The theory of bounded sets on metric spaces has been studied by many authors with different motivations. For instance, Kubrusly and Willard proved that a metric space ( X , d ) is totally bounded if and only if every sequence in X has a Cauchy subsequence. In 2012, Olela Otafudu investigated total boundedness of the u-injective hull of a totally bounded T₀-ultra-quasi-metric space. He first defined a set to be bounded if it is contained in a double ball and total bounded if it is contained in the union of finite number of τ ( q s ) -open balls. He then proved that total boundedness is preserved by the ultra-quasimetrically injective hull of a T₀-ultra-quasi-metric space (see ( [1], Proposition 5.4.1)).

According to Cobzas ( [2], p. 63), a quasi-pseudometric space ( X , q ) is said to be totally bounded if for each ε > 0 there exists a finite subset M ε = { x 1 , x 2 , x 3 , ⋯ , x k } of X such that X ⊆ ∪ j = 1 k B q s ( x j , ε ) . As it is known, in metric spaces precompactness and total boundedness are equivalent notions, a result that is not true in quasi-metric spaces (see ( [2], Proposition 1.2.21)). In quasi metric spaces, Mukonda and Otafudu have defined a set to be Bourbaki bounded if for each ε > 0 and a nutural number n, there exists a finite subset M ε = { x 1 , x 2 , x 3 , ⋯ , x k } of X such that X ⊆ ∪ j = 1 k B q n ( x j , ε ) .

Morever, our recent work [3] has extended the concept of bornology from metric settings to the framework of quasi-metrics. Naturally, this has led to the speculation of what is the relationship between the bornology of bounded sets and other types of bornologies on quasi-metric spaces. Toachieve this, a careful study of bornologyof bounded sets, bornology of totally bounded sets and bornolgies of bourbaki bounded sets in quasi-pseudometric spaces is required.

In this present work, we intend to generalize some classical bornological results of Garrido and Meroño [4] on classes of bounded sets from metric spaces to the category of quasi-metric spaces. For instance, given a compatible quasi-metric, we intend to give some necessary and sufficient conditions for which a bornology of totally bounded sets and bornology of bourbaki bounded sets coincide with our quasi-metric bornology studied in [5].

This section recalls and introduces the terminology and notation for quasi-metric spaces we will use in the sequel. Further details about theory of asymmetric topology can be found in [2] [6] [7].

Definition 2.1. Let X be a set and let q : X × X → [ 0, ∞ ) be a function mapping into the set [ 0, ∞ ) of the nonnegative reals. Then, q is called a quasi-pseudometric on X if

1) q ( x , x ) = 0 whenever x ∈ X .

2) q ( x , z ) ≤ q ( x , y ) + q ( y , z ) whenever x , y , z ∈ X .

We say q is a T₀-quasi-metric provided that q also satisfies the following condition:

q ( x , y ) = 0 = q ( y , x )   implies   x = y .

If q is a quasi-pseudometric on a set X, then q − 1 : X × X → [ 0, ∞ ) defined by q − 1 ( x , y ) = q ( y , x ) for every x , y ∈ X , often called the conjugate quasi-pseudometric, is also quasi-pseudometric on X. The quasi-pseudometric on a set X such that q = q − 1 is a pseudometric. Note that if ( X , q ) is a quasi-metric space, then q s = max { q , q − 1 } = q ∨ q − 1 is also a metric.

Remark 2.2. [2] Let ( X , q ) be a quasi-pseudometric space. The open ball of radius ε > 0 centred at x ∈ X is the set D q ( x , ε ) = { y ∈ X : q ( x , y ) < ε } . The collection of open balls yields a base for the topology τ ( q ) and it is called the topology induced by q on X. Similarly, the closed ball of radius ε ≥ 0 centred at x ∈ X is the set D q [ x , ε ] = { y ∈ X : q ( x , y ) ≤ ε } . If ( X , q ) is a quasi-pseudometric space, then the pair { D q [ x , r ] ; D q t [ x , s ] } where x ∈ X and r , s ∈ [ 0, ∞ ) is called a double ball. In general, { ( D q ( x i , r i ) ) i ∈ I ; ( D q t ( x i , s i ) ) i ∈ I } , with x i ∈ X and r i , s i ∈ [ 0, ∞ ) , is called the family of double balls.

Note that the set D q ( x , ε ) = { y ∈ X : q ( x , y ) < ε } is a τ ( q t ) -closed set, but not τ ( q ) -closed in general. The following inclusions holds:

D q s ( x , ε ) ⊂ D q ( x , ε )     and     D q s ( x , ε ) ⊂ D q t ( x , ε ) .

Definition 2.3. ( [3], Definition 4.1) Let ( X , q ) be a quasi-pseudometric. An arbitrary subset A is called q-bounded if only if there exists x ∈ X , r > 0 and s > 0 such that A ⊆ D q ( x , r ) ∩ D q − 1 ( x , s ) .

Definition 2.4. Let ( X , q ) be a quasi-pseudometric space and F ⊆ X . We say that F is totally bounded, if for any δ > 0 there exists a finite subset { f 1 , f 2 , ⋯ , f k } of X such that

F ⊆ ∪ i = 1 k     D q ( f i , δ ) .

