For example, R, the set of real numbers is
Dedekind-complete, but not lattice-complete.) An ordered field F is said to be
archimedean if Z, the set of integers is cofinal, i.e., for every x ∈ F, there exists
n ∈ Z such that n ≥ x. For instance, Q(

√
n) := {a + b

√
n : a, b ∈ Q,

√
n /∈ Q}

is an archimedean field.

Theorem 1.1 ([3, Theorem 0.21]). An ordered field is archimedean if and only
if it is isomorphic to a subfield of the ordered field R.

A brief outline of this paper is as follows. In Section 2, we introduce the
G-metric spaces related to L-groups (i.e., partially ordered groups which are
lattices) and study them further. In Section 3, the basic topological prop-
erties based on the notion of g-disk are studied. We prove that if G is a
densely ordered group or an infinite cyclic group, then every G-metric space
is Hausdorff. It is shown that if G is a Dedekind-complete densely ordered
group, (X, d) a G-metric space, d is bounded and A ⊆ X, then f : X → G
given by f(x) = d(x, A) := inf{d(x, a) : a ∈ A} is continuous and further
x ∈ clXA if and only if f(x) = e (the identity element of G). Moreover, let

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Metric spaces related to Abelian groups

F(X) be the family of nonempty closed sets in X, e < g ∈ G, A, B ∈ F(X)
and dA(B) = sup{d(b, A) : b ∈ B}. Then for the G-metric space (F(X), d′)
(note, d′(A, B) = sup{dA(B), dB(A)}), we have d′(A, B) ≤ g if and only if
A ⊆ Nd(B, ḡ) and B ⊆ Nd(A, ḡ), where Nd(A, ḡ) = {x ∈ X : d(x, A) ≤ g}.
Particularly, if G is a densely ordered group and further a closed subset of R,
X is a G-metric space and K(X) is the family of nonempty compact sets in
X, then d′(A, B) < g if and only if A ⊆ Nd(B, g) and B ⊆ Nd(A, g), where
Nd(A, g) = {x ∈ X : d(x, A) < g}.

2. G-metric spaces

Definition 2.1. A group G with a partial ordering relation ≤ is called a
partially ordered group (in brief, a poset group) if the binary operation of G
preserves the order, i.e.,

g1 ≥ g2 implies g1g3 ≥ g2g3 for all g1, g2, g3 ∈ G. (R1)
Moreover, if a poset group G is a lattice then G is called an L-group.

From the above definition, the following facts are evident: g1 ≥ g2 if and
only if g1g

−1
2 ≥ e; g ≥ e if and only if g−1 ≤ e; if g1 ≥ g3 and g2 ≥ g4, then

g1g2 ≥ g3g4. For example, every archimedean field with the addition is an
L-group. But Zn with the addition of modulo n is not a poset group yet, since
this addition does not preserve the order. For an L-group G and g ∈ G, we let
|g| = sup{g, g−1} = g ∨ g−1 = |g−1|.
Example 2.2. Consider the group G := Z×Z×. . .×Z (k-times) with the usual
addition and the identity element e = (0, 0, . . . , 0). Let g1 = (m1, m2, . . . , mk),
g2 = (n1, n2, . . . , nk) ∈ G. Define

g1 ≤ g2 if and only if mi ≤ ni for all i = 1, 2, . . . , k.
We see that ≤ is a partial ordering relation on G. Also, the condition (R1) in
the above definition is satisfied, i.e., G is a poset group. Let zi = max{mi, ni}
and z′i = min{mi, ni}, where i = 1, 2, . . . , k. Let g3 = (z1, z2, . . . , zk) and
g4 = (z

′
1, z

′
2, . . . , z

′
k). Then we obtain g1 ∨ g2 = g3 and g1 ∧ g2 = g4. Hence, G

is an L-group.

By an ordered group, we mean a poset group which is a totally ordered set
by its partial ordering relation. It is clear that an ordered group is an L-group.
Finally, by Dedekind-complete group, we mean an ordered group which is a
Dedekind-complete set with its partial ordering relation, i.e., every nonempty
subset with an upper bound has a supremum, or equivalently, every nonempty
subset with a lower bound has an infimum. For example, every archimedean
field with the addition is a Dedekind-complete group.

