Communications on Applied Nonlinear Analysis 

ISSN: 1074-133X 

Vol 32 No. 9s (2025) 

 

199 https://internationalpubls.com 

Quasinormed Cones and Bicompletion Isometries 

 

Manal Yagoub Ahmed Juma  

Department of mathematic, College of Science, Qassim University, Buraidah, Saudi Arabia.  
M.juma@qu.edu.sa 

 

Article History: 

Received: 13-11-2024 

Revised:25-12-2024 

Accepted:09-01-2025 

Abstract:  

Since (X, e_(p_j )) is an extended quasi-metric cone, we demonstrate how any quasi-

norm p_jon an actual cancellative cone X naturally implies an extended quasi-metric 

e_(p_j ) on that cone. We demonstrate that bicompletion respects the structure of a 

quasi-normalized cone under bijective isometries. In fact, we find that isometries are 

not generally injective in this case. In addition, a few situations are shown. 

Key words:  Bicompletion, quasi-cone, calculative, bijective isometry, injective. 

 

1. Introduction and Preliminary Information 

The letters ℝ+, 𝜔, and N will be used for the sets of nonnegative real numbers, nonnegative 

integer numbers, and positive integer numbers, respectively, throughout this essay. The letters 

ℝ+, 𝜔, and N will be used for the sets of nonnegative real numbers, nonnegative integer 

numbers, and positive integer numbers, respectively, throughout this essay. Remind that a 

semigroup (𝑋, ℝ+) is a monoid if its neutral element is 0 [2]. For each 𝑥, 𝑦 𝑖𝑛 𝑋 and 𝑟, 𝑠 𝑖𝑛 ℝ+ 

,the cone (on ℝ+) is defined as a triple ((𝑋, ℝ+,⋅) where (𝑋, +) is an Abelian monoid and ⋅ is 

a function from ℝ+ × 𝑋 to 𝑋. 

       (𝑎) 𝑟 ⋅ (𝑠 ⋅ 𝑥) = (𝑟𝑠) ⋅ 𝑥; 

       (b) 𝑟 ⋅ (𝑥 + 𝑦) = (𝑟 ⋅ 𝑥) + (𝑟 ⋅ 𝑦); 

       (c)  (𝑟 + 𝑠) ⋅ 𝑥 = (𝑟 ⋅ 𝑥) + (𝑠 ⋅ 𝑥); 

       (𝑑)1 ⋅ 𝑥 = 𝑥. 

Every element 𝑥 ∈ 𝑋 that allows an inverse is distinct and is denoted by -x, as is normal.  

When Y is a part of X and +|𝑌 and ⋅|𝑌are the limits of + and  to 𝑌, respectively, the cone 

(𝑌, +|𝑌, ⋅|𝑌)) is said to be a subcone of a cone (𝑋, +,⋅).Assume that (𝑋, +,⋅) is a cone. as long 

as 𝑓𝑗(𝑥 + 𝑦) ≤ 𝑓𝑗(𝑥) + 𝑓𝑗(𝑦) 

Definition (1.1):   The function 𝑓𝑗 ∶ 𝑋 → ℝ  is considered subadditive for any 𝑥, 𝑦 ∈ 𝑋. For any 

x in 𝑋 and 𝑟 𝑖𝑛 , ℝ+, a quasi-norm on a cone (X,+,⋅) is a subadditive function 𝑝𝑗 ∶ 𝑋 → ℝ+ such 

that :  

(a) 𝑥 = 0 if and only if −𝑥 ∈ 𝑋 and 𝑝𝑗(𝑥) = 𝑝𝑗(−𝑥) = 0,and 

(b) 𝑝𝑗(𝑟 ⋅ 𝑥) = 𝑟𝑝𝑗(𝑥). 

Definition (2.1):  A quasi-norm p on 𝑋 is a norm on a cone (