Definition 2.5. Let ( X , q ) be a quasi-pseudometric space and F ⊆ X . We say that F is q-Bourbaki-bounded, if for any δ > 0 there exists a finite subset { f 1 , f 2 , ⋯ , f k } of X and for some positive integer n such that

F ⊆ ∪ i = 1 k     D q n ( f i , δ ) .

Definition 2.6. A bornology on a set X is a collection B of subsets of X which satisfies the following conditions:

1) B forms a cover of X, i.e. X = ∪ B ;

2) for any B ∈ B , and A ⊆ B , then A ∈ B ;

3) B is stable under finite unions, i.e. if X 1 , X 2 , ⋯ , X n ∈ B , then

∪ i = 1 n     X i ∈ B .

If we take a nonempty set X and a bornology B on X, then the pair ( X , B ) is called a bornological universe. For every nonempty set X, the family B = { B ⊂ X : B   isfinite } is the smallest bornology on X.

Recall from [3] that the bornology of quasi-pseudometric bounded sets is denoted by B q ( X ) . However, in [8], the family of totally bounded subsets and boubark bounded sets their bornologies are denoted by T B q ( X ) and B B q ( X ) respectively. We will compare these bornologies in the next sections.

Let ( X , q ) be a T₀-quasi-metric space. Then ( X , q ) is called bicomplete provided that the metric space ( X , q s ) is complete. A mapping f between two quasi-metric spaces ( X , q ) and ( Y , ρ ) is said to be quasi-isometry if q ( f ( x ) , f ( y ) ) = ρ ( x , y ) for all x , y in X.

A bicompletion of a quasi-metric space ( X , q ) is a bicomplete quasi-metric space ( X ˜ , q ˜ ) in which ( X , q ) can be quasi-isometrically embedded as a τ ( q ˜ s ) -dense subspace.

We recall the concepts of asymmetric norms and semi-Lipschitz functions in quasi-metric spaces.

Definition 2.7. [2] An asymmetric norm on a real vector space X is a function ‖     ⋅     | : X → [ 0, ∞ ) satisfying the conditions:

1) ‖     x     | = ‖     −   x     | = 0 then x = 0 ;

2) ‖     a x     | = a ‖     x     | ;

3) ‖     x + y     | ≤ ‖     x     | + ‖     y     | ,

for all x , y ∈ X and a ≥ 0 . Then the pair ( X , ‖     ⋅     | ) is called an asymmetric normed space.

The conjugate asymmetric norm |     ⋅     ‖ of ‖     ⋅     | and the symmetrized norm ‖   ⋅   ‖ of ‖     ⋅     | are defined respectively by

|     x     ‖ : = ‖     −   x     |     and     ‖   x   ‖ : = max { |     x     ‖ , ‖     x     | }     for   any   x ∈ X .

An asymmetric norm ‖     ⋅     | on X induces a quasi-metric q ∥ ⋅ | on X defined by

q ∥ ⋅ | ( x , y ) = ‖   x − y   ‖       for   any     x , y ∈ X .

If ( X , ‖   ⋅   ‖ ) is a normed lattice space, then the function ‖     x     | : = ‖ x + ‖ with x + = max { x ,0 } is an asymmetric norm on X.

Definition 2.8. Let ( X , q ) be a quasi-metric space and ( Y , ‖     ⋅     | ) be an asymmetric normed space. Then a function φ : ( X , q ) → ( Y , ‖     ⋅     | ) is called k-semi-Lipschitz (or semi-Lipschitz) if there exists k ≥ 0 such that

‖   φ ( x ) − φ ( y )   | ≤ k q ( x , y )     for   all   x , y ∈ X . (1)

A number k satisfying inquality (1) is called semi-Lipschitz constant for φ .

This section is as a result of the distinction that we gave in [3] about the bornologies B q ( X ) and B q s ( X ) . We will investigate further the connection between the bornologies B q s ( X ) , B q ( X ) , T B q ( X ) and B B q ( X ) .

Lemma 3.1. If ( X , q ) is a quasi-metric space. Then the following statement is true:

B q s ( X ) ⊆ B q ( X ) (2)

and the quasi-metric bornologies B q ( X ) and B q t ( X ) are equivalent.

Proof. Let A ∈ B q s ( X ) , then A is q s -bounded. By Remark 2.2, A is q-bounded too. Thus A ∈ B q ( X ) . The equivalence of B q ( X ) and B q t ( X ) comes from the fact that any subset A of X is q-bounded if and only if it is q t -bounded. □

The converse of Lemma 3.1 above does not holds. i.e., a set on a quasi-metric can be q-bounded but not q s -bounded (check ( [3], Remark 4.2)).

Definition 3.2. ( [6], p.85) Let ( X , q ) be a T₀-quasi-metric space. Then ( X , q ) is called joincompact provided that the metric space ( X , q s ) is compact.

Theorem 3.3. (Compare ( [9], Theorem 3.78).) Let ( X , q ) be a T₀-quasi-metric space. A set B ⊆ X is joincompact if and only if B is both bicomplete and totally bounded.

Proof. We leave this proof to the reader. □

We rephrase the above theorem in the following Corrolary as proved by Fletcher and Lindgreen in quasi-uniform spaces (see ( [7], p. 65)).