Corollary 2.3. If G is an L-group and g1, g2 ∈ G, then |g1g2| ≤ |g1||g2|.
Proof. We note that g1, g

−1
1 ≤ |g1| and g2, g

−1
2 ≤ |g2|. Definition 2.1 now

gives g1g2 ≤ |g1||g2| and also g−11 g
−1
2 ≤ |g1||g2|. So we have that |g1g2| =

sup{g1g2, (g1g2)−1} ≤ |g1g2|, and the result holds. �

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A. Veisi and A. Delbaznasab

Definition 2.4. Let G be a poset group and X a nonempty set. We say the
function d : X ×X → G is a G-metric on X, whenever the following conditions
hold, for every x, y, z ∈ X.

(i) d(x, y) ≥ e (e is the identity element in G),
(ii) d(x, y) = e if and only if x = y,
(iii) d(x, y) = d(y, x),
(iv) d(x, y) ≤ d(x, z)d(z, y) (triangle inequality).

The pair (X, d) (briefly, X) is called a G-metric space. Evidently, every
metric space is a G-metric space, when (G, ∗) = (R, +). If all axioms but the
second part of Definition 2.4 are satisfied, we call d a G-pseudometric and then
X a G-pseudometric space. Defining d(x, y) = e for all x and y in X, gives
a G-pseudometric on X, called the trivial G-pseudometric, in this case, d is a
G-metric if and only if X is the singleton set {x}. Although all the material of
this section is developed for G-metric spaces, the basic results remain true for
G-pseudometric spaces as well. If (X, d) is a G-metric on X and A is a subset
of X, then A inherits a G-metric structure from X in an obvious way, making
A a G-metric space.

In the following example, we will present some examples of G-metric spaces.
Before it, let X = Rn, G1 = (R, +), G2 = ((0, +∞), .), G3 = (R − {−1}, ∗)

and G4 = (Z2, ⊕), where +, . are usual addition and multiplication, the symbol
⊕ is the addition of modulo 2 and ∗ is defined as follows: x ∗ y = x + y + xy.
In G3, the identity element is 0 and the inverse of x is x

−1 = −x
1+x

. Checking
of the associative property of ∗ is easy.
Moreover, let ϕi : Gi ×Gi → Gi, where i = 1, 2, 3, 4, such that ϕ1(x, y) = x−y,
ϕ2(x, y) =

x
y
, ϕ3(x, y) =

−xy
1+x

and ϕ4(x, y) =

{

0 if x = y
1 if x 6= y. Then, since each

ϕi is continuous, each Gi is a topological group as subspaces of R with the usual
topology.

Example 2.5. Let X = Rn, Gi, where i = 1, 2, 3, 4, be as defined in the
previous discussion. For x = (x1, x2, . . . , xn), y = (y1, y2, . . . , yn) ∈ X, ‖x − y‖
is the usual norm, i.e., ‖x − y‖ =

(
∑n

i=1(xi − yi)2
)

1

2 . We claim that each di,
the functions below, is a Gi-metric and therefore (X, di) is a Gi-metric space.
We only check that d2 and d3 satisfy (iv) of Definition 2.4. Other conditions
are routine.

(1) Let d1 : X × X → G1 such that d1(x, y) = ‖x − y‖.
(2) Let d2 : X × X → G2 such that d2(x, y) = e‖x−y‖.
(3) Let d3 : X × X → G3 such that d3(x, y) = e‖x−y‖ − 1.
(4) Let d4 : X × X → G4 such that d(x, y) = 0 if x = y; and 1 if x 6= y. d4

is called a discrete G-metric.

We notice that the identity elements in G2 and G3 are 1 and 0 respectively.
Moreover, d2(x, y) ≥ 1 and d3(x, y) ≥ 0. Now,

d2(x, y) = e
‖x−y‖ ≤ e(‖x−z‖+‖z−y‖) = e‖x−z‖e‖z−y‖ = d2(x, z)d2(z, y).

© AGT, UPV, 2021 Appl. Gen. Topol. 22, no. 1 172



Metric spaces related to Abelian groups

Also, we have

d3(x, y) = e
‖x−y‖ − 1 ≤ e‖x−z‖e‖z−y‖ − 1

= (e‖x−z‖ − 1) + (e‖z−y‖ − 1) + (e‖x−z‖ − 1)(e‖z−y‖ − 1)
= d3(x, z) + d3(z, y) + d3(x, z)d3(z, y)

= d3(x, z)d3(z, y).

So d2 and d3 satisfy the triangle inequality of Definition 2.4.

