Functional Analysis In Asymmetric Normed Spaces 2013th Edition Stefan Cobzas
Functional Analysis In Asymmetric Normed Spaces 2013th Edition Stefan Cobzas
Functional Analysis In Asymmetric Normed Spaces 2013th Edition Stefan Cobzas
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12. Introduction
The main goal of this book is to present the basic results on asymmetric normed
spaces. Since the basic topological tools come from quasi-metric spaces and quasi-
uniform spaces, the first chapter contains a thorough presentation of some funda-
mental results from the theory of these spaces. The focus is on those which are
most used in functional analysis – completeness, compactness and Baire category.
For a good presentation of the general theory of quasi-uniform and quasi-metric
spaces, a well-established and thickly developed branch of general topology, one
can consult the classical monograph by Fletcher and Lindgren [80] and some sub-
sequent survey papers by Künzi (see the bibliography at the end of the book). The
survey paper [45] may be viewed as a skeleton of this book.
A quasi-metric is a function 𝜌 on 𝑋 × 𝑋 satisfying all the axioms of a metric
with the exception of the symmetry: it is possible that 𝜌(𝑦, 𝑥) ∕= 𝜌(𝑥, 𝑦) for some
𝑥, 𝑦 ∈ 𝑋. In this case ¯
𝜌(𝑥, 𝑦) = 𝜌(𝑦, 𝑥) is another quasi-metric on 𝑋, called the
conjugate of 𝜌, and 𝜌𝑠
= 𝜌 ∨ ¯
𝜌 is a metric on 𝑋. Asymmetric metric spaces are
called quasi-metric spaces. The term quasi-metric was proposed as early as 1931 by
Wilson [239], see also [1]. Quasi-metric spaces were considered also by Niemytzki
[168] in connection with the axioms defining a metric space and metrizability. In
[27], [186] they are called oriented metric spaces and in [187] spaces with weak
metric.
This apparently innocent modification of the axioms of a metric space drasti-
cally changes the whole theory, mainly with respect to completeness, compactness
and total boundedness. There are a lot of completeness notions in quasi-metric
and quasi-uniform spaces, all agreeing with the usual notion of completeness in
the case of metric or uniform spaces, each of them having its advantages and
weaknesses.
Also, concerning compactness, the situation is totally different in quasi-metric
spaces – for instance, sequential compactness does not agree with compactness,
in contrast to the case of metric spaces. In spite of these peculiarities there are
a lot of positive results relating compactness with various kinds of completeness
and total boundedness. Baire category also needs a special treatment, including
some bitopological results.
Quasi-uniform spaces form a natural extension of both quasi-metric spaces
and uniform spaces. A quasi-uniformity is a family 𝒰 of subsets of 𝑋 × 𝑋, called
vii
13. viii Introduction
entourages, satisfying all the axioms of a uniformity excepting symmetry: one does
not suppose that 𝒰 has a base formed of symmetric entourages. Again, 𝒰−1
=
{𝑈−1
: 𝑈 ∈ 𝒰} is another quasi-uniformity on 𝑋, called the conjugate of 𝒰, and
𝒰𝑠
= 𝒰 ∨𝒰−1
is a uniformity. The notions of completeness can be transposed from
quasi-metric spaces to quasi-uniform spaces by replacing sequences with nets or
filters. Again the focus is on the relations between compactness, completeness and
total boundedness within this framework.
On a quasi-metric space (𝑋, 𝜌) there are two natural topologies generated
by the quasi-metric 𝜌 and its conjugate ¯
𝜌, respectively by the quasi-uniformity 𝒰
and its conjugate 𝒰−1
, making quasi-metric and quasi-uniform spaces bitopologi-
cal spaces. For this reason the first chapter of the book contains a quite detailed
introduction to bitopological spaces, including Urysohn and Tietze type theorems
for semi-continuous functions on bitopological spaces, compactness and Baire cat-
egory.
Following the advice of Einar Hille [105] that “a functional analyst is an
analyst, first and foremost, and not a degenerate species of a topologist”, after
this detour in topology we turn to functional analysis. Functional analysis in the
asymmetric case, meaning the study of asymmetric normed spaces, asymmetric
locally convex spaces and of operators acting between them, with emphasis on
linear functionals and dual spaces, is treated in the second chapter.
An asymmetric norm is a positive definite sublinear functional 𝑝 on a real
vector space 𝑋. Since the possibility that 𝑝(𝑥) = 𝑝(−𝑥) for some 𝑥 ∈ 𝑋 is not
excluded, ¯
𝑝(𝑥) = 𝑝(−𝑥), 𝑥 ∈ 𝑋, is another asymmetric norm on 𝑋 called the con-
jugate of 𝑝, and 𝑝𝑠
= 𝑝 ∨ ¯
𝑝 is a norm on 𝑋. The topological notion are considered
with respect to the attached metric 𝜌𝑝(𝑥, 𝑦) = 𝑝(𝑦 − 𝑥), 𝑥, 𝑦 ∈ 𝑋. Any asym-
metric norm can be obtained as the Minkowski gauge functional of an absorbing
convex subset of 𝑋. Asymmetric locally convex spaces are defined as vector spaces
equipped with a topology generated by a family of asymmetric seminorms.
Of great importance is the asymmetric norm 𝑢 on ℝ given by 𝑢(𝑡) = 𝑡+
, 𝑡 ∈
ℝ, with conjugate ¯
𝑢(𝑡) = 𝑡−
and 𝑢𝑠
= ∣⋅∣. The topology generated by 𝑢 is called the
upper topology of ℝ, while that generated by its conjugate ¯
𝑢, the lower topology.
If (𝑇, 𝜏) is a topological space, then a real-valued function 𝑓 on 𝑇 is upper semi-
continuous as a function from 𝑇 to (ℝ, ∣ ⋅ ∣) if and only if it is continuous from 𝑇
to (ℝ, 𝑢). Similarly, 𝑓 is (𝜏, ∣ ⋅ ∣)-lower semi-continuous if and only if it is (𝜏, ¯
𝑢)-
continuous.
The main differences with respect to the classical functional analysis (mean-
ing analysis over the fields ℝ or ℂ) come from the fact that the asymmetric norm 𝑝
does not generate a vector topology on 𝑋: the addition is continuous with respect
to the product topology on 𝑋, but the multiplication by scalars is continuous only
when restricted to (0; ∞) × 𝑋. Also, for each fixed 𝑥, the function 𝑓 : ℝ → 𝑋
given by 𝑓(𝑡) = 𝑡𝑥, 𝑡 ∈ ℝ, is continuous. The dual space of an asymmetric normed
space (𝑋, 𝑝), denoted by 𝑋♭
𝑝, formed by all linear and ∣ ⋅ ∣-upper semi-continuous
functions, or, equivalently, linear continuous functionals from (𝑋, 𝑝) to (ℝ, 𝑢), is
14. Introduction ix
not a linear space but merely a cone contained in the dual space 𝑋∗
= (𝑋, 𝑝𝑠
)∗
of the associated normed space (𝑋, 𝑝𝑠
). The situation is similar for the set of all
continuous linear operators between two asymmetric normed spaces, as well as for
asymmetric locally convex spaces.
In spite of these differences, many results from classical functional analysis
have their counterparts in the asymmetric case, by taking care of the interplay
between the asymmetric norm 𝑝 and its conjugate ¯
𝑝. Among the positive results
we mention: Hahn-Banach type theorems and separation results for convex sets,
Krein-Milman type theorems, analogs of the fundamental principles – open map-
ping and closed graph theorems – an analog of the Schauder theorem on the com-
pactness of a conjugate mapping. Applications are given to best approximation
problems and, as relevant examples, one considers normed lattices equipped with
asymmetric norms and spaces of semi-Lipschitz functions on quasi-metric spaces.
It is difficult to localize the first moment when asymmetric norms were used,
but it goes back as early as 1968 in a paper by Duffin and Karlovitz (1968) [70],
who proposed the term asymmetric norm. Krein and Nudelman (1973) [129] used
also asymmetric norms in their study of some extremal problems related to the
Markov moment problem. Remark that the relevance of sublinear functionals for
some problems of convex analysis and of mathematical analysis was emphasized
also by H. König in the 1970s. A systematic study of the properties of asymmetric
normed spaces started with the papers of S. Romaguera, from the Polytechnic
University of Valencia, and his collaborators from the same university and from
other universities in Spain: Alegre, Ferrer, Garcı́a-Raffi, Sánchez Pérez, Sánchez
Álvarez, Sanchis, Valero (see the bibliography). Besides its intrinsic interest, their
study was motivated also by applications in Computer Science, namely to the
complexity analysis of programs, results obtained in cooperation with Professor
Schellekens from the National University of Ireland.
Containing very recent results, some of them appearing for the first time in
print, in the focus of current research, on quasi-metric, quasi-uniform, asymmetric
normed and asymmetric locally convex spaces, the book can be used as a reference
by researchers in this domain. Due to the detailed exposition of the subject, the
book can be also used as an introductory text for newcomers.
Acknowledgement. The author expresses his gratitude to the staff of Birkhäuser-
Springer, particularly to the Editors Anna Mätzener and Sylvia Lotrovsky, for the
professional work and excellent cooperation during the publication process, and
to Ute McCrory from Springer DE for support.
I want to mention also that this research was supported by
Grant CNCSIS 2261, ID 543.
Notation. We present here, for the convenience of the reader, some symbols that
are used throughout the text, which could differ from the standard ones. Other
notations are standard or explained in the text, some of them being included in
the index at the end of the book.
15. x Introduction
∙ ℕ = {1, 2, . . .} – the set of natural numbers (positive integers);
∙ [𝑎; 𝑏], (𝑎; 𝑏), (𝑎; 𝑏], [𝑎; 𝑏) – intervals;
∙ (𝑎, 𝑏) – an ordered pair;
∙ 𝐵𝜌[𝑥, 𝑟] = {𝑦 ∈ 𝑋 : 𝜌(𝑥, 𝑦) ≤ 𝑟} – a closed ball in a quasi-metric space (𝑋, 𝜌);
∙ 𝐵𝜌(𝑥, 𝑟) = {𝑦 ∈ 𝑋 : 𝜌(𝑥, 𝑦) < 𝑟} – an open ball;
∙ 𝜌𝑝(𝑥, 𝑦) = 𝑝(𝑦 − 𝑥) – the quasi-metric associated to an asymmetric norm 𝑝;
∙ 𝐵𝑝 = {𝑥 ∈ 𝑋 : 𝑝(𝑥) ≤ 1} – the closed unit ball of an asymmetric normed
space (𝑋, 𝑝);
∙ 𝐵′
𝑝 = {𝑥 ∈ 𝑋 : 𝑝(𝑥) < 1} – the open unit ball;
∙ 𝑆𝑝 = {𝑥 ∈ 𝑋 : 𝑝(𝑥) = 1} – the unit sphere;
∙ 𝑢 is the standard asymmetric norm 𝑢(𝑡) = 𝑡+
on ℝ.
Cluj-Napoca, July 2012 Ştefan Cobzaş
17. 2 Chapter 1. Quasi-metric and Quasi-uniform Spaces
1.1 Topological properties of quasi-metric
and quasi-uniform spaces
In this section we shall present the basic topological properties of quasi-metric and
quasi-uniform spaces, with emphasis on asymmetric normed spaces and asymmet-
ric locally convex spaces. Since quasi-metric and quasi-uniform spaces are par-
ticular cases of bitopological spaces, a quite detailed presentation of bitopological
spaces is also given, including compactness, normality, regularity, a Tikhonov type
theorem on the existence of semi-continuous functions and Tietze-Urysohn type
theorems on the extension of semi-continuous functions.
1.1.1 Quasi-metric spaces and asymmetric normed spaces
A quasi-semimetric on a set 𝑋 is a mapping 𝜌 : 𝑋 × 𝑋 → [0; ∞) satisfying the
following conditions:
(QM1) 𝜌(𝑥, 𝑦) ≥ 0, 𝑎𝑛𝑑 𝜌(𝑥, 𝑥) = 0;
(QM2) 𝜌(𝑥, 𝑧) ≤ 𝜌(𝑥, 𝑦) + 𝜌(𝑦, 𝑧) ,
for all 𝑥, 𝑦, 𝑧 ∈ 𝑋. If, further,
(QM3) 𝜌(𝑥, 𝑦) = 𝜌(𝑦, 𝑥) = 0 ⇒ 𝑥 = 𝑦 ,
for all 𝑥, 𝑦 ∈ 𝑋, then 𝜌 is called a quasi-metric. The pair (𝑋, 𝜌) is called a quasi-
semimetric space, respectively a quasi-metric space. The conjugate of the quasi-
semimetric 𝜌 is the quasi-semimetric ¯
𝜌(𝑥, 𝑦) = 𝜌(𝑦, 𝑥), 𝑥, 𝑦 ∈ 𝑋. The mapping
𝜌𝑠
(𝑥, 𝑦) = max{𝜌(𝑥, 𝑦), ¯
𝜌(𝑥, 𝑦)}, 𝑥, 𝑦 ∈ 𝑋, is a semimetric on 𝑋 which is a metric
if and only if 𝜌 is a quasi-metric. Sometimes one works with extended quasi-
semimetrics, meaning that the quasi-semimetric 𝜌 can take the value +∞ for
some 𝑥, 𝑦 ∈ 𝑋. The following inequalities hold for these quasi-semimetrics for all
𝑥, 𝑦 ∈ 𝑋:
𝜌(𝑥, 𝑦) ≤ 𝜌𝑠
(𝑥, 𝑦) and ¯
𝜌(𝑥, 𝑦) ≤ 𝜌𝑠
(𝑥, 𝑦) . (1.1.1)
An asymmetric norm on a real vector space 𝑋 is a functional 𝑝 : 𝑋 → [0, ∞)
satisfying the conditions
(AN1) 𝑝(𝑥) = 𝑝(−𝑥) = 0 ⇒ 𝑥 = 0;
(AN2) 𝑝(𝛼𝑥) = 𝛼𝑝(𝑥);
(AN3) 𝑝(𝑥 + 𝑦) ≤ 𝑝(𝑥) + 𝑝(𝑦) ,
for all 𝑥, 𝑦 ∈ 𝑋 and 𝛼 ≥ 0.
If 𝑝 satisfies only the conditions (AN2) and (AN3), then it is called an asym-
metric seminorm. The pair (𝑋, 𝑝) is called an asymmetric normed (respectively
seminormed) space. Again, in some instances, the value +∞ will be allowed for
18. 1.1. Topological properties of quasi-metric and quasi-uniform spaces 3
𝑝 in which case we shall call it an extended asymmetric norm (or seminorm). An
asymmetric seminorm 𝑝 defines a quasi-semimetric 𝜌𝑝 on 𝑋 through the formula
𝜌𝑝(𝑥, 𝑦) = 𝑝(𝑦 − 𝑥), 𝑥, 𝑦 ∈ 𝑋 . (1.1.2)
Defining the conjugate asymmetric seminorm ¯
𝑝 and the seminorm 𝑝𝑠
by
¯
𝑝(𝑥) = 𝑝(−𝑥) and 𝑝𝑠
(𝑥) = max{𝑝(𝑥), 𝑝(−𝑥)} , (1.1.3)
for 𝑥 ∈ 𝑋, the inequalities (1.1.1) become
𝑝(𝑥) ≤ 𝑝𝑠
(𝑥) and ¯
𝑝(𝑥) ≤ 𝑝𝑠
(𝑥) , (1.1.4)
for all 𝑥 ∈ 𝑋. Obviously, 𝑝𝑠
is a norm when 𝑝 is an asymmetric norm and (𝑋, 𝑝𝑠
)
is a normed space.
The conjugates of 𝜌 and 𝑝 are denoted also by 𝜌−1
and 𝑝−1
, a notation that
we shall use occasionally.
If (𝑋, 𝜌) is a quasi-semimetric space, then for 𝑥 ∈ 𝑋 and 𝑟 > 0 we define the
balls in 𝑋 by the formulae
𝐵𝜌(𝑥, 𝑟) = {𝑦 ∈ 𝑋 : 𝜌(𝑥, 𝑦) < 𝑟} – the open ball, and
𝐵𝜌[𝑥, 𝑟] = {𝑦 ∈ 𝑋 : 𝜌(𝑥, 𝑦) ≤ 𝑟} – the closed ball.
In the case of an asymmetric seminormed space (𝑋, 𝑝) the balls are given by
𝐵𝑝(𝑥, 𝑟) = {𝑦 ∈ 𝑋 : 𝑝(𝑦 −𝑥) < 𝑟}, respectively 𝐵𝑝[𝑥, 𝑟] = {𝑦 ∈ 𝑋 : 𝑝(𝑦 −𝑥) ≤ 𝑟} .
The closed unit ball of 𝑋 is 𝐵𝑝 = 𝐵𝑝[0, 1] and the open unit ball is 𝐵′
𝑝 =
𝐵𝑝(0, 1). In this case the following formulae hold true:
𝐵𝑝[𝑥, 𝑟] = 𝑥 + 𝑟𝐵𝑝 and 𝐵𝑝(𝑥, 𝑟) = 𝑥 + 𝑟𝐵′
𝑝 , (1.1.5)
that is, any of the unit balls of 𝑋 completely determines its quasi-metric structure.
If necessary, these balls will be denoted by 𝐵𝑝,𝑋 and 𝐵′
𝑝,𝑋, respectively.
The conjugate ¯
𝑝 of 𝑝 is defined by 𝑝(𝑥) = 𝑝(−𝑥), 𝑥 ∈ 𝑋, and the associate
seminorm is 𝑝𝑠
(𝑥) = max{𝑝(𝑥), ¯
𝑝(𝑥)}, 𝑥 ∈ 𝑋. The seminorm 𝑝 is an asymmetric
norm if and only if 𝑝𝑠
is a norm on 𝑋. Sometimes an asymmetric norm will be
denoted by the symbol ∥ ⋅ ∣, a notation proposed by Krein and Nudelman, [129,
Ch. IX, §5], in their book on the theory of moments.
Remark 1.1.1. Since the terms “quasi-norm”, “quasi-normed space” and “quasi-
Banach space” are already “registered trademarks” (see, for instance, the survey
by Kalton [106]), we can not use these terms to designate an asymmetric norm,
an asymmetric normed space or an asymmetric biBanach space. A quasi-normed
space is a vector space 𝑋 equipped with a functional ∥ ⋅ ∥ : 𝑋 → [0; ∞), satisfying
all the axioms of a norm, excepting the triangle inequality which is replaced by
∥𝑥 + 𝑦∥ ≤ 𝐶(∥𝑥∥ + ∥𝑦∥), 𝑥, 𝑦 ∈ 𝑋 ,
19. 4 Chapter 1. Quasi-metric and Quasi-uniform Spaces
for some constant 𝐶 ≥ 1. It is obvious that for 𝐶 = 1 the functional ∥ ⋅ ∥ is a
norm. The reverse situation is also encountered: in [232] a quasi-metric space is
a metric space (𝑋, 𝜌) in which the triangle inequality is replaced by 𝜌(𝑥, 𝑧) ≤
𝐶 (𝜌(𝑥, 𝑦) + 𝜌(𝑦, 𝑧)) , for some 𝐶 > 0.
Note also that the alternative terms “quasi-pseudometric” is used by many
topologists instead of “quasi-semimetric”. We have preferred the term semimetric
to be in concordance with the notion of seminorm – in this way an asymmetric
seminorm induces a quasi-semimetric.
1.1.2 The topology of a quasi-semimetric space
The topology 𝜏(𝜌) of a quasi-semimetric space (𝑋, 𝜌) can be defined starting from
the family 𝒱𝜌(𝑥) of neighborhoods of an arbitrary point 𝑥 ∈ 𝑋:
𝑉 ∈ 𝒱𝜌(𝑥) ⇐⇒ ∃𝑟 > 0 such that 𝐵𝜌(𝑥, 𝑟) ⊂ 𝑉
⇐⇒ ∃𝑟′
> 0 such that 𝐵𝜌[𝑥, 𝑟′
] ⊂ 𝑉 .
To see the equivalence in the above definition, we can take, for instance,
𝑟′
= 𝑟/2.
A set 𝐺 ⊂ 𝑋 is 𝜏(𝜌)-open if and only if for every 𝑥 ∈ 𝐺 there exists 𝑟 = 𝑟𝑥 > 0
such that 𝐵𝜌(𝑥, 𝑟) ⊂ 𝐺. Sometimes we shall say that 𝑉 is a 𝜌-neighborhood of 𝑥
or that the set 𝐺 is 𝜌-open.
The convergence of a sequence (𝑥𝑛) to 𝑥 with respect to 𝜏(𝜌), called 𝜌-
convergence and denoted by 𝑥𝑛
𝜌
−
→ 𝑥, can be characterized in the following way:
𝑥𝑛
𝜌
−
→ 𝑥 ⇐⇒ 𝜌(𝑥, 𝑥𝑛) → 0 . (1.1.6)
Also
𝑥𝑛
¯
𝜌
−
→ 𝑥 ⇐⇒ ¯
𝜌(𝑥, 𝑥𝑛) → 0 ⇐⇒ 𝜌(𝑥𝑛, 𝑥) → 0 . (1.1.7)
The following proposition contains some simple properties of convergent se-
quences.
Proposition 1.1.2. Let (𝑥𝑛) be a sequence in a quasi-semimetric space (𝑋, 𝜌).
1. If (𝑥𝑛) is 𝜏𝜌-convergent to 𝑥 and 𝜏¯
𝜌-convergent to 𝑦, then 𝜌(𝑥, 𝑦) = 0.
2. If (𝑥𝑛) is 𝜏𝜌-convergent to 𝑥 and 𝜌(𝑦, 𝑥) = 0, then (𝑥𝑛) is also 𝜏𝜌-convergent
to 𝑦.
Proof. 1. Letting 𝑛 → ∞ in the inequality 𝜌(𝑥, 𝑦) ≤ 𝜌(𝑥, 𝑥𝑛)+𝜌(𝑥𝑛, 𝑦), one obtains
𝜌(𝑥, 𝑦) = 0.
2. Follows from the relations 𝜌(𝑦, 𝑥𝑛) ≤ 𝜌(𝑦, 𝑥) + 𝜌(𝑥, 𝑥𝑛) = 𝜌(𝑥, 𝑥𝑛) → 0 as
𝑛 → ∞. □
20. 1.1. Topological properties of quasi-metric and quasi-uniform spaces 5
Using the conjugate quasi-semimetric ¯
𝜌 one obtains another topology 𝜏(¯
𝜌). A
third one is the topology 𝜏(𝜌𝑠
) generated by the semimetric 𝜌𝑠
. Sometimes, (see,
for instance, Menucci [151] and Collins and Zimmer [50]) the balls with respect
to 𝜌 are called forward balls and the topology 𝜏(𝜌) is called the forward topology,
while the balls with respect to ¯
𝜌 are called backward balls and the topology 𝜏(¯
𝜌)
the backward topology. We shall use sometimes the alternative notation 𝜏𝜌, 𝜏¯
𝜌, 𝜏𝜌𝑠
to designate these topologies.
As a space with two topologies, 𝜏𝜌 and 𝜏¯
𝜌, a quasi-semimetric space can be
viewed as a bitopological space in the sense of Kelly [111] (see also the book [72])
and so, all the results valid for bitopological spaces apply to a quasi-semimetric
space. A bitopological space is simply a set 𝑇 endowed with two topologies 𝜏 and
𝜈. A bitopological space is denoted by (𝑇, 𝜏, 𝜈).
The following example is very important in what follows.
Example 1.1.3. On the field ℝ of real numbers consider the asymmetric norm
𝑢(𝛼) = 𝛼+
:= max{𝛼, 0}. Then, for 𝛼 ∈ ℝ, ¯
𝑢(𝛼) = 𝛼−
:= max{−𝛼, 0} and
𝑢𝑠
(𝛼) = ∣𝛼∣. The topology 𝜏(𝑢) generated by 𝑢 is called the upper topology of
ℝ, while the topology 𝜏(¯
𝑢) generated by ¯
𝑢 is called the lower topology of ℝ. A
basis of open 𝜏(𝑢)-neighborhoods of a point 𝛼 ∈ ℝ is formed of the intervals
(−∞; 𝛼 + 𝜀), 𝜀 > 0. A basis of open 𝜏(¯
𝑢)-neighborhoods is formed of the intervals
(𝛼 − 𝜀; ∞), 𝜀 > 0.
In this space the addition is continuous from (ℝ × ℝ, 𝜏𝑢 × 𝜏𝑢) to (ℝ, 𝜏𝑢), but
the multiplication need not be continuous at every point (𝛼, 𝛽) ∈ ℝ × ℝ.
The continuity property can be directly verified. To see the last assertion,
let 𝑉 = (−∞; 𝛼𝛽 + 𝜀), be a 𝜏𝑢-neighborhood of 𝛼𝛽, for some 𝜀 > 0. Since the
𝜏𝑢-neighborhoods of 𝛼 and 𝛽 contain −𝑛, for 𝑛 ∈ ℕ sufficiently large, it follows
that 𝑛2
= (−𝑛)(−𝑛) does not belong to 𝑉, for 𝑛 large enough.
Remark 1.1.4 (communicated by M.D. Mabula). In the asymmetric normed space
(ℝ, 𝑢) the sequence 𝛼𝑛 = (−1)𝑛
, 𝑛 ∈ ℕ, is 𝑢-convergent to 1 and ¯
𝑢-convergent to
−1, but it is not convergent in (ℝ, ∣ ⋅ ∣).
For a sequence (𝑥𝑛) in a quasi-semimetric space (𝑋, 𝜌), denote by 𝐿((𝑥𝑛))
the set of all 𝜌-limits of the sequence (𝑥𝑛), that is
𝐿𝜌((𝑥𝑛)) = {𝑥 ∈ 𝑋 : lim
𝑛
𝜌(𝑥, 𝑥𝑛) = 0}. (1.1.8)
The following proposition gives a characterization of this set in (ℝ, 𝑢).
Proposition 1.1.5 ([51]). Let (𝛼𝑛) be a sequence of real numbers. Then
𝐿𝑢((𝛼𝑛)) =
{
[lim sup𝑛 𝛼𝑛; ∞) if lim sup𝑛 𝛼𝑛 ∈ ℝ,
ℝ if lim sup𝑛 𝛼𝑛 = −∞.
(1.1.9)
If lim sup𝑛 𝛼𝑛 = ∞, then the sequence (𝛼𝑛) is not 𝑢-convergent.
21. 6 Chapter 1. Quasi-metric and Quasi-uniform Spaces
Similarly,
𝐿𝑢((𝛼𝑛)) =
{
(−∞; lim inf𝑛 𝛼𝑛] if lim inf𝑛 𝛼𝑛 ∈ ℝ,
ℝ if lim sup𝑛 𝛼𝑛 = ∞.
(1.1.10)
If lim sup𝑛 𝛼𝑛 = −∞, then the sequence (𝛼𝑛) is not ¯
𝑢-convergent.
Another important topological example is the so-called Sorgenfrey topology
on ℝ.