Corollary 3.4. ( [7], Proposition 3.36) Let ( X , q ) be a T₀-quasi-metric space. Then ( X , q ) is totally bounded if and only ( X ˜ , q ˜ s ) is compact.

Definition 3.5. ( [2], Definition 1.44) Let ( X , q ) be a T₀-quasi-metric space. Then ( X , q ) is called supseparable provided that the metric space ( X , q s ) is separable.

Proposition 3.6. (Compare ( [9], Proposition 3.72)) A totally bounded quasi-pseudometric space ( X , q ) is supseparable.

Proof. Suppose ( X , q ) is totally bounded, for any positive interge n, we can

find a finite set A n ⊆ X such that for all x ∈ X , q s ( x , A n ) < 1 n . Now let

B = ∪ n ∈ ℕ A n . The set B is either finite or infinitely countable, thus countable. To show the τ ( q s ) -density of B, let us pick x ∈ X , then we have

q s ( x , B ) ≤ q s ( x , A n ) < 1 n implying that q s ( x , B ) = 0 and x ∈ cl τ ( q s ) ( B ) . This

proves that x is a q s -limit point of B and hence B is a τ ( q s ) -dense subset of X. Consequently, ( X , q s ) separable and by Definition 3.5, ( X , q ) is supseparable. □

The next example shows that for finite dimension spaces, total boundedness coincide with boundedness.

Example 3.7. If we equip a real unit interval X = [ 0,1 ] with the T₀-quasi-metric q ( x , y ) = max { x − y ,0 } , then the pair ( X , q ) is both q-bounded and totally bounded space.

Proof. It can be seen that X is q-bounded. Now If we pick { 0,1 } to be a finite subset of X = [ 0,1 ] and ε = 1 / 2 , then

X ⊂ B q s ( 0, 1 / 2 ) ∪ B q s ( 1, 1 / 2 ) . □

The next Lemma proves that for infinite dimension spaces, total boundedness and quasi-metric boundedness are two different notions.

Lemma 3.8. Let ( X , q ) be a quasi-metric space, then T B q ( X ) ⊆ B q ( X ) .

Proof. Let B ∈ T B q ( X ) . For ε > 0 , there exists a finite subset

F ε = { x 1 , x 2 , x 3 , ⋯ , x k } of B such that B ⊆ ∪ j = 1 k B q s ( x j , ε ) . The set B is a finite family of q s -bounded subsets thus its is q s -bounded. Hence B ∈ B q ( X ) by Lemma 3.1. □

The following example illustrates the converse of Lemma 3.8 above.

Example 3.9. Let us equip the set of natural numbers ℕ with the T₀-quasi-metric

q ( x , y ) = { x − y if   x ≥ y 1 if   x < y

The T₀-quasi-metric space ( ℕ , q ) is q-bounded but not q-totally bounded.

Proof. For all x , y ∈ ℕ we can find k ≥ 0 such that q ( x , y ) ≤ k . But any finite set { x 1 , x 2 , x 3 , ⋯ , x n } ⊂ ℕ with the discrete metric q s , the set ℕ can not be covered by D q s ( x i , ε ) for 1 ≤ i ≤ n . Hence, ( ℕ , q ) is not q-totally bounded. □

It is important to note that T B q s ( X ) is a metric bornology in the sense of Beer et al. [10].

Definition 3.10. Let ( X , q ) be a quasi-pseudometric space and δ > 0 . For any ∅ ≠ F ⊂ X , we define the δ -enlargement D q ( F , δ ) of F by

D q ( F , δ ) : = { x ∈ X : dist ( F , x ) < δ } = ∪ f ∈ F     D q ( f , x )

and

D q t ( F , δ ) : = { x ∈ X : dist t ( F , x ) < δ } = ∪ f ∈ F     D q t ( f , x ) .

Furthermore,

D q s ( F , δ ) = max { D q ( F , δ ) , D q t ( F , δ ) } = ∪ f ∈ F     D q s ( f , x ) .

Remark 3.11. For a given quasi-pseudometric space ( X , q ) . For any δ > 0 and x , y ∈ X . It is easy to see that if ( x i ) i = 0 n is a δ -chain in ( X , q s ) of length n from x to y, then ( x i ) i = 0 n is also a δ -chain in ( X , q ) and in ( X , q t ) of length n from x to y. We have

D q s n ( x , δ ) ⊆ D q n ( x , δ ) (3)

and

D q s n ( x , δ ) ⊆ D q t n ( x , δ ) . (4)

Lemma 3.12. Let ( X , q ) be a quasi-pseudometric space and for any ε , δ > 0 . We have D q ( D q ( F , ε ) , δ ) ⊂ D q ( F , ε + δ ) .

Lemma 3.13. Let ( X , q ) be a quasi-pseudometric space and δ > 0 . For any x ∈ X and n = 0 , 1 , 2 , ⋯ , we have D q n ( x , δ ) ⊆ D q n + 1 ( x , δ ) .

Corollary 3.14. Let ( X , q ) be a quasi-pseudometric space and δ > 0 . If there exists a δ -chain of length n from x to y in ( X , q t ) , then there exists a δ -chain of length n from y to x in ( X , q ) whenever x , y ∈ X .

