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@ Appl. Gen. Topol. 19, no. 2 (2018), 253-260doi:10.4995/agt.2018.9009
c© AGT, UPV, 2018

τ-metrizable spaces

A. C. Megaritis

Technological Educational Institute of Peloponnese, Department of Computer Engineering, 23100,

Sparta, Greece (thanasismeg13@gmail.com)

Communicated by D. Georgiou

Abstract

In [1], A. A. Borubaev introduced the concept of τ-metric space, where
τ is an arbitrary cardinal number. The class of τ-metric spaces as τ
runs through the cardinal numbers contains all ordinary metric spaces
(for τ = 1) and thus these spaces are a generalization of metric spaces.
In this paper the notion of τ-metrizable space is considered.

2010 MSC: 54A05; 54E35.

Keywords: τ-metric space; τ-metrizable space; τ-metrization theorem.

1. Preliminaries and notations

Our notation and terminology is standard and generally follows [2]. The
cardinality of a set X is denoted by |X|. Throughout, we denote by τ an
arbitrary nonzero cardinal number. The cardinalities of the natural numbers
and of the real numbers are denoted by ℵ0 and c, respectively. The character,
the weight and the density of a topological space X are denoted by χ(X), w(X)
and d(X), respectively. As usual I denotes the closed unit interval [0, 1] with
the Euclidean metric topology.

By Rτ+ we denote the topological product of τ copies of the space R+ =
[0, +∞) (with the natural topology). On the space Rτ+, the operations of
addition, multiplication, and multiplication by a scalar, as well as a partial
ordering, are defined in a natural way (coordinatewise).

Now, we present the notion of τ-metric space [1]. Let X be a nonempty set.
A mapping ρτ : X × X → R

τ
+ is called a τ-metric on X if the following axioms

hold:

Received 28 November 2017 – Accepted 02 April 2018

http://dx.doi.org/10.4995/agt.2018.9009


A. C. Megaritis

(1) ρτ(x, y) = θ if and only if x = y, where θ is the point of the space R
τ
+

whose all coordinates are zeros.

(2) ρτ(x, y) = ρτ(y, x) for all x, y ∈ X.

(3) ρτ(x, z) 6 ρτ (x, y) + ρτ(y, z) for all x, y, z ∈ X.

The pair (X, ρτ) is called a τ-metric space and the elements of X are called
points.

Every τ-metric space (X, ρτ ) generates a Tychonoff (that is, completely
regular and Hausdorff) topological space (X, Tρτ ). The topology Tρτ on X
defined by the local basis consisting of the sets of the form

G(x) = {y ∈ X : ρτ (x, y) ∈ O(θ)},

where O(θ) runs through all open neighbourhoods of the point θ in the space
R

τ
+, of each point x ∈ X is called the topology induced by the τ-metric ρτ.
In this paper the notion of τ-metrizable space is introduced. The paper

is organized as follows. Section 2 contains the basic concepts of τ-metrizable
spaces. Generally, τ-metrizable spaces may be not metrizable. We prove that
if τ 6 ℵ0, then every τ-metrizable space is metrizable. In section 3 we obtain a
generalization of the classical metrization theorem of Urysohn. More precisely,
we prove that every Tychonoff space of weight τ > ℵ0 is τ-metrizable. Finally,
in section 4 we prove that every compact τ-metrizable space has density less
than or equal to τ.

2. Basic concepts

The notion of a τ-metric space leads to the notion of a τ-metrizable space
which is inserted in the following definition.

Definition 2.1. A topological space (X, T ) is called τ-metrizable if there exists
a τ-metric ρτ on the set X such that the topology Tρτ induced by the τ-metric
ρτ coincides with the original topology T of X. τ-metrics on the set X which
induce the original topology of X will be called τ-metrics on the space X.

Note that τ-metrizable spaces are useful because only such spaces can be
presented as limits of τ-long projective systems of metric spaces [1, Theorem
3].

Proposition 2.2. A metric space is τ-metrizable.

Proof. Let (X, ρ) be a metric space, tρ be the topology induced by the metric
ρ, and let τ be a cardinal number. Consider a set Λ such that |Λ| = τ and
set ρλ = ρ for each λ ∈ Λ. The mapping ρτ : X × X → R

τ
+ defined by

ρτ(x, y) = {ρλ(x, y)}λ∈Λ for every x, y ∈ X is a τ-metric on X. It is easy to
see that tρ = Tρτ . �

Proposition 2.3. A τ-metrizable space is τ′-metrizable for every cardinal
number τ′ > τ.