Example 1.1.6 (The Sorgenfrey line). For 𝑥, 𝑦 ∈ ℝ define a quasi-metric 𝜌 by
𝜌(𝑥, 𝑦) = 𝑦 − 𝑥, if 𝑥 ≤ 𝑦 and 𝜌(𝑥, 𝑦) = 1 if 𝑥 > 𝑦. A basis of open 𝜏𝜌-open neigh-
borhoods of a point 𝑥 ∈ ℝ is formed by the family [𝑥; 𝑥+𝜀), 0 < 𝜀 < 1. The family
of intervals (𝑥 − 𝜀; 𝑥], 0 < 𝜀 < 1, forms a basis of open 𝜏¯
𝜌 -open neighborhoods of
𝑥. Obviously, the topologies 𝜏𝜌 and 𝜏¯
𝜌 are Hausdorff and 𝜌𝑠
(𝑥, 𝑦) = 1 for 𝑥 ∕= 𝑦,
so that 𝜏(𝜌𝑠
) is the discrete topology of ℝ.
We shall present, for the convenience of the reader, the separation axioms.
A topological space (𝑇, 𝜏) is called
∙ 𝑇0 if for any pair 𝑠, 𝑡 of distinct points in 𝑇 , at least one of them has a
neighborhood not containing the other;
∙ 𝑇1 if for any pair 𝑠, 𝑡 of distinct points in 𝑇 , each of them has a neighborhood
not containing the other (this is equivalent to the fact that the set {𝑡} is closed
for every 𝑡 ∈ 𝑇 );
∙ Hausdorff or 𝑇2 if for any pair 𝑠, 𝑡 of distinct points in 𝑇 , there exist neigh-
borhoods 𝑈 of 𝑠 and 𝑉 of 𝑡 such that 𝑈 ∩ 𝑉 = ∅;
∙ regular if for each 𝑡 ∈ 𝑇 and each closed subset 𝑆 of 𝑇, not containing 𝑡,
there are disjoint open subsets 𝑈, 𝑉 of 𝑇 such that 𝑡 ∈ 𝑈 and 𝑆 ⊂ 𝑉. In other
words a point and a closed set not containing it can be separated by open
sets. This is equivalent to the fact that every point in 𝑇 has a neighborhood
base formed of closed sets. If 𝑇 is regular and 𝑇1, then it is called a 𝑇3 space.
∙ completely regular, or Tikhonov, or 𝑇3 1
2
, if for every 𝑡 ∈ 𝑇 and every closed
subset 𝑆 of 𝑇 not containing 𝑡 there is a continuous function 𝑓 : 𝑇 → [0; 1]
such that 𝑓(𝑡) = 1 and 𝑓(𝑠) = 0 for each 𝑠 ∈ 𝑆.
∙ normal if any pair 𝑆1, 𝑆2 of disjoint closed sets can be separated by open sets,
that is there exist two disjoint open sets 𝐺1 ⊃ 𝑆1 and 𝐺2 ⊃ 𝑆2. A normal 𝑇1
space is called a 𝑇4 space.
In general, we shall say that a bitopological space (𝑇, 𝜏, 𝜈) has a property 𝑃
if both the topologies 𝜏 and 𝜈 have the property 𝑃.
Now we introduce, following Kelly [111], some separation properties specific
to a bitopological space (𝑇, 𝜏, 𝜈).
The bitopological space (𝑇, 𝜏, 𝜈) is called pairwise Hausdorff if for each pair of
distinct points 𝑠, 𝑡 ∈ 𝑇 there exists a 𝜏-neighborhood 𝑈 of 𝑠 and a 𝜈-neighborhood
22. 1.1. Topological properties of quasi-metric and quasi-uniform spaces 7
𝑉 of 𝑡 such that 𝑈 ∩𝑉 = ∅. It is obvious that if 𝑇 is pairwise Hausdorff, then both
of the topologies 𝜏 and 𝜈 are 𝑇1.
Remark 1.1.7. Taking into account the symmetry (𝑥 ∕= 𝑦 ⇐⇒ 𝑦 ∕= 𝑥) it follows
that a bitopological space (𝑇, 𝜏, 𝜈) is pairwise Hausdorff if and only if for every
pair of distinct points 𝑥, 𝑦 from 𝑋 the following condition holds:
(∃𝑈 ∈ 𝜏, ∃𝑉 ∈ 𝜈, 𝑥 ∈ 𝑈 ∧ 𝑦 ∈ 𝑉 ∧ 𝑈 ∩ 𝑉 = ∅)
∧ (∃𝑈1 ∈ 𝜏, ∃𝑉1 ∈ 𝜈, 𝑦 ∈ 𝑈1 ∧ 𝑥 ∈ 𝑉1 ∧ 𝑈1 ∩ 𝑉1 = ∅) .
(1.1.11)
The topology 𝜏 is called regular with respect to 𝜈 if every 𝑡 ∈ 𝑇 has a 𝜏-
neighborhood base formed of 𝜈-closed sets or, equivalently, if for every 𝑡 ∈ 𝑇 and
every 𝜏-closed subset 𝑆 of 𝑇 not containing 𝑡, there exist a 𝜏-open set 𝑈 and a
𝜈-open set 𝑉 such that 𝑡 ∈ 𝑈, 𝑆 ⊂ 𝑉 and 𝑈 ∩ 𝑉 = ∅.
One says that the bitopological space (𝑇, 𝜏, 𝜈) is pairwise regular if 𝜏 is regular
with respect to 𝜈 and 𝜈 is regular with respect to 𝜏.
The bitopological space (𝑇, 𝜏, 𝜈) is called pairwise normal if given a 𝜏-closed
subset 𝐴 of 𝑇 and a 𝜈-closed subset 𝐵 of 𝑇 with 𝐴 ∩ 𝐵 = ∅, there exist a 𝜈-
open subset 𝑈 of 𝑇 and a 𝜏-open subset 𝑉 of 𝑇 such that 𝐴 ⊂ 𝑈, 𝐵 ⊂ 𝑉, and
𝑈 ∩ 𝑉 = ∅. Equivalently, the bitopological space (𝑇, 𝜏, 𝜈) is pairwise normal if
given an 𝜏-closed set 𝐶 and a 𝜈-open set 𝐷 with 𝐶 ⊂ 𝐷 there exist a 𝜈-open set
𝐺 and a 𝜏-closed set 𝐹 such that
𝐶 ⊂ 𝐺 ⊂ 𝐹 ⊂ 𝐷 , (1.1.12)
or, equivalently, there exists a 𝜈-open set 𝐺 such that
𝐶 ⊂ 𝐺 ⊂ 𝐺
𝜏
⊂ 𝐷 , (1.1.13)
where 𝐺
𝜏
denotes the closure of 𝐺 with respect to 𝜏. A bitopological space (𝑇, 𝜏, 𝜈)
is called quasi-semimetrizable if there exists a quasi-semimetric 𝜌 on 𝑇 such that
𝜏 = 𝜏𝜌 and 𝜈 = 𝜏¯
𝜌. If 𝜌 is a semimetric, then 𝜏 = 𝜈.
The following topological properties are true for quasi-semimetric spaces. We
use the abbreviation lsc for lower semicontinuous and usc for upper semicontinu-
ous.
Proposition 1.1.8. If (𝑋, 𝜌) is a quasi-semimetric space, then
1. Any ball 𝐵𝜌(𝑥, 𝑟) is 𝜏(𝜌)-open and a ball 𝐵𝜌[𝑥, 𝑟] is 𝜏(¯
𝜌)-closed. The ball
𝐵𝜌[𝑥, 𝑟] need not be 𝜏(𝜌)-closed.
Also, the following inclusions hold:
𝐵𝜌𝑠 (𝑥, 𝑟) ⊂ 𝐵𝜌(𝑥, 𝑟) and 𝐵𝜌𝑠 (𝑥, 𝑟) ⊂ 𝐵¯
𝜌(𝑥, 𝑟) ,
with similar inclusions for the closed balls.
2. The topology 𝜏(𝜌𝑠
) is finer than the topologies 𝜏(𝜌) and 𝜏(¯
𝜌). This means
that:
23. 8 Chapter 1. Quasi-metric and Quasi-uniform Spaces
∙ any 𝜏(𝜌)-open (closed) set is 𝜏(𝜌𝑠
)-open (closed); similar results hold
for the topology 𝜏(¯
𝜌);
∙ the identity mappings from (𝑋, 𝜏(𝜌𝑠
)) to (𝑋, 𝜏(𝜌)) and to (𝑋, 𝜏(¯
𝜌)) are
continuous;
∙ a sequence (𝑥𝑛) in 𝑋 is 𝜏(𝜌𝑠
)-convergent to 𝑥 ∈ 𝑋 if and only if it is
𝜏(𝜌)-convergent and 𝜏(¯
𝜌)-convergent to 𝑥.
3. If 𝜌 is a quasi-metric, then the topologies 𝜏(𝜌) and 𝜏(¯
𝜌) are 𝑇0, but not
necessarily 𝑇1 (and so nor 𝑇2, in contrast to the case of metric spaces).
The topology 𝜏(𝜌) is 𝑇1 if and only if 𝜌(𝑥, 𝑦) > 0 whenever 𝑥 ∕= 𝑦. In this
case, 𝜏(¯
𝜌) is also 𝑇1 and, as a bitopological space, 𝑋 is pairwise Hausdorff.
4. For every fixed 𝑥 ∈ 𝑋, the mapping 𝜌(𝑥, ⋅) : 𝑋 → (ℝ, ∣⋅∣) is 𝜏𝜌-usc and 𝜏¯
𝜌-lsc.
For every fixed 𝑦 ∈ 𝑋, the mapping 𝜌(⋅, 𝑦) : 𝑋 → (ℝ, ∣⋅∣) is 𝜏𝜌-lsc and 𝜏¯
𝜌-usc.
5. ([145]) The mapping 𝜌(𝑥, ⋅) : 𝑋 → (ℝ, ∣⋅∣) is 𝜏𝜌-continuous at 𝑥 ∈ 𝑋, if and
only if 𝜏𝜌-cl(𝐵𝜌(𝑥, 𝑟)) ⊂ 𝐵𝜌[𝑥, 𝑟] for all 𝑟 > 0.
Similar results hold for an asymmetric seminorm 𝑝, its conjugate ¯
𝑝 and the
associated seminorm 𝑝𝑠
.
Proof. 1. For 𝑦 ∈ 𝐵𝜌(𝑥, 𝑟) we have 𝐵𝜌(𝑥, 𝑟′
) ⊂ 𝐵𝜌(𝑥, 𝑟), where 𝑟′
:= 𝑟−𝜌(𝑥, 𝑦) > 0.
Also, if 𝑦 ∈ 𝐵𝜌[𝑥, 𝑟] and 𝑟′
:= 𝜌(𝑥, 𝑦) − 𝑟 > 0, then 𝐵¯
𝜌(𝑥, 𝑟′
) ∩ 𝐵𝜌[𝑥, 𝑟] = ∅, or,
equivalently, 𝐵¯
𝜌(𝑥, 𝑟′
) ⊂ 𝑋 ∖ 𝐵𝜌[𝑥, 𝑟]. Indeed, if 𝑧 ∈ 𝐵¯
𝜌(𝑥, 𝑟′
) ∩ 𝐵𝜌[𝑥, 𝑟], then
𝜌(𝑥, 𝑦) ≤ 𝜌(𝑥, 𝑧) + 𝜌(𝑧, 𝑦) = 𝜌(𝑥, 𝑧) + ¯
𝜌(𝑦, 𝑧) < 𝑟 + 𝑟′
= 𝜌(𝑥, 𝑦) ,
a contradiction.
The inclusions from 1 follows from the inequalities (1.1.1) and, in their turn,
they imply the assertions from the second point of the proposition.
The assertions from 2 are obvious.
3. If 𝑥, 𝑦 are distinct points in the quasi-metric space (𝑋, 𝜌), then max{𝜌(𝑥, 𝑦),
𝜌(𝑦, 𝑥)} > 0. If 𝜌(𝑥, 𝑦) > 0, then 𝑦 /
∈ 𝐵𝜌(𝑥, 𝑟), where 𝑟 = 𝜌(𝑥, 𝑦). Similarly, if
𝜌(𝑦, 𝑥) > 0, then 𝑥 /
∈ 𝐵𝜌(𝑦, 𝑟′
), where 𝑟′
= 𝜌(𝑦, 𝑥). Consequently, 𝜏(𝜌) is 𝑇0 and
𝜏(¯
𝜌) as well.
Suppose that 𝜌(𝑥, 𝑦) > 0 for every 𝑥 ∕= 𝑦. Then 𝑦 /
∈ 𝐵𝜌(𝑥, 𝜌(𝑥, 𝑦)). Since
𝜌(𝑦, 𝑥) > 0 too, 𝑥 /
∈ 𝐵𝜌(𝑦, 𝜌(𝑦, 𝑥)), showing that the topology 𝜏𝜌 is 𝑇1. Similarly
𝜏¯
𝜌 is 𝑇1.
Also, 𝐵𝜌(𝑥, 𝑟) ∩ 𝐵¯
𝜌(𝑦, 𝑟) = ∅, where 𝑟 > 0 is given by 2𝑟 := 𝜌(𝑥, 𝑦) > 0.
Indeed, if 𝑧 ∈ 𝐵𝜌(𝑥, 𝑟) ∩ 𝐵¯
𝜌(𝑦, 𝑟), then
𝜌(𝑥, 𝑦) ≤ 𝜌(𝑥, 𝑧) + 𝜌(𝑧, 𝑦) < 𝑟 + 𝑟 = 𝜌(𝑥, 𝑦) ,
a contradiction which shows that the bitopological space (𝑋, 𝜏𝜌, 𝜏¯
𝜌) is pairwise
Hausdorff.
It is easy to check that if 𝜏𝜌 is 𝑇1, then 𝜌(𝑥, 𝑦) > 0 for every pair of distinct
elements 𝑥, 𝑦 ∈ 𝑋.
24. 1.1. Topological properties of quasi-metric and quasi-uniform spaces 9
4. To prove that 𝜌(𝑥, ⋅) is 𝜏𝜌-usc and 𝜏¯
𝜌-lsc, we have to show that the set
{𝑦 ∈ 𝑋 : 𝜌(𝑥, 𝑦) < 𝛼} is 𝜏𝜌-open and {𝑦 ∈ 𝑋 : 𝜌(𝑥, 𝑦) > 𝛼} is 𝜏¯
𝜌-open, for every
𝛼 ∈ ℝ, properties that are easy to check.
Indeed, for 𝑦 ∈ 𝑋 such that 𝜌(𝑥, 𝑦) < 𝛼, let 𝑟 := 𝛼 − 𝜌(𝑥, 𝑦) > 0. If 𝑧 ∈ 𝑋 is
such that 𝜌(𝑦, 𝑧) < 𝑟, then
𝜌(𝑥, 𝑧) ≤ 𝜌(𝑥, 𝑦) + 𝜌(𝑦, 𝑧) < 𝜌(𝑥, 𝑦) + 𝑟 = 𝛼,
showing that 𝐵𝜌(𝑦, 𝑟) ⊂ {𝑦 ∈ 𝑋 : 𝜌(𝑥, 𝑦) < 𝛼} .
Similarly, for 𝑦 ∈ 𝑋 with 𝜌(𝑥, 𝑦) > 𝛼 take 𝑟 := 𝜌(𝑥, 𝑦) − 𝛼 > 0. If𝑧 ∈ 𝑋
satisfies 𝜌(𝑧, 𝑦) = ¯
𝜌(𝑦, 𝑧) < 𝑟, then
𝜌(𝑥, 𝑦) ≤ 𝜌(𝑥, 𝑧) + 𝜌(𝑧, 𝑦) < 𝜌(𝑥, 𝑧) + 𝑟 ,
so that 𝜌(𝑥, 𝑧) > 𝜌(𝑥, 𝑦) − 𝑟 = 𝛼. Consequently, 𝐵¯
𝜌(𝑦, 𝑟) ⊂ {𝑦 ∈ 𝑋 : 𝜌(𝑥, 𝑦) > 𝛼}.
5. Suppose that 𝜌(𝑥, ⋅) is continuous on 𝑋. If 𝑦 ∈ 𝜏𝜌-cl(𝐵𝜌(𝑥, 𝑟)), then there
exists a sequence (𝑦𝑛) in 𝐵𝜌(𝑥, 𝑟) such that 𝑦𝑛
𝜌
−
→ 𝑦 as 𝑛 → ∞. The continuity of
𝜌(𝑥, ⋅) implies 𝜌(𝑥, 𝑦) = lim𝑛→∞ 𝜌(𝑥, 𝑦𝑛) ≤ 𝑟, that is 𝑦 ∈ 𝐵𝜌[𝑥, 𝑟].
To prove the converse, suppose that there exists 𝑥 ∈ 𝑋 such that 𝜌(𝑥, ⋅) is
discontinuous at some 𝑦 ∈ 𝑋. Then there exist 𝑟 > 0 and a sequence (𝑦𝑛) in 𝑋
such that 𝑦𝑛
𝜌
−
→ 𝑦 and 𝜌(𝑥, 𝑦𝑛) ∈ (−∞, 𝜌(𝑥, 𝑦) − 𝑟) ∪ (𝜌(𝑥, 𝑦) + 𝑟, ∞) for all 𝑛 ∈ ℕ.
If there exists an infinity of 𝑛 ∈ ℕ such that 𝜌(𝑥, 𝑦𝑛) > 𝜌(𝑥, 𝑦) + 𝑟, then, by the
𝜏𝜌-usc of the function 𝜌(𝑥, ⋅), 𝜌(𝑥, 𝑦) + 𝑟 ≤ lim sup𝑛 𝜌(𝑥, 𝑦𝑛) ≤ 𝜌(𝑥, 𝑦), leading to
the contradiction 𝑟 ≤ 0.
Consequently 𝜌(𝑥, 𝑦𝑛) < 𝜌(𝑥, 𝑦) − 𝑟, that is 𝑦𝑛 ∈ 𝐵𝜌(𝑥, 𝜌(𝑥, 𝑦) − 𝑟) for all
𝑛 ∈ ℕ excepting a finitely many. By hypothesis, 𝑦 ∈ 𝐵𝜌[𝑥, 𝜌(𝑥, 𝑦) − 𝑟], that is
𝜌(𝑥, 𝑦) ≤ 𝜌(𝑥, 𝑦) − 𝑟, leading again to the contradiction 𝑟 ≤ 0. □
One can define other pairwise separation axioms. Call a bitopological space
(𝑇, 𝜏, 𝜈) pairwise 𝑇0 if for any pair 𝑥, 𝑦 of distinct points in 𝑇 either there exists
a 𝜏-open set 𝑈 such that 𝑥 ∈ 𝑈 and 𝑦 /
∈ 𝑈 or there exists 𝑉 ∈ 𝜈 such that 𝑦 ∈ 𝑉
and 𝑥 /
∈ 𝑉. Again, by symmetry, it follows that the space (𝑇, 𝜏, 𝜈) is pairwise 𝑇0 if
and only if
[(∃𝑈 ∈ 𝜏, 𝑥 ∈ 𝑈 ∧ 𝑦 /
∈ 𝑈) ∨ (∃𝑉 ∈ 𝜈, 𝑦 ∈ 𝑉 ∧ 𝑥 /
∈ 𝑉 )]
∧ [(∃𝑈1 ∈ 𝜏, 𝑦 ∈ 𝑈1 ∧ 𝑥 /
∈ 𝑈1) ∨ (∃𝑉1 ∈ 𝜈, 𝑥 ∈ 𝑉1 ∧ 𝑦 /
∈ 𝑉1)] .
(1.1.14)
Taking into account the mutual distributivity of the operators ∧ and ∨ it
follows that this condition is equivalent to
[(∃𝑈 ∈ 𝜏, 𝑥 ∈ 𝑈 ∧ 𝑦 /
∈ 𝑈) ∧ (∃𝑈1 ∈ 𝜏, 𝑦 ∈ 𝑈1 ∧ 𝑥 /
∈ 𝑈1)]
∨ [(∃𝑈 ∈ 𝜏, 𝑥 ∈ 𝑈 ∧ 𝑦 /
∈ 𝑈) ∧ (∃𝑉1 ∈ 𝜈, 𝑥 ∈ 𝑉1 ∧ 𝑦 /
∈ 𝑉1)]
∨ [(∃𝑉 ∈ 𝜏, 𝑦 ∈ 𝑉 ∧ 𝑥 /
∈ 𝑉 ) ∧ (∃𝑈1 ∈ 𝜏, 𝑦 ∈ 𝑈1 ∧ 𝑥 /
∈ 𝑈1)]
∨ [(∃𝑉 ∈ 𝜏, 𝑦 ∈ 𝑉 ∧ 𝑥 /
∈ 𝑉 ) ∧ (∃𝑉1 ∈ 𝜈, 𝑥 ∈ 𝑉1 ∧ 𝑦 /
∈ 𝑉1)] .
(1.1.15)
25. 10 Chapter 1. Quasi-metric and Quasi-uniform Spaces
Similar conditions hold for the notion of pairwise 𝑇1. A bitopological space
(𝑇, 𝜏, 𝜈) is called pairwise 𝑇1 if for any pair 𝑥, 𝑦 of distinct points in 𝑇 the following
condition holds:
[(∃𝑈 ∈ 𝜏, 𝑥 ∈ 𝑈 ∧ 𝑦 /
∈ 𝑈) ∧ (∃𝑉 ∈ 𝜈, 𝑦 ∈ 𝑉 ∧ 𝑥 /
∈ 𝑉 )]
∧ [(∃𝑈1 ∈ 𝜏, 𝑦 ∈ 𝑈1 ∧ 𝑥 /
∈ 𝑈1) ∧ (∃𝑉1 ∈ 𝜈, 𝑥 ∈ 𝑉1 ∧ 𝑦 /
∈ 𝑉1)] .
(1.1.16)
In the case of a quasi-semimetric space the following proposition holds.
Proposition 1.1.9. Let (𝑋, 𝜌) be a quasi-semimetric space. If the associated bitopo-
logical space (𝑋, 𝜏𝜌, 𝜏¯
𝜌) is pairwise 𝑇0, then 𝜌(𝑥, 𝑦) > 0 for any pair of distinct
points 𝑥, 𝑦 ∈ 𝑋.
Proof. Suppose that there exists 𝑥, 𝑦 ∈ 𝑋 with 𝑥 ∕= 𝑦 and 𝜌(𝑥, 𝑦) = 0. Then
𝑦 ∈ 𝐵𝜌(𝑥, 𝑟) and 𝑥 ∈ 𝐵¯
𝜌(𝑥, 𝑟) for every 𝑟 > 0, so that the condition (1.1.14) does
not hold. □
Taking into account Proposition 1.1.8.3, one obtains the following corollary.
Corollary 1.1.10. For a quasi-semimetric space (𝑋, 𝜌) the following are equivalent.
1. The bitopological space (𝑋, 𝜏𝜌, 𝜏¯
𝜌) is pairwise 𝑇0.
2. The bitopological space (𝑋, 𝜏𝜌, 𝜏¯
𝜌) is pairwise 𝑇1.
3. The bitopological space (𝑋, 𝜏𝜌, 𝜏¯
𝜌) is pairwise Hausdorff.
Better continuity properties of the distance function holds in the class of
the so-called balanced quasi-metric spaces. Following Doitchinov [63], a 𝑇1 quasi-
metric space (𝑋, 𝜌) is called balanced if the following condition holds:
[lim
𝑚,𝑛
𝜌(𝑣𝑚, 𝑢𝑛) = 0 ∧ ∀𝑛, 𝜌(𝑢, 𝑢𝑛) ≤ 𝑟 ∧ ∀𝑚, 𝜌(𝑣𝑚, 𝑣) ≤ 𝑠] ⇒ 𝜌(𝑢, 𝑣) ≤ 𝑟 + 𝑠,
(1.1.17)
for all sequences (𝑢𝑛), (𝑣𝑚) in 𝑋 and all 𝑢, 𝑣 ∈ 𝑋.
A pair (𝑋, 𝜌) where 𝜌 is a balanced quasi-metric is called a balanced quasi-
metric space or a 𝐵-quasi-metric space.
In the following proposition we collect some consequences of this definition.
Proposition 1.1.11 ([63]). Let (𝑋, 𝜌) be a 𝐵-quasi-metric space.
1. The following assertions hold for all sequences (𝑥𝑚), (𝑦𝑛) in 𝑋 and all
𝑥, 𝑦 ∈ 𝑋.
(i) [lim𝑛 𝜌(𝑥, 𝑥𝑛) = 0 ∧ ∀𝑛, 𝜌(𝑦, 𝑥𝑛) ≤ 𝑟] ⇒ 𝜌(𝑦, 𝑥) ≤ 𝑟;
(1.1.18)
(ii) [lim𝑛 𝜌(𝑥𝑛, 𝑥) = 0 ∧ ∀𝑛, 𝜌(𝑥𝑛, 𝑦) ≤ 𝑟] ⇒ 𝜌(𝑥, 𝑦) ≤ 𝑟;
(i) [lim𝑛 𝜌(𝑥, 𝑥𝑛) = 0 ∧ lim𝑛 𝜌(𝑦, 𝑥𝑛) = 0] ⇒ 𝑥 = 𝑦;
(1.1.19)
(ii) [lim𝑛 𝜌(𝑥𝑛, 𝑥) = 0 ∧ lim𝑛 𝜌(𝑥𝑛, 𝑦) = 0] ⇒ 𝑥 = 𝑦 ;
26. 1.1. Topological properties of quasi-metric and quasi-uniform spaces 11
[lim𝑛 𝜌(𝑥𝑛,𝑥)=0 ∧ lim𝑛 𝜌(𝑦,𝑦𝑚)=0 ∧ ∀𝑚,𝑛, 𝜌(𝑥𝑛,𝑦𝑚)≤𝑟] ⇒𝜌(𝑥,𝑦)≤𝑟;
(1.1.20)
(i) [lim𝑛𝜌(𝑥𝑛,𝑥)=0 ∧ lim𝑚 𝜌(𝑦,𝑦𝑚)=0] ⇒lim𝑚,𝑛𝜌(𝑥𝑚,𝑦𝑛)=𝜌(𝑥,𝑦);
(ii) lim𝑛 𝜌(𝑥𝑛, 𝑥) = 0 ⇒ lim𝑛 𝜌(𝑥𝑛, 𝑦) = 𝜌(𝑥, 𝑦);
(iii) lim𝑛 𝜌(𝑥, 𝑥𝑛) = 0 ⇒ lim𝑛 𝜌(𝑦, 𝑥𝑛) = 𝜌(𝑦, 𝑥) . (1.1.21)
2. The topologies 𝜏(𝜌) and 𝜏(¯
𝜌) are 𝑇2 (Hausdorff). For every fixed 𝑦 ∈ 𝑋 the
function 𝜌(⋅, 𝑦) is 𝜏(¯
𝜌) continuous on 𝑋 and 𝜌(𝑦, ⋅) is 𝜏(𝜌)-continuous on 𝑋.
Proof. 1. To prove (1.1.18).(i) and (ii), take in (1.1.17) 𝑢𝑛 = 𝑥𝑛, 𝑣𝑚 = 𝑥, 𝑢 =
𝑦, 𝑣 = 𝑥, respectively, 𝑢𝑚 = 𝑥, 𝑣𝑛 = 𝑥𝑛, 𝑢 = 𝑥, 𝑣 = 𝑦 (with 𝑚, 𝑛 having
interchanged roles).
To prove (1.1.19).(i) let 𝜀 > 0. Since 𝜌(𝑦, 𝑥𝑛) → 0, there exists 𝑛𝜀 ∈ ℕ such
that 𝜌(𝑦, 𝑥𝑛) ≤ 𝜀 for all 𝑛 ≥ 𝑛𝜀. Since 𝜌(𝑥, 𝑥𝑛) → 0, an application of (1.1.18).(i)
yields 𝜌(𝑦, 𝑥) ≤ 𝜀. Since 𝜀 > 0 was arbitrarily chosen, this implies 𝜌(𝑦, 𝑥) = 0 and
so 𝑥 = 𝑦 (in the definition of a balanced qm space we have required that the
topology 𝜏(𝜌) is 𝑇1). The assertion (ii) follows similarly.