Lemma 3.15. If ( X , q ) is a quasi-metric space. Then the following statement is true:

B B q s ( X ) ⊆ B B q ( X ) (5)

and the quasi-metric bornologies B B q ( X ) and B B q t ( X ) are equivalent.

Proof. Let δ > 0 . Suppose that F ∈ B B q s ( X ) . Then there exists a finite set { f 1 , f 2 , ⋯ , f k } ⊂ X such that

F ⊆ ∪ i = 1 k     D q s n ( f i , δ )

for some positive integer n. By inclusion (3) we have F ⊆ ∪ i = 1 k     D q n ( f i , δ ) for some positive integer n. Hence F ∈ B B q ( X ) . Note that Corollary 3.14 confirms the equivalence of B B q ( X ) and B B q t ( X ) . □

The converse of the above lemma does not always hold. Let us determine this from the following example.

Example 3.16. Consider the four point set X = { 1 , 2 , 3 , 4 } . If we equip X with T₀-quasi-metric q defined by the distance matrix

Q = ( 0 1 2 1 1 0 1 2 2 1 0 1 2 1 1 0 )

that is, q ( i , j ) = q i , j whenever i , j ∈ X . The symmetrized metric q s of q is induced by the matrix

Q s = ( 0 1 2 2 1 0 1 2 2 1 0 1 2 2 1 0 )

Let δ = 1 , 5 > 0 . If we consider the sequence ( f i ) i = 0 2 : = ( 4 , 2 , 1 ) . Then we have

q ( f 0 , f 1 ) = ( 4 , 2 ) = 1 = q ( f 1 , f 2 ) = q ( 2 , 1 ) < δ .

Hence the sequence ( f i ) i = 0 2 : = ( 4 , 2 , 1 ) is a δ -chain in ( X , q ) of length 2 from 4 to 1. But the same sequence ( f i ) i = 0 2 : = ( 4 , 2 , 1 ) is not a δ -chain in ( X , q s ) of length 2 from 4 to 1 because q s ( f 0 , f 1 ) = q s ( 4 , 2 ) = 2 > δ .

We state the following lemma that we will use in our next proposition.

Lemma 3.17. Let ( X , q ) be a quasi-pseudometric space. For some positive integer n, δ > 0 and x ∈ X , we have

∪ i = 1 k     D q n ( x i , δ ) ⊆ ∪ i = 1 k     D q ( x i , n δ ) .

Proof. Let y ∈ ∪ i = 1 k     D q n ( x i , δ ) , then for some j with 1 ≤ j ≤ k , y ∈ D q n ( x j , δ ) . Moreover, for some j with 1 ≤ j ≤ k , there exists { f 0 , f 1 , ⋯ , f n } a δ -chain of length n from x j to y such that f 0 = x j , f n = y and q ( f i − 1 , f i ) < δ for all i with 1 ≤ i ≤ n . Furthermore, we have

q ( x j , y ) = q ( f 0 , f n ) ≤ q ( f 0 , f 1 ) + q ( f 1 , f 2 ) + ⋯ + q ( f n − 1 , f n ) < δ + δ + ⋯ + δ < n δ .

Thus, for some j with 1 ≤ j ≤ k , y ∈ D q ( x j , n δ ) . Hence, y ∈ ∪ i = i k     D q ( x i , n δ ) . □

Proposition 3.18. Given a quasi-pseudometric space ( X , q ) . If F is a subset of X and δ > 0 , then we have the following conditions:

1) T B q ( X ) ⊆ B B q ( X ) .

2) B B q ( X ) ⊆ B q ( X ) .

Proof.

1) Let δ > 0 . Suppose F ∈ T B q ( X ) then there exists a set { f 1 , f 2 , ⋯ , f k } ⊆ X such that

F ⊆ ∪ i = 1 k     D q s ( f i , δ ) = ∪ i = 1 k     D q s 1 ( f i , δ ) ⊆ ∪ i = 1 k     D q 1 ( f i , δ )

for some positive integer n = 1 . Therefore, F ∈ B B q ( X ) .

2) Since F ∈ B B q ( X ) there exists a set { x 1 , x 2 , ⋯ , x k } ⊆ X and some positive integer n such that for δ > 0 we have F ⊆ ∪ i = 1 k     D q n ( x i , δ ) . By Lemma 3.17, F ⊆ ∪ i = 1 k     D q n ( x i , δ ) ⊆ ∪ i = 1 k     D q ( x i , n δ ) . Hence, F ∈ B q ( X ) . □

Let us provide the summary of the connections between these bornologies in the following remark.

Remark 3.19. If ( X , q ) is a quasi-pseudometric space, then we have the following inlusions:

T B q ( X ) ⊆ B B q s ( X ) ⊆ B B q ( X ) ⊆ B q ( X )

But if ( X , ‖     ⋅     | ) is an asymmetric normed space, then we have

B B q ∥ ⋅ | ( X ) = B q ∥ ⋅ | ( X ) .

We have provided the proof in Proposition 4.1.

One would still wonder, if is it indeed posible to find a quasi-metric metric q ′ equivalent to q such that B q ′ ( X ) = B B q ( X ) or B B q ′ ( X ) = T B q ( X ) .