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τ-metrizable spaces

Proof. Let X be a τ-metrizable space, ρτ be a τ-metric on the space X and
τ′ be a cardinal number such that τ′ > τ. Consider two sets K and Λ such
that K ⊂ Λ, |K| = τ and |Λ| = τ′, and set ρτ(x, y) = {ρ

k
τ(x, y)}k∈K for every

x, y ∈ X. Let k0 be one fixed element of K. The mapping ρτ′ : X × X → R
τ′

+

defined by ρτ′(x, y) = {ρ
λ
τ′(x, y)}λ∈Λ for every x, y ∈ X, where

ρλτ′(x, y) =

{

ρλτ (x, y), if λ ∈ K

ρk0τ (x, y), if λ ∈ Λ \ K,

is a τ′-metric on X such that Tρ
τ
′
= Tρτ . �

The following examples show that τ-metrizable spaces may be not metri-
zable.

Example 2.4. The product Rc =
∏

λ∈Λ Xλ, where Xλ = R for every λ ∈ Λ and
|Λ| = c, of uncountably many copies of the real line R is not metrizable, since
it is not first-countable. However, the space Rc is c-metrizable. Assuming each
copy Xλ of R has its usual metric dλ, the mapping ρc : R

c×Rc → Rc+ defined by
ρc(x, y) = {dλ(xλ, yλ)}λ∈Λ for every x = {xλ}λ∈Λ ∈ R

c and y = {yλ}λ∈Λ ∈ R
c

is a c-metric on Rc and the topology induced by ρc coincides with the product
topology.

Example 2.5. Let R be the set of real numbers with the discrete topology
D and (R∞, D∞) be the Alexandroff’s one-point compactification of the space
(R, D), that is R∞ = R∪{∞} and D∞ = D∪{R∞\K : K is a finite subset of R}.
The space (R∞, D∞) is not metrizable (because it is not separable). We prove
that the space (R∞, D∞) is c-metrizable. Let Fin(R) be the collection of all
the nonempty finite subsets of R with |Fin(R)| = c. For every F ∈ Fin(R) we
define:

(1) ρF (x, x) = 0 for each x ∈ R∞.

(2) ρF (x, ∞) = ρF (∞, x) =

{

0, if x /∈ F

1, otherwise
for each x ∈ R.

(3) ρF (x, y) =

{

0, if x /∈ F and y /∈ F

1, otherwise
for each x, y ∈ R with x 6= y.

The mapping ρc : R∞ ×R∞ → R
c

+ defined by ρc(x, y) = {ρF (x, y)}F ∈F in(R) for
every x, y ∈ R∞ is a c-metric on R∞. We prove that the topology Tρ

c

induced
by the c-metric ρc coincides with the topology D∞.

Let x ∈ R. If G(x) = {y ∈ R∞ : ρc(x, y) ∈ O(θ)}, where O(θ) is an open
neighbourhood of the point θ in the space Rc+, then {x} ∈ D∞ and {x} ⊆ G(x).
Moreover, for the open neighbourhood

∏

F ∈F in(R) WF of the point θ, where

WF =

{

[0, 1
2
), if F = {x}

R+, otherwise

we have G(x) = {y ∈ R∞ : ρc(x, y) ∈
∏

F ∈F in(R) WF } ⊆ {x}.

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A. C. Megaritis

Now, we consider the point ∞ of R∞. If {∞} ∪ (R \ K), where K ∈ Fin(R)
is an open neighbourhood of the point ∞ in the space R∞, then for the open
neighbourhood

∏

F ∈F in(R) WF of the point θ, where

WF =

{

[0, 1
2
), if F = K

R+, otherwise

we have G(∞) = {y ∈ R∞ : ρc(∞, y) ∈
∏

F ∈F in(R) WF } ⊆ {∞} ∪ (R \ K).