The proof of (1.1.20). By (1.1.18).(ii) lim𝑛 𝜌(𝑥𝑛, 𝑥) = 0 and ∀𝑛, 𝜌(𝑥𝑛, 𝑦𝑚) ≤
𝑟, imply 𝜌(𝑥, 𝑦𝑚) ≤ 𝑟, for every 𝑚 ∈ ℕ. Since lim𝑚 𝜌(𝑦, 𝑦𝑚) = 0, an application of
(1.1.18).(i) yields 𝜌(𝑥, 𝑦) ≤ 𝑟.
To prove (1.1.21).(i) let 𝜀 > 0. Observe first that the inequality
𝜌(𝑥𝑚, 𝑦𝑛) ≤ 𝜌(𝑥𝑚, 𝑥) + 𝜌(𝑥, 𝑦) + 𝜌(𝑦, 𝑦𝑛) ,
and the hypotheses imply the existence of 𝑘0 ∈ ℕ such that
𝜌(𝑥𝑚, 𝑦𝑛) ≤ 𝜌(𝑥, 𝑦) + 𝜀 ,
for all 𝑚, 𝑛 ≥ 𝑘0. The proof will be complete if we prove the existence of 𝑙0 ∈ ℕ
such that
𝜌(𝑥𝑚, 𝑦𝑛) > 𝜌(𝑥, 𝑦) − 𝜀 ,
for all 𝑚, 𝑛 ≥ 𝑙0.
If contrary, then there exists the subsequences (𝑥𝑚𝑖 ) of (𝑥𝑚) and (𝑦𝑛𝑗 ) of
(𝑦𝑛) such that
𝜌(𝑥𝑚𝑖 , 𝑦𝑛𝑗 ) ≤ 𝜌(𝑥, 𝑦) − 𝜀 ,
for all 𝑖, 𝑗 ∈ ℕ. Since lim𝑖 𝜌(𝑥𝑚𝑖 , 𝑥) = 0 = lim𝑗 𝜌(𝑦, 𝑦𝑛𝑗 ), an application of (1.1.20)
yields the contradiction 𝜌(𝑥, 𝑦) ≤ 𝜌(𝑥, 𝑦) − 𝜀.
The assertions from (ii) and (iii) follow from (i).
2. The fact that the topologies 𝜏(𝜌) and 𝜏(¯
𝜌) are 𝑇2 follows from (1.1.19).(i)
and (ii), respectively.
The 𝜏(¯
𝜌)-continuity of the function 𝜌(⋅, 𝑦) follows from (1.1.21).(ii) and the
𝜏(𝜌)-continuity of the function 𝜌(𝑦, ⋅) follows from (1.1.21).(iii). □
27. 12 Chapter 1. Quasi-metric and Quasi-uniform Spaces
As it is well known the distance function to a subset plays a key role in the
study of metric spaces. As we shall see, the same is true in the asymmetric case
where, due to the asymmetry, we have two kinds of distance functions.
Let (𝑋, 𝜌) be a quasi-semimetric space. For a nonempty subset 𝐴 of 𝑋 and
𝑥 ∈ 𝑋 put
𝜌(𝑥, 𝐴) = inf{𝜌(𝑥, 𝑦) : 𝑦 ∈ 𝐴} and 𝜌(𝐴, 𝑥) = inf{𝜌(𝑦, 𝑥) : 𝑦 ∈ 𝐴} . (1.1.22)
It is obvious that 𝜌(𝐴, 𝑥) = ¯
𝜌(𝑥, 𝐴). A sequence (𝑦𝑛) in 𝐴 such that lim𝑛 𝜌(𝑥, 𝑦𝑛) =
𝜌(𝑥, 𝐴) is called a minimizing sequence for 𝜌(𝑥, 𝐴), with a similar definition for
minimizing sequences for 𝜌(𝐴, 𝑥). Since 𝐴 ∕= ∅, minimizing sequences always exist.
In the following proposition we collect the basic properties of the distance
functions.
Proposition 1.1.12. Let (𝑋, 𝜌) be a quasi-semimetric space, 𝐴 a nonempty subset
of 𝑋 and 𝑥, 𝑥′
∈ 𝑋. The following are true.
1. 𝜌(𝑥, 𝐴) ≤ 𝜌(𝑥, 𝑥′
) + 𝜌(𝑥′
, 𝐴) and 𝜌(𝐴, 𝑥) ≤ 𝜌(𝐴, 𝑥′
) + 𝜌(𝑥′
, 𝑥).
2. 𝜌(𝑥, 𝐴) = 0 ⇐⇒ 𝑥 ∈ 𝜏𝜌-cl(𝐴) and 𝜌(𝐴, 𝑥) = 0 ⇐⇒ 𝑥 ∈ 𝜏¯
𝜌-cl(𝐴).
3. The function 𝜌(⋅, 𝐴) : 𝑋 → ℝ is 𝜏𝜌-lsc and 𝜏¯
𝜌-usc, and the function 𝜌(𝐴, ⋅) :
𝑋 → ℝ is 𝜏𝜌-usc and 𝜏¯
𝜌-lsc.
Proof. 1. For any 𝑦 ∈ 𝐴,
𝜌(𝑥, 𝐴) ≤ 𝜌(𝑥, 𝑦) ≤ 𝜌(𝑥, 𝑥′
) + 𝜌(𝑥′
, 𝑦) .
Taking the infimum with respect to 𝑦 ∈ 𝐴 one obtains 𝜌(𝑥, 𝐴) ≤ 𝜌(𝑥, 𝑥′
) +
𝜌(𝑥′
, 𝐴). The second inequality from 1 can be proved similarly.
2. The proof of the first assertion follows from the equivalences
𝜌(𝑥, 𝐴) = 0 ⇐⇒ ∃(𝑦𝑛) in 𝐴, lim
𝑛→∞
𝜌(𝑥, 𝑦𝑛) = 0
⇐⇒ ∃(𝑦𝑛) in 𝐴, 𝑦𝑛
𝜌
−
→ 𝑥 ⇐⇒ 𝑥 ∈ 𝜏𝜌- cl(𝐴) .
The second equivalence from 2 can be proved similarly.
3. Suppose that for some 𝛼 ∈ ℝ, 𝜌(𝑥, 𝐴) > 𝛼. Taking 𝑟 := 𝜌(𝑥, 𝐴) − 𝛼 > 0,
it follows that
𝜌(𝑥, 𝐴) ≤ 𝜌(𝑥, 𝑥′
) + 𝜌(𝑥′
, 𝐴) < 𝜌(𝑥, 𝐴) − 𝛼 + 𝜌(𝑥′
, 𝐴) ,
for every 𝑥′
∈ 𝑋 with 𝜌(𝑥, 𝑥′
) < 𝑟. Consequently, 𝜌(𝑥′
, 𝐴) > 𝛼 for every 𝑥′
∈
𝐵𝜌(𝑥, 𝑟), proving the 𝜏𝜌-lsc of the mapping 𝜌(⋅, 𝐴) at 𝑥.
Similarly, if 𝜌(𝑥, 𝐴) < 𝛼,
𝜌(𝑥′
, 𝐴) ≤ 𝜌(𝑥′
, 𝑥) + 𝜌(𝑥, 𝐴) ,
28. 1.1. Topological properties of quasi-metric and quasi-uniform spaces 13
implies 𝜌(𝑥′
, 𝐴) < 𝛼 for every 𝑥′
∈ 𝑋 with 𝜌(𝑥′
, 𝑥) < 𝑟, where 𝑟 := 𝛼−𝜌(𝑥, 𝐴) > 0.
So 𝜌(𝑥′
, 𝐴) < 𝛼 for every 𝑥′
∈ 𝐵¯
𝜌(𝑥, 𝑟), proving the 𝜏¯
𝜌-usc of the mapping 𝜌(⋅, 𝐴)
at 𝑥.
The case of the distance function 𝜌(𝐴, ⋅) can be treated similarly. □
Based on the properties of the distance mapping we can prove further bitopo-
logical properties of quasi-semimetric spaces.
Proposition 1.1.13 (Kelly [111]). Let (𝑋, 𝜌) be a quasi-semimetric space.
1. The topology of a quasi-metric space is pairwise regular and pairwise normal.
2. If 𝜏𝜌 ⊂ 𝜏¯
𝜌, then the topology 𝜏¯
𝜌 is semimetrizable.
3. If the mapping 𝜌(𝑥, ⋅) : 𝑋 → (ℝ, ∣ ⋅ ∣) is 𝜏𝜌-continuous for every 𝑥 ∈ 𝑋, then
the topology 𝜏𝜌 is regular.
If 𝜌(𝑥, ⋅) : 𝑋 → (ℝ, ∣ ⋅ ∣) is 𝜏¯
𝜌-continuous for every 𝑥 ∈ 𝑋, then the topology
𝜏¯
𝜌 is semi-metrizable.
Similar results hold for an asymmetric seminormed space (𝑋, 𝑝).
Proof. 1. Since {𝐵𝜌[𝑥, 𝑟] : 𝑟 > 0} is a 𝜏(𝜌)-neighborhood base of the point 𝑥 formed
of 𝜏(¯
𝜌)-closed sets and {𝐵¯
𝜌[𝑥, 𝑟] : 𝑟 > 0} is a 𝜏(¯
𝜌)-neighborhood base of the point
𝑥 formed of 𝜏(𝜌)-closed sets, it follows that the bitopological space (𝑋, 𝜏(𝜌), 𝜏(¯
𝜌))
is pairwise regular.
To prove the normality of 𝑋, let 𝐴, 𝐵 ⊂ 𝑋, , such that 𝐴 is 𝜏(𝜌)-closed, 𝐵
is 𝜏(¯
𝜌)-closed and 𝐴 ∩ 𝐵 = ∅.
By the assertions 3 and 4 of Proposition 1.1.12,
𝐴 = {𝑥 ∈ 𝑋 : 𝜌(𝑥, 𝐴) = 0} and 𝐵 = {𝑥 ∈ 𝑋 : ¯
𝜌(𝑥, 𝐵) = 0} .
Let
𝑈 = {𝑥 ∈ 𝑋 : 𝜌(𝑥, 𝐴) < ¯
𝜌(𝑥, 𝐵)} and 𝑉 = {𝑥 ∈ 𝑋 : ¯
𝜌(𝑥, 𝐵) < 𝜌(𝑥, 𝐴)} .
It is obvious that 𝑈 ∩ 𝑉 = ∅. By Proposition 1.1.12 the mapping 𝜌(⋅, 𝐴) is
𝜏(¯
𝜌)-usc, so that the set 𝑈 is 𝜏(¯
𝜌)-open. By the same proposition, the mapping
𝜌(𝐴, ⋅) is 𝜏(𝜌)-usc, so that the set 𝑉 is 𝜏(𝜌)-open.
If 𝑥 ∈ 𝐴, then 𝑥 /
∈ 𝐵, so that 𝜌(𝑥, 𝐴) = 0 < ¯
𝜌(𝑥, 𝐵), showing that 𝐴 ⊂ 𝑈.
Similarly, 𝐵 ⊂ 𝑉.
2. The semimetric topology 𝜏𝜌𝑠 is the smallest topology finer than 𝜏𝜌 and 𝜏 ¯
𝜌,
so that 𝜏 ¯
𝜌 = 𝜏𝜌𝑠 .
3. If the mapping 𝜌(𝑥, ⋅) : 𝑋 → (ℝ, ∣ ⋅ ∣) is 𝜏(𝜌)-continuous, then the balls
𝐵𝜌[𝑥, 𝑟] = {𝑦 ∈ 𝑋 : 𝜌(𝑥, 𝑦) ≤ 𝑟} are 𝜏(𝜌)-closed for every 𝑟 > 0. Since they form a
𝜏(𝜌)-neighborhood base of the point 𝑥, it follows that the topology 𝜏(𝜌) is regular.
Suppose now that the mapping 𝜌(𝑥, ⋅) : 𝑋 → (ℝ, ∣ ⋅ ∣) is 𝜏(¯
𝜌)-continuous for
every 𝑥 ∈ 𝑋. The 𝜏(¯
𝜌)-continuity of 𝜌(𝑥, ⋅) implies that the spheres 𝐵𝜌(𝑥, 𝑟) =
29. 14 Chapter 1. Quasi-metric and Quasi-uniform Spaces
{𝑦 ∈ 𝑋 : 𝜌(𝑥, 𝑦) < 𝑟} are 𝜏(¯
𝜌)-open for every 𝑟 > 0, showing that the topology
𝜏(¯
𝜌) is finer than 𝜏(𝜌).
The topology 𝜏𝑠
generated by the semimetric 𝜌𝑠
= max{𝜌, ¯
𝜌} is the smallest
topology finer that both 𝜏(𝜌) and 𝜏(¯
𝜌), that is 𝜏𝑠
= 𝜏(𝜌) ∨ 𝜏(¯
𝜌) = 𝜏(¯
𝜌), implying
the semimetrizability of 𝜏(¯
𝜌). □
Remark 1.1.14. 1. The properties from the assertions 3, 4, 5 of Proposition 1.1.8
are taken from Kelly [111].
2. The lower and upper continuity properties from the assertion 4 of Propo-
sition 1.1.8 are equivalent to the fact that the mapping 𝜌(𝑥, ⋅) : 𝑋 → ℝ is (𝜏𝜌, 𝜏𝑢)-
continuous, respectively (𝜏¯
𝜌, 𝜏¯
𝑢)-continuous. Similar equivalences hold for the semi-
continuity properties of the mapping 𝜌(⋅, 𝑦).
3. The quasi-metric space (ℝ, 𝑢) from Example 1.1.3 is not 𝑇1 because, for
instance, any neighborhood of 1 contains 0.
1.1.3 More on bitopological spaces
Kelly [111] proved several basic results on bitopological spaces including extensions
of the classical theorems of Uryson and Tietze to semi-continuous functions defined
on bitopological spaces. The Urysohn type theorem proved in [111] is the following.
Theorem 1.1.15. Let (𝑇, 𝜏, 𝜈) be a pairwise normal bitopological space, 𝐴 ⊂ 𝑇 𝜏-
closed and 𝐵 ⊂ 𝑇 𝜈-closed with 𝐴 ∩ 𝐵 = ∅. Then there exists a 𝜏-lsc and 𝜈-usc
function 𝑓 : 𝑇 → [0; 1] such that
∀𝑡 ∈ 𝐴, 𝑓(𝑡) = 0 and ∀𝑡 ∈ 𝐵, 𝑓(𝑡) = 1 . (1.1.23)
Proof. The proof follows the line of the proof of the classical Urysohn theorem as
given, for instance, in Pedersen [172]. Put 𝐴0 = 𝐴 and 𝐶1 = 𝑇 ∖ 𝐵. Then 𝐴0 is
𝜏-closed, 𝐶1 is 𝜈-open and 𝐴0 ⊂ 𝐶1. By (1.1.12) there exists a 𝜏-closed set 𝐴1/2
and a 𝜈-open set 𝐶1/2 such that
𝐴0 ⊂ 𝐶1/2 ⊂ 𝐴1/2 ⊂ 𝐶1 .
Applying the same condition to the pairs 𝐴0 ⊂ 𝐶1/2 and 𝐴1/2 ⊂ 𝐶1, we affirm
the existence of two 𝜏-closed sets 𝐴1/4, 𝐴3/4 and of two 𝜈-open sets 𝐶1/4, 𝐺3/4 such
that
𝐴0 ⊂ 𝐶1/4 ⊂ 𝐴1/4 ⊂ 𝐶1/2 ⊂ 𝐴1/2 ⊂ 𝐶3/4 ⊂ 𝐴3/4 ⊂ 𝐶1 .
Continuing in this manner, for any dyadic rational number 𝑝 ∈ {𝑖 ⋅ 2−𝑘
: 1 =
1, 2, . . ., 2𝑘
− 1}, 𝑘 ∈ ℕ, one finds a 𝜏-closed set 𝐴𝑝 and a 𝜈-open set 𝐶𝑝. Putting,
for convenience, 𝐴𝑝 = ∅ for 𝑝 < 0, 𝐴𝑝 = 𝑇 for 𝑝 ≥ 1, 𝐶𝑝 = ∅ for 𝑝 ≤ 0 and 𝐶𝑝 = 𝑇
for 𝑝 > 1, it follows that these sets satisfy the relations
𝐶𝑝 ⊂ 𝐶𝑞 ⊂ 𝐴𝑞 ⊂ 𝐴𝑟 for all dyadic rational numbers 𝑝 ≤ 𝑞 ≤ 𝑟, and
𝐴𝑝 ⊂ 𝐶𝑞 for all dyadic rational numbers 𝑝 < 𝑞 .
30. 1.1. Topological properties of quasi-metric and quasi-uniform spaces 15
Define the function 𝑓 : 𝑇 → ℝ by
𝑓(𝑡) = inf{𝑝 : 𝑡 ∈ 𝐶𝑝}, 𝑡 ∈ 𝑇 .
It follows that
𝑓(𝑡) = inf{𝑝 : 𝑡 ∈ 𝐴𝑝}, 0 ≤ 𝑓(𝑡) ≤ 1, ∀𝑡 ∈ 𝑇,
𝑓(𝑡) = 0, 𝑡 ∈ 𝐴, and 𝑓(𝑡) = 1, 𝑡 ∈ 𝐵 = 𝑇 ∖ 𝐶1 .
To prove that 𝑓 is 𝜈-usc we have to show that for every number 𝛼, 0 < 𝛼 ≤ 1,
the set 𝑓−1
([0; 𝛼)) is 𝜈-open. Let 0 < 𝛼 ≤ 1 and 𝑡 ∈ 𝑇 such that 𝑓(𝑡) < 𝛼. Using
the definition of 𝑓 in terms of the 𝜈-open sets 𝐶𝑝, it follows that there exists a
dyadic number 𝑝 such that 𝑝 < 𝛼 and 𝑡 ∈ 𝐶𝑝, that is
𝑓(𝑡) < 𝛼 ⇐⇒ ∃𝑝, 𝑝 < 𝛼, 𝑡 ∈ 𝐶𝑝 ⇐⇒ 𝑡 ∈ ∪𝑝<𝛼𝐶𝑝 .
Consequently, 𝑓−1
([0; 𝛼)) = ∪𝑝<𝛼𝐶𝑝 showing that the set 𝑓−1
([0; 𝛼)) is 𝜈-
open.
To show that 𝑓 is also 𝜏-lsc we shall use the expression of 𝑓 in terms of
the 𝜏-closed sets 𝐴𝑝, 𝑓(𝑡) = inf{𝑝 : 𝑡 ∈ 𝐴𝑝}. We have to show that for every
𝛽, 0 ≤ 𝛽 < 1, the set 𝑓−1
((𝛽, 1]) is 𝜏-open. Let 𝑡 ∈ 𝑇 such that 𝑓(𝑡) > 𝛽. If 𝑡 ∈ 𝐴𝑝
for every 𝑝 > 𝛽, then, by the definition of 𝑓, 𝑓(𝑡) ≤ 𝑝. Consequently, 𝑓(𝑡) ≤ 𝑝 for
every 𝑝 > 𝛽, implying 𝑓(𝑡) ≤ 𝛽. This shows that
𝑓(𝑡) > 𝛽 ⇐⇒ ∃𝑝 > 𝛽, 𝑡 ∈ 𝑇 ∖ 𝐴𝑝 ⇐⇒ 𝑡 ∈ ∪𝑝>𝛽(𝑇 ∖ 𝐴𝑝) .
It follows that the set 𝑓−1
((𝛽, 1]) = ∪𝑝>𝛽(𝑇 ∖ 𝐴𝑝) is 𝜏-open. □
The following theorem extends to bitopological spaces a result of Katětov
[107, 108] asserting that a topological space 𝑇 is normal if and only if for any pair
of functions 𝑓, 𝑔 : 𝑇 → ℝ such that 𝑓 is usc, 𝑔 is lsc and 𝑓 ≤ 𝑔, there exists a
continuous function ℎ : 𝑇 → ℝ such that 𝑓 ≤ ℎ ≤ 𝑔. The proof given here follows
the ideas suggested in Engelking [73, Exercise 2.7.2] for normal spaces.
Theorem 1.1.16 (Lane [144]). A bitopological space (𝑇, 𝜏, 𝜈) is pairwise normal if
and only if for every pair 𝑓, 𝑔 : 𝑇 → ℝ of functions such that 𝑓 is 𝜈-usc, 𝑔 is 𝜏-lsc,
and 𝑓 ≤ 𝑔, there exists a 𝜏-lsc and 𝜈-usc function ℎ : 𝑇 → ℝ such that 𝑔 ≤ 𝑓 ≤ ℎ.
The following lemma extends to pairwise normal spaces a separation result
valid in normal spaces.
Lemma 1.1.17. Let (𝑇, 𝜏, 𝜈) be a pairwise normal bitopological space. Let 𝐴, 𝐵 ⊂ 𝑇
such that 𝐴 is 𝜏-𝐹𝜎, 𝐵 is 𝜈-𝐹𝜎 and
𝐴
𝜏
∩ 𝐵 = ∅ = 𝐴 ∩ 𝐵
𝜈
. (1.1.24)
Then there exist a 𝜈-open set 𝑈 ⊃ 𝐴 and a 𝜏-open set 𝑉 ⊃ 𝐵 such that
𝑈 ∩ 𝑉 = ∅.
31. 16 Chapter 1. Quasi-metric and Quasi-uniform Spaces
The sets 𝑈 and 𝑉 satisfy also the relations
𝐴 ⊂ 𝑈 ⊂ 𝑈
𝜏
⊂ 𝑇 ∖ 𝐵 and 𝐵 ⊂ 𝑉 ⊂ 𝑉
𝜈
⊂ 𝑇 ∖ 𝐴 . (1.1.25)
Proof. Let 𝐴 = ∪∞
𝑛=1𝐴𝑛 and 𝐵 = ∪∞
𝑛=1𝐵𝑛, with 𝐴𝑛 𝜏-closed and 𝐵𝑛 𝜈-closed.
Applying (1.1.13) to the sets 𝐴1 ⊂ 𝑇 ∖ 𝐵
𝜈
, there exists a 𝜈-open set 𝑈1 such that
𝐴1 ⊂ 𝑈1 ⊂ 𝑈
𝜏
1 ⊂ 𝑇 ∖ 𝐵
𝜈
.
By the same condition (with 𝜏 and 𝜈 interchanged), there exists a 𝜏-open set
𝑉1 such that
𝐵1 ⊂ 𝑉1 ⊂ 𝑉
𝜈
1 ⊂ 𝑇 ∖ (𝐴
𝜏
∪ 𝑈
𝜏
1) .
Continuing in this manner, we find inductively the 𝜈-open sets 𝑈𝑛 and the
𝜏-open sets 𝑉𝑛 such that
𝐴𝑛 ∪ 𝑈
𝜏
1 ∪ ⋅ ⋅ ⋅ ∪ 𝑈
𝜏
𝑛−1 ⊂ 𝑈𝑛 ⊂ 𝑈
𝜏
𝑛 ⊂ 𝑇 ∖ (𝐵
𝜈
∪ 𝑉
𝜈
1 ∪ ⋅ ⋅ ⋅ ∪ 𝑉
𝜈
𝑛−1),
𝐵𝑛 ∪ 𝑉
𝜈
1 ∪ ⋅ ⋅ ⋅ ∪ 𝑣𝜈
𝑛−1 ⊂ 𝑉𝑛 ⊂ 𝑈
𝜈
𝑛 ⊂ 𝑇 ∖ (𝐴
𝜏
∪ 𝑈
𝜏
1 ∪ ⋅ ⋅ ⋅ ∪ 𝑈
𝜈
𝑛) ,
for all 𝑛 ∈ ℕ, where 𝑈0 = 𝑉0 = ∅.
It follows that the set 𝑈 = ∪∞
𝑛=1𝑈𝑛 is 𝜈-open, 𝑉 = ∪∞
𝑛=1𝑉𝑛 is 𝜏-open, 𝐴 ⊂
𝑈, 𝐵 ⊂ 𝑉 and 𝑈 ∩ 𝑉 = ∅.
Since 𝑉 is 𝜏-open and 𝑈 ∩ 𝑉 = ∅, it follows that 𝑈
𝜏
∩ 𝑉 = ∅, so that
𝑈
𝜏
∩ 𝐵 = ∅, i.e., 𝑈
𝜏
⊂ 𝑇 ∖ 𝐵. The second set of inclusions in (1.1.25) follows
similarly. □
Proof of Theorem 1.1.16. Suppose first that 𝑓, 𝑔 : 𝑇 → [0; 1]. For 𝑟 ∈ ℚ ∩ [0; 1] let
𝐴𝑟 = 𝑔−1
([0; 𝑟)) (𝐴0 = ∅), 𝐵𝑟 = 𝑓−1
([0; 𝑟]) and 𝐶𝑟 = 𝑇 ∖𝑓−1
([0; 𝑟]) = 𝑓−1
((𝑟; 1]).
Since 𝑔 is 𝜏-lsc the set 𝑔−1
([0; 𝑟′
]) is 𝜏-closed, so that the set 𝐴𝑟 = ∪{𝑔−1
([0; 𝑟′
]) :
𝑟′
∈ ℚ, 0 < 𝑟′
< 𝑟} is 𝜏-𝐹𝜎. Since 𝑓 is 𝜈-usc, the set 𝑓−1
([𝑟′
; 1]) is 𝜈-closed, so
that 𝐶𝑟 = ∪{𝑓−1
([𝑟′
; 1]) : 𝑟′
∈ ℚ, 𝑟 < 𝑟′
< 1} is 𝜈-𝐹𝜎 and 𝐵𝑟 = 𝑇 ∖ 𝐶𝑟 is 𝜈-𝐺𝛿. It
is easy to check that
𝐴
𝜏
𝑟 ∩ 𝐶𝑟 = ∅ = 𝐴𝑟 ∩ 𝐶
𝜈
𝑟 . (1.1.26)
Indeed, if (𝑡𝑖 : 𝑖 ∈ 𝐼) is a net in 𝐴𝑟 that is 𝜏-convergent to 𝑡 ∈ 𝑇, then, by
the 𝜏-lsc of the function 𝑔, 𝑔(𝑡) ≤ lim inf𝑖 𝑔(𝑡𝑖) ≤ 𝑟, because 𝑔(𝑡𝑖) ≤ 𝑟, 𝑖 ∈ 𝐼. It
follows that 𝑡 /
∈ 𝐶𝑟. Similarly, using the 𝜈-usc of the function 𝑓 one obtains that
𝑡 /
∈ 𝐴 for every 𝑡 ∈ 𝐶
𝜈
𝑟 .
As in the proof of the bitopological Urysohn theorem (Theorem 1.1.15), we
shall work with the set Λ2 = {𝑖 ⋅ 2−𝑘
: 𝑖 = 0, 1, 2, . . ., 2𝑘
, 𝑘 ∈ ℕ} of dyadic rational
numbers in [0; 1] (including 0 and 1).