Proposition 4.1. Suppose that ( X , ‖     ⋅     | ) is an asymmetric normed space. Then we have the following:

B B q ∥ ⋅ | ( X ) = B q ∥ ⋅ | ( X ) .

Proof. For B B q ∥ ⋅ | ( X ) ⊆ B q ∥ ⋅ | ( X ) follows from Proposition 3.18 (b).

For B B q ∥ ⋅ | ( X ) ⊇ B q ∥ ⋅ | ( X ) , suppose that F is q ∥ ⋅ | -bounded then F ⊆ D q ∥ ⋅ | ( x 0 , ε ) for some x 0 ∈ X and ε > 0 . For any δ > 0 , there exists

n ∈ ℕ such that ε n < δ .

Let f ∈ F . We define z i : = x 0 + i n ( f − x 0 ) whenever i with 1 ≤ i ≤ n and z 0 = x 0 . Then

‖     q ∥ ⋅ | ( z i − 1 , z i )     | = ‖     z i − 1 − z i     | = ‖   [ x 0 + i − 1 n ( f − x 0 ) ] − [ x 0 + i n ( f − x 0 ) ]   | = ‖ x 0 n − f n | = ‖ 1 n ( x 0 − f ) | < ε n < δ .

Thus, for any f ∈ F we have obtained a δ -chain of length n on ( X , q ∥ ⋅ | ) from z 0 to f. Therefore, f ∈ ∪ k = 0 n     D q ∥ ⋅ | 1 ( z k , δ ) . □

Definition 4.2. [11] Given a Hilbert cube H = [ 0,1 ] ℕ , the product topology is defined in a usual way by a quasi-pseudometric

ρ q ( x , y ) = ∑ n = 1 ∞ u ( x n , y n ) 2 n

where u ( x n , y n ) = max { x n − y n , 0 } .

Theorem 4.3. ( [11], Theorem 3.10) Every supseparable quasi-metric space is embeddable as subspace of the Hilbert cube H = [ 0,1 ] ℕ .

Theorem 4.4 (Tychonoff’s Theorem). The topological product of a family of compact spaces is compact.

Theorem 4.5. (Compare ( [10], Theorem 3.1).) Let ( X , q ) be a quasi-metric space and let x 0 ∈ X . The following conditions are equivalent:

1) There exists an equivalent quasi-metric ρ such that B q ( X ) = T B ρ ( X ) .

2) The quasi-metric space ( X , q ) is supseparable.

3) There is an embedding Φ of X into some quasi-metrizable space Y such that the family { cl τ ( q Y s ) [ Φ ( C q ( x 0 , n ) ∩ C q − 1 ( x 0 , s ) ) ] : n , s ∈ ℕ } is cofinal in K 0 ( Y ) .

4) There exists an equivalent quasi-metric ρ with B q ( X ) = T B ρ ( X ) = B ρ ( X ) .

Proof.

1 ⇒ 2: If there exists an equivalent quasi-metric space ρ such that B q ( X ) = T B ρ ( X ) , then X = ∪ i = 1 n     B i where B i are ρ -totally bounded subsets. This means that X is a countable union of ρ -totally bounded sets, thus its ρ -totally bounded and by Proposition 3.6, the quasi-metric space ( X , q ) is supseparable.

2 ⇒ 3: First case: If q is bounded, then by Theorem 4.3, we can find an embed

ding Φ : ( X , q ) → ( [ 0,1 ] ℕ , ρ q ) . Let Y = cl τ ( ρ q s ) ( Φ ( X ) ) and choose n , s ∈ ℕ so that Y = cl τ ( ρ q s ) [ Φ ( C q ( x 0 , n ) ∩ C q − 1 ( x 0 , s ) ) ] . Since [ 0,1 ] ℕ is joincompact with

respect to product topology, its subset Y is joincompact and confinal in K 0 ( Y ) .

Second case: If q is unbounded, consider { x i : i ∈ ℕ } as a τ ( q s ) -dense subset in X. For each i in ℕ , Let us define f i : X ⇒ ℝ by f i ( x ) = q ( x , x i ) . Now if A is a nonempty τ ( q s ) -closed subset of X and x ∉ A then we can choose x i with q s ( x , x i ) < q s ( A , x i ) and f i ( x ) ∉ cl τ ( q ) ( f i ( C ) ) . From the choice of x i , the set { f i : i ∈ ℕ } separates points from τ ( q s ) -closed sets and we can define an embedding Φ : X → ℝ ℕ by Φ ( x ) = { f i ( x ) } i = 1 ∞ equipped with the product topology.

Now let p be a quasi-metric compatible with the product topology on ℝ ℕ , we

now prove that Y = cl τ ( p s ) Φ ( X ) ⊆ ℝ ℕ equipped with the relative topology is

cofinal in K 0 ( Y ) . If n ∈ ℕ is chosen arbitrary then for each i ∈ ℕ ,

f i ( C q ( x 0 , n ) ∩ C q − 1 ( x 0 , s ) ) is q-bounded, so by the Theorem 4.4,

Y = cl τ ( p s ) [ Φ ( C q ( x 0 , n ) ∩ C q − 1 ( x 0 , s ) ) ] is joincompact as it is contained in a product ℝ ℕ . Suppose Y = cl τ ( p s ) [ Φ ( C q ( x 0 , n ) ∩ C q − 1 ( x 0 , s ) ) ] is not confinal

in K 0 ( Y ) . Let B ∈ K 0 ( Y ) \ Y then for each n ∈ ℕ , take y n ∈ B and pick

x n ∈ X with q ( x n , x 0 ) > n and p s ( y n ; Φ ( x n ) ) < 1 n .