Finally, let
∏

F ∈F in(R) UF be an open neighbourhood of the point θ in the

space Rc+ and suppose that {F ∈ Fin(R) : UF 6= R+} = {K1, . . . , Km}. Then,

{∞} ∪ (R \ (K1 ∪ . . . ∪ Km)) ⊆ G(∞) = {y ∈ R∞ : ρc(∞, y) ∈
∏

F ∈F in(R)

UF }.

However, a τ-metrizable space may be metrizable considering addition con-
ditions as the following assertions show.

Proposition 2.6. A n-metric space is metrizable for every finite cardinal num-
ber n.

Proof. Let (X, ρn) be a n-metric space and Tρn be the topology induced by ρn.
Consider a vector expression of the form ρn(x, y) = (ρ

1
n(x, y), . . . , ρ

n
n(x, y)) for

every x, y ∈ X. The mapping ρ : X × X → R+ defined by

ρ(x, y) = max{ρ1n(x, y), . . . , ρ
n
n(x, y)}

for every x, y ∈ X is a metric on X. It is easy to see that the metric topology
is the same as Tρn. �

Definition 2.7. Two τ-metrics ρ1τ and ρ2τ on a set X are called equivalent
if they induce the same topology on X, that is Tρ1τ = Tρ2τ .

Example 2.8. Let ρτ be a τ-metric on X. Consider a set Λ such that |Λ| = τ
and let us set ρτ(x, y) = {ρ

λ
τ (x, y)}λ∈Λ for every x, y ∈ X. The mapping

ρ∗τ : X × X → R
τ
+ defined by ρ

∗
τ (x, y) =

{

min{1, ρλτ (x, y)}
}

λ∈Λ
for every

x, y ∈ X is a τ-metric on X equivalent to ρτ.

Proposition 2.9. An ℵ0-metric space is metrizable.

Proof. Let (X, ρ
ℵ0
) be an ℵ0-metric space. Consider the equivalent ℵ0-metric

ρ∗
ℵ0

to ρ
ℵ0

of Example 2.8. Let ρ∗
ℵ0
(x, y) = (ρ∗1

ℵ0
(x, y), ρ∗2

ℵ0
(x, y), . . .) for every

x, y ∈ X. The mapping ρ : X × X → R+ defined by

ρ(x, y) =

∞
∑

i=1

1

2i
ρ∗i
ℵ0
(x, y)

for every x, y ∈ X is a metric on X. The process of proving that the topology
induced by the metric ρ coincides with the topology Tρ

ℵ0

is similar to the proof

of the Theorem 4.2.2 of [2]. �

Corollary 2.10. If τ 6 ℵ0, then every τ-metrizable space is metrizable.

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τ-metrizable spaces

Proof. Follows directly from Propositions 2.6 and 2.9. �

Proposition 2.11. For each τ > ℵ0 there is a τ-metrizable space Xτ with
w(Xτ ) = τ, which is not metrizable.

Proof. Let Xτ be the Alexandroff’s one-point compactification of a discrete
space X of cardinality τ, where τ > ℵ0. The space Xτ is not metrizable
(because it is not separable). It is known that |Fin(X)| = |X| = τ. Therefore,
in the same manner as in Example 2.5, we can prove that the space Xτ is
τ-metrizable. Let us note that w(Xτ ) = τ. �

Proposition 2.12. For every τ > ℵ0 and every τ-metrizable space X, we have
χ(X) 6 τ.

Proof. Let X be a τ-metrizable space and ρτ be a τ-metric on the space X
with τ > ℵ0. Consider a set Λ such that |Λ| = τ. The family Bθ of all products
∏

λ∈Λ Wλ, where finitely many Wλ are intervals of the form [0, b) with rational
b and the remaining Wλ = R+, form a local basis of the point θ in the space
R

τ
+. Hence, for every x ∈ X, the family

B(x) = {G(x) = {y ∈ X : ρτ(x, y) ∈ B} : B ∈ Bθ}

is a local basis of the point x in the space X. Since |Bθ| = τ, we have |B(x)| 6
τ. �

3. A τ-metrization theorem

Metrization theorems are theorems that give sufficient conditions for a topo-
logical space to be metrizable (see [2,5]). In this section we obtain a general-
ization of the classical metrization theorem of Urysohn.

Lemma 3.1. If (X, ρτ) is a τ-metric space and A is a subspace of X, then the
topology induced by the restriction of the τ-metric ρτ to A × A is the same as
the subspace topology of A in X.