The equalities (1.1.26) show that the sets 𝐴1 and 𝐶1 satisfy the hypotheses
of Lemma 1.1.17, so that, by (1.1.25) there exists a 𝜈-open set 𝑈1 such that
𝐴1 ⊂ 𝑈1 ⊂ 𝑈
𝜏
1 ⊂ 𝐵1 . (1.1.27)
32. 1.1. Topological properties of quasi-metric and quasi-uniform spaces 17
Similarly, there exists a 𝜈-open set 𝑈1/2 such that
𝐴1/2 ⊂ 𝑈1/2 ⊂ 𝑈
𝜏
1/2 ⊂ 𝐵1/2 ∩ 𝑈1 .
At the next step we find two 𝜈-open sets 𝑈1/4, 𝑈3/4 such that
𝐴1/4 ⊂ 𝑈1/4 ⊂ 𝑈
𝜏
1/4 ⊂ 𝐵1/4 ∩ 𝑈1/2, and
𝐴3/4 ∪ 𝑈
𝜏
1/2 ⊂ 𝑈3/4 ⊂ 𝑈
𝜏
3/4 ⊂ 𝐵3/4 ∩ 𝑈1 .
Continuing in this manner, one obtains by induction, the 𝜈-open sets 𝑈𝑟
satisfying (1.1.27) and
𝐴(2𝑖−1)⋅2−𝑚 ∪𝑈
𝜏
(𝑖−1)⋅2−𝑚+1 ⊂ 𝑈(2𝑖−1)⋅2−𝑚 ⊂ 𝑈
𝜏
(2𝑖−1)⋅2−𝑚 ⊂ 𝐵(2𝑖−1)⋅2−𝑚 ∩𝑈𝑖⋅2−𝑚+1 ,
for 𝑖 = 1, 2, . . ., 2𝑚−1
, 𝑚 ∈ ℕ, where 𝑈0 = ∅. It follows that the inclusions
(i) 𝐴𝑟 ⊂ 𝑈𝑟 ⊂ 𝑈
𝜏
𝑟 ⊂ 𝐵𝑟
(1.1.28)
(ii) 𝑈𝑟 ⊂ 𝑈
𝜏
𝑟 ⊂ 𝑈𝑟′ ,
hold for all 𝑟, 𝑟′
∈ Λ2 with 𝑟 < 𝑟′
.
Put, for convenience, 𝑈𝑟 = 𝑇 for 𝑟 > 1 and define the function ℎ : 𝑇 →
[0; 1] by
ℎ(𝑡) = inf{𝑟 : 𝑟 ∈ Λ2 ∪ (1, ∞) such that 𝑡 ∈ 𝑈𝑟}, 𝑡 ∈ 𝑇 .
Since the sets 𝑈𝑟 are 𝜈-open, reasoning as in the proof of the 𝜈-usc of the function
𝑓 in Theorem 1.1.15, one can show that the function ℎ is 𝜈-usc.
By (1.1.28).(ii),
ℎ(𝑡) = inf{𝑟 ∈ Λ2 : 𝑡 ∈ 𝑈
𝜏
𝑟 } ,
and the sets 𝑈
𝜏
𝑟 are 𝜏-closed, so that, following again the ideas of the proof of the
𝜏-lsc of the function 𝑓 in Theorem 1.1.15, one can show that the function ℎ is
𝜏-lsc.
It remains to show that 𝑓 ≤ ℎ ≤ 𝑔. Let 𝑡 ∈ 𝑇. If 𝑟 ∈ Λ2, 𝑟 < 𝑓(𝑡), then,
for every 𝑟′
∈ Λ2, 0 ≤ 𝑟′
< 𝑟, 𝑡 /
∈ 𝑓−1
([𝑜, 𝑟′
]) = 𝐵𝑟′ , and so 𝑡 /
∈ 𝑈𝑟′ , implying
ℎ(𝑡) ≥ 𝑟. That is ℎ(𝑡) ≥ 𝑟 for every 𝑟 ∈ Λ2 with 𝑟 < 𝑓(𝑡), and so 𝑓(𝑡) ≤ ℎ(𝑡).
If 𝑟 ∈ Λ2 is such that 𝑔(𝑡) < 𝑟, then 𝑡 ∈ 𝑔−1
([0; 𝑟)) = 𝐴𝑟 ⊂ 𝑈𝑟, and so
ℎ(𝑡) ≤ 𝑟. Since ℎ(𝑡) ≤ 𝑟 for every 𝑟 ∈ Λ2 with 𝑔(𝑡) < 𝑟, it follows that ℎ(𝑡) ≤ 𝑔(𝑡).
The general case, when 𝑓, 𝑔 : 𝑇 → ℝ, 𝑓 is 𝜈-usc, 𝑔 is 𝜏-lsc and 𝑓 ≤ 𝑔, can be
reduced to the preceding one by composing with the function𝜓 : ℝ → (0; 1),
𝜓(𝑡) =
∣𝑡∣ + 𝑡 + 1
2(∣𝑡∣ + 1)
, , 𝑡 ∈ ℝ .
The function 𝜓 is a strictly increasing homeomorphism between ℝ and (0; 1),
and its inverse 𝜓−1
: (0; 1) → ℝ is strictly increasing and continuous. Consequently,
33. 18 Chapter 1. Quasi-metric and Quasi-uniform Spaces
applying the proved result to the functions ˜
𝑓 = 𝜓 ∘ 𝑓 and ˜
𝑔 = 𝜓 ∘ 𝐺 we confirm
the existence of a 𝜏-lsc and 𝜈-usc function ℎ̃ : 𝑇 → (0; 1) such that ˜
𝑓 ≤ ℎ̃ ≤ ˜
𝑔.
But then, the function ℎ = 𝜓−1
∘ ℎ̃ : 𝑇 → ℝ is 𝜏-lsc, 𝜈-usc and 𝑓 = 𝜓−1
∘ ˜
𝑓 ≤
ℎ ≤ 𝜓−1
∘ ˜
𝑔 = 𝑔.
To prove the converse, suppose that 𝐴, 𝐵 are two disjoint subsets of 𝑇 such
that 𝐴 is 𝜈-closed and 𝐵 is 𝜏-closed. Then the characteristic function 𝜒𝐴 of the
set 𝐴 is 𝜈-usc and 𝜒𝑇 ∖𝐵 is 𝜏-lsc. The inclusion 𝐴 ⊂ 𝑇 ∖ 𝐵 implies 𝜒𝐴 ≤ 𝜒𝑇 ∖𝐵,
so that, by hypothesis, there exists a 𝜈-usc and 𝜏-lsc function ℎ : 𝑇 → [0; 1] such
that 𝜒𝐴 ≤ ℎ ≤ 𝜒𝑇 ∖𝐵. By the 𝜏-lsc of ℎ, the set 𝑈 = ℎ−1
((1
2 ; 1]) is 𝜏-open and
𝐴 ⊂ 𝑈. By the 𝜈-usc of the function ℎ, the set 𝑉 = ℎ−1
([1
2 ; 1]) is 𝜈-closed and
𝐴 ⊂ 𝑈 ⊂ 𝑉 ⊂ 𝑇 ∖ 𝐵. □
The proof of the following result follows that in topological spaces, see En-
gelking [73, Theorem 1.5.15].
Proposition 1.1.18 (Kelly [111]). A pairwise regular bitopological space (𝑇, 𝜏, 𝜈)
satisfying the second countability axiom (i.e., both 𝜏 and 𝜈 satisfy this axiom) is
pairwise normal.
As it is well known, a relatively easy consequence of the Urysohn theorem
is the Tietze extension theorem, but the proof given for topological spaces can-
not be adapted to bitopological spaces. Kelly [111] proved a Tietze type theorem
asserting that any real-valued 𝜏-usc and 𝜈-lsc function defined on 𝜏-closed and
𝜈-closed subset of a pairwise normal bitopological space 𝑇 admits a 𝜏-usc and
𝜈-lsc extension to the whole space 𝑇. Lane [144] showed by a counterexample that
the result is false in this form, and gave a proper formulation.
The example is the following.
Example 1.1.19. Let 𝑇 be an uncountable set and 𝐴 = {𝑡1, 𝑡2, . . . } be an infinite
countable subset of 𝑇. Let 𝜏 be formed by the empty set and the complements
of finite or countable subsets of 𝑇 , and let 𝜈 be the discrete topology. Then the
bitopological space (𝑇, 𝜏, 𝜈) is pairwise normal, the function 𝑓 : 𝐴 → ℝ, 𝑓(𝑡𝑖) =
𝑖, 𝑖 ∈ ℕ, is 𝜏-lsc and 𝜈-usc, but has no 𝜏-lsc and 𝜈-usc extension to 𝑇 .
Indeed, if 𝐴, 𝐵 are disjoint subsets of 𝑇 such that 𝐴 is 𝜏-closed and 𝐵 is
𝜈-closed, then 𝐴 is also 𝜈-open and 𝑇 ∖ 𝐴 is 𝜏-open, so that (𝑇, 𝜏, 𝜈) is pairwise
normal. The set 𝐴 = {𝑡1, 𝑡2, . . . } is 𝜏- and 𝜈-closed and the function 𝑓 : 𝐴 →
ℝ, 𝑓(𝑡𝑖) = 𝑖, 𝑖 ∈ ℕ, is 𝜏-lsc and 𝜈-usc. Suppose that 𝐹 : 𝑇 → ℝ is a 𝜏-lsc and
𝜈-usc extension of 𝑓. Then the sets 𝐴𝑖 = {𝑡 ∈ 𝑇 : 𝐹(𝑡) ≤ 𝑖}, 𝑖 ∈ ℕ, are 𝜏-closed.
Because 𝐴𝑖 ∕= 𝑇, 𝐴𝑖 must be at most countable, implying that 𝑇 = ∪𝑖∈ℕ𝐴𝑖 is
countable, in contradiction to the hypothesis.
A general Tietze type theorem holds only for bounded semi-continuous func-
tions.
Theorem 1.1.20. Let (𝑇, 𝜏, 𝜈) be a pairwise normal bitopological space and 𝐴 a
𝜏-closed and 𝜈-closed subset of 𝑇 . Then every bounded, 𝜏-lsc and 𝜈-usc function
34. 1.1. Topological properties of quasi-metric and quasi-uniform spaces 19
𝑓 : 𝐴 → ℝ admits a 𝜏-lsc and 𝜈-usc extension 𝐹 : 𝑇 → ℝ such that
inf 𝐹(𝑇 ) = inf 𝑓(𝐴) and sup 𝐹(𝑇 ) = sup 𝑓(𝐴) . (1.1.29)
Proof. Let
𝛼 = inf 𝑓(𝐴) and 𝛽 = sup 𝑓(𝐴) .
Define the functions 𝑔, ℎ : 𝑇 → ℝ by 𝑔(𝑡) = ℎ(𝑡) = 𝑓(𝑡) for 𝑡 ∈ 𝐴, 𝑔(𝑡) = 𝛼
and ℎ(𝑡) = 𝛽 for 𝑡 ∈ 𝑇 ∖ 𝐴. It is easily seen that 𝑔 is 𝜈-usc, ℎ is 𝜏-lsc and 𝑔 ≤ ℎ.
By Theorem 1.1.16 there exists a 𝜏-lsc and 𝜈-usc function 𝐹 : 𝑇 → ℝ such that
𝑔 ≤ 𝐹 ≤ ℎ. It follows that 𝐹 is the desired extension of 𝑓. □
Lane [144] gave some sufficient conditions on the subset 𝐴 of the bitopological
space 𝑇 in order for any 𝜏-lsc and 𝜈-usc real-valued function to have a 𝜏-lsc and
𝜈-usc extension to 𝑇. In particular this is true for quasi-metric spaces.
Proposition 1.1.21. Let (𝑋, 𝜌) be a quasi-semimetric space and 𝐴 a nonempty 𝜏𝜌-
and 𝜏¯
𝜌-closed subset of 𝑋. Then any 𝜏𝜌-lsc and 𝜏¯
𝜌-usc function 𝑓 : 𝐴 → (ℝ, ∣ ⋅ ∣)
admits a 𝜏𝜌-lsc and 𝜏𝜌-usc extension to the whole space 𝑋.
Proof. By Proposition 1.1.12, the function ℎ1(𝑥) = 𝜌(𝐴, 𝑥), 𝑥 ∈ 𝑋, is 𝜏𝜌-usc and
𝜏¯
𝜌-lsc, while the function ℎ2(𝑥) = 𝜌(𝑥, 𝐴), 𝑥 ∈ 𝑋, is 𝜏𝜌-lsc and 𝜏¯
𝜌-usc.
The function 𝑔 = 𝑓/(1 + ∣𝑓∣) is 𝜏𝜌-lsc and 𝜏¯
𝜌-usc and −1 < 𝑔(𝑥) < 1 for all
𝑥 ∈ 𝑋. By Theorem 1.1.20 𝑔 has a 𝜏𝜌-lsc and 𝜏¯
𝜌-usc extension 𝐺 : 𝑋 → [−1, 1].
The function 𝐺+
= 𝐺 ∨ 0 is 𝜏𝜌-lsc, 𝜏¯
𝜌-usc and 0 ≤ 𝐺+
≤ 1, while the function
𝐺−
= −(𝐺 ∧ 0) is 𝜏𝜌-usc, 𝜏¯
𝜌-lsc and 0 ≤ 𝐺−
≤ 1.
The function 𝐺1 = 𝐺+
/(1 + ℎ1) is 𝜏𝜌-lsc, 𝜏¯
𝜌-usc and 0 ≤ 𝐺1(𝑥) < 1, 𝑥 ∈ 𝑋,
while the function 𝐺2 = 𝐺+
/(1 + ℎ2) is 𝜏𝜌-usc, 𝜏¯
𝜌-lsc and 0 ≤ 𝐺2(𝑥) < 1, 𝑥 ∈ 𝑋.
Also, for every 𝑥 ∈ 𝐴,
𝐺1(𝑥) = 𝐺+
(𝑥) and 𝐺2(𝑥) = 𝐺−
(𝑥) .
It follows that the function 𝐺 = 𝐺1 − 𝐺2 is 𝜏𝜌-lsc, 𝜏¯
𝜌-usc and −1 < 𝐺(𝑥) <
1, 𝑥 ∈ 𝑋. Also, for every 𝑥 ∈ 𝐴,
𝐺(𝑥) = 𝐺1(𝑥) − 𝐺2(𝑥) = 𝐺+
(𝑥) − 𝐺−
(𝑥) = 𝐺(𝑥) = 𝑔(𝑥) ,
that is 𝐺 is a 𝜏𝜌-lsc and 𝜏¯
𝜌-usc extension of 𝑔. But then the function 𝐹 = 𝐺/(1−∣𝐺∣)
is 𝜏𝜌-lsc and 𝜏¯
𝜌-usc extension of the function 𝑓. □
Remark 1.1.22. The proof of Theorem 1.1.20, based on Theorem 1.1.16, is taken
from [144]. Another proof, based on the Urysohn theorem for bitopological spaces
(Theorem 1.1.15) is given in [78]. In [78] some conditions on the set 𝐴 ⊂ 𝑇 ensuring
the existence of a 𝜏-usc and 𝜈-lsc extension 𝐹 of an arbitrary 𝜏-usc and 𝜈-lsc 𝐹
on 𝐴 are given too.
35. 20 Chapter 1. Quasi-metric and Quasi-uniform Spaces
A bitopological space (𝑇, 𝜏, 𝜈) is called pairwise perfectly normal if it is pair-
wise normal and every 𝜏-closed (𝜈-closed) subset of 𝑇 is 𝜈-𝐺𝛿 (𝜏-𝐺𝛿).
The following characterization of pairwise perfectly normal bitopological
spaces extends a well-known result in topology, see, e.g., Engelking [73, Theorem
1.5.19].
Theorem 1.1.23 ([78], [144]). A bitopological space (𝑇, 𝜏, 𝜈) is pairwise perfectly
normal if and only if for any pair 𝐴, 𝐵 of subsets of 𝑇 such that 𝐴 is 𝜏-closed, 𝐵
is 𝜈-closed and 𝐴 ∩ 𝐵 = ∅, there exists a 𝜏-lsc and 𝜈-usc function 𝑓 : 𝑇 → [0; 1]
such that 𝐴 = 𝑓−1
(0) and 𝐵 = 𝑓−1
(1).
Bı̂rsan studied in the paper [24] pairwise completely regular bitopological
spaces and extended to this setting Hausdorff’s embedding theorem. A bitopo-
logical space (𝑋, 𝜏, 𝜈) is called 𝜏-completely regular with respect to 𝜈 if for every
𝑥 ∈ 𝑋 and every 𝜏-open neighborhood 𝑈 of 𝑥 there exits a 𝜏-lsc and 𝜈-usc function
𝑓 : 𝑋 → [0; 1] such that 𝑓𝑥) = 1 and 𝑓(𝑋 ∖ 𝑈) = {0}. Equivalently, (𝑋, 𝜏, 𝜈) is
𝜏-completely regular with respect to 𝜈 if and only if for every 𝜏-closed set 𝑍 and
every point 𝑥 ∈ 𝑋 ∖ 𝑍 there exists a 𝜏-lsc and 𝜈-usc function 𝑓 : 𝑋 → [0; 1] such
that 𝑓(𝑥) = 1 and 𝑓(𝑍) = {0}.
Consider the bitopological space (ℝ, 𝜏𝑢, 𝜏¯
𝑢) from Example 1.1.3 and let 𝐼 =
[0; 1] with the induced induced topologies.
By definition a bitopological cube is a product 𝐼𝐴
of 𝐴 copies of the bitopo-
logical space 𝐼, where 𝐴 an arbitrary nonempty set. Any bitopological cube is
pairwise completely regular and pairwise compact.
A bitopological space (𝑋, 𝜏, 𝜈) is called weakly pairwise Hausdorff if for every
pair 𝑥, 𝑦 of distinct points from 𝑋 there exist a 𝜏-open neighborhood 𝑈 of one of
these points and a 𝜈-open neighborhood 𝑉 of the other one with 𝑈 ∩ 𝑉 = ∅. The
space (ℝ, 𝜏𝑢, 𝜏¯
𝑢) is weakly pairwise Hausdorff but not pairwise Hausdorff.
Theorem 1.1.24 ([24]). A weakly pairwise Hausdorff bitopological space is pair-
wise completely regular if and only if it is bi-homeomorphic to a subspace of a
bitopological cube.
A mapping 𝑓 : (𝑋, 𝜏1, 𝜏2) → (𝑌, 𝜈1, 𝜈2) between two bitopological spaces
is called bi-continuous if it is both (𝜏1, 𝜈1)-continuous and (𝜏2, 𝜈2)-continuous. If
𝑓 is a bijection such that both 𝑓 and 𝑓−1
are bi-continuous, then 𝑓 is called a
bi-homeomorphism.
The paper [25] is concerned with bitopological groups and [26] with some
properties relating quasi-uniform and bitopological spaces.
An important problem in topology is that of metrizability of topological
spaces. As we have seen, quasi-semimetric spaces are bitopological spaces, and so
the corresponding problem in this setting would be that of quasi-metrizability of
bitopological spaces. The following result is the analog of Urysohn metrizability
theorem and improves Proposition 1.1.18.
36. 1.1. Topological properties of quasi-metric and quasi-uniform spaces 21
Theorem 1.1.25 (Kelly [111]). If (𝑇, 𝜏, 𝜈) is a pairwise regular bitopological space
such that both 𝜏 and 𝜈 satisfy the second axiom of countability, then it is quasi-
semimetrizable. If further, 𝑇 is pairwise Hausdorff, then it is quasi-metrizable.
Lane [144] proved a bitopological version of the Nagata-Smirnov metrizability
theorem.
1.1.4 Compactness in bitopological spaces
We shall briefly present some results on compactness in bitopological spaces ob-
tained by Bı̂rsan [23].
If 𝒜, ℬ are two covers of a set 𝑋 one says that ℬ refines 𝒜 (or that ℬ is a
refinement of 𝒜) if
∀𝐵 ∈ ℬ, ∃𝐴 ∈ 𝒜, 𝐵 ⊂ 𝐴 . (1.1.30)
Following [23], a bitopological space (𝑋, 𝜏, 𝜈) is called:
∙ 𝜏-compact with respect to 𝜈 if every 𝜏-open cover of 𝑋 admits a finite 𝜈-open
refinement that covers 𝑋;
∙ pairwise compact if it is 𝜏-compact with respect to 𝜈 and 𝜈-compact with
respect to 𝜏.
It is obvious that a topological space (𝑋, 𝜏) is compact if and only if for
every open cover of 𝑋 there exists a finite refinement of 𝑋 which is an open cover
of 𝑋, so that the above definitions are natural extensions of the usual notion
of compactness. The bitopological compactness can be characterized in terms of
closed sets.
Proposition 1.1.26 ([23]). Let (𝑋, 𝜏, 𝜈) be a bitopological space. The following are
equivalent.
1. The space 𝑋 is 𝜏-compact with respect to 𝜈.
2. If a family 𝐹𝑖, 𝑖 ∈ 𝐼, of 𝜏-closed sets has empty intersection, then there
exists a finite family 𝐻𝑗, 𝑗 ∈ 𝐽, of 𝜈-closed sets having empty intersection
and such that for every 𝑗 ∈ 𝐽 there exists 𝑖 ∈ 𝐼 with 𝐻𝑗 ⊃ 𝐹𝑖.
Proof. 1 ⇒ 2. Let 𝐹𝑖, 𝑖 ∈ 𝐼, be a family of 𝜏-closed sets with empty intersection.
Then their complements 𝐺𝑖 = ∁(𝐹𝑖), 𝑖 ∈ 𝐼, are 𝜏-open and cover 𝑋, so there exists
a finite refinement 𝑍𝑗, 𝑗 ∈ 𝐽, of {𝐺𝑖} with 𝜈-open sets that covers 𝑋. Putting
𝐻𝑗 = ∁(𝑍𝑗), 𝑗 ∈ 𝐽, it follows that each 𝐻𝑗 is 𝜈-closed, ∩𝑗∈𝐽 𝐻𝑗 = ∁ (∪𝑗∈𝐽 𝑍𝑗) = ∅.
Also, for every 𝑗 ∈ 𝐽 there exists 𝑖 ∈ 𝐼 such that 𝑍𝑗 ⊂ 𝐺𝑖 ⇐⇒ 𝐻𝑗 ⊃ 𝐹𝑖.
The implication 2 ⇒ 1 is proved similarly. □
Proposition 1.1.27 ([23]). If the bitopological space (𝑋, 𝜏, 𝜈) is pairwise Hausdorff
and (𝑋, 𝜏) is compact, then 𝜏 ⊂ 𝜈.
37. 22 Chapter 1. Quasi-metric and Quasi-uniform Spaces
Proof. It is sufficient to show that every 𝜏-closed subset of 𝑋 is 𝜈-closed. Let
𝑍 ⊂ 𝑋 be 𝜏-closed and 𝑦 /
∈ 𝑍. Then for every 𝑧 ∈ 𝑍 there exists a 𝜏-open set
𝑈𝑧 containing 𝑧 and a 𝜈-open set 𝑉𝑧 containing 𝑦 such that 𝑈𝑧 ∩ 𝑉𝑧 = ∅. By the
𝜏-compactness of 𝑋, the 𝜏-open cover 𝑈𝑧, 𝑧 ∈ 𝑍, of 𝑍 contains a finite subcover
𝑈𝑧1 , . . . , 𝑈𝑧𝑛 . It follows that 𝑉 = ∩𝑛
𝑘=1𝑉𝑧𝑘
is a 𝜈-open neighborhood of 𝑦 and
𝑉 ∩ 𝑍 ⊂ 𝑉 ∩ (∪𝑛
𝑘=1𝑈𝑧𝑘
) = ∅, so that 𝑉 ⊂ 𝑋 ∖ 𝑍. Consequently 𝑋 ∖ 𝑍 is 𝜈-open
and 𝑍 is 𝜈-closed. □
This proposition has the following corollary.
Corollary 1.1.28. Let (𝑋, 𝜏, 𝜈) be a pairwise Hausdorff bitopological space.
1. If both the topologies 𝜏 and 𝜈 are compact, then 𝜏 = 𝜈.
2. If the space 𝑋 is 𝜏-compact with respect to 𝜈, then 𝜏 ⊂ 𝜈.
3. If (𝑋, 𝜏, 𝜈) is pairwise compact, then 𝜏 = 𝜈.
It is known that a subset of a topological space is compact if and only if it
is compact with respect to the induced topology, a result that does not hold in
bitopological spaces.
Call a subset 𝑌 of a bitopological space (𝑋, 𝜏, 𝜈) 𝜏-compact with respect to 𝜈
if it is 𝜏∣𝑌 -compact with respect to 𝜈∣𝑌 as a bitopological subspace of 𝑋.
Proposition 1.1.29 ([23]). Let (𝑋, 𝜏, 𝜈) be a bitopological space and 𝑌 ⊂ 𝑋.
1. If every 𝜏-open cover of 𝑌 admits a finite refinement that is a 𝜈-open cover
of 𝑌 , then 𝑌 is 𝜏-compact with respect to 𝜈.
2. If the set 𝑌 is 𝜈-open, then the converse is also true.
It is shown by an example [23, Example 8] that the second assertion from
the above proposition is not true for arbitrary subsets.
The preservation of compactness by continuous mappings takes the following
form.
Proposition 1.1.30 ([23]). Let (𝑋, 𝜏1, 𝜏2) and (𝑋, 𝜈1, 𝜈2) be bitopological spaces.
Suppose that 𝑋 is 𝜏1-compact with respect to 𝜏2. If the mapping 𝑓 : 𝑋 → 𝑌 is
(𝜏1, 𝜈1)-continuous and (𝜏2, 𝜈2)-open, then 𝑓(𝑋) is 𝜈1-compact with respect to 𝜈2.
Proof. One shows that the hypotheses from Proposition 1.1.29.1 are fulfilled. □
In the same paper [23], Bı̂rsan extended to this setting Alexander’s subbase
theorem and Tikhonov’s theorem on the compactness of the product of compact
topological spaces.
Theorem 1.1.31 (Alexander’s subbase theorem, [23]). Let (𝑋, 𝜏, 𝜈) be a bitopolog-
ical space, 𝒜 a subbase of the topology 𝜏 and ℬ a subbase of the topology 𝜈.
1. If every cover of 𝑋 with sets in 𝒜 admits a finite refinement with elements
from 𝜈 that covers 𝑋, then 𝑋 is 𝜏-compact with respect to 𝜈.
38. 1.1. Topological properties of quasi-metric and quasi-uniform spaces 23
2. If every cover of 𝑋 with sets in 𝒜 admits a finite refinement with elements
from ℬ that covers 𝑋, then the bitopological space 𝑋 is pairwise compact.
The bitopological product of a family (𝑋𝑖, 𝜏𝑖, 𝜈𝑖), 𝑖 ∈ 𝐼, of bitopological spaces
is the bitopological space (𝑋, 𝜏, 𝜈), where 𝑋 =
∏
𝑖∈𝐼 𝑋𝑖, 𝜏 =
∏
𝑖∈𝐼 𝜏𝑖 and 𝜈 =
∏
𝑖∈𝐼 𝜈𝑖. The so-defined bitopological product has properties similar to that of the
usual topological product.
Proposition 1.1.32. Let (𝑋𝑖, 𝜏𝑖, 𝜈𝑖), 𝑖 ∈ 𝐼, be a family of bitopological spaces,
(𝑋, 𝜏, 𝜈) their bitopological product.
1. For each 𝑖 ∈ 𝐼 the projection 𝑝𝑖 : 𝑋 → 𝑋𝑖 is bi-continuous and bi-open, i.e.,
(𝜏, 𝜏𝑖)-continuous, (𝜈, 𝜈𝑖)-continuous and (𝜏, 𝜏𝑖)-open and (𝜈, 𝜈𝑖)-open.