A G-metric d on a set X is called bounded if d(x, y) ≤ g0, for all x, y ∈ X
and some g0 ∈ G. Thus, the next result is now immediate.
Corollary 2.6. Let G be an L-group, d a G-metric on X, e < g1 a fixed
element in G and d1(x, y) = inf{d(x, y), g1}. Then d1 is a bounded G-metric.
Lemma 2.7. Let G be an L-group; A and B are finite subsets of G such that
A ≥ e (i.e., a ≥ e, for all a ∈ A) and B ≥ e. Then

(i) if A ≤ B (i.e., for each a ∈ A there is b ∈ B such that a ≤ b) and
e ≤ g ∈ G, then sup(gA) = g sup A ≤ g sup B = sup(gB), where
gA = {ga : a ∈ A}.

(ii) if A ≥ B (i.e., for each a ∈ A there is b ∈ B such that b ≤ a) and
e ≤ g ∈ G, then inf(gB) = g inf B ≤ g inf A = inf(gA).

(iii) sup(AB) = sup A sup B, and also inf(AB) = inf A inf B, where AB =
{ab : a ∈ A, b ∈ B}.

Proof. First, we note that by definition of an L-group, each of the finite sets
A, B and AB has a supremum and an infimum in G. The proofs of (i) and
(ii) are routine. (iii). Let sup A = α, sup B = β and sup(AB) = γ. Since G
is an L-group, it is a poset group. So by Definition 2.1, we have ab ≤ αβ, for
all a ∈ A and b ∈ B. Evidently, γ ≤ αβ. Now, we are ready to show that
γ = αβ. For the reverse inclusion, let a ∈ A be fixed. Then ab ≤ γ implies
b ≤ a−1γ. Therefore, B is bounded by a−1γ. So β = sup B ≤ a−1γ, in other
words, a ≤ β−1γ. Since a ∈ A is arbitrary, we deduce that A is bounded by
β−1γ. Thus, α ≤ β−1γ. This yields αβ ≤ γ, and we are through. The proof of
another assertion (infimum) is done similarly. �

Proposition 2.8. Let G be an ordered group. Then defining

d : G × G → G given by d(g1, g2) = |g1g−12 |,
turns G into a G-metric space.

Proof. We claim that d is a G-metric on G. First, we note that since G is an
ordered group, it is an L-group and further g ∈ G gives g ≥ e or g−1 ≥ e. So
|g| = |g−1| = sup{g, g−1} = g or g−1. Hence, |g| ≥ e and therefore conditions
(i)-(iii) of Definition 2.4 hold. Moreover, if g1, g2, g3 ∈ G, then Corollary 2.3
implies

d(g1, g2) = |g1g−12 | = |(g1g
−1
3 )(g3g

−1
2 )| ≤ |g1g

−1
3 ||g3g

−1
2 |

= d(g1, g3)d(g3, g2)

© AGT, UPV, 2021 Appl. Gen. Topol. 22, no. 1 173



A. Veisi and A. Delbaznasab

This gives d satisfies the triangle inequality, i.e., it is a G-metric on G and
hence G is a G-metric space. �

Corollary 2.9. Let G be an ordered group, X a nonempty set and f : X → G
a function. Then d : X × X → G given by d(x, y) = |f(x)f−1(y)| is a G-
pseudometric on X. Moreover, d is a G-metric on X if and only if f is one-
one.

The next proposition generalizes Proposition 2.8.

Proposition 2.10. Let G be an ordered group. Then each of the following
binary operations on Gn (the n product of G), turns it into a G-metric space,
where g = (g1, g2, . . . , gn) and g

′ = (g′1, g
′
2, . . . , g

′
n) are arbitrary elements of

Gn.

(i) d1 : G
n × Gn → G defined by d1(g, g′) = |g−11 g′1||g

−1
2 g

′
2| . . . |g−1n g′n|.

(ii) d2 : G
n × Gn → G defined by

d2(g, g
′) = sup{|g−11 g′1|, |g

−1
2 g

′
2|, . . . , |g−1n g′n|}.