By the joincompactness of B and the quasi-metrizability of Y, we can find some q s -subsequence { y n k } k = 1 ∞ of { y n } n = 1 ∞ such that p s ( y n k , y 0 ) = 0 . This implies that q ( Φ ( x n k ) , y 0 ) = 0 . But this is not possible, since q is unbounded.

3 ⇒ 4: If ( X , ρ ) is an quasi-metric equivalent to q then B q ( X ) = B ρ ( X ) by ( [3], Theorem 5.4). To prove that B q ( X ) = T B q ( X ) , let B ∈ T B ρ ( X ) and

( Y , ρ ˜ ) be a bicompletion of ρ . Since ρ ˜ is bicomplete by, the set cl τ ( ρ ˜ s ) ( B )

is compact. Given the cofinality of K 0 ( Y ) , let us choose n ∈ ℕ with cl τ ( ρ ˜ s ) ( B ) ⊆ cl τ ( ρ ˜ s ) ( C q ( x 0 , n ) ∩ C q − 1 ( x 0 , s ) ) . But this means that

B ⊆ cl q s ( B ) ⊆ cl q s ( C q ( x 0 , n ) ∩ C q − 1 ( x 0 , s ) ) = C q ( x 0 , n ) ∩ C q − 1 ( x 0 , s ) .

Thus B ∈ B q ( X ) and it follows that T B ρ ( X ) ⊆ B q ( X ) . For the reverse inclusion. If B ∈ B q ( X ) , we can choose n ∈ ℕ with B ⊆ C q ( x 0 , n ) ∩ C q − 1 ( x 0 , s ) .

The B ⊆ cl τ ( ρ ˜ s ) ( C q ( x 0 , n ) ∩ C q − 1 ( x 0 , s ) ) is compact and ρ ˜ -totally bounded.

Therefore, B ∈ T B ρ ( X ) . The equivalence 4 ⇒ 1 follows from ( [3], Theorem 5.4). □

Definition 4.6. (Compare ( [10], Definition 3)). Let ( X , q ) be a T₀-quasi-metric space. Given the point p ∉ X and a quasi-metric bornology B q ( X ) on X we can form the one-point extension of X associated with B q ( X ) by a X ′ = X ∪ { p } .

If τ ( q ) is the topology X, then the corresponding topology on X ′ is defined by

τ ( q ) ∪ { { p } ∪ X \ B : B = cl τ ( q ) ( B ) ∈ B q ( X ) } .

The quasi-metric bornology associated with X ′ is denoted by B q ( X ′ ) .

Remark 4.7. If B 0 is a τ ( q ) -closed base of the bornology then { { p } ∪ X \ B : B ∈ B 0 } forms a τ ( q ) -neighbourhood base at the point p.

Lemma 4.8. Let ( X , q ) be a T₀-quasi-metric space. If the bornology B q ( X ) is quasi-metrizable then the associated bornolgy B q ( X ′ ) on X ′ is quasi-metrizable.

Theorem 4.9. (Compare ( [10], Theorem 3.4)) Let ( X , q ) be a quasi-metric space The following conditions are equivalent:

1) T B q ( X ) has a countable base;

2) There exists an equivalent quasi-metric q ′ such that T B q ( X ) = B q ′ ( X )

3) The one-point extension of X associated with T B q ( X ) is quasi-metrizable.

4) The one-point extension of X associated with T B q ( X ) has a τ ( q ) -neighborhood base at the ideal point.

Proof.

1 ⇒ 2: Since T B q ( X ) has a countable base by Hu’s theorem (see ( [3], Theorem 4.18)) there exists an equivalent quasi-metric q ′ such that T B q ( X ) = B q ′ ( X ) .

2 ⇒ 3: By (2), T B q ( X ) = B q ′ ( X ) . From Lemma 4.8 B q ( X ′ ) on X ′ is quasi-metrizable thus T B q ( X ′ ) is quasi-metrizable.

3 ⇒ 4 Since the bornology T B q ( X ′ ) has a τ ( q s ) -closed base, thus by the Remark 4.7 T B q ( X ′ ) has a τ ( q s ) -neighborhood base at the ideal point.

4 ⇒ 1: If T B q ( X ′ ) have a τ ( q s ) -neighborhood base at each point, then T B q ( X ) has countable base. □

Definition 4.10. Let ( X , q ) be a quasi-metric space and ( Y , ‖     ⋅     | ) be an asymmetric normed space. A function φ : ( X , q ) → ( Y , ‖     ⋅     | ) is called semi-Lipschitz in the small if there exists δ > 0 and k ≥ 0 such that if q ( x , y ) < δ then ‖   φ ( x ) − φ ( y )   | ≤ k q ( x , y ) .