2. If (𝑋, 𝜏, 𝜈) is 𝜏-compact with respect to 𝜈 (pairwise compact), then for each
𝑖 ∈ 𝐼 the bitopological space (𝑋𝑖, 𝜏𝑖, 𝜈𝑖) is 𝜏𝑖-compact with respect to 𝜈𝑖
(respectively, pairwise compact).
Theorem 1.1.33 (Tikhonov compactness theorem, [23]). Let (𝑋𝑖, 𝜏𝑖, 𝜈𝑖), 𝑖 ∈ 𝐼,
be a family of bitopological spaces. If each bitopological space 𝑋𝑖 is 𝜏𝑖-compact
with respect to 𝜈𝑖 (pairwise compact), then their bitopological product (𝑋, 𝜏, 𝜈) is
𝜏-compact with respect to 𝜈 (respectively, pairwise compact).
In spite of the good results holding for this notion of compactness, it has
a serious drawback – a finite set need not be compact – as it is shown by the
following example.
Example 1.1.34 ([23]). Consider a three point set 𝑋 = {𝑎, 𝑏, 𝑐} with the topologies
𝜏 = {∅, {𝑎}, {𝑏, 𝑐}, 𝑋} and 𝜈 = {∅, {𝑐}, {𝑎, 𝑏}, 𝑋}. Then the topologies 𝜏 and 𝜈 are
compact, but 𝜏 is not compact with respect to 𝜈, nor 𝜈 is compact with respect
to 𝜏.
The paper [23] contains also examples of
∙ a pairwise compact, not pairwise normal, bitopological space (𝑋, 𝜏, 𝜈) with
𝜏 ∕= 𝜈;
∙ a pairwise Hausdorff bitopological space (𝑋, 𝜏, 𝜈) which is 𝜏-compact with
respect to 𝜈, but not pairwise normal;
∙ a pairwise normal, pairwise compact bitopological space (𝑋, 𝜏, 𝜈) which is
not pairwise regular.
There are also other notions of compactness in bitopological spaces, a good
analysis of various relations holding between them is done in the paper [52]. Note
that our terminology differs from that in [52]. For convenience we shall call a
bitopological space pairwise 𝐵-compact if it is pairwise compact in the sense con-
sidered by Bı̂rsan.
Let (𝑋, 𝜏, 𝜈) be a bitopological space. A cover 𝒜 ⊂ 𝜏 ∪𝜈 of 𝑋 is called (𝜏, 𝜈)-
open. If, in addition, 𝒜 contains a nonempty set from 𝜏 and a nonempty set from
𝜈, then 𝒜 is called pairwise open.
39. 24 Chapter 1. Quasi-metric and Quasi-uniform Spaces
The bitopological space (𝑋, 𝜏, 𝜈) is called:
∙ pairwise S-compact provided that every pairwise open cover of 𝑋 contains a
finite subcover (Swart [231]);
∙ semi-compact provided that every (𝜏, 𝜈)-open cover of 𝑋 contains a finite
subcover (Datta [53, 54]);
∙ pseudo-compact provided that every bi-continuous function 𝑓 : (𝑋, 𝜏, 𝜈) →
(ℝ, 𝜏𝑢, 𝜏¯
𝑢) is bounded (Saegrove [213]);
∙ pairwise real-compact provided it is bi-homeomorphic to the intersection of
a
∏
𝑖∈𝐼 𝜏𝑢-closed subset and a
∏
𝑖∈𝐼 𝜏¯
𝑢-closed subset of a product of 𝐼 copies
of (ℝ, 𝜏𝑢, 𝜏¯
𝑢) (Saegrove [213]);
∙ bicompact provided it is both pseudo-compact and pairwise real-compact
(Saegrove [213]).
In the following theorem we collect some results from [52].
Theorem 1.1.35. Let (𝑋, 𝜏, 𝑛𝑢) be a bitopological space.
1. The bitopological space (𝑋, 𝜏, 𝜈) is semi-compact if and only if it is compact
with respect to the topology 𝜏 ∨ 𝜈.
2. The bitopological space (𝑋, 𝜏, 𝜈) is semi-compact if and only if it is pairwise
compact, 𝜏-compact and 𝜈-compact.
3. The bitopological space (𝑋, 𝜏, 𝜈) is pairwise 𝑆-compact if and only if each
𝜏-closed proper subset of 𝑋 is 𝜈-compact and each 𝜈-closed proper subset of
𝑋 is 𝜏-compact.
4. If the bitopological space (𝑋, 𝜏, 𝜈) is semi-compact or pairwise 𝐵-compact,
then it is pseudo-compact.
5. If the bitopological space (𝑋, 𝜏, 𝜈) is pairwise Hausdorff and either semi-
compact, bicompact or pairwise 𝐵-compact, then 𝜏 = 𝜈.
5. If the bitopological space (𝑋, 𝜏, 𝜈) is either semi-compact, bicompact or pair-
wise 𝐵-compact, then both of the topologies 𝜏 and 𝜈 are compact.
Remark 1.1.36.
1. Any finite bitopological space is semi-compact and pairwise 𝑆-compact. Con-
sequently, Example 1.1.34 furnishes also an example of a semi-compact and
pairwise 𝑆-compact bitopological space which is not pairwise 𝐵-compact.
2. Example 3 in [52] gives a pairwise 𝐵-compact bitopological space which is
neither pairwise 𝑆-compact nor semi-compact.
3. Example 1 in [52] gives a pairwise 𝑆-compact bitopological space (𝑋, 𝜏, 𝜈)
such that (𝑋, 𝜈) is not compact.
Paracompactness is even a more delicate matter to be treated within the
framework of bitopological spaces. For some attempts and discussions see the
papers [29], [55], [124], [178] and [179].
40. 1.1. Topological properties of quasi-metric and quasi-uniform spaces 25
We mention also the following metrizability conditions.
Theorem 1.1.37. Let (𝑇, 𝜏, 𝜈) be a bitopological space and (𝑋, 𝑞) a quasi-metric
space.
1. If the bitopological space 𝑇 is quasi-metrizable and 𝜏 is locally (countably)
compact with respect to 𝜈, then (𝑇, 𝜈) is metrizable. The same conclusion
holds if locally compact is replaced by paracompact.
2. If the topology 𝜏(¯
𝜌) is sequentially compact, then the topology 𝜏(𝜌) is metriz-
able. The same is true if 𝜏(¯
𝜌) is compact.
3. If (𝑋, 𝜌) is sequentially compact and Hausdorff, then the topology 𝜏(𝜌) is
metrizable.
Remark 1.1.38. Assertions 1 and 2 are taken from [178] while the third one is
from [227].
1.1.5 Topological properties of asymmetric seminormed spaces
In general the topology generated by an asymmetric norm is 𝑇0 but not 𝑇1. Indeed,
it is easy to check that the space (ℝ, 𝑢) from Example 1.1.3 is 𝑇0 but −1 belongs to
every neighborhood of 1. A condition that the topology of an asymmetric normed
space be Hausdorff was found by Garcı́a-Raffi, Romaguera and Sánchez-Pérez [91],
in terms of a functional 𝑝⋄
: 𝑋 → [0; ∞) associated to an asymmetric seminorm
𝑝 defined on a real vector space 𝑋, a result that was extended to asymmetric
locally convex spaces in [41], see Subsection 1.1.7. The separation properties of an
asymmetric seminormed space will be presented in Proposition 1.1.40.
The functional 𝑝⋄
is defined by the formula
𝑝⋄
(𝑥) = inf{𝑝(𝑥′
) + 𝑝(𝑥′
− 𝑥) : 𝑥′
∈ 𝑋}, 𝑥 ∈ 𝑋. (1.1.31)
In the following proposition we present the properties of 𝑝⋄
.
Proposition 1.1.39. The functional 𝑝⋄
is a (symmetric) seminorm on 𝑋, 𝑝⋄
≤ 𝑝,
and 𝑝⋄
is the greatest seminorm on 𝑋 majorized by 𝑝.
Proof. First observe that, replacing 𝑥′
by 𝑥′
− 𝑥 in (1.1.31), we get
𝑝⋄
(−𝑥) = inf{𝑝(𝑥′
) + 𝑝(𝑥′
+ 𝑥) : 𝑥′
∈ 𝑋}
= inf{𝑝(𝑥′
− 𝑥) + 𝑝((𝑥′
− 𝑥) + 𝑥) : 𝑥′
∈ 𝑋} = 𝑝⋄
(𝑥) ,
so that 𝑝⋄
is symmetric. The positive homogeneity of 𝑝⋄
, 𝑝⋄
(𝛼𝑥) = 𝛼𝑝⋄
(𝑥), 𝑥 ∈
𝑋, 𝛼 ≥ 0, is obvious. For 𝑥, 𝑦 ∈ 𝑋 and arbitrary 𝑥′
, 𝑦′
∈ 𝑋 we have
𝑝⋄
(𝑥 + 𝑦) ≤ 𝑝(𝑥′
+ 𝑦′
) + 𝑝(𝑥′
+ 𝑦′
− 𝑥 − 𝑦) ≤ 𝑝(𝑥′
) + 𝑝(𝑥′
− 𝑥) + 𝑝(𝑦′
) + 𝑝(𝑦′
− 𝑦) ,
so that, passing to infimum with respect to 𝑥′
, 𝑦′
∈ 𝑋, we obtain the subadditivity
of 𝑝⋄
,
𝑝⋄
(𝑥 + 𝑦) ≤ 𝑝⋄
(𝑥) + 𝑝⋄
(𝑦) .
41. 26 Chapter 1. Quasi-metric and Quasi-uniform Spaces
Suppose now that there exists a seminorm 𝑞 on 𝑋 such that 𝑞 ≤ 𝑝, i.e.,
∀𝑧 ∈ 𝑋, 𝑞(𝑧) ≤ 𝑝(𝑧), and 𝑝⋄
(𝑥) < 𝑞(𝑥) ≤ 𝑝(𝑥), for some 𝑥 ∈ 𝑋. Then, by the
definition of 𝑝⋄
, there exists 𝑥′
∈ 𝑋 such that 𝑝⋄
(𝑥) < 𝑝(𝑥′
) + 𝑝(𝑥′
− 𝑥) < 𝑞(𝑥),
leading to the contradiction
𝑞(𝑥) ≤ 𝑞(𝑥′
) + 𝑞(𝑥 − 𝑥′
) = 𝑞(𝑥′
) + 𝑞(𝑥′
− 𝑥) ≤ 𝑝(𝑥′
) + 𝑝(𝑥′
− 𝑥) < 𝑞(𝑥) . □
In the following proposition we collect the separation properties of an asym-
metric seminormed space.
Proposition 1.1.40 ([91]). Let (𝑋, 𝑝) be an asymmetric seminormed space.
1. The topology 𝜏𝑝 is 𝑇0 if and only if for every 𝑥 ∈ 𝑋, 𝑥 ∕= 0, 𝑝(𝑥) > 0 or
𝑝(−𝑥) > 0, in other words, if and only if 𝑝 is an asymmetric norm.
2. The topology 𝜏𝑝 is 𝑇1 if and only if 𝑝(𝑥) > 0 for every 𝑥 ∈ 𝑋, 𝑥 ∕= 0.
3. The topology 𝜏𝑝 is Hausdorff if and only if 𝑝⋄
(𝑥) > 0 for every 𝑥 ∕= 0.
Proof. The assertion from 1 follows from Proposition 1.1.8.3.
2. If 𝑝⋄
(𝑥) > 0 whenever 𝑥 ∕= 0, then 𝑝⋄
is a norm on 𝑋, so that the topology
𝜏𝑝⋄ generated by 𝑝⋄
is Hausdorff. The inequality 𝑝⋄
≤ 𝑝 implies that the topology
𝜏𝑝 is finer than 𝜏𝑝⋄ , so it is Hausdorff, too.
Suppose 𝑝⋄
(𝑥) = 0 for some 𝑥 ∕= 0. By the definition (1.1.31) of 𝑝⋄
, there
exists a sequence (𝑥𝑛) in 𝑋 such that lim𝑛[𝑝(𝑥𝑛) + 𝑝(𝑥𝑛 − 𝑥)] = 𝑝⋄
(𝑥) = 0. This
implies lim𝑛 𝑝(𝑥𝑛) = 0 and lim𝑛 𝑝(𝑥𝑛 −𝑥) = 0, showing that the sequence (𝑥𝑛) has
two limits with respect to 𝜏𝜌. Consequently, the topology 𝜏𝑝 is not Hausdorff. □
Remark 1.1.41. It is known that a 𝑇0 topological vector space (TVS) is Hausdorff
and completely regular (see [149, Theorem 2.2.14]), a result that is no longer true
in asymmetric normed spaces. An easy example illustrating this situation is that
of the space (ℝ, 𝑢) from Example 1.1.3, which is 𝑇0 but not 𝑇1.
The following proposition shows that an asymmetric normed space is not
necessarily a topological vector space.
Proposition 1.1.42. If (𝑋, 𝑝) is an asymmetric normed space, then the topology
𝜏𝑝 is translation invariant, so that the addition + : 𝑋 × 𝑋 → 𝑋 is continuous.
Also any additive mapping between two asymmetric normed spaces (𝑋, 𝑝), (𝑌, 𝑞)
is continuous if and only if it is continuous at 0 ∈ 𝑋 (or at an arbitrary point
𝑥0 ∈ 𝑋).
The scalar multiplication is not always continuous, so that an asymmetric
normed space need not be a topological vector space.
Proof. The fact that the topology 𝜏𝑝 is translation invariant follows from the
formulae (1.1.5). Example 1.1.3 shows that the multiplication by scalars need not
be continuous. □
Another example was given by Borodin [28].
42. 1.1. Topological properties of quasi-metric and quasi-uniform spaces 27
Example 1.1.43. In the space 𝑋 = 𝐶0[0; 1], where
𝐶0[0; 1] =
{
𝑓 ∈ 𝐶[0; 1] :
∫ 1
0
𝑓(𝑡)𝑑𝑡 = 0
}
,
with the asymmetric seminorm 𝑝(𝑓) = max 𝑓([0; 1]), the multiplication by scalars
is not continuous at the point (−1, 0) ∈ ℝ × 𝑋.
To prove this, we show that (−1)𝐵𝑝[0, 𝑟] is not contained in 𝐵𝑝[0, 1] for any
𝑟 > 0. Indeed, let 𝑡𝑛 = 1/𝑛 and
𝑓𝑛(𝑡) =
{
𝑟(𝑛 − 1)(𝑛𝑡 − 1), 0 ≤ 𝑡 ≤ 𝑡𝑛,
𝑟 𝑛
𝑛−1
(
𝑡 − 1
𝑛
)
, 𝑡𝑛 < 𝑡 ≤ 1 ,
for 𝑛 ∈ ℕ. Then 𝑓𝑛 ∈ 𝐶0[0; 1], 𝑝(𝑓𝑛) = 𝑟, −𝑓𝑛 ∈ 𝐶0[0; 1] and 𝑝(−𝑓𝑛) = (𝑛−1)𝑟 >
1 for sufficiently large 𝑛.
Remark 1.1.44. It is known (and easy to check) that if a topology is separable, then
any coarser topology is separable. In particular, if the semimetric space (𝑋, 𝜌𝑠
) is
separable, then the quasi-metric space (𝑋, 𝜌) is also separable (with respect to the
topology 𝜏𝜌).
The following example shows that the converse is not true, in general.
Example 1.1.45 (Borodin [28]). There exists an asymmetric normed space (𝑋, 𝑝)
which is 𝜏𝑝-separable but not 𝜏𝑝𝑠 -separable.
Take
𝑋 =
{
𝑥 ∈ ℓ∞ : 𝑥 = (𝑥𝑘),
∞
∑
𝑘=1
𝑥𝑘
2𝑘
= 0
}
, (1.1.32)
with the asymmetric norm 𝑝(𝑥) = sup𝑘 𝑥𝑘. It is clear that 𝑝 is an asymmetric
norm on 𝑋 satisfying 𝑝(𝑥) > 0 whenever 𝑥 ∕= 0 and 𝑝𝑠
(𝑥) = ∥𝑥∥∞ = sup𝑘 ∣𝑥𝑘∣
is the usual sup-norm on ℓ∞. Because 𝜑(𝑥) =
∑∞
𝑘=1 2−𝑘
𝑥𝑘, 𝑥 = (𝑥𝑘) ∈ ℓ∞, is
a continuous linear functional on ℓ∞, it follows that 𝑋 = ker 𝜑 is a codimension
one closed subspace of ℓ∞. Since ℓ∞ is nonseparable with respect to 𝑝𝑠
, 𝑋 is also
nonseparable with respect to 𝑝𝑠
.
Let us show that 𝑋 is 𝑝-separable. Consider the set 𝑌 formed of all 𝑦 = (𝑦𝑘)
such that 𝑦𝑘 ∈ ℚ for all 𝑘 and there exists 𝑛 = 𝑛(𝑦) such that 𝑦𝑘 = 𝑦𝑛+1 for all
𝑘 > 𝑛. It is clear that 𝑌 is contained in 𝑋 and that 𝑌 is countable. To show that
𝑌 is 𝑝-dense in 𝑋, let 𝑥 ∈ 𝑋 and 𝜀 > 0.
Choose 𝑛 ∈ ℕ such that
𝑛
∑
𝑖=1
2−𝑖
𝜀 −
𝑛
∑
𝑖=1
2−𝑖
𝑥𝑖 > ∥𝑥∥∞
∞
∑
𝑗=𝑛+1
2−𝑗
. (1.1.33)
43. 28 Chapter 1. Quasi-metric and Quasi-uniform Spaces
This is possible because the left-hand side of (1.1.33) tends to 𝜀 for 𝑛 → ∞,
while the right-hand side tends to 0. Choose 𝑦𝑘 ∈ ℚ ∩ (𝑥𝑘 − 2𝜀; 𝑥𝑘 − 𝜀) for 𝑘 =
1, . . . , 𝑛, and let
𝑦𝑘 = 𝛼 := −
( 𝑛
∑
𝑖=1
2−𝑖
𝑦𝑖
)
:
⎛
⎝
∞
∑
𝑗=𝑛+1
2−𝑗
⎞
⎠ ,
for 𝑘 > 𝑛. Then 𝑦 = (𝑦𝑘) ∈ 𝑌, 𝜀 < 𝑦𝑘 − 𝑥𝑘 < 2𝜀, for 𝑘 = 1, . . . , 𝑛 and, by (1.1.33),
𝑦𝑘 = 𝛼 >
( 𝑛
∑
𝑖=1
2−𝑖
(𝜀 − 𝑥𝑖)
)
:
⎛
⎝
∞
∑
𝑗=𝑛+1
2−𝑗
⎞
⎠ > ∥𝑥∥∞ ,
for 𝑘 > 𝑛. It follows that 𝑥𝑘 − 𝑦𝑘 ≤ ∥𝑥∥∞ − 𝑦𝑘 < 0 for 𝑘 > 𝑛, so that 𝑝(𝑥 − 𝑦) =
max{𝑥𝑘 − 𝑦𝑘 : 1 ≤ 𝑘 ≤ 𝑛} < 2𝜀.
As it is well known, by the classical Banach-Mazur theorem, any separable
real Banach space can be linearly and isometrically embedded in the Banach space
𝐶[0; 1] of all continuous real-valued functions on [0; 1] with the sup-norm, see, for
instance, [74, Theorem 5.17]. In other words, 𝐶[0; 1] is a universal space in the
category of separable real Banach spaces. The validity of this result in the case of
asymmetric normed spaces was discussed by Alimov [11] and Borodin [28]. The
above example shows that some attention must be paid to the notion of separability
that we are using.
Denote by (𝐶[0; 1], 1, 0) the space of all real-valued continuous functions on
[0; 1] functions with the asymmetric norm
∥𝑓∣ = max{𝑓+(𝑡) : 𝑡 ∈ [0; 1]}, (1.1.34)
for 𝑓 ∈ 𝐶[0; 1], where 𝑓+(𝑡) = max{𝑓(𝑡), 0}, 𝑡 ∈ [0; 1] .
Theorem 1.1.46 ([28]). Any 𝑇1 asymmetric normed space (𝑋, 𝑝) such that the asso-
ciated normed space (𝑋, 𝑝𝑠
) is separable is isometrically isomorphic to a subspace
of the asymmetric normed space (𝐶[0; 1], 1, 0).
Since any linear isometry from (𝑋, 𝑝) to (𝐶[0; 1], 1, 0) induces a linear isome-
try from (𝑋, 𝑝𝑠
) to the usual Banach space 𝐶[0; 1], the separability condition with
respect to the symmetric norm 𝑝𝑠
is necessary for the validity of the Banach-Mazur
theorem.
Some complements to this result were given by Alimov [11].
Theorem 1.1.47 ([11]). A 𝑇1 asymmetric normed space (𝑋, 𝑝) is isometrically iso-
morphic to an affine variety 𝑍 of the usual Banach space 𝐶[0; 1] if and only if the
topology 𝜏𝑝 is metrizable and separable.
Proof. Note that the topology 𝜏𝑝 of a 𝑇1 asymmetric normed space (𝑋, 𝑝) is metriz-
able if and only if 𝜏𝑝 = 𝜏¯
𝑝 = 𝜏𝑝𝑠 , so that the topology 𝜏𝑝 is generated by the
44. 1.1. Topological properties of quasi-metric and quasi-uniform spaces 29
associated norm 𝑝𝑠
. Also, in this case, it does not matter with respect to which of
the topologies 𝜏𝑝, 𝜏¯
𝑝, or 𝜏𝑝𝑠 , is considered the separability.
Suppose that the topology 𝜏𝑝 is metrizable and separable. Then the topology
generated by the norm 𝑝𝑠
agrees with that generated by the asymmetric norm 𝑝,
implying the existence of two numbers 0 < 𝑅 < 𝑅′
such that 𝑅𝐵𝑝𝑠 ⊂ 𝐵𝑝 ⊂ 𝑅′
𝐵𝑝𝑠 ,
where 𝐵𝑝𝑠 , 𝐵𝑝 denote the corresponding closed unit balls. Since 𝛼𝛿𝐵𝛿𝑝𝑠 = 𝛿 𝐵𝑝𝑠 ,
we can suppose, replacing, if necessary, 𝑝𝑠
by 𝑞 = 𝛼𝑝𝑠
for some 0 < 𝛼 < 1, that
there exist a norm 𝑞 on 𝑋 and the numbers 0 < 𝑟 < 𝑟′
< 1 such that
𝑟𝐵𝑞 ⊂ 𝐵𝑝 ⊂ 𝑟′
𝐵𝑞 ⊂ 𝐵𝑞.
In the space 𝑋 × ℝ consider the sets 𝑈 = co [(𝐵𝑝 × {1}) ∪ (𝐵𝑞 × {0})], 𝑉 =
𝑈 ∪ co({𝑣} ∪ 𝐵𝑝 × {1}), where 𝑣 = (0, 1 + 𝑟), and 𝑊 = 𝑉 ∪ (−𝑉 ). It follows that
the sets 𝑈, 𝑉 are convex and 𝑊 is a bounded absolutely convex body, so that it
generates a norm ∥ ⋅ ∥ on 𝑋 × ℝ. Since 𝑊 ∩ (𝑋 × {0}) = 𝐵𝑞 × {0}, it follows that
∥(𝑥, 0)∥ = 𝑞(𝑥) for every 𝑥 ∈ 𝑋. If 𝑌 is a countable dense subset of (𝑋, 𝑞), then
𝑌 × ℚ is a countable dense subset of (𝑋 × ℝ, ∥ ⋅∥). By the classical Banach-Mazur
theorem there exists an isometric linear mapping 𝜑 : (𝑋×ℝ, ∥⋅∥) → (𝐶[0; 1], ∥⋅∥∞).
The image 𝑍 = 𝜑(𝑋 × {1}) of the hyperplane 𝑋 × {1} by 𝜑 is an affine variety
in 𝐶[0; 1]. Since 𝑊 ∩ (𝑋 × {1}) = 𝐵𝑝 × {1} and 𝜑 is an isometry, it follows that
the set 𝐾 = 𝑍 ∩ 𝐵𝐶[0;1] is isometric to 𝐵. In particular, the Minkowski functional
𝑃𝐾,𝜉 of the set 𝐾 with respect to the point 𝜉 = 𝜑(0, 1) agrees with the Minkowski
functional of the set 𝐵, that is with the asymmetric norm 𝑝. □
Remark 1.1.48. Let 𝐾 be a convex subset of a linear space 𝑋. The set 𝐾 is called
absorbing with respect to a point 𝜉 ∈ 𝐾 if for every 𝑥 ∈ 𝑋 there exists 𝑡 > 0 such
that 𝑥 ∈ 𝜉 + 𝑡(𝐾 − 𝜉). The Minkowski functional 𝑝𝐾,𝜉 of the set 𝐾 with respect
to 𝜉 is defined by
𝑝𝐾,𝜉(𝑥) = 𝑝𝐾−𝜉(𝑥 − 𝜉) = inf{𝑡 > 0 : 𝑥 ∈ 𝜉 + 𝑡(𝐾 − 𝜉)}. (1.1.35)
The isometry mentioned at the end of the proof of Theorem 1.1.46 is given
by the formula
𝑝𝐾,𝜉(𝜑(𝑥)) = 𝑝𝐾−𝜉(𝜑(𝑥 − 𝑣)) = 𝑝(𝑥), 𝑥 ∈ 𝑋.
A corollary of the Banach-Mazur theorem asserts that any separable metric
space can be isometrically embedded in 𝐶[0; 1] (see [74, Corollary 5.18]). Kleiber
and Pervin [117] extended this result to metric spaces of arbitrary density character
𝛼, where 𝛼 is an uncountable cardinal number, by proving that such a space can
be isometrically embedded in the space 𝐶([0; 1]𝛼
). The density character of a
topological space 𝑇 is the smallest cardinal number 𝛼 such that 𝑇 contains a
dense subset of cardinality 𝛼. As remarked by Alimov [11], these results hold also
for a quasi-metric space (𝑋, 𝜌), the density character being that of the associated
metric space (𝑋, 𝜌𝑠
).
45. 30 Chapter 1. Quasi-metric and Quasi-uniform Spaces
1.1.6 Quasi-uniform spaces
Quasi-semimetric spaces are particular cases of quasi-uniform spaces. A quasi-
uniformity on a set 𝑋 is a filter 𝒰 on 𝑋 × 𝑋 such that
(QU1) Δ(𝑋) ⊂ 𝑈, ∀𝑈 ∈ 𝒰;
(QU1) ∀𝑈 ∈ 𝒰, ∃𝑉 ∈ 𝒰, such that 𝑉 ∘ 𝑉 ⊂ 𝑈 ,
where Δ(𝑋) = {(𝑥, 𝑥) : 𝑥 ∈ 𝑋} denotes the diagonal of 𝑋 and, for 𝑀, 𝑁 ⊂ 𝑋 ×𝑋,
𝑀 ∘ 𝑁 = {(𝑥, 𝑧) ∈ 𝑋 × 𝑋 : ∃𝑦 ∈ 𝑋, (𝑥, 𝑦) ∈ 𝑀 and (𝑦, 𝑧) ∈ 𝑁} .
The composition 𝑉 ∘ 𝑉 is denoted sometimes simply by 𝑉 2
. Since every
entourage contains the diagonal Δ(𝑋), the inclusion 𝑉 2
⊂ 𝑈 implies 𝑉 ⊂ 𝑈.