Proof. We only check that the triangle inequality for d1 and d2. Other condi-
tions are routine. (i). Let g

′′

= (g
′′

1 , g
′′

2 , . . . , g
′′

n) ∈ Gn. Then

d1(g, g
′′

) = |g−11 g
′′

1 ||g−12 g
′′

2 | . . . |g−1n g
′′

n|
= |(g−11 g′1)(g′1

−1
g

′′

1 )||(g−12 g′2)(g′2
−1

g
′′

2 )| . . . |(g−1n g′n)(g′n
−1

g
′′

n)|
≤ |g−11 g′1||g′1

−1
g

′′

1 ||g−12 g′2||g′2
−1

g
′′

2 | . . . |g−1n g′n||g′n
−1

g
′′

n|
=

(

|g−11 g
′
1|g−12 g

′
2| . . . |g−1n g′n|

)(

|g′1
−1

g
′′

1 ||g′2
−1

g
′′

2 | . . . |g′n
−1

g
′′

n|
)

= d1(g, g
′)d1(g

′, g
′′

).

Notice that the above inequality is obtained by Corollary 2.3. (ii). Let g
′′

=

(g
′′

1 , g
′′

2 , . . . , g
′′

n) ∈ Gn and let

A = {|g−11 g
′′

1 |, |g−12 g
′′

2 |, . . . , |g−1n g
′′

n|},

B = {|g−11 g
′
1||g′1

−1
g

′′

1 |, |g−12 g
′
2||g′2

−1
g

′′

2 |, . . . , |g−1n g′n||g′n
−1

g
′′

n|},

B1 = {|g−11 g′1|, |g
−1
2 g

′
2|, . . . , |g−1n g′n|}, and

B2 = {|g′1
−1

g
′′

1 |, |g′2
−1

g
′′

2 |, . . . , |g′n
−1

g
′′

n|}.

We notice that G is an L-group. Now, according to Lemma 2.7, we have
A ≤ B ≤ B1B2. Therefore,

d2(g, g
′′

) = sup A ≤ sup B ≤ sup(B1B2) = sup B1 sup B2 = d2(g, g′)d2(g′, g
′′

),

which completes the proof. �

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Metric spaces related to Abelian groups

3. Basic topological concepts in G-metric spaces and some
related results

We begin with the following definition.

Definition 3.1. Let G be a poset group, (X, d) a G-metric space and x a point
of X. Given e < g ∈ G, we let

Nd(x, g) = {y ∈ X : d(x, y) < g},
and call it the g-disk centered at x. Also, we put Nd(x, g) = {y ∈ X, d(x, y) ≤
g}.

A subset U of X is said to be open in X if either U = ∅ or for every x ∈ U
there is a g ∈ G such that Nd(x, g) ⊆ U. Here, x is called an interior point
of U. The set of all interior points of U is called the interior of U, denoted
by U◦ (or intXU). Also, a set F is called closed if and only if its set-theoretic
complement is an open set in X. Evidently, a set F is closed if and if every
g-disk centered at x meets F , then x ∈ F .
Corollary 3.2. Every g-disk Nd(x, g) is an open set in X (and hence X r
Nd(x, g) = {y ∈ X : d(x, y) ≥ g} is a closed set in X).
Proof. Let y ∈ Nd(x, g). Then g1 = d(x, y) < g. We claim that Nd(y, gg−11 ) ⊆
Nd(x, g) (note, gg

−1
1 > e). To see this, assume that z ∈ Nd(y, gg

−1
1 ). Hence,

d(z, x) ≤ d(z, y)d(y, x) < gg−11 g1 = g. This yields z ∈ Nd(x, g), i.e., Nd(y, gg
−1
1 )

⊆ Nd(x, g), and we are done. �
Definition 3.3. Let X be a G-metric space and A ⊆ X. The closure of A in
X is denoted by clXA (or briefly clA) and defined by the set

clA = ∩{F ⊆ X : F is closed in X and A ⊆ F}.
By the above definition, A is closed if and only if A = clA.

Corollary 3.4. Let G be a poset group, (X, d) a G-metric space and Ā = {x ∈
X : Nd(x, g) ∩ A 6= ∅ for all e < g ∈ G}, where A ⊆ X. Then

(1) A = clA.

(2) If x ∈ X and g ∈ G, then Nd(x, g) ⊆ Nd(x, g).
Proof. (1). Let x /∈ clA. Then x /∈ F , for some closed set F containing A. Now,
since XrF is open, there exists e < g ∈ G, such that x ∈ Nd(x, g) ⊆ XrF . So
x /∈ Ā. Conversely, suppose that x /∈ Ā. So for some e < g ∈ G, Nd(x, g)∩A =
∅. Therefore, the closed set X r Nd(x, g) contains A but not x. This gives x /∈
clA, and we are done. (2). Suppose that y /∈ Nd(x, g). So d(x, y) > g and hence
g1 = d(x, y)g

−1 > e. We claim that Nd(y, g1) ∩ Nd(x, g) = ∅. Otherwise, for
some z ∈ Nd(y, g1) ∩ Nd(x, g), we have d(x, y) ≤ d(x, z)d(z, y) < gg1 = d(x, y),
a contradiction. Hence, y /∈ Nd(x, g) and we are done. �
Proposition 3.5. Let G be an ordered group and X a G-metric space. Then
the open sets in X have the following properties:

(i) X and ∅ are both open.