The following lemma follows directly from the definitions of semi-Lipschitz in the small function and uniformly continuous.

Lemma 4.11. Let ( X , q ) be a quasi-metric space and ( Y , ‖     ⋯     | ) be an asymmetric normed space. If a function φ : ( X , q ) → ( Y , ‖     ⋅     | ) is semi-Lipschitz in the small, then φ : ( X , q ) → ( Y , ‖     ⋅     | ) is uniformly continuous.

Theorem 4.12. (Compare ( [12], Theorem 3.4)) Let ( X , q ) be a quasi-metric space and ∅ ≠ F ⊆ X . Then the following conditions are equivalent:

1) F ∈ B B q ( X ) ;

2) if ( Y , ‖     ⋅     | ) is an asymmetric normed space and φ : ( X , q ) → ( Y , ‖     ⋅     | ) is uniformly continuous, then φ ( F ) ∈ B q ∥ ⋅ | ( Y ) ;

3) if ( Y , ‖     ⋅     | ) is an asymmetric normed space and φ : ( X , q ) → ( Y , ‖     ⋅     | ) is semi-Lipschitz in the small function, then φ ( F ) ∈ B q ∥ ⋅ | ( Y ) ;

4) if φ : ( X , q ) → ( ℝ , u ) is semi-Lipschitz in the small function, then φ ( F ) ∈ B u ( ℝ ) .

Proof.

(1) ⇒ (2) If φ : ( X , q ) → ( Y , ‖     ⋅     | ) is uniformly continuous then there exists δ > 0 such that whenever x , y ∈ X with q ( x , y ) < δ , we have

q ∥ ⋅ | ( φ ( x ) , φ ( y ) ) = ‖   φ ( x ) − φ ( y )   | < 1. (6)

By the q-Bourbaki-boundedness of F, there exists A : = { a 1 , a 2 , ⋯ , a m } ⊆ X such that

F ⊆ ∪ i = 1 m     D q n ( a i , δ )

for some positive integer n. If we take f artbitrary in F, then there exists k with 1 ≤ k ≤ m such that f ∈ D q n ( a k , δ ) . Then for some k with 1 ≤ k ≤ m , there exists a δ -chain { f 0 , f 1 , ⋯ , f n } with f 0 = a k , f n = f and

q ( f i − 1 , f i ) < δ     whenever   i   with     1 ≤ i ≤ m . (7)

It follows from the uniform continuity of φ and inequality (6) that

q ∥ ⋅ | ( φ ( f i − 1 ) , φ ( f i ) ) < 1     whenever   i     with     1 ≤ i ≤ m . (8)

Hence, for some k with 1 ≤ k ≤ m , we have

q ∥ . | ( φ ( a k ) , φ ( f ) ) = q ∥ . | ( f 0 , f n ) ≤ q ∥ . | ( f 0 , f 1 ) + q ∥ . | ( f 1 , f 2 ) + ⋯ + q ∥ . | ( f n − 1 , f n ) < n .

Thus, φ ( f ) ∈ ∪ i = 1 m     D q ∥ . | ( φ ( a i ) , n ) for any f ∈ F and φ ( F ) ⊆ D q ( φ ( A ) , n ) . Therefore, φ ( F ) is q ∥ . | -bounded.

(2) ⇒ (3) Follows from Lemma 4.11.

(3) ⇒ (4) Follows directly by replacing ( Y , ‖     ⋅     | ) with ( ℝ , u ) in (3).

(4) ⇒ (1). Suppose that F is not q-Bourbaki-bounded. Then there exists a δ > 0 such that if { f 1 , f 2 , ⋯ , f k } ⊆ X and a positive integer n, we have F ⊆ ∪ i = 1 k     D q n ( f i , δ ) . We have two cases on the structure of F.

Case 1: If f ∈ F , then there exists a positive integer n such that for all j ∈ ℕ

F ∩ D q n ( f , δ ) = F ∩ f ≍ δ .

Let f 1 be an arbitrary point of F. We choose a positive integer n 1 such that

F ∩ D q n 1 ( f 1 , δ ) = F ∩ f 1 ≍ δ .

Since F is not q-Bourbaki-bounded, there exists f 2 ∈ F such that f 2 ∉ D q n 1 ( f 1 , δ ) . It follows that f 1 ≍ δ ≠ f 2 ≍ δ by the choice of n 1 .

One chooses another n 2 ∈ ℤ + such that n 2 > n 1 and F ∩ D q n 2 ( f 2 , δ ) = F ∩ f 2 ≍ δ . Moreover, since F ⊆ ∪ j = 1 2     D q n 2 ( f j , δ ) , we can find f 3 ∈ F \ ( f 3 ≍ δ ∪ f 2 ≍ δ ) . Continuing this procedure by induction, we can find a sequence ( f j ) with distinct terms in F such that for any i ≠ j we have f i ≍ δ ≠ f j ≍ δ . Therefore, we define a function : ( X , q ) → ( ℝ , u ) by

φ ( x ) = { j if     x ≍ δ f j   for   some   j 0 otherwise .

It follows that the function φ is constant on D q ( x , δ ) and it is unbounded on F since φ ( f j ) = j . Therefore, the function φ is semi-Lipschitz in the small function.