If the filter 𝒰 satisfies also the condition
(U3) ∀𝑈, 𝑈 ∈ 𝒰 ⇒ 𝑈−1
∈ 𝒰 ,
where
𝑈−1
= {(𝑦, 𝑥) ∈ 𝑋 × 𝑋 : (𝑥, 𝑦) ∈ 𝑈} ,
then 𝒰 is called a uniformity on 𝑋. The sets in 𝒰 are called entourages. The pair
(𝑋, 𝒰) is called a quasi-uniform space.
For 𝑈 ∈ 𝒰, 𝑥 ∈ 𝑋 and 𝑍 ⊂ 𝑋 put
𝑈(𝑥) = {𝑦 ∈ 𝑋 : (𝑥, 𝑦) ∈ 𝑈} and 𝑈[𝑍] =
∪
{𝑈(𝑧) : 𝑧 ∈ 𝑍} .
A quasi-uniformity 𝒰 generates a topology 𝜏(𝒰) on 𝑋 for which the family of sets
{𝑈(𝑥) : 𝑈 ∈ 𝒰}
is a base of neighborhoods of the point 𝑥 ∈ 𝑋. A mapping 𝑓 between two quasi-
uniform spaces (𝑋, 𝒰), (𝑌, 𝒲) is called quasi-uniformly continuous if for every
𝑊 ∈ 𝒲 there exists 𝑈 ∈ 𝒰 such that (𝑓(𝑥), 𝑓(𝑦)) ∈ 𝑊 for all (𝑥, 𝑦) ∈ 𝑈. By
the definition of the topology generated by a quasi-uniformity, it is clear that a
quasi-uniformly continuous mapping is continuous with respect to the topologies
𝜏(𝒰), 𝜏(𝒲).
The family of sets
𝒰−1
= {𝑈−1
: 𝑈 ∈ 𝒰} (1.1.36)
is another quasi-uniformity on 𝑋 called the conjugate quasi-uniformity. With re-
spect to the topologies 𝜏(𝒰) and 𝜏(𝒰−1
), 𝑋 is a bitopological space. The equiva-
lences of the separation axioms from Corollary 1.1.10 holds in this case too.
Proposition 1.1.49. For a quasi-uniform space (𝑋, 𝒰) the following are equivalent.
1. The bitopological space (𝑋, 𝜏(𝒰), 𝜏(𝒰−1
) is pairwise 𝑇0.
2. The bitopological space (𝑋, 𝜏(𝒰), 𝜏(𝒰−1
) is pairwise 𝑇1.
3. The bitopological space (𝑋, 𝜏(𝒰), 𝜏(𝒰−1
) is pairwise Hausdorff.
46. 1.1. Topological properties of quasi-metric and quasi-uniform spaces 31
Proof. Obviously, it is sufficient to prove the implication 1 ⇒ 3. Suppose, for
instance, that for 𝑥 ∕= 𝑦 in 𝑋, there exists 𝑈 ∈ 𝒰 such that 𝑦 /
∈ 𝑈(𝑥). Taking
𝑉 ∈ 𝒰 such that 𝑉 2
⊂ 𝑈, it follows that 𝑉 (𝑥) ∩ 𝑉 −1
(𝑦) = ∅. Indeed
𝑧 ∈ 𝑉 (𝑥)∩𝑉 −1
(𝑦) ⇐⇒ (𝑥, 𝑧) ∈ 𝑉 ∧(𝑧, 𝑦) ∈ 𝑉 ⇒ (𝑥, 𝑦) ∈ 𝑉 2
⊂ 𝑈 ⇒ 𝑦 ∈ 𝑈(𝑥) ,
a contradiction. As the other cases (see (1.1.15)) can be treated similarly, it follows
that the bitopological space (𝑋, 𝜏(𝒰), 𝜏(𝒰−1
) is pairwise Hausdorff. □
If (𝑋, 𝜌) is a quasi-semimetric space, then
𝑉𝜀 = {(𝑥, 𝑦) ∈ 𝑋 × 𝑋 : 𝜌(𝑥, 𝑦) < 𝜀}, 𝜀 > 0 ,
is a basis for a quasi-uniformity 𝒰𝜌 on 𝑋. The family
𝑉 −
𝜀 = {(𝑥, 𝑦) ∈ 𝑋 × 𝑋 : 𝜌(𝑥, 𝑦) ≤ 𝜀}, 𝜀 > 0 ,
generates the same quasi-uniformity. Since 𝑉𝜀(𝑥) = 𝐵𝜌(𝑥, 𝜀) and 𝑉 −
𝜀 (𝑥) = 𝐵𝜌[𝑥, 𝜀],
it follows that the topologies generated by the quasi-semimetric 𝜌 and by the quasi-
uniformity 𝒰𝜌 agree, i.e., 𝜏𝜌 = 𝜏(𝒰𝜌).
If 𝑃 is a family of quasi-semimetrics, then the family of sets 𝑉𝑑,𝜀, 𝑑 ∈ 𝑃, 𝜀 >
0, generates a quasi-uniformity on 𝑋, the sets
𝑉𝑑,𝜀 = {(𝑥, 𝑦) ∈ 𝑋 × 𝑋 : 𝑑(𝑥, 𝑦) < 𝜀}, 𝜀 > 0, 𝑑 ∈ 𝑃 ,
being a base for this quasi-uniformity.
The following proposition is an adaptation of [110, Ch. 6, Thm. 11].
Proposition 1.1.50. Let (𝑋, 𝒰) be a quasi-uniform space and 𝑑 a quasi-semimetric
on 𝑋. Then the function 𝑑 is quasi-uniformly continuous on 𝑋 if and only if
𝑉𝑑,𝜀 ∈ 𝒰 for every 𝜀 > 0.
Proof. The space 𝑋 × 𝑋 is considered equipped with the quasi-uniformity gen-
erated by the base {𝑈 × 𝑈 : 𝑈 ∈ 𝒰}. Consequently, due to the symmetry in the
definition of product quasi-uniformity, the function 𝑑 : (𝑋 × 𝑋, 𝒰 × 𝒰) → (ℝ, 𝒰𝑢)
is quasi-uniformly continuous if and only if for every 𝜀 > 0 there exists 𝑈 ∈ 𝒰
such that for every (𝑥, 𝑦), (𝑢, 𝑣) ∈ 𝑈, ∣𝑑(𝑥, 𝑦) − 𝑑(𝑢, 𝑣)∣ < 𝜀.
Now, if 𝑑 is supposed quasi-uniformly continuous, then for every 𝜀 > 0 there
exists 𝑈 ∈ 𝒰 such that the above condition is satisfied. Taking (𝑢, 𝑣) = (𝑦, 𝑦) ∈ 𝑈
it follows that (𝑥, 𝑦) ∈ 𝑉𝑑,𝜀 for every (𝑥, 𝑦) ∈ 𝑈, that is 𝑈 ⊂ 𝑉𝑑,𝜀, and so 𝑉𝑑,𝜀 ∈ 𝒰.
Conversely, suppose that 𝑉𝑑,𝜀 ∈ 𝒰 for every 𝜀 > 0 and prove that the quasi-
semimetric 𝑑 is quasi-uniformly continuous on 𝑋. Given 𝜀 > 0, (𝑥, 𝑦), (𝑢, 𝑣) ∈ 𝑉𝑑,𝜀
imply 𝑑(𝑥, 𝑦) < 𝜀 and 𝑑(𝑢, 𝑣) < 𝜀, so that 𝑑(𝑥, 𝑦) − 𝑑(𝑢, 𝑣) < 𝜀 and 𝑑(𝑢, 𝑣) −
𝑑(𝑥, 𝑦) < 𝜀, that is ∣𝑑(𝑥, 𝑦) − 𝑑(𝑢, 𝑣)∣ < 𝜀, proving the quasi-uniformity of the
function 𝑑 on 𝑋 × 𝑋. □
47. 32 Chapter 1. Quasi-metric and Quasi-uniform Spaces
To prove that every quasi-uniformity is generated by a family of quasi-
seminorms in the way described above, we need the following result in general
topology.
Proposition 1.1.51 (Kelley [110]). Let 𝑋 be a nonempty set and {𝑈𝑛 : 𝑛 ∈ ℕ} a
family of nonempty subsets of 𝑋 × 𝑋 such that 𝑈0 = 𝑋 × 𝑋, each 𝑈𝑛 contains
the diagonal Δ(𝑋) and
𝑈𝑛+1 ∘ 𝑈𝑛+1 ∘ 𝑈𝑛+1 ⊂ 𝑈𝑛 , (1.1.37)
for all 𝑛 ∈ ℕ ∪ {0}. Then there exists a quasi-semimetric 𝑑 on 𝑋 such that
𝑈𝑛+1 ⊂ {(𝑥, 𝑦) ∈ 𝑋 × 𝑋 : 𝑑(𝑥, 𝑦) < 2−𝑛
} ⊂ 𝑈𝑛 , 𝑛 ∈ ℕ ∪ {0} . (1.1.38)
If, in addition, each 𝑈𝑛 is symmetric, then there exists a semimetric 𝑑 satisfying
(1.1.38).
Proof. For the convenience of the reader, we include a proof of this result. Observe
that the fact that each 𝑈𝑛 contains the diagonal and (1.1.37) implies that 𝑈𝑛+1 ⊂
𝑈𝑛 for all 𝑛 ∈ ℕ ∪ {0}.
Define a function 𝑓 : 𝑋 × 𝑋 → [0; ∞) by
𝑓(𝑥, 𝑦) = 2−𝑛
if (𝑥, 𝑦) ∈ 𝑈𝑛 ∖ 𝑈𝑛+1, for some 𝑛 ∈ ℕ ∪ {0}, and
𝑓(𝑥, 𝑦) = 0 if (𝑥, 𝑦) ∈ ∩𝑛𝑈𝑛 .
(1.1.39)
Observe that there exists at most one 𝑛 ∈ ℕ∪{0} such that (𝑥, 𝑦) ∈ 𝑈𝑛∖𝑈𝑛+1,
so that the function 𝑓 is well defined.
From the definition of the function 𝑓 it is clear that
𝑓(𝑥, 𝑦) ≤ 2−𝑛
⇐⇒ (𝑥, 𝑦) ∈ 𝑈𝑛 . (1.1.40)
A sequence 𝑥0, 𝑥1, . . . , 𝑥𝑛+1 of points in 𝑋 with 𝑥0 = 𝑥 and 𝑥𝑛+1 = 𝑦 is
called a chain connecting 𝑥 and 𝑦 and 𝑛 is called the length of the chain. Define
𝑑 : 𝑋 × 𝑋 → [0; ∞) by
𝑑(𝑥, 𝑦) = inf
𝑛
∑
𝑖=0
𝑓(𝑥𝑖, 𝑥𝑖+1) , (1.1.41)
where the infimum is taken over all chains connecting 𝑥 and 𝑦. From the definition
it is clear that 𝑑(𝑥, 𝑦) ≤ 𝑓(𝑥, 𝑦), so that, taking into account (1.1.40),
𝑈𝑛+1 ⊂ {(𝑥, 𝑦) ∈ 𝑋 × 𝑋 : 𝑑(𝑥, 𝑦) < 2−𝑛
} .
Indeed, (𝑥, 𝑦) ∈ 𝑈𝑛+1 implies 𝑓(𝑥, 𝑦) ≤ 2−𝑛−1
, so that 𝑑(𝑥, 𝑦) ≤ 2−𝑛−1
< 2−𝑛
.
Let us prove that
𝑓(𝑥0, 𝑥𝑛+1) ≤ 2
𝑛
∑
𝑖=0
𝑓(𝑥𝑖, 𝑥𝑖+1) , (1.1.42)
for all chains 𝑥0, 𝑥1, . . . , 𝑥𝑛+1 in 𝑋.
48. 1.1. Topological properties of quasi-metric and quasi-uniform spaces 33
The proof will be done by induction over the length 𝑛 of the chain. For 𝑛 = 0
the inequality (1.1.42) is trivial. Suppose that it holds for every 𝑚, 0 ≤ 𝑚 < 𝑛,
and prove it for 𝑛. Let 𝑎 =
∑𝑛
𝑖=0 𝑓(𝑥𝑖, 𝑥𝑖+1). The case 𝑎 = 0 is trivial, as well as
the case when 𝑓(𝑥𝑖, 𝑥𝑖+1) = 𝑎 for some 𝑖 ∈ {0, . . ., 𝑛}.
Excluding these situations, let 𝑘 be the greatest number between 0 and 𝑛
such that
∑𝑘
𝑖=0 𝑓(𝑥𝑖, 𝑥𝑖+1) ≤ 𝑎/2.
If 0 ≤ 𝑘 < 𝑛 − 1, then
∑𝑛
𝑖=𝑘+2 𝑓(𝑥𝑖, 𝑥𝑖+1) ≤ 𝑎/2 so that, by the induction
hypothesis,
𝑓(𝑥0, 𝑥𝑘+1) ≤ 𝑎, 𝑓(𝑥𝑘+2, 𝑥𝑛+1) ≤ 𝑎 ,
and, obviously, 𝑓(𝑥𝑘+1, 𝑥𝑘+2) ≤ 𝑎.
Let 𝑚 ≥ 0 be the smallest integer such that 2−𝑚
≤ 𝑎. Then
𝑓(𝑥0, 𝑥𝑘+1) ≤ 2−𝑚
, 𝑓(𝑥𝑘+1, 𝑥𝑘+2) ≤ 2−𝑚
and 𝑓(𝑥𝑘+2, 𝑥𝑛+1) ≤ 2−𝑚
,
so that, by (1.1.40), (𝑥0, 𝑥𝑘+1), (𝑥𝑘+1, 𝑥𝑘+2), (𝑥𝑘+2, 𝑥𝑛+1) ∈ 𝑈𝑚, implying
(𝑥0, 𝑥𝑛+1) ∈ 𝑈𝑚 ∘ 𝑈𝑚 ∘ 𝑈𝑚 ⊂ 𝑈𝑚−1,
which, by the same equivalence, yields
𝑓(𝑥0, 𝑥𝑛+1) ≤ 2−𝑚+1
= 2 ⋅ 2−𝑚
≤ 2 𝑎 .
If 𝑘 = 𝑛 − 1, then, reasoning as above,
𝑓(𝑥0, 𝑥𝑛) ≤ 2−𝑚
and 𝑓(𝑥𝑛, 𝑥𝑛+1) ≤ 2−𝑚
,
implying (𝑥0, 𝑥𝑛), (𝑥𝑛, 𝑥𝑛+1) ∈ 𝑈𝑚, so that
(𝑥0, 𝑥𝑛+1) ∈ 𝑈𝑚 ∘ 𝑈𝑚 ⊂ 𝑈𝑚 ∘ 𝑈𝑚 ∘ 𝑈𝑚 ⊂ 𝑈𝑚−1.
(The inclusion 𝑈𝑚 ∘ 𝑈𝑚 ⊂ 𝑈𝑚 ∘ 𝑈𝑚 ∘ 𝑈𝑚 follows from the fact that Δ(𝑋) ⊂ 𝑈𝑚.)
Now, (1.1.42) and the definition (1.1.41) of 𝑑 imply 𝑓(𝑥, 𝑦) ≤ 2𝑑(𝑥, 𝑦).
Consequently, if 𝑑(𝑥, 𝑦) < 2−𝑛
, then
𝑓(𝑥, 𝑦) < 2−𝑛+1
⇐⇒ 𝑓(𝑥, 𝑦) ≤ 2−𝑛
⇐⇒ (𝑥, 𝑦) ∈ 𝑈𝑛 ,
so that the second inclusion in (1.1.38) holds too.
If, in addition, each 𝑈𝑛 is symmetric, then, obviously, the function 𝑓 and also
𝑑, are symmetric. □
Theorem 1.1.52. Any quasi-uniform space is generated by a family of quasi-uni-
formly continuous quasi-semimetrics in the way described above.
Proof. It is clear that a family 𝑃 of quasi-semimetrics generates a quasi-uniformity
𝒰𝑃 and that each quasi-semimetric in 𝑃 is quasi-uniformly continuous with respect
to 𝒰𝑃 .
49. 34 Chapter 1. Quasi-metric and Quasi-uniform Spaces
Given a quasi-uniform space (𝑋, 𝒰), let 𝑃 denote the family of all quasi-
uniformly continuous quasi-semimetrics on 𝑋. By Proposition 1.1.50, 𝑉𝑑,𝜀 ∈ 𝒰 for
every 𝑑 ∈ 𝑃 and every 𝜀 > 0, implying 𝒰𝑃 ⊂ 𝒰.
Conversely, for 𝑈 ∈ 𝒰, choose a family 𝑈𝑛 ∈ 𝒰, 𝑛 ∈ ℕ, of entourages
such that 𝑈1 = 𝑈 and 𝑈𝑛+1 ∘ 𝑈𝑛+1 ∘ 𝑈𝑛+1 ⊂ 𝑈𝑛 for all 𝑛 ∈ ℕ. By Proposition
1.1.51 there exists a quasi-semimetric 𝑑 satisfying (1.1.38). The first inclusions
in (1.1.38) show that 𝑑 is quasi-uniformly continuous, while {(𝑥, 𝑦) ∈ 𝑋 × 𝑋 :
𝑑(𝑥, 𝑦) < 2−2
} ⊂ 𝑈1 = 𝑈 shows that 𝑈 belongs to 𝒰𝑃 , and so 𝒰 ⊂ 𝒰𝑃 . □
The following quasi-metrizability criterium for a quasi-uniform space is the
analog of a well-known one for uniform spaces.
Corollary 1.1.53. A quasi-uniform space (𝑋, 𝒰) is quasi-semimetrizable if and only
if the quasi-uniformity 𝒰 has a countable basis.
Proof. The necessity is clear. Suppose now that the quasi-uniformity 𝒰 has a
countable basis 𝐵𝑘, 𝑘 ∈ ℕ. Apply Proposition 1.1.51 for 𝑈𝑘
1 = 𝐵𝑘 to find a family
𝑑𝑘 of quasi-uniformly continuous semimetrics 𝑑𝑘, 𝑘 ∈ ℕ, satisfying (1.1.38). It
follows that the family {𝑑𝑘 : 𝑘 ∈ ℕ} generates the quasi-uniformity 𝒰. It is a
standard procedure to check that 𝑑 =
∑
𝑘 2−𝑘
𝑑𝑘/(1 + 𝑑𝑘) is a quasi-semimetric on
𝑋 which generates 𝒰. □
Remark 1.1.54. For other quasi-metrizability and metrizability results for quasi-
uniform spaces see Künzi [137] and the references given therein.
As it was shown by Pervin [175], every topological space is quasi-uniformizable.
Theorem 1.1.55 ([175]). Let (𝑋, 𝜏) be a topological space. Then the family of subsets
ℬ = {(𝐺 × 𝐶) ∪ ((𝑋 ∖ 𝐺) × 𝑋) : 𝐺 ∈ 𝜏} (1.1.43)
is a subbase of a quasi-uniformity on 𝑋 that generates the topology 𝑋.
As Pervin remarked in [175] the quasi-uniformity generating the topology is
not unique.
Example 1.1.56. Let ℝ with the natural quasi-uniformity 𝒰𝑑 be generated by the
quasi-metric 𝑑(𝑥, 𝑦) = max{𝑦 − 𝑥, 0}. Then the quasi-uniformity 𝒰 generated by
the subbase (1.1.43) is not even comparable with the quasi-uniformity 𝒰𝑑, although
both generate the same topology 𝜏𝑑 = 𝜏𝑢 on ℝ (see Example 1.1.3).
An account of the theory of quasi-uniform and quasi-metric spaces up to
1982 is given in the book by Fletcher and Lindgren [80]. The survey papers by
Künzi [132, 133, 134, 135, 136, 137] are good guides for subsequent developments.
Another book on quasi-uniform spaces is [153]. The properties of bitopologies and
quasi-uniformities on function spaces were studied in [197].
50. 1.1. Topological properties of quasi-metric and quasi-uniform spaces 35
A function 𝑓 : (𝑋, 𝒰) → (𝑌, 𝒱) between two quasi-uniform spaces is called
quasi-uniformly continuous on a subset 𝐴 of 𝑋 if
∀𝑉 ∈ 𝒱, ∃𝑈 ∈ 𝒰, ∀(𝑥, 𝑦), (𝑥, 𝑦) ∈ (𝐴 × 𝐴) ∩ 𝑈 ⇒ (𝑓(𝑥), 𝑓(𝑦)) ∈ 𝑉 . (1.1.44)
A quasi-uniform isomorphism is a bijective quasi-uniformly continuous function
𝑓 such that the inverse function 𝑓−1
is also quasi-uniformly continuous.
As it is well known, if (𝑋, 𝒰), (𝑌, 𝒱) are uniform spaces, then any continuous
functions 𝑓 : 𝑋 → 𝑌 is uniformly continuous on every compact subset of 𝑋. This
result is no longer true in the case of quasi-uniform spaces. Lambrinos [141] gives
an example of a continuous function from a compact quasi-uniform space (𝑋, 𝒰)
to a a quasi-uniform space (𝑋, 𝒱) that is not quasi-uniformly continuous and gives
a proper formulation to this result.
Theorem 1.1.57. A continuous function from a compact quasi-uniform space to a
uniform space is uniformly continuous.
This result corrects an example given without proof in [153, p. 5] of a con-
tinuous function from a compact quasi-uniform space to a uniform space which is
not quasi-uniformly continuous.
In fact Lambrinos proves a slightly more general result concerning continuous
functions on topologically bounded sets, a notion considered by him in the paper
[140]. A subset 𝐴 of a topological space (𝑋, 𝜏) is called 𝜏-bounded (or topologically
bounded) if any family of open sets covering 𝑋 contains a finite subfamily covering
𝐴. Obviously, a compact set is topologically bounded and the space 𝑋 is topolog-
ically bounded if and only if it is compact. Lambrinos [140] gives an example of a
topologically bounded set that is not relatively compact.
Theorem 1.1.57 will be a consequence of the following more general result.
Theorem 1.1.58. A continuous function from a quasi-uniform space (𝑋, 𝒰) to a
uniform space (𝑋, 𝒱) is uniformly continuous on every 𝜏(𝒰)-bounded subset of 𝑋.
Proof. Suppose that 𝐴 is a 𝜏(𝒰)-bounded subset of 𝑋. For an arbitrary 𝑉 ∈ 𝒱 let
𝑊 be a symmetric entourage in 𝒱 such that 𝑊2
⊂ 𝑉.
By the continuity of 𝑓, for every 𝑥 ∈ 𝑋 there exists 𝑈𝑥 ∈ 𝒰 such that
𝑓(𝑈𝑥(𝑥)) ⊂ 𝑊(𝑓(𝑥)). Let 𝑆𝑥 ∈ 𝒰 such that 𝑆2
𝑥 ⊂ 𝑈𝑥, 𝑥 ∈ 𝑋. Since the family
{𝜏(𝒰)- int 𝑆𝑥(𝑥) : 𝑥 ∈ 𝑋} is an open cover of 𝑋, there exists 𝑥1, . . . , 𝑥𝑛 ∈ 𝑋 such
that 𝐴 ⊂ ∪𝑛
𝑘=1𝑆𝑥𝑘
(𝑥𝑘).
Put 𝑈 := ∩𝑛
𝑘=1𝑆𝑥𝑘
and show that the condition (1.1.44) is satisfied by 𝑈
and 𝑉 .
For (𝑥, 𝑦) ∈ 𝑈∩(𝐴×𝐴) there exists 𝑗 ∈ {1, . . . , 𝑛} such that 𝑥 ∈ 𝑆𝑥𝑗 (𝑥𝑗) ⇐⇒
(𝑥𝑗, 𝑥) ∈ 𝑆𝑥𝑗 . Since 𝑆2
𝑥𝑗
⊂ 𝑈𝑥𝑗 , it follows that 𝑆𝑥𝑗 ⊂ 𝑈𝑥𝑗 and 𝑥 ∈ 𝑈𝑥𝑗 (𝑥𝑗).
Since (𝑥, 𝑦) ∈ 𝑈 = ∩𝑛
𝑘=1𝑆𝑥𝑘
, it follows that (𝑥, 𝑦) ∈ 𝑆𝑥𝑗 , so that (𝑥𝑗, 𝑦) ∈
51. 36 Chapter 1. Quasi-metric and Quasi-uniform Spaces
𝑆2
𝑥𝑗
⊂ 𝑈𝑥𝑗 . Consequently, 𝑥, 𝑦 ∈ 𝑈𝑥𝑗 (𝑥𝑗), implying
𝑓(𝑥) ∈ 𝑊(𝑓(𝑥𝑗)) ⇐⇒ (𝑓(𝑥𝑗, 𝑓(𝑥)) ∈ 𝑊
⇐⇒ (𝑓(𝑥), 𝑓(𝑥𝑗)) ∈ 𝑊 (𝑊 is symmetric),
𝑓(𝑦) ∈ 𝑊(𝑓(𝑥𝑗)) ⇐⇒ (𝑓(𝑥𝑗), 𝑓(𝑦)) ∈ 𝑊 ,
so that (𝑓(𝑥), 𝑓(𝑦)) ∈ 𝑊2
⊂ 𝑈. □
1.1.7 Asymmetric locally convex spaces
In this subsection we shall present some properties of asymmetric locally convex
spaces. Note that they were considered as early as 1997 in the paper [5]. In our
presentation we shall follow the paper [41]. A more general approach, based on
the theory of ordered locally convex cones, is considered in [109] and [212].
Let 𝑃 be a family of asymmetric seminorms on a real vector space 𝑋. Denote
by ℱ(𝑃) the family of all nonempty finite subsets of 𝑃 and, for 𝐹 ∈ ℱ(𝑃), 𝑥 ∈ 𝑋,
and 𝑟 > 0 let
𝐵𝐹 [𝑥, 𝑟] = {𝑦 ∈ 𝑋 : 𝑝(𝑦 − 𝑥) ≤ 𝑟, 𝑝 ∈ 𝐹} = ∩{𝐵𝑝[𝑥, 𝑟] : 𝑝 ∈ 𝐹}
and
𝐵𝐹 (𝑥, 𝑟) = {𝑦 ∈ 𝑋 : 𝑝(𝑦 − 𝑥) < 𝑟, 𝑝 ∈ 𝐹} = ∩{𝐵𝑝(𝑥, 𝑟) : 𝑝 ∈ 𝐹}
denote the closed, respectively open multiball of center 𝑥 and radius 𝑟. It is im-
mediate that these multiballs are convex absorbing subsets of 𝑋.
The asymmetric locally convex topology associated to the family 𝑃 of asym-
metric seminorms on a real vector space 𝑋 is the topology 𝜏𝑃 having as basis of
neighborhoods of any point 𝑥 ∈ 𝑋 the family 𝒩(𝑥) = {𝐵𝐹 (𝑥, 𝑟) : 𝑟 > 0, 𝐹 ∈
ℱ(𝑃)} of open multiballs. The family {𝐵𝐹 [𝑥, 𝑟] : 𝑟 > 0, 𝐹 ∈ ℱ(𝑃)} of closed
multiballs is also a neighborhood basis at 𝑥 for 𝜏𝑃 .
We shall abbreviate locally convex space as LCS.
It is easy to check that for every 𝑥 ∈ 𝑋 the family 𝒩(𝑥) fulfills the require-
ments of a neighborhood basis at 𝑥 so that it defines a topology 𝜏𝑃 on 𝑋 (or
𝜏(𝑃)).