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A. Veisi and A. Delbaznasab

(ii) Every union of open sets is open.
(iii) Every finite intersection of open sets is open.

Proof. (i) and (ii) are clear. (iii). Let x ∈
⋂n

i=1 Ui, where Ui is an open set in
X. Take gi ∈ G such that x ∈ Nd(x, gi) ⊆ Ui. Since G is an ordered group,
there exists g ∈ G such that g = inf{gi}ni=1 (note, the elements gi form a chain
and hence g is one of them). Thus, x ∈ Nd(x, g) ⊆

⋂n
i=1 Nd(x, gi) ⊆

⋂n
i=1 Ui,

which completes the proof. �

By the above proposition, every G-metric d on a set X defines a topology
τd on X; members of τd, or, open subsets of X are unions of g-disks. Clearly,
the family of all g-disks is a base for (X, τd). We call τd the topology induced
by the G-metric d (or G-metric topology).

Remark 3.6. Even if G is a Dedekind-complete group, a countable intersection
of open sets in a G-metric space need not be an open set necessarily. To see
this, consider R as a G-metric space, where G is (R, +) or ((0, ∞), .). Also,
recall the fact that every point a of R is a Gδ-set, i.e., {a} =

⋂∞
n=1(a−

1
n
, a+ 1

n
).

In [2, I.3], an ordered set X is called a densely ordered set, if no cut of X is
a jump, or equivalently, for every pair x, y of elements of X satisfying x < y,
there exists a z ∈ X, such that x < z < y.

Definition 3.7. An ordered group G is called a densely ordered group if it is
a densely ordered set with its total ordering relation.

It is clear that densely ordered groups are infinite. For example, every
archimedean field like as R and Q(

√
n) := {a + b

√
n : a, b ∈ Q,

√
n /∈ Q} with

the addition is a densely ordered group. Z, the group of integers is an ordered
group which is not a densely ordered group while Zn with the addition of mod-
ulo n, is not a poset group yet.

From now on, the group G is assumed to be a densely ordered group.

Proposition 3.8. Let G be a densely ordered group, (X, d) a G-metric space,
x ∈ X and g ∈ G be fixed, and A = {y ∈ X : d(x, y) > g}. Then A is an open
set in X (and hence Nd(x, g) := {y ∈ G : d(x, y) ≤ g} is closed).

Proof. Let y ∈ A be fixed. Then d(x, y) > g. We must show that y is an
interior point of A. Let g1 = d(x, y)g

−1. Then g1 > e and d(x, y) = gg1. Since
G is a densely ordered group, we take e < g2 < g1 and claim that Nd(y, g2) is
contained in A entirely. To see this, let z ∈ Nd(y, g2). Then we have d(y, z) < g2
and so g−12 < d

−1(y, z). Now, the inequality d(x, y) ≤ d(x, z)d(z, y) yields

g < gg1g
−1
2 < d(x, y)d

−1(y, z) ≤ d(x, z).

(Notice that g < gg1g
−1
2 if and only if g2 < g1.) Therefore, g < d(x, z), i.e., y

is an interior point of A. So A is an open set in X, and we are done. �

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Metric spaces related to Abelian groups

Proposition 3.9. Let G be an ordered group and (X, d) a G-metric space.
Then the following statements hold.

(i) If G is a densely ordered group, then X is Hausdorff.
(ii) If G is an infinite cyclic group, then X is discrete (and so it is first

countable).

Proof. (i). Let x, y ∈ X and d(x, y) = g > e. By assumption, since G is a
densely ordered set, we can take g1, g2 ∈ G such that e < g1 < g2 < g. Now,
we claim that two disks Nd(x, g1) and Nd(y, gg

−1
2 ) are disjoint (note, gg

−1
2 > e).