Case 2: If there exists f ∈ F and for all positive integer n, there exists j ∈ ℕ such that

F ∩ D q n ( f , δ ) ⊂ F ∩ D q n + j ( f , δ ) .

For x ≍ δ f , let n ( x ) be the smallest positive integer n such that

x ∈ F ∩ D q n ( f , δ ) . (9)

We then define the function φ : ( X , q ) → ( ℝ , u ) by

φ ( x ) = { ( n ( x ) − 1 ) δ + dist q ( x , D q n ( x ) − 1 ( f , δ ) ) if   x ≠ f   and   x ≍ δ f 0 otherwise .

By definition, the function φ is unbounded on F. We now have to show that if x ≠ y and q ( x , y ) < δ , then for k = 2

u ( φ ( x ) , φ ( y ) ) ≤ k q ( x , y ) .

If either x or y is not related to f with respect to ≍ δ , then since x ≠ y , both x and y are not related to f with respect to ≍ δ and

u ( φ ( x ) , φ ( y ) ) = 0 < 2 q ( x , y ) .

If x ≍ δ f and y ≍ δ f , then we have some cases on n ( x ) and n ( y ) :

If n ( x ) > n ( y ) . Suppose that n ( y ) = 0 then y = f and 0 < q ( x , y ) < δ which implies that y ∈ D q ( x , δ ) hence n ( x ) = 1 .

Furthermore,

u ( φ ( x ) , φ ( y ) ) = u [ ( 1 − 1 ) δ + dist q ( x , D q 0 ( f , δ ) ) ,0 ] = dist q ( x , { y } ) = q ( x , y ) < 2 q ( x , y ) .

If n ( y ) ≥ 1 and n ( x ) = n ( y ) , then

u ( φ ( x ) , φ ( y ) ) = max { [ dist q ( x , D q n ( x ) − 1 ( f , δ ) ) − dist q ( y , D q n ( x ) − 1 ( f , δ ) ) ] ,0 } ≤ q ( x , y ) < 2 q ( x , y ) .

If n ( y ) ≥ 1 and n ( x ) > n ( y ) (i.e., n ( x ) = n ( y ) + 1 ) with φ ( x ) ≤ φ ( y ) , then there is nothing to prove since u ( φ ( x ) , φ ( y ) ) = 0 < 2 q ( x , y ) .

If φ ( x ) > φ ( y ) , then

u ( φ ( x ) , φ ( y ) ) = φ ( x ) − φ ( y ) = [ ( n ( x ) − 1 ) δ + dist q ( x , D q n ( x ) − 1 ( f , δ ) ) ]     − [ ( n ( y ) − 1 ) δ + dist q ( y , D q n ( y ) − 1 ( f , δ ) ) ] = ( n ( y ) + 1 − 1 ) δ − ( n ( y ) − 1 ) δ − dist q ( x , D q n ( y ) + 1 − 1 ( f , δ ) )     − [ ( n ( y ) − 1 ) δ − dist q ( y , D q n ( y ) − 1 ( f , δ ) ) ] .

Furthermore,

u ( φ ( x ) , φ ( y ) ) = δ + dist q ( x , D q n ( y ) ( f , δ ) ) − [ ( n ( y ) − 1 ) δ − dist q ( y , D q n ( y ) − 1 ( f , δ ) ) ] ≤ δ + q ( x , y ) + dist q ( y , D q n ( y ) ( f , δ ) ) − dist q ( y , D q n ( y ) − 1 ( f , δ ) ) .

Since n ( w ) is the smallest n such that y ∈ F ∩ D q n ( f , δ ) , it therefore means

dist q ( y , D q n ( y ) ( f , δ ) ) = 0.

Thus, we have

u ( φ ( x ) , φ ( y ) ) ≤ δ + q ( x , y ) − dist q ( y , D q n ( y ) − 1 ( f , δ ) ) . (10)

We claim that,

δ − q ( x , y ) ≤ dist q ( y , D q n ( y ) − 1 ( f , δ ) ) . (11)

Suppose otherwise, i.e., dist q ( y , D q n ( y ) − 1 ( f , δ ) ) < δ − q ( x , y ) , then

dist q ( x , D q n ( y ) − 1 ( f , δ ) ) ≤ q ( x , y ) + dist q ( y , D q n ( y ) − 1 ( f , δ ) ) < q ( x , y ) + δ − q ( x , w ) < δ .

So x ∈ D q n ( y ) − 1 ( f , δ ) which implies that n ( x ) ≤ n ( y ) − 1 + 1 but this is a contradiction since n ( x ) > n ( y ) .

Combining (10) and (11) we have

u ( φ ( x ) , φ ( y ) ) ≤ δ + q ( x , y ) − δ + q ( x , y ) ≤ 2 q ( x , y ) .

Therefore, the proof is complete. □

The authors declare no conflicts of interest regarding the publication of this paper.

Mukonda, D., Matindih, L.K. and Moyo, E. (2022) Some Classes of Bounded Sets in Quasi-Metric Spaces. Advances in Pure Mathematics, 12, 701-714. https://doi.org/10.4236/apm.2022.1212053