Obviously, for 𝑃 = {𝑝} we obtain the topology 𝜏𝑝 of an asymmetric semi-
normed space (X,p) considered above, i.e., 𝜏{𝑝} = 𝜏𝑝.
The sets
𝑈𝐹,𝜀 = {(𝑥, 𝑦) ∈ 𝑋 × 𝑋 : 𝑝(𝑦 − 𝑥) < 𝜀, 𝑝 ∈ 𝐹} , 𝐹 ∈ ℱ(𝑃), 𝜀 > 0 , (1.1.45)
form a subbasis of a quasi-uniformity 𝒰𝑃 on 𝑋 generating the topology 𝜏𝑃 .
We say that the family 𝑃 is directed if for any 𝑝1, 𝑝2 ∈ 𝑃 there exists 𝑝 ∈ 𝑃
such that 𝑝 ≥ 𝑝𝑖, 𝑖 = 1, 2, where 𝑝 ≥ 𝑞 stands for the pointwise ordering: 𝑝(𝑥) ≥
𝑞(𝑥) for all 𝑥 ∈ 𝑋. If the family 𝑃 is directed then for any 𝜏𝑃 -neighborhood of
52. 1.1. Topological properties of quasi-metric and quasi-uniform spaces 37
a point 𝑥 ∈ 𝑋 there exist 𝑝 ∈ 𝑃 and 𝑟 > 0 such that 𝐵𝑝(𝑥, 𝑟) ⊂ 𝑉 (respectively
𝐵𝑝[𝑥, 𝑟] ⊂ 𝑉 ). Indeed, if 𝐵𝐹 (𝑥, 𝑟) ⊂ 𝑉 then there exists 𝑝 ∈ 𝑃 such that 𝑝 ≥ 𝑞 for
all 𝑞 ∈ 𝐹 so that 𝐵𝑝(𝑥, 𝑟) ⊂ 𝐵𝐹 (𝑥, 𝑟) ⊂ 𝑉. Also 𝑊 ∈ 𝒰𝑃 if and only if there exist
𝑝 ∈ 𝑃 and 𝜀 > 0 such that 𝑈𝑝,𝜀 ⊂ 𝑊, that is the family of sets (1.1.45) is a basis
for the quasi-uniformity 𝒰𝑃 .
For 𝐹 ∈ ℱ(𝑃) let
𝑝𝐹 (𝑥) = max{𝑝(𝑥) : 𝑝 ∈ 𝐹}, 𝑥 ∈ 𝑋. (1.1.46)
Then 𝑝𝐹 is an asymmetric seminorm on 𝑋 and
𝐵𝐹 [𝑥, 𝑟] = 𝐵𝑝𝐹 [𝑥, 𝑟] and 𝐵𝐹 (𝑥, 𝑟) = 𝐵𝑝𝐹 (𝑥, 𝑟). (1.1.47)
The family
𝑃𝑑 = {𝑝𝐹 : 𝐹 ∈ ℱ(𝑃)}, (1.1.48)
where 𝑝𝐹 is defined by (1.1.46), is a directed family of asymmetric seminorms
generating the same topology as 𝑃, i.e., 𝜏𝑃𝑑
= 𝜏𝑃 . Therefore, without restricting
the generality, we can always suppose that the family 𝑃 of asymmetric seminorms
is directed.
Because 𝐵𝐹 [𝑥, 𝑟] = 𝑥 + 𝐵𝐹 [0, 𝑟] and 𝐵𝐹 (𝑥, 𝑟) = 𝑥 + 𝐵𝐹 (0, 𝑟), the topology
𝜏𝑃 is translation invariant
𝒱(𝑥) = {𝑥 + 𝑉 : 𝑉 ∈ 𝒱(0)} ,
where by 𝒱(𝑥) we have denoted the family of all neighborhoods with respect to
𝜏𝑃 of a point 𝑥 ∈ 𝑋.
The addition + : 𝑋 × 𝑋 → 𝑋 is continuous. Indeed, for 𝑥, 𝑦 ∈ 𝑋 and the
neighborhood 𝐵𝐹 (𝑥+𝑦, 𝑟) of 𝑥+𝑦 we have 𝐵𝐹 (𝑥, 𝑟/2)+𝐵𝐹 (𝑦, 𝑟/2) ⊂ 𝐵𝐹 (𝑥+𝑦, 𝑟).
As we have seen in Proposition 1.1.42, the multiplication by scalars need not
be continuous, even in asymmetric seminormed spaces.
For an asymmetric seminorm 𝑝 on a vector space put
¯
𝑝(𝑥) = 𝑝(−𝑥), and 𝑝𝑠
(𝑥) = max{𝑝(𝑥), ¯
𝑝(𝑥)} ,
for all 𝑥 ∈ 𝑋.
If 𝑃 is a family of asymmetric seminorms on 𝑋, then
¯
𝑃 = {¯
𝑝 : 𝑝 ∈ 𝑃} and 𝑃𝑠
= {𝑝𝑠
: 𝑝 ∈ 𝑃} .
The following proposition contains some simple properties of asymmetric
LCS.
Proposition 1.1.59. Let 𝑃 be a directed family of asymmetric seminorms on a real
vector space 𝑋. Then
53. 38 Chapter 1. Quasi-metric and Quasi-uniform Spaces
1. For every 𝑝 ∈ 𝑃, 𝑥 ∈ 𝑋 and 𝑟 > 0,
𝐵¯
𝑝(0, 𝑟) = −𝐵𝑝(0, 𝑟), 𝐵¯
𝑝[0, 𝑟] = −𝐵𝑝[0, 𝑟],
𝐵𝑝𝑠 (𝑥, 𝑟) = 𝐵𝑝(𝑥, 𝑟) ∩ 𝐵¯
𝑝(𝑥, 𝑟) and 𝐵𝑝𝑠 [𝑥, 𝑟] = 𝐵𝑝[𝑥, 𝑟] ∩ 𝐵¯
𝑝[𝑥, 𝑟] .
2. Any 𝜏(𝑃)-open set is 𝜏(𝑃𝑠
)-open and any 𝜏( ¯
𝑃)-open set is 𝜏(𝑃𝑠
)-open, that
is 𝜏(𝑃) ⊂ 𝜏(𝑃𝑠
) and 𝜏( ¯
𝑃 ) ⊂ 𝜏(𝑃𝑠
). The same inclusions hold for the corre-
sponding closed sets.
3. Any 𝜏(𝑃)-continuous (or 𝜏( ¯
𝑃 )-continuous) mapping 𝑓 from 𝑋 to a topolog-
ical space 𝑇 is 𝜏(𝑃𝑠
)-continuous.
4. A ball 𝐵𝑝(𝑥, 𝑟) is 𝜏(𝑃)-open. A ball 𝐵𝑝[𝑥, 𝑟] is 𝜏( ¯
𝑃 )-closed and it need not
be 𝜏(𝑃)-closed.
Proof. The assertions 1 and 2 are immediate and 3 is a consequence of 2, so we
only need to prove 4.
For 𝑦 ∈ 𝐵𝑝(𝑥, 𝑟) let 𝑟′
:= 𝑟 − 𝑝(𝑦 − 𝑥) > 0. Since 𝑝(𝑧 − 𝑦) < 𝑟′
implies
𝑝(𝑧−𝑥) ≤ 𝑝(𝑧−𝑦)+𝑝(𝑦−𝑥) < 𝑟′
+𝑝(𝑦−𝑥) = 𝑟 it follows that 𝐵𝑝(𝑦, 𝑟′
) ⊂ 𝐵𝑝(𝑥, 𝑟).
To prove that 𝐵𝑝[𝑥, 𝑟] is 𝜏( ¯
𝑃 )-closed let 𝑦 ∈ 𝑋 ∖ 𝐵𝑝[𝑥, 𝑟]. Then 𝑟′
:= 𝑝(𝑦 −
𝑥) − 𝑟 > 0 and 𝐵¯
𝑝(𝑦, 𝑟′
) ⊂ 𝑋 ∖ 𝐵𝑝[𝑥, 𝑟]. Indeed, if there exists an element 𝑧 ∈
𝐵𝑝[𝑥, 𝑟] ∩ 𝐵¯
𝑝(𝑦, 𝑟′
), then one obtains the contradiction
𝑝(𝑦 − 𝑥) ≤ 𝑝(𝑦 − 𝑧) + 𝑝(𝑧 − 𝑥) = ¯
𝑝(𝑦 − 𝑧) + 𝑝(𝑧 − 𝑥) < 𝑟′
+ 𝑟 = 𝑝(𝑦 − 𝑥) .
Consequently, 𝑋 ∖ 𝐵𝑝[𝑥, 𝑟] is 𝜏( ¯
𝑃 )-open and so 𝐵𝑝[𝑥, 𝑟] is 𝜏( ¯
𝑃 )-closed. □
Example 1.1.60. In ℝ with the upper topology 𝜏𝑢, where 𝑢(𝑥) = max{𝑥, 0}, 𝑥 ∈
ℝ, we have 𝐵𝑢[0, 1] = (−∞; 1] and ℝ ∖ 𝐵𝑢[0, 1] = (1; +∞) is 𝜏¯
𝑢-open, but not
𝜏𝑢-open.
The following property is easy to check.
Proposition 1.1.61. Let (𝑋, 𝑃) be an asymmetric LCS. A net (𝑥𝑖 : 𝑖 ∈ 𝐼) is 𝜏𝑃 -
convergent to 𝑥 if and only if
∀𝑝 ∈ 𝑃, lim
𝑖
𝑝(𝑥𝑖 − 𝑥) = 0 . (1.1.49)
The topology 𝜏𝑝 generated by an asymmetric norm is not always Hausdorff.
A necessary and sufficient condition in order that 𝜏𝑝 be Hausdorff is given in
Proposition 1.1.40.
The following characterization of the Hausdorff separation property for lo-
cally convex spaces is well known, see, e.g., [238, Lemma VIII.1.4].
Proposition 1.1.62. Let (𝑋, 𝑄) be a locally convex space, where 𝑄 is a family of
seminorms generating the topology 𝜏𝑄 of 𝑋.
The topology 𝜏𝑄 is Hausdorff separated if and only if for every 𝑥 ∈ 𝑋, 𝑥 ∕= 0,
there exists 𝑞 ∈ 𝑄 such that 𝑞(𝑥) > 0.
54. 1.1. Topological properties of quasi-metric and quasi-uniform spaces 39
In the case of asymmetric locally convex spaces we have the following sepa-
ration properties.
Proposition 1.1.63. Let 𝑃 be a family of asymmetric seminorms on a real vector
space 𝑋.
1. The topology 𝜏𝑃 is 𝑇0 if and only if for every 𝑥 ∕= 0 in 𝑋 there exists 𝑝 ∈ 𝑃
such that 𝑝(𝑥) > 0 or 𝑝(−𝑥) > 0.
2. The topology 𝜏𝑃 is 𝑇1 if and only if for every 𝑥 ∕= 0 in 𝑋 there exists 𝑝 ∈ 𝑃
such that 𝑝(𝑥) > 0.
3. The topology 𝜏𝑃 is 𝑇2 if and only if for every 𝑥 ∕= 0 in 𝑋 there exists 𝑝 ∈ 𝑃
such that 𝑝⋄
(𝑥) > 0, where 𝑝⋄
is given by (1.1.31).
Proof. 1. Let 𝑥 ∕= 𝑦 in 𝑋. Then 𝑥 − 𝑦 ∕= 0 and 𝑦 − 𝑥 ∕= 0, so that, by hypothesis,
there exists 𝑝 ∈ 𝑃 such that 𝑝(𝑥−𝑦) > 0 or 𝑝(𝑦 −𝑥) > 0. If, say 𝑟 := 𝑝(𝑦 −𝑥) > 0,
then 𝑦 /
∈ 𝐵𝑝(𝑥, 𝑟). Similarly, 𝑟′
:= 𝑝(𝑥 − 𝑦) > 0 implies 𝑥 /
∈ 𝐵𝑝(𝑦, 𝑟′
).
Conversely, if 𝜏𝑃 is 𝑇0, then for 𝑥 ∕= 0 there exists 𝑈 ∈ 𝒱(0) such that 𝑥 /
∈ 𝑈,
or there exists 𝑈′
∈ 𝒱(𝑥) such that 0 /
∈ 𝑈′
. Let 𝐹 ∈ ℱ(𝑃) and 𝑟 > 0 such that
𝐵𝐹 (𝑜, 𝑟) ⊂ 𝑈. Since 𝑥 /
∈ 𝑈, there exists 𝑝 ∈ 𝐹 such that 𝑝(𝑥) ≥ 𝑟 > 0. In the
second case, let 𝐹′
∈ ℱ(𝑃) and 𝑟′
> 0 such that 𝐵𝐹 ′ (𝑥, 𝑟) ⊂ 𝑈′
. Since 𝑥 /
∈ 𝑈′
,
there exists 𝑞 ∈ 𝐹′
such that 𝑞(−𝑥) = 𝑞(0 − 𝑥) ≥ 𝑟′
> 0.
2. If 𝑥 ∕= 𝑦, then there exists 𝑝1, 𝑝2 ∈ 𝑃 such that 𝑟1 := 𝑝1(𝑦 − 𝑥) > 0
and 𝑟2 := 𝑝2(𝑥 − 𝑦) > 0, implying 𝑦 /
∈ 𝐵𝑝1 (𝑥, 𝑟1) and 𝑥 /
∈ 𝐵𝑝2 (𝑥, 𝑟2), that is the
topology 𝜏𝑃 is 𝑇1.
Conversely, if 𝜏𝑃 is 𝑇1, then for every 𝑥 ∕= 0 in 𝑋 there exists 𝑈 ∈ 𝒱(0) such
that 𝑥 /
∈ 𝑈. If 𝐹 ∈ ℱ(𝑃) and 𝑟 > 0 are such that 𝐵𝐹 (0, 𝑟) ⊂ 𝑈, then 𝑥 /
∈ 𝐵𝐹 (0, 𝑟),
so there exists 𝑝 ∈ 𝐹 with 𝑝(𝑥) = 𝑝(𝑥 − 0) ≥ 𝑟 > 0.
3. Suppose that 𝑃 is directed and let
𝑃⋄
= {𝑝⋄
: 𝑝 ∈ 𝑃} ,
where for 𝑝 ∈ 𝑃, 𝑝⋄
is defined by (1.1.31).
By Proposition 1.1.39, 𝑝⋄
is a seminorm on 𝑋. Denote by 𝜏𝑃 ⋄ the locally
convex topology on 𝑋 generated by the family 𝑃⋄
of seminorms. The topology 𝜏𝑃
is finer than 𝜏𝑃 ⋄ . Indeed, 𝐺 ∈ 𝜏𝑃 ⋄ is equivalent to the fact that for every 𝑥 ∈ 𝐺
there exist 𝑝 ∈ 𝑃 and 𝑟 > 0 such that 𝐵𝑝⋄ (𝑥, 𝑟) ⊂ 𝐺. Because, by Proposition
1.1.39, 𝑝⋄
(𝑦 − 𝑥) ≤ 𝑝(𝑦 − 𝑥) < 𝑟 it follows that 𝐵𝑝(𝑥, 𝑟) ⊂ 𝐵𝑝⋄ (𝑥, 𝑟) ⊂ 𝐺, so that
𝐺 ∈ 𝜏𝑃 . If for every 𝑥 ∈ 𝑋, 𝑥 ∕= 0, there exists 𝑝 ∈ 𝑃 such that 𝑝⋄
(𝑥) > 0, then,
by Proposition 1.1.62, the locally convex topology 𝜏𝑃 ⋄ is separated Hausdorff, and
so will be the finer topology 𝜏𝑃 .
Conversely, suppose that the topology 𝜏𝑃 is Hausdorff and show that 𝑝⋄
(𝑥) =
0 for all 𝑝 ∈ 𝑃 implies 𝑥 = 0.
Let 𝑥 ∈ 𝑃 be such that 𝑝⋄
(𝑥) = 0 for all 𝑝 ∈ 𝑃. By the definition (1.1.31) of
the seminorm 𝑝⋄
, for every 𝑝 ∈ 𝑃 and 𝑛 ∈ ℕ there exists an element 𝑥(𝑝,𝑛) ∈ 𝑋
56. It was well they had hastened, for the rascally landlord of the place
had, by this time, aroused all the half-castes in the place and as,
headed by Captain Sparhawk, they set off into the jungle, there was
a scattering firing of shots behind them. Nobody was hurt, however,
and they hastened forward to the place where Muldoon told them
the capture had taken place.
Salloo was consulted and he made a careful examination of the
surroundings. It was considered quite safe to make this halt, as the
tumult behind them had died out and was probably only incited by
the hotel owner in order to get them out of the village.
“Must wait till light come,” decided Salloo at last, “no can make out
trail in dark.”
It seemed a whole eternity till dawn, but at last it grew light and the
Malay darted hither and thither in the vicinity. At last he announced
to Captain Sparhawk that he thought he knew, from the direction
the trail took, the place to which the prisoners had been conveyed.
“Me think they take um to old fort on river,” he declared.
“Then let us go there at once,” said Captain Sparhawk eagerly. “Is it
far?”
“No velly far through jungle. But Salloo no know trail. Velly bad
swamp in there and if no know trail get in tlubble plenty quick.”
“Then we can’t reach them,” said Billy with a groan.
“Salloo know other way,” was the reply, “we go round by ribber.
Then climbee cliff, find fort at top.”
“Then let’s start at once,” said Captain Sparhawk. “I don’t want to
lose a second of time.”
“No, begorry, those spalpeens may have taken them further on by
the time we git there if we don’t put a good foot forward,” said
Muldoon.
Salloo glanced up at the sky. A light, fleecy haze overspread it.
57. “Nuther reason we hully,” he said. “Salloo think big storm come to-
mollow. Rain washee out the tlacks.”
They set off along a narrow track that Salloo said would bring them
to the river, whose course they must follow to the deserted fort. The
jungle contained every kind of tropical growth, and huge ferns as big
as trees waved over the path. But the atmosphere was close and
feverish, with a humid heat that was very tiring. At times they
encountered vines which had grown across the trail and had to be
cut. Some of these were thin and wiry and could cut like a knife;
others were as thick as a man’s arm and bore brilliant, though
poisonous-looking blossoms of every color.
“Bad traveling,” remarked Captain Sparhawk, “still I suppose we
must expect that on a seldomly frequented trail.”
“Him get velly bad further on,” was all the comfort Salloo could offer,
“but not velly far to ribber once we strike udder trail.”
Before long they came to the track he had referred to which
branched off at right angles to the one along which they had been
traveling.
Several miles were covered, however, when it became time to halt
for lunch. They made a hasty meal of canned goods instead of
stopping to light fires, as Salloo thought it would be inadvisable to
advertise their whereabouts by smoke columns in case the “enemy”
had scouts out. They had hardly resumed their wearisome journey
when they were startled by hearing a cry from a distance. Salloo
came to an instant halt.
“Keep out ob sight, all of you,” he said, “Salloo go see what makee
noise.”
He glided off into the dense vegetation with the silent, undulatory
movements of a snake.
“Begorry, I wonder what that critter was?” said Muldoon in a low
voice.
58. “I don’t know. I only hope it wasn’t a band of natives who might
prove unfriendly,” muttered Billy.
“Well, so far we have had more trouble with white men than with
natives,” said Captain Sparhawk, a remark of which they all felt the
truth.
“It might have been monkeys chattering,” suggested Raynor, after a
pause, during which they all listened for some sign of Salloo.
“And spaking of the divil,” exclaimed Muldoon, “look, there’s a
monkey looking at us now. See those two black oys back in the
threes?”
He pointed with his forefinger and they all gazed in that direction. It
was Billy who first discovered the nature of Muldoon’s monkey.
“That’s not a monkey. It’s a big snake! Look out for yourselves!” he
yelled.
“A python!” cried Captain Sparhawk.
He started back and the others did the same. But Muldoon tripped
over a bow and fell sprawling headlong. As he scrambled to his feet
a serpent’s form appeared above him as it swung from a big tree.
The next instant there was a cry of horror from them all.
The serpent had made a sudden lunge and a cry broke from
Muldoon as, before he could make a move to help himself, he was
enwrapped in the spiral folds of the great python.
Captain Sparhawk seized his revolver from his belt and leveled the
weapon. But the next moment he lowered it. To have fired would
have been to imperil Muldoon’s life, and there might still be a chance
of saving him.
The monstrous reptile that had the unfortunate boatswain in its grip
was large, even judged by the standards of the immense pythons of
the New Guinea and Borneo forests. It must have been fully thirty-
five feet long.
59. Billy could not endure the sight and put his hands in front of his
eyes. When he removed them it was to behold a stirring sight.
61. CHAPTER XXX.—IN THE COILS OF A PYTHON.
From the jungle there had darted a lithe figure. It was Salloo. He
had traced the source of the mysterious cries to a troop of monkeys.
He was returning when Muldoon’s despairing cry broke on his ears.
The Malay, guessing that there was serious trouble, glided through
the jungle at the best speed of which he was capable, making his
way swiftly through thickets that a white man could not have passed
at all. There is one weapon with which a Malay is always armed—his
kriss, a razor-edged sword about two feet long, with a “wavy”
outline. This kriss Salloo now drew from under his single garment.
One instant it flashed in the sunlight and the next, during which it
was impossible to follow its movements, so swift were they, the
python’s head was severed. But instantly, by a convulsive
movement, its coils tightened and Muldoon emitted another pitiful
cry. But, fortunately, the life of the snake had departed and soon its
coils relaxed and its gaudily-colored body slipped in a heap to the
ground.
They all sprang forward to Muldoon’s aid, for the man, powerful and
rugged, was almost in a state of collapse as the result of his terrible
experience. An examination by Captain Sparhawk soon showed that
no bones had been broken, as they had at first feared, and after
restoratives had been administered, and after a short rest, Muldoon
announced that he was ready to march on again.
“That was a close shave, Muldoon,” remarked Raynor, as they
pressed onward, after Muldoon had nearly wrung the hand off Salloo
in expressing his thanks for the Malay’s courageous act, which had
undoubtedly saved the boatswain’s life.
“Ouch! Don’t say a wurrud,” groaned the Irishman, “I thought I was
a goner sure. Divil a bit more of snakes is it I want to see.”
62. That evening they reached the river, and leaving them camped,
Salloo set off on a scouting expedition. It was a long time before he
returned, but when he came in he brought good news. He had
located the old fort and reported that the ruffians who had carried
off Mr. Jukes and Jack were all there enjoying themselves round a
big fire and apparently in no fear of an attack.
“Me see um white boy there, too,” he added. “Same boy hang round
hotel at Bomobori all time.”
“Donald Judson!” exclaimed Billy. “How can that be possible? I can’t
fit him into this at all.”
“Well, the question is, now that we have tracked the rascals, what’s
the next move,” said Captain Sparhawk.
“Me think now good time attack,” counseled Salloo. “They no think
anyone near. Give ’em heap big surprise.”
“Begorry, that’s well said, naygur,” approved Muldoon, “I’m aching to
git a good crack at thim.”
After some consultation it was decided to make the attack at once. If
they delayed they would have to wait till the next night in order to
surprise Broom’s band and there was no telling what might happen
during the twenty-four hours that would elapse.
Luckily, there was a moon, though it was somewhat obscured by the
haze which Salloo had drawn attention to as presaging a storm. The
party, piloted by Salloo, started off up the river, which was low, as
the weather had been dry and there was plenty of room for them to
pass between the bank and the water’s edge.
At last they arrived in sight of the cliff and Raynor’s heart gave a
bound. At the top they could see the red glare of the camp fire,
though they could not see any of the men.
“There’s one good thing, the ascent of the cliff will be easy,” said
Billy, in a whisper, as he drew attention to the knotted and twisted
vines that hung down it.
63. “Yes, we’ll need no scaling ladders,” rejoined Captain Sparhawk.
“No need for usum vines,” declared the Malay. “Salloo know a path
to top.”
Telling them to remain where they were, the faithful fellow set off on
another scouting expedition. His kriss glittered menacingly in the
moonlight as he went on, trying to keep in the shadow of the cliff.
Arrived at the path he knew of, he glided noiselessly up it, although
it was a steep and tortuous one, and soon was at the top of the cliff.
Through the gloom he made out a solitary figure sitting on a rock far
removed from the campfire, about which the rest were gathered.
The Malay guessed it was a sentry, although the fellow was not
keeping a very careful watch and appeared to be half asleep.
“Me fixee you one minute,” grinned the Malay to himself.
He cast himself on his stomach in the long grass that grew on the
cliff-top and began worming his way round the sentry so as to
approach him from the rear. He scarcely made a sound as he moved
with wonderful rapidity.
The sentry appeared to shake off his drowsiness suddenly and rose
to his feet just as Salloo was within a few inches of him. But he left
his rifle leaning against the rock on which he had been seated.
Instantly Salloo leaped from the grass and the next instant the kriss
was at the thunderstruck sentry’s throat.
“You no speak or me killee,” grated out the Malay, and one glance
convinced the sentry that Salloo would carry out his threat.
Salloo stooped and picking up a small pebble cast it over the cliff. It
fell almost at the feet of Captain Sparhawk and Billy, who were
anxiously on the look-out for this signal, which had been
prearranged.
“Forward,” ordered Captain Sparhawk, who was in the lead. Next
came Billy, then Muldoon and last the natives, some of whom had
spears, and others the peculiar blow-pipes used by the Papuans to
shoot poisoned darts.
64. The advance was made in silence, and at the top of the cliff they
found Salloo waiting for them. He was garmentless, having used his
single cloak to tie up the sentry with. Grass stuffed in the man’s
mouth had effectually gagged him.
“Good for you, Salloo,” said the captain approvingly, to which the
native replied with a grin.
“Now we take him down below and find out some things from him,”
said the Malay.
The helpless prisoner was bundled back down the trail and brought
to the camp at the foot of the cliff. Here he was roped to a tree and
the gag taken out of his mouth. But the sight of Salloo’s ever-ready
kriss kept him from making any outcry.
Yes, he said, the old, fat man and the boy were all right. They had
not been fed though, and wouldn’t be till a ransom was forthcoming.
This made the whites boil with indignation. Questioned as to how
many were in the band, he said he did not know, and as he stuck to
this it was thought best not to waste any more time questioning
him.
After a consultation the gag was replaced, but the ropes were
loosened so that with a little exertion the man could set himself free.
“If, for any reason, we couldn’t come back, and we left the ropes
tight, he would perish,” said Captain Sparhawk, “and we want no
human lives to our account.”
“Me leave him there starve to death plitty quick,” growled Salloo,
with a scowl at the crestfallen prisoner.
At the foot of the cliff all was now dark and silent as the grave. The
moon was obscured by a cloud and it was an ideal moment for the
dash on the camp to begin.
“We go plenty slow or maybe take big tumble,” advised Salloo.
He was in advance but Billy was close at his heels. Cautiously they
ascended, taking great care not to dislodge loose stones which
65. might have been fatal to their plans. At last the stream was far
below them and the summit of the cliff within reach.
It was at this moment that a torch flashed above them, glaring into
their upturned faces.
“What’s all this, who are you?” a voice demanded.
“Silence if you value your life,” came from Captain Sparhawk.
“It’s Donald Judson!” exclaimed Billy.
“Billy Raynor,” cried the other in his turn. “How did you——?”
“Don’t utter another word,” ordered Captain Sparhawk. “Put your
hands above your head, you young rascal.”
“Not much I won’t!” exclaimed Judson.