Otherwise, for some x′ ∈ Nd(x, g1) ∩ Nd(y, gg−12 ), we have d(x, x′) < g1 and
d(x′, y) < gg−12 . Hence, g = d(x, y) ≤ d(x, x′)d(x′, y) < g1gg

−1
2 . Therefore,

e < g1g
−1
2 , or equivalently, g2 < g1, a contradiction. So we are done.

(ii). Let e < g ∈ G be the generator of G. Then G = {gn : n ∈ Z}, in fact,
we have G ∼= Z, the additive group of integers with the generator 1 (or −1).
We note that the elements of G form a chain. So we obtain

. . . < g−3 < g−2 < g−1 < e < g < g2 < g3 < . . .

Therefore, for each x ∈ X, Nd(x, g) = {y ∈ X : d(x, y) < g} = {y ∈ X :
d(x, y) = e} = {x}. This yields X is discrete (note, in this case G is not a
densely ordered group). �

Remark 3.10. By Proposition 3.9(ii), if G is an infinite cyclic group then X
is first countable. But the converse of that result may be false, since every
metric space is first countable, whereas the additive group (R, +) is not even
countably generated.

In general, the converse of the above proposition does not need to be true.
In the next example, we give examples of Hausdorff G-metric spaces such that
the group G is neither a densely ordered group nor an infinite cyclic group.

Example 3.11. (i) Let d1 : Z × Z → Z defined by d1(m, n) = |m − n|.
Then, since Nd1(m, 1) = {m}, we obtain Z is a discrete Z-metric space.
So it is Hausdorff, whereas Z is not a densely ordered group. But if Z
is considered as a Q-metric space with the same definition, d1(m, n) =
|m − n|, it is a discrete Q-metric space while Q is a densely ordered
group.

(ii) Let G := Z×Z with the identity element e = (0, 0). Define d2 : Z×Z →
G with d2(m, n) = (|m−n|, |m−n|). By example 2.2, G is an L-group.
It is easy to see that d2 is a G-metric on Z. Now, let g = (1, 1). Then

Nd2
(

m, g
)

= {n ∈ Z : d2(m, n) < g} = {m}.
This yields Z is a discrete G-metric space, whereas G = Z × Z is not a
cyclic group (note, it is a finitely generated group which is generated
by the set {(0, 1), (1, 0)}).

Definition 3.12. If (X, dX) (resp. (Y, dY )) is a G1- (resp. G2-) metric space,
a function f : X → Y is called continuous at x0 ∈ X if and only if for each
e2 < g2 ∈ G2 there is some e1 < g1 ∈ G1 such that dY (f(x0), f(y)) < g2,

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A. Veisi and A. Delbaznasab

whenever dX(x0, y) < g1. f is called continuous on X, if it is continuous at
every x ∈ X.

A simple translation of the above definition is:

Corollary 3.13. A function f : X → Y is continuous at x0 ∈ X if and
only if for each g2-disk NdY (f(x0), g2) centered at f(x0), there is some g1-disk
NdX (x0, g1) centered at x0, such that f(NdX (x0, g1)) ⊆ NdY (f(x0), g2).

Theorem 3.14. If (X, dX) and (Y, dY ) are G1- and G2-metric spaces respec-
tively, a function f : X → Y is continuous at x0 ∈ X if and only if for each
open set V of Y containing f(x0), there exists an open set U of X containing
x0 such that f(U) ⊆ V.

Proof. (⇒) : Suppose that f is continuous at x0 and V is an open set in Y
containing f(x0). By definition of open sets, there is g2 ∈ G2 such that f(x0) ∈
NdY (f(x0), g2) ⊆ V . By Corollary 3.13, there exists a g1-disk NdX (x0, g1)
centered at x0 such that f(NdX (x0, g1)) ⊆ NdY (f(x0), g2) ⊆ V, where g1 ∈ G1.
It now suffices to choose U = NdX (x0, g1).

(⇐) : Consider e2 < g2 ∈ G2 and NdY (f(x0), g2) as an open set in Y
containing f(x0). By hypothesis, there exists an open set U in X containing
x0 such that f(U) ⊆ NdY (f(x0), g2). Also, we can take e1 < g1 ∈ G1 such
that NdX (x0, g1) ⊆ U. So f(NdX (x0, g1)) ⊆ f(U) ⊆ NdY (f(x0), g2), and we
are done. �

The following lemma is the counterpart of Lemma 2.7 for a Dedekind-
complete group G. The only difference is that there A and B were finite
subsets of G but here these sets must be bounded.