Theorem 3.2. Every Tychonoff space of weight τ > ℵ0 is τ-metrizable.

Proof. Let X be a Tychonoff space such that w(X) = τ > ℵ0. The space
Iτ =

∏

λ∈Λ Xλ, where Xλ = I for every λ ∈ Λ and |Λ| = τ is τ-metrizable
(see Example 2.4). Assuming each copy Xλ of I has its usual metric dλ, the
mapping dτ : I

τ × Iτ → Rτ+ defined by dτ(x, y) = {dλ(xλ, yλ)}λ∈Λ for every
x = {xλ}λ∈Λ ∈ I

τ and y = {yλ}λ∈Λ ∈ I
τ is a τ-metric on Iτ. We shall

prove that X is τ-metrizable by imbedding X into the τ-metrizable space Iτ,
i.e. by showing that X is homeomorphic with a subspace of Iτ . But this
follows immediately from the fact that the Tychonoff cube Iτ is universal for
all Tychonoff spaces of weight τ (see [2, Theorem 2.3.23]). By Lemma 3.1, the
space X is τ-metrizable. �

As every τ-metrizable space is Tychonoff (see [1]), we get the following result.

Corollary 3.3. A space of weight τ > ℵ0 is τ-metrizable if and only if it is
Tychonoff.

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A. C. Megaritis

Remark 3.4. We can use Theorem 3.2 to find τ-metrizable spaces, where τ >
ℵ0, that are not metrizable. Below we consider some examples. Example
3.5 is a c-metrizable space which is not second-countable, Example 3.6 is a c-
metrizable space which is not normal and Example 3.7 is a 2τ-metrizable space,
where τ > c, which is not metrizable.

Example 3.5. Let S be the Sorgenfrey line, that is the real line with the
topology in which local basis of x are the sets [x, y) for y > x. Since X is
separable but not second-countable, it cannot be metrizable. Furthermore, S
is Tychonoff and w(S) = c. From Theorem 3.2 it follows that the Sorgenfrey
line is a c-metrizable space.

Example 3.6. Let P = {(α, β) ∈ R2 : β > 0} be the open upper half-plane
with the Euclidean topology and L = {(α, β) ∈ R2 : β = 0}. We set X = P ∪L.
For every x ∈ P let B(x) be the family of all open discs in P centered at x. For
every x ∈ L let B(x) be the family of all sets of the form {x}∪D, where D is an
open disc in P which is tangent to L at the point x. The family T of all subsets
of X that are unions of subfamilies of ∪{B(x) : x ∈ X} is a topology on X and
the family {B(x) : x ∈ X} is a neighbourhood system for the topological space
(X, T ). The space X is called the Niemytzki plane (see, for example, [2, 4]).
X is a Tychonoff space with w(X) = c, which is not normal. Therefore, by
Theorem 3.2, X is a c-metrizable space, but not metrizable.

Example 3.7. Let βD(τ) be the Čech-Stone compactification of the discrete
space D(τ) of cardinality τ > c. Then, w(βD(τ)) = 2τ (see [2, Theorem
3.6.11]). Since βD(τ) is zero-dimensional (see [2, Theorem 3.6.13]), it is Ty-
chonoff. The space D(τ) is not compact. Therefore, βD(τ) is not metrizable
(see [3, Exercise 9, §38, Ch.5]). From Theorem 3.2 it follows that βD(τ) is 2τ-
metrizable. Particularly, if one assumes the continuum hypothesis, the Čech-
Stone compactification βω of the discrete space of the non-negative integers
ω = {0, 1, 2, . . .} is c-metrizable.

Remark 3.8. A space X may be τ-metrizable for some infinite cardinal number
τ < w(X), as shown in the following example.

Example 3.9. Let Λ be a set of cardinality τ > ℵ0, D(κ) the discrete space of
cardinality κ > τ, and F =

∏

λ∈Λ Xλ, where Xλ = D(κ) for every λ ∈ Λ, with
the Tychonoff product topology. We note that the points of F are functions
from Λ to D(κ). The space F is not metrizable for χ(F) = τ (see [2, Exercise
2.3.F(b)]). Moreover, w(F) = κ (see [2, Exercise 2.3.F(a)]). We prove that the
space F is τ-metrizable. For every λ ∈ Λ we define:

(1) ρλ(f, f) = 0 for each f ∈ F .