He flung his torch full in Billy’s face and then started at top speed for
the camp fire, yelling the alarm at the top of his lungs.
For a minute Billy was in peril of losing his balance as the torch
struck him. But Salloo caught and held him firmly. The torch dropped
with a splash and hiss into the waters of the river below.
By this time Salloo scrambled to the cliff summit and made off after
young Judson. Both reached the camp fire at about the same time.
The others, following close on Salloo’s heels, saw Donald turn, catch
sight of the glittering kriss, and then, with a yell of dismay, tumble
headlong. He lay quite still and had apparently been stunned by the
violence of the fall.
“Help! Help!” It was Jack’s voice from the fort and was instantly
recognised by Billy.
But by this time the men about the fire, headed by ‘Bully’ Broom,
were on their feet. There was no time for them to get their
weapons, which had been left inside the fort so that they would not
rust in the damp night air. The battle was a brief one, although some
shots were fired, none of which, in the excitement, took effect.
66. Billy, by a clever ruse, brought the engagement to a speedy
termination. In the midst of the fight he turned toward the cliff and
then raising his voice as if summoning help, he shouted:
“This way, captain. Bring that company up here. Let the others
guard the river.”
“Get out of here, boys,” roared Broom, completely taken in. “I’ll
settle with you later on,” he cried, shaking his fist as he turned and
followed the rout of his followers, who, imagining they were being
pursued by great numbers, made off at top speed for the jungle,
which soon swallowed them up.
68. CHAPTER XXXI.—THE JOURNEY RESUMED.
“Thank Heaven that is over,” said Mr. Jukes, as he sat on an old
bench in the fort after he and Jack had been released. “You may
depend upon it that I shall not forget the part that Salloo and all of
you played in our rescue.”
It was some two hours after the “battle,” if the rout of the rascals
who had captured Mr. Jukes and Jack could be termed such. The
kidnappers’ larder had been ransacked and a good meal enjoyed by
all hands, especially, as may be imagined, by the two captives who
had been without food for almost twenty-four hours.
Donald Judson, looking hang-dog and abject, was huddled on a
bench in a corner of the room. He had been picked up after the fray,
having shammed insensibility to avoid being injured, and was easily
captured by the victors.
“You certainly came in the nick of time,” said Jack. “From what I
could hear them saying, that scoundrel Broom was actually
contemplating torturing us if that check was not signed by Mr.
Jukes.”
The millionaire shuddered. His experiences had greatly affected him.
“That young ruffian over yonder,” he nodded his head toward
Judson, “was the instigator of the idea to get money out of me, I
believe,” he said. “He ought to be punished severely.”
“I didn’t,” whined Judson miserably, “I—I—that fellow Broom did it
all.”
“What’s the use of your lying, Judson,” exclaimed Jack, “you met
Broom at Bomobori. It’s as plain as day now, and furnished him with
an account of as much of our plans as we had confided to you.”
69. “Well, maybe I did,” mumbled Judson sullenly, “but I didn’t put him
up to getting money out of you.”
“Nonsense,” said Captain Sparhawk, “you are as bad as Broom is—
worse, in fact, for you are a lad of decent upbringing.”
No more was said to Judson that night, and they retired to catch a
few hours sleep, leaving the “carriers” under Salloo on guard. The
Malay amused himself by making hideous faces at the unfortunate
Donald and flourishing his kriss under his nose. By daylight the
wretched prisoner was half dead from fear. Captain Sparhawk sternly
warned Salloo not to tease him any more, at which the Malay
appeared to be much surprised.
“Him enemy,” he said, “why no can do what like with him?”
Breakfast, of which Donald was given his share, was eaten in the
fort, and after that meal the natives were sent down to the river to
bring up all the supplies which had been left there. They reported
that the prisoner Salloo had made had succeeded, as they intended
he should, in loosening his bonds during the night and had vanished.
As soon as the boxes containing the wireless apparatus and the
hand-generator arrived, Jack lost no time in setting them up and as
soon as he raised the yacht sent a full account of Broom’s rascally
conduct to her. The first officer at once left to notify the authorities
and ask that a keen lookout be kept for Broom’s schooner.
“Broom will never guess that we have any means of communicating
with Bomobori,” the boy explained, “and if he returns there, will
bungle into a fine trap.”
“Begorry, I hope he does,” commented Muldoon, “shure that wireless
is an illigant invintion entirely.”
“If Broom is captured, as many other criminals have been, by its aid,
it will have proved its splendid usefulness once more,” declared Mr.
Jukes. “Ready, you might flash another message saying that I will
give $1,000 to anyone who captures ‘Bully’ Broom.”
70. After this had been done, the question arose of what to do with
Donald Judson. They had no desire to have the young rascal as a
traveling companion, but at the same time they did not see how
they could very well turn him loose in the jungle in which he might
starve to death. It was a problem that they were still discussing
when Donald himself spoke up in the timid, fawning voice he
affected when in trouble.
“See here,” he said, “if you won’t make trouble for me maybe I can
help you out.”
“In what way?” sharply asked Mr. Jukes.
“Why I saw Broom put a map or something that looked like one in a
cupboard in the room that door opens into,” said the boy, pointing to
the end of the room. “I thought maybe it might have something to
do with your brother, Mr. Jukes.”
“Come here at once and show me,” ordered the millionaire. “I don’t
suppose it was anything of great importance,” he added.
“Perhaps not,” whimpered Donald, “but if it is will you let it count in
my favor?”
“I shall consider that later,” said Mr. Jukes sternly, as they all
followed the boy into the room he indicated. In one corner was a
rough cupboard. Mr. Jukes opened this and took out a rolled-up
paper. He spread it out on the table and they all pressed about him.
“It’s a map!” cried Billy.
“Yes, and of this part of the country, too,” cried Jack. “See, there’s
that village, Taroo, where we stopped two nights ago.”
“And what’s this leading along the river from this place marked 'Fort’
on the map?” asked Mr. Judson, his eyes shining as his forefinger
traced a red ink line that zig-zagged along till it left the river and
struck inland to what appeared to be intended to show a range of
mountains. “The Kini-Balu Mountains,” he read out.
71. “The Kini-Balu Mountains!” echoed Salloo, “me know them. Me bet
your brother up there. One time ‘Bully’ Bloom he helpee Kini-Balu
men fight big battle 'gainst Tariani tribe. Kini-Balus win and now
heap like ‘Bully’ Bloom hide your brother up there.”
“It is possible,” mused the millionaire, “and—yes, by jove! Look
here.”
Indicated on the map in red letters, at a spot in the heart of the Kini-
Balu country, was a place marked “Cave.”
“Do you think it possible that that can be ‘Bully’ Broom’s hiding place
for the other Mr. Jukes?” asked Jack.
“I don’t know, but it appears probable,” rejoined the millionaire.
“Me membel now sometime ‘Bully’ Bloom go way from Bomobori
long time,” said Salloo, “nobody know where he go. That time when
cruiser come look for him. Maybe he hide up there.”
“It seems worth trying at any rate,” said Mr. Jukes, in the manner of
one who has reached a decision.
“It seems reasonable to suppose that if Broom had taken your
brother and his men anywhere on the island it would have been to
some such inaccessible spot as that,” said Captain Sparhawk.
“Well thin, what’s to privint us going up among the 'balloon’ men, or
whativer they call thimsilves?” asked Muldoon.
“It may be attended by some danger,” said Mr. Jukes. “From what
Salloo said the Kini-Balu men are a very war-like tribe. They might
attack us. How about that, Salloo?”
The Malay’s reply was not one calculated to reassure them.
“Kini-Balu men head hunters,” he said, “Maybe they no hurt us. But
maybe take our heads. Salloo no 'fraid, though.”
“Then, by golly, neither are we,” declared Muldoon.
After more discussion, it was decided to advance cautiously into the
Kini-Balu country and then do some scouting to see how matters lay.
72. If the natives were hostile, and if they were convinced that Mr. Jukes
was really a captive among them, guarded by their warriors at ‘Bully’
Broom’s orders, then they would return to Bomobori without risking
their lives and come back with a strong force. If everything appeared
to be pacific, then they would seek out the place indicated on the
map and settle the question of whether or no it was actually the
place of the pearl hunter’s confinement.
74. CHAPTER XXXII.—A STORM IN THE JUNGLE.
Two days later, before they turned away from the river, they heard
some news of the Kini-Balus from a party of natives bound down-
stream in dug-outs. Salloo learned from them that the tribe was at
war, at least so it was supposed by the canoeists from the fact that
they had heard that the chief of the Kini-Balus had been making
levies of cattle and corn among his subjects.
“That sounds bad,” said Mr. Jukes, when this news had been
interpreted to the party.
“No, him good,” asserted Salloo positively.
“How do you make that out?” asked Jack.
“If Kini-Balus makee war, they leave only women and old men at
home. They no fight us,” argued the Malay, and they had to admit
that there was a good deal of truth in what he said.
“We’re all going to get killed anyhow,” whimpered Donald, who had
been taken along by the party, much against their will, in
consideration of the services he had rendered in showing them the
hiding place of the map.
“Him heap big coward,” muttered Salloo. “Boy’s body, girl’s heart.”
It was on the afternoon of the second day that the storm that Salloo
had predicted overtook them. They were passing through a dense
forest of magnificent trees when the eternal twilight that reigned
under the great branches deepened till it was almost totally dark.
Astonished at this phenomenon, for it was long before the proper
hour for night to descend, they questioned Salloo.
“Big storm come,” he said, “me thinkee we better get out of here.
Lightning hit a tlee maybe he killee us.”
75. The birds of the jungle screamed discordantly, as if warning each
other of what was coming. Troops of monkeys swung through the
trees as if seeking refuge, and the almost deafening chorus of
insects and lizards gave way to total silence. It seemed as if nature
was holding her breath preparatory to some great crisis.
“We had better look for some safe place to stay before it breaks,”
counseled Captain Sparhawk. “A hurricane in the jungle is a serious
matter. Trees are rooted up and struck by lightning and in the forest
it is very dangerous for anyone to be caught by such a storm.”
“Me findee place,” said Salloo, and struck off down a dim trail
leading toward the river. “Follow me, evelybody, and hully up.”
They needed no urging. The gloom and quiet of the forest was
overawing. It had begun to get on their nerves. Under Salloo’s
guidance they soon found themselves at a great mass of rocks on a
high bank overlooking the river. The great masses of stone were
piled in such a way that the crevices among them formed regular
caves.
“We getee in here,” said Salloo, indicating the largest of them. “I
send my men in annuder one.”
“I’m not going in there,” declared Donald, “there might be snakes or
wild beasts inside.”
“You’d better come in or be blown away,” said Captain Sparhawk.
He had hardly spoken, before the storm broke in all its fury. Donald,
with a cry of alarm, followed the others into shelter.
“Gracious, this beats anything I ever saw, even that storm off the
Pamatous,” shouted Jack, above the shrieking of the wind.
“Him blow more big bimeby,” said Salloo, “him big storm this. You
see.”
The trees swayed violently, and before long, from their shelter, they
saw a big one torn up by the roots and hurtled from the bank into
the river. The wind grew more violent. The dark air was filled with
76. flying branches, leaves and sticks. Birds, large and small, were swept
by, powerless to contend with the furious gale. Donald was crouched
back in a far corner of their shelter, too frightened to do anything
more than mumble and whimper.
The river began to rise and add its mighty voice to the other sounds,
although no rain had yet fallen where they were. The darkness
increased, but suddenly everything was lit up in a livid glare that
made them all blink.
“Lightning,” exclaimed Salloo, “now him comin’.”
Then down came the rain. It literally fell in sheets, blotting out
everything like a fog even when the constant flashes illuminated the
scene. The water began to pour into their shelter from above,
making it a very uncomfortable place. Soon the water was up to
their knees and in the cave occupied by the carriers the men stood
upright with their burdens on their heads to keep them out of the
water.
“Gracious, I never saw so much water come down in my life,”
exclaimed Jack. “It’s a regular—my!”
There had come a flash, a red ribbon of flame, so blinding that for
an instant they could not see. It was followed by a crack of thunder
that seemed to have split the sky. Donald gave a yell of alarm.
“Him hittee something close by for sure,” declared Salloo. He was
right. Presently they saw a tall ceiba tree burst into flame like a
torch. Fanned by the wind, it blazed fiercely even in the downpour.
Its red glare lit up their faces in a ghostly manner, for it was not
more than a few feet from their place of refuge.
“My, this is awful,” muttered Raynor. “Thank goodness we got out
from under those trees in time.”
“Amen to that,” said Captain Sparhawk solemnly.
It rained for the rest of that night and in the morning they were
sorry-looking objects. Everything was wet, and although they had
tried to light a fire during the night, after the first violence of the
77. storm had abated, they had not succeeded. But when, shortly before
noon, the sun did come out, it shone down with a heat that made
the whole wet earth steam. Clothes were spread out on the rocks to
dry, as was the rest of the outfit. Fortunately, the bags the carriers
bore were mostly of waterproof material, so not much damage was
done to the contents.
It was a scene of havoc on which they gazed. The river ran high and
its surface was littered with the bodies of dead monkeys, snakes,
great trees torn up bodily, and other debris eloquent of the violence
of the hurricane.
All round them lay big trees and the bodies of countless birds that
had been dashed to death. It was some time before Salloo could
persuade a fire to burn, but among the rocks, in crevices the rain
had not penetrated, he found old dried leaves and sticks which made
capital kindling and at last they cooked a hot meal, in need of which
they all stood badly.
Then it was off on the long trail again. Late that afternoon, just as
they were making camp, a party of natives came along the trail.
They carried the skins of numerous beautiful birds that they had
brought down with their blow-pipes. They were friendly and the
boys bought some of the skins. Afterward Salloo had a long talk with
them and, this being concluded, they kept on their way while our
party went on with its preparations for spending the night.
Salloo had some news to disclose, he said. The natives he had been
talking to knew the Kini-Balu Mountains well and told him, after he
had described the cave they were looking for, that it was a very bad
place. Nobody liked to go near it.
“On account of the Kini-Balus?” asked Mr. Jukes.
“No, on account um ghosts,” rejoined Salloo; “ghost of Taratao, old-
time chief of Kini-Balus haunt him.”
“Begorry, so long as the ghosts ain’t got a punch it’s sorra a bit I
care for ’em,” declared Muldoon valiantly.
78. That evening Salloo had a novelty for supper in the form of the flesh
of a huge lizard, or iguana. At first the boys and their companions
did not want to touch it, for in life it had been a hideous looking
monster. But being pressed by Salloo, they consented, and found it
very good eating. Its flesh tasted like chicken, though even more
delicate.
It was about an hour after the meal when they were preparing for
bed that Jack complained that he was feeling poorly. He said he had
a headache and a feeling of vertigo. The others then admitted
experiencing the same symptoms. Nausea soon succeeded these
and ere long they were all convinced that they had been poisoned
by eating the iguana. The natives, who camped some distance off
with Salloo, experienced no such illness but then they had eaten
none of the iguana which, to Captain Sparhawk’s mind, made it all
the more certain that it was the giant lizard’s flesh that had made
them ill.
Salloo was called from the native camp and bitterly reproached for
inducing them to eat it. He protested that it could not have been the
iguana that had made them ill. Had he not himself eaten it? But in
the end he returned to the native camp with his head hung down,
completely crushed by what he deemed the injustice of his white
friends in blaming him for their illness. At first they were not greatly
alarmed, not deeming it possible that they had actually been
poisoned, and Captain Sparhawk administered remedies from the
medicine chest. But, to their alarm, instead of decreasing in severity,
their sufferings grew more acute as the night wore on.
Their ideas became confused, and as in sea-sickness in an acute
stage, they lay about, not caring whether they lived or died. If they
tried to rise, their heads swam, their feet tottered. Thus it was that
Salloo found them in the morning when he came from the native
camp.
The faithful fellow was seriously alarmed and set up a mighty wailing
which soon brought his followers running over. But the sufferers only
turned dull eyes upon them and moaned in their pain. Plainly they
79. were in such a serious condition that unless something was done
soon to relieve them, death itself might put an end to their misery.
Salloo looked about him wildly, hoping to catch some solution to the
mystery of this sudden illness. He raised his eyes upward and his lips
moved as if he were invoking the aid of some heathen deity.
But suddenly the expression on his countenance changed. His eyes
were fixed on the leaves of a tree under which the sufferers had
passed the night. For the first time, too, he became aware of a
peculiarly sickening odor in the air. It smelled like carrion. As some
huge scarlet flowers which grew on the tree began to open to the
daylight (they had been closed at night) this terrible stench became
stronger. Salloo uttered a single shout of comprehension.
“Upas!” It was echoed by his companions, whom Salloo at once
directed to pick up the sufferers and carry them to some distance.
When the last had been transported, Salloo got water from a forest
pool and poured it over them. One by one they began to revive.
Jack, who was one of the first to come to, rose dizzily to his feet and
tried to walk. But Salloo gently made him lie down again. After an
hour or so all felt better and partook of some soup and weak tea.
“Salloo, you are forgiven,” said Captain Sparhawk, “but never
persuade us to eat lizard again. You came near being the death of
us all.”
“Faith, oi was niver so near the Pearly Gates before,” declared
Muldoon emphatically.
“Him no lizard hurt you,” declared Salloo vehemently; “lizard heap
good. Upas he hurt you. If I no see it and have you moved away you
plitty soon have died.”
“What do you mean, Salloo?” asked Mr. Jukes. “Do you mean our
sickness had anything to do with the tree we camped under?”
“Ebblyting,” was the reply; “him tree was the upas.”
“I see it all now,” exclaimed Captain Sparhawk. “That tree was the
deadly upas of which you may have heard. Every one in the Indian
80. Archipelago knows of it. Within its great red blossoms are the
sepulchre of birds and insects whose bodies, lying rotting there, give
out that terrible odor which ought to warn all travelers against it. But
we camped when it was getting dark and the flowers were closed,
keeping the noxious reek from escaping and warning us. Salloo is
right, and if he had not had us dragged from under it we should
have perished miserably.”
“I remember reading somewhere of the upas,” said Jack, “but I
always thought its deadly qualities were exaggerated. After our last
night’s experience I’ll know better.”
“I suppose the heat of our camp fire under the branches had
something to do with it, too,” said Billy.
“Undoubtedly,” declared the captain. “And then as we sat around
after supper we were, unknown to ourselves, inhaling the deadly
vapor till we grew sick. Instead of moving away before we grew
worse, as we certainly would have if we had known the cause of our
malady, we made ourselves worse by lying down to sleep with that
poisonous breath as our only atmosphere. Salloo, your lizard is
vindicated, and to show you it is, the next one you shoot I’ll
volunteer to eat.”
But although recovered, they still felt weak from the effects of their
terrible night under the upas, whose Latin name, if any one wishes
to know it, is antiaris toxicaria. In fact, their feelings were very like
those of persons just getting over sea-sickness. They felt buoyantly
well and happy, but not yet quite strong enough for the hard work of
the trail. So they remained where they were till the next day and
then pushed on once more on their quest.
82. CHAPTER XXXIII.—THE GIANT SPIDERS.
When they resumed their journey the next morning they
encountered a new form of obstacle in the form of the webs of huge
red bird-catching spiders, whose nets stretched from tree to tree in
the forest, looked like seine nets in a fisherman’s village hung out to
dry, or to make another comparison, miles of mosquito netting hung
between the tree trunks. Through these webs they had to make
their way for a long distance.
The boys did not like it at all, and Donald Judson, who was
particularly averse to spiders, slunk in the rear till the natives, with
shouts and yells, cut down the webs that hung across the trail. The
soft silky substance of the webs struck them in the face and clung
glutinously and covered their clothes with a coating of white fleece.
As they forced their way through this repulsive feature of New
Guinea forest travel, they could, from time to time, see the hideous
forms of the huge and venomous spiders that had spread the webs
peering at them from dark retreats in the crevices of trees or else
scuttling off on long, hairy legs to safety. It did not require much
imagination to picture their anger at this ruthless destruction of their
homes. That night they camped near the edge of a big swamp, and
the two boys, weary of the monotony of the long march and tired of
canned stuff and preserved goods, volunteered to set out with rifles
and see if they could not bring in something more palatable.
As they had camped early when the swamp crossed their path, there
was plenty of time for them to go quite a distance in search of
game. In a short time they had brought down two birds that looked
something like partridges, as well as shooting an odd-looking bird
like a huge parrot, with a gigantic bill and horny head. They were
some distance apart, separated by a brake of reeds, when Jack
heard a sudden cry of alarm from Billy.
83. Disregarding the danger of snakes, he pushed his way through the
brake at once. As he came in sight of Billy, who was standing staring
into the forest as if petrified, Jack, too, received a shock. Not far
from Billy was what he at first thought was a man. But such a man!
Not even in a nightmare had the boy ever beheld such a hideous
form.
This man, if such he was, was covered all over with red hair, thick
and shaggy, except on the face, which was darker and bereft of hair,
but from which two yellow eyes glared malevolently. In an instant
the true nature of this creature flashed upon Jack. It was an orang-
outang, and a monster, too, that stood facing them, its long arms
trailing in front of it. But even though stooped over, it was as large
as the average man, with a massive chest and shoulders.
“Take a shot at it, Jack,” urged Billy.
But Jack shook his head.
“It looks too horribly human,” he said. “Besides, it doesn’t look as if
it would attack us. It seems to be more possessed by curiosity than
anything else.”
Perhaps the boy was right, for after eyeing them for a few seconds
more the monstrous creature shuffled off for the edge of a big sheet
of water on whose margin they stood, and began tearing up some
sort of water plants and eating their roots with many grunts of
satisfaction. He waded in almost knee deep, stuffing his bag-like
cheeks full and chewing with huge satisfaction. The boys gazed at
this strange picture with fascination.
But suddenly the monster stopped eating and stood erect. Its hair
began to bristle and it uttered an angry sort of growl. Apparently it
was not fear but anger that possessed this colossus of the forests as
it glanced angrily about it. The cause of its emotion was not long in
appearing. From the stagnant waters was approaching an antagonist
formidable indeed—a giant saurian—a crocodile larger than any the
boys had ever seen in any zoo.
84. The boys naturally expected to see the orang-outang beat a hasty
retreat. But instead it stood its ground, merely drawing back a few
inches as the crocodile’s hideous snout and scaly body were
successively protruded from the water. Jack now recalled what Salloo
had told him one night in camp about the orang. The Malay had said
it was the king of the New Guinea forests, fearing no man, beast or
reptile, and this certainly appeared to be the case in this instance.
Had it wished to beat a retreat to safety, the mias, as the natives
called the red gorilla, might easily have done so. One leap and he
could have grasped a tree trunk, up which he could have scrambled
in a jiffy. On the contrary, after its first backward steps, which
brought it almost out of the water, the creature stood upright and,
uttering savage growls, beat on its hairy chest with its huge arms,
producing a sound like the reverberations of a savage “tom-tom.”
The scaly reptile continued to advance. Perhaps, to its eyes, the red
gorilla was simply a native, a poor weak human being, such as
possibly had fallen victim to the great crocodile before. However that
may have been, the saurian, without undue hurry, could be seen to
be making straight for the red ape and, maneuvering so as to get its
monstrous armor-plated tail in position to give a fatal flail-like
sweep, which would fling the orang-outang into the water, stunning
it and making it an easy prey.
It appeared to flatten itself as it reached shallow water, its ugly
lizard-like legs spread out on each side of its scaly body almost
horizontally. Then, with a suddenness that made the boys catch their
breaths in a quick gasp, the monster gave a sudden leap, aiding this
maneuver by its tail, which it suddenly stiffened as if it had been a
spring.
Its whole length was launched into the air as it sprang, and for a
flash its wide-opened jaws with their hideous rows of triangular
teeth, appeared to engulf the red ape. But while the boys were still
held spell-bound by this spectacle, such a one as perhaps no human
being but a lone native hunter had ever beheld before, the red
gorilla gave a mighty leap. It was partly straight up and partly to one
85. side. As the great jaws of the saurian came together with a snap like
that of a titanic steel trap, the red ape landed fair and square on the
scaled monster’s back.
Straddling the plated hide, the great hairy legs gripped the
crocodile’s sides as a bronco buster grips his fractious mount. And
now commenced a struggle between these two denizens of the
deepest New Guinea forests such as the two young spectators
remembered with photographic vividness to the end of their lives.
On the part of the crocodile the battle was simply a series of leaps
and wild tail threshings in an effort to dislodge his nimble foe. The
grass and weeds were mown down as if by a scythe by the sweeps
of the great tail, but the ape held firm, his little eyes twinkling
wickedly. With one arm it clutched the rough hide firmly, but the
other was waving about like a tentacle seeking something to grasp.
During the struggle the jaws of the crocodile had been frequently
snapped, but they only closed on empty air. As in all the saurian
tribe, during this process the upper jaw had pointed nearly vertically
upward, making an opening big enough to swallow a canoe.
Suddenly the watchers saw the orang’s purpose. All at once the
disengaged arm made a swift sweep forward and grasped the
extended upper jaw.
“Great Scott! he’s done for now,” cried Billy. “That jaw will close and
cut his fingers off.”
“Hold on,” warned Jack. “Watch. I’ve heard these creatures can bend
rifle barrels as if they were made of lead. Perhaps—look!”
The orang suddenly shifted his position. He was now kneeling on the
crocodile’s back, his knees braced firmly on its armor-plated neck
and his second arm aiding the first in the task of keeping those jaws,
once apart, from ever coming together again. Then summoning
every ounce of that strength that has made the orang the most
dreaded of all the forest animals in that part of the world, even the
Bornean tiger owning his supremacy, the red gorilla gave one grand
wrench.
86. There was a tearing sound as of a tree being torn from its roots, and
the alligator’s body writhed and threshed about convulsively. The
great ape sprang free from the scaly monster and with hoarse
laughter that sounded like the merriment of a maniac, it gazed on
the saurian’s struggles. But it was not destined to see the end of
them. In its agony the great crocodile instinctively made for the
water and was soon out of sight, threshing and writhing until a
clump of water-cane hid it from sight.
Then, and not till then, did the orang take its eyes from its
conquered enemy. But when it had seen the last of it, the hairy
creature turned and appeared to be contemplating fresh victory. The
lust of battle was in its wicked little eyes.
“Down, Billy, down with you quick,” warned Jack, pulling his chum
aside in the thicket. “If it comes this way, shoot at once. I wouldn’t
want to come to close quarters with a creature like that. I thought
Salloo was drawing the long bow when he told me about the mias,
as he called it, but he didn’t put it on thick enough.”
“If only we’d had a camera,” was Billy’s regret. But for the next few
moments there were more important things to think about. The
orang stood upright, looking about him in a truculent manner. It
almost appeared as if, now that his battle with the saurian was over,
he had recollected the human figure he had seen not long before,
but had paid little heed to it in his haste to make his evening meal
among the water plants.
In fact, he started shamblingly toward the brake where the boys
were concealed with leveled rifles and fingers on triggers. But the
great creature’s life was spared, for that time at least, for had the
boys fired he must have fallen at the first bullets from the high-
powered rifles. After advancing a few paces, he changed his mind
and, grumbling to himself, he shuffled off and was soon lost in the
gloom of the forest.
“We ought to have shot him, Jack,” muttered Billy as they started
back to camp with what game they bagged.
87. “What, kill a fine old warrior like that without cause? Could you have
done it, Billy?”
“Um—well—er—no, I don’t believe I could,” rejoined his chum. “After
all, that crocodile started the scrap and—and I guess every American
likes a good fighter.”
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