Lemma 3.15. Let G be a Dedekind-complete group and; A and B are bounded
subsets of G such that A ≥ e (i.e., a ≥ e, for all a ∈ A) and B ≥ e. Then

(i) if A ≤ B (i.e., for each a ∈ A there is b ∈ B such that a ≤ b) and
e ≤ g ∈ G, then sup(gA) = g sup A ≤ g sup B = sup(gB), where
gA = {ga : a ∈ A}.

(ii) if A ≥ B (i.e., for each a ∈ A there is b ∈ B such that b ≤ a) and
e ≤ g ∈ G, then inf(gB) = g inf B ≤ g inf A = inf(gA).

(iii) sup(AB) = sup A sup B, and also inf(AB) = inf A inf B, where AB =
{ab : a ∈ A, b ∈ B}.

In the remainder of this article, G is assumed to be a Dedekind-complete
densely ordered group (i.e., a densely ordered group in which every bounded
nonempty subset has a supremum and an infimum in G), (X, d) a G-metric
space, and d is bounded. The distance of a point x to a set A (⊆ X) is defined
by d(x, A) = inf{d(x, a) : a ∈ A}, if A 6= ∅, and d(x, ∅) = e.

Theorem 3.16.

(i) The mapping f : X → G defined by f(x) = d(x, A) is continuous.
(ii) x ∈ clXA if and only if f(x) = d(x, A) = e, in fact, clXA = f−1(e).

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Metric spaces related to Abelian groups

Proof. (i). First, by Proposition 2.8, we have (G, d′) is a G-metric space, where
d′(g1, g2) = |g1g−12 |. Let x0 ∈ X, g0 ∈ G and Nd′(f(x0), g0) be an open set
containing f(x0). Then

d(x0, a) ≤ d(x0, x)d(x, a), and d(x, a) ≤ d(x, x0)d(x0, a). (R2)
Now, if we let G1 = {d(x0, a) : a ∈ A} and G2 = {d(x0, x)d(x, a) : a ∈ A},
then G1 and G2 are two subsets of G with the same cardinality and G2 ≥ G1.
By Lemma 3.15 (ii), we have infa∈A G1 ≤ infa∈A G2. In other words, taking
infimum on both sides of each of the inequalities in (R2) with respect to a ∈ A,
we obtain

inf
a∈A

d(x0, a) ≤ d(x0, x) inf
a∈A

d(x, a), and inf
a∈A

d(x, a) ≤ d(x, x0) inf
a∈A

d(x0, a).

Thus, f(x0) ≤ d(x, x0)f(x) and f(x) ≤ d(x, x0)f(x0). Hence, f(x0)f−1(x) ≤
d(x, x0) and also f(x)f

−1(x0) ≤ d(x, x0), i.e., d(x, x0) is a common upper
bound for f(x0)f

−1(x) and f−1(x0)f(x). Therefore,

d′(f(x0), f(x)) = |f(x)f−1(x0)| = sup{f(x0)f−1(x), f−1(x0)f(x)} ≤ d(x, x0).
Now, for the g0-disk Nd(x0, g0) we have f(Nd(x0, g0)) ⊆ Nd′(f(x0), g0), and
we are through.

(ii). Necessity: First, we note that by Corollary 3.4, clA = Ā = {x ∈ X :
Nd(x, g) ∩ A 6= ∅, for all g > e}. If d(x, A) = g > e then d(x, a) ≥ g > e, for all
a ∈ A. By assumption, since G is a densely ordered group, we can take g1 ∈ G
such that g > g1 > e. Now, we observe that Nd(x, g1) ∩ A = ∅. Hence, x /∈ A.

Sufficiency: Let x /∈ A. Then Nd(x, g) ∩ A = ∅, for some e < g ∈ G. Hence,
d(x, a) ≥ g, for all a ∈ A. Therefore, d(x, A) ≥ g > e. So d(x, A) 6= e, and we
are done. �

Theorem 3.17. Let G be a Dedekind-complete densely ordered group, (X, d)
a G-metric space, d is bounded, g ∈ G, and let F(X) be the family of all
nonempty closed subsets of X. For A, B ∈ F(X) define

dB(A) = sup{d(a, B) : a ∈ A}, and d′(A, B) = sup{dA(B), dB(A)}.
Then the following statements hold.