(2) ρλ(f, g) =

{

0, if f(λ) = g(λ)

1, otherwise
for each f, g ∈ F with f 6= g.

The mapping ρτ : F × F → R
τ
+ defined by ρτ (f, g) = {ρλ(f, g)}λ∈Λ for every

f, g ∈ F is a τ-metric on F . We prove that the topology Tρ
τ

induced by the
τ-metric ρτ coincides with the Tychonoff product topology.

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τ-metrizable spaces

Let f ∈ F ,
∏

λ∈Λ Uλ be an open neighbourhood of the point θ in the space
R

τ
+, and suppose that {λ ∈ Λ : Uλ 6= R+} = {λ1, . . . , λm}. For the open

neighbourhood
∏

λ∈Λ Wλ of the point f, where

Wλ =

{

{f(λ)}, if λ ∈ {λ1, . . . , λm}

D(κ), otherwise

we have
∏

λ∈Λ Wλ ⊆ G(f) = {g ∈ F : ρτ(f, g) ∈
∏

λ∈Λ Uλ}.
Now, let f ∈ F and

∏

λ∈Λ Wλ be an open neighbourhood of the point f in
the space F , and suppose that {λ ∈ Λ : Wλ 6= D(κ)} = {λ1, . . . , λm}. For the
open neighbourhood

∏

λ∈Λ Uλ of the point θ, where

Uλ =

{

[0, 1
2
), if λ ∈ {λ1, . . . , λm}

R+, otherwise

we have G(f) = {g ∈ F : ρτ(f, g) ∈
∏

λ∈Λ Uλ} ⊆
∏

λ∈Λ Wλ.

4. Compact τ-metrizable spaces

It is well known that every compact metrizable space is separable. An ana-
logous result for τ-metrizable spaces is stated in this section.

Let us consider a set Λ such that |Λ| = τ > ℵ0 and let Bε be the family
of all open subsets

∏

λ∈Λ Wλ of the product R
τ
+, where finitely many Wλ are

intervals of the form [0, ε) and the remaining Wλ = R+.

Definition 4.1. Let (X, ρτ ) be a τ-metric space. A subset A of X is called
Oε-dense in (X, ρτ), where Oε ∈ Bε, if for every x ∈ X there exists a ∈ A such
that ρτ (x, a) ∈ Oε.

Definition 4.2. A τ-metric space (X, ρτ) is called ε-totally bounded if for
every Oε ∈ Bε there exists a finite subset A of X which is Oε-dense in (X, ρτ).
The τ-metric space (X, ρτ ) is called totally bounded if it is ε-totally bounded
for every ε > 0.

Recall that the density d(X) of a topological space X, is defined to be
d(X) = min{|D| : D is a dense subset of X}.

Proposition 4.3. For every totally bounded τ-metric space X, the inequality
d(X) 6 τ holds.

Proof. Let n ∈ {1, 2, . . .}. For each O1/n ∈ B1/n, let A(O1/n) be a finite O1/n-
dense subset of X and consider the subset An = ∪{A(O1/n) : O1/n ∈ B1/n} of
X with |An| 6 τ. The subset A = ∪

∞
n=1An of X is dense and |A| 6 τ. �

Proposition 4.4. Every compact τ-metric space X is totally bounded.

Proof. Let ε > 0. For every Oε ∈ Bε the family

{G(x) = {y ∈ X : ρτ (x, y) ∈ Oε} : x ∈ X}

forms an open cover of X. By compactness of X, there exists a finite subset
A of X such that

⋃

a∈A G(a) = X. For every x ∈ X there exists a ∈ A with

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A. C. Megaritis

x ∈ G(a). Therefore, ρτ(x, a) ∈ Oε and the subset A of X is Oε-dense in
(X, ρτ). �

Theorem 4.5. For every compact τ-metrizable space X we have d(X) 6 τ.

Proof. Let X be a compact τ-metrizable space. According to Proposition 4.4,
the space X is totally bounded. Therefore, by virtue of Proposition 4.3, d(X) 6
τ. �

Acknowledgements. The author would like to thank both referees for their

valuable comments and suggestions.

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