(1) d′ is a G-metric on F(X). We call it the Hausdorff G-metric on F(X).
(2) d′(A, B) ≤ g if and only if A ⊆ Nd(B, ḡ) and B ⊆ Nd(A, ḡ), where

Nd(A, ḡ) = {x ∈ X : d(x, A) ≤ g}.
Proof. (1). (i) and (iii) of Definition 2.4 are evident. Let d′(A, B) = e. Then
dB(A) = e = dA(B). So d(a, B) = e for all a ∈ A. By Theorem 3.16 (ii),
a ∈ clB = B, i,e., A ⊆ B. Similarly, B ⊆ A. This proves (ii) of Definition 2.4.
For the proof of triangle inequality, let A, B, C ∈ F(X) and a ∈ A, b ∈ B, c ∈ C.
We notice that d(a, B) ≤ d(a, b) and d(b, C) ≤ dC(B). Thus,

d(a, B) ≤ d(a, b) ≤ d(a, c)d(c, b).
Taking infimum on both sides of the above inequality with respect to c ∈ C
plus Lemma 3.15 yield

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A. Veisi and A. Delbaznasab

d(a, B) ≤ infc∈C{d(a, c)d(c, b)} = infc∈C d(a, c) infc∈C d(c, b).
Therefore, d(a, B) ≤ d(a, C)d(b, C). Since d(b, C) ≤ dC(B), we have d(a, B) ≤
d(a, C)dC(B). Taking supremum on both sides of the latter inequality with
respect to a ∈ A, we obtain

dB(A) ≤ dC(A)dC(B). (R3)
On the other hand, taking infimum over c ∈ C on both sides of the inequalities
d(b, A) ≤ d(a, b) ≤ d(a, c)d(c, b) we obtain d(b, A) ≤ d(a, C)d(b, C) (Lemma
3.15). Furthermore, d(a, C) ≤ dC(A) gives d(b, A) ≤ dC(A)d(b, C). Now, take
supremum on both sides of the latter inequality respect to b ∈ B. Thus,

dA(B) ≤ dC(A)dC(B). (R4)
Combining (R3) and (R4) we get

d′(A, B) ≤ dC(A)dC(B) ≤ d′(A, C)d′(C, B).
Hence, d′ satisfies (iv) of Definition 2.4, and we are done.

(2). (⇒) : Let d′(A, B) ≤ g. Then dB(A) ≤ g and dA(B) ≤ g. Hence,
d(a, B) ≤ g, for all a in A. So A ⊆ Nd(B, ḡ). Similarly, B ⊆ Nd(A, ḡ).

(⇐) : Since A ⊆ Nd(B, ḡ), it gives d(a, B) ≤ g, for all a in A, and therefore
dB(A) = supa∈A d(a, B) ≤ g. The assertion dA(B) ≤ g is deduced similarly.
So d′(A, B) ≤ g, and we are through. �
Corollary 3.18. Let G be a densely ordered group and further a closed subset
of R, K(X) the family of nonempty compact subsets of X and A, B ∈ K(X)
such that X, d, g, dA and d

′ be as defined in Theorem 3.17. Then d′(A, B) < g
if and only if A ⊆ Nd(B, g) and B ⊆ Nd(A, g), where Nd(A, g) = {x ∈ X :
d(x, A) < g}.
Proof. We first recall the fact that a nonempty subset of R has the least-upper-
bound property (equivalently, the greatest-lower-bound property) if and only
if it is closed in R. So G has the least-upper-bound property and hence it is a
Dedekind-complete densely ordered group. Moreover, by Proposition 3.9, X is
Hausdorff and therefore every compact set in X is closed. Thus, the conditions
of Theorem 3.17 are satisfied. The necessary condition is obvious. To prove
the sufficiency, let us define

f1, f2 : X → G with f1(x) = d(x, A) and f2(x) = d(x, B).
Now, since A and B are compact subsets of X and further; f1 and f2 are
continuous functions on X (Theorem 3.16), f1(B) and f2(A) are compact sets
in G (note, since G is closed, f1 and f2 are well defined). Therefore, sup f1(B) ∈
f1(B) and also sup f2(A) ∈ f2(A). So we have

dA(B) = sup f1(B) = f1(b1) = d(b1, A), for some b1 ∈ B,
and also

dB(A) = sup f2(A) = f2(a2) = d(a2, B), for some a2 ∈ A.

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Metric spaces related to Abelian groups

By assumption, we now get dA(B) < g and dB(A) < g. Hence, d
′(A, B) < g,

and we are through. �

Acknowledgements. The authors are grateful to the referee for providing

helpful comments and recommendations to improve the quality of the paper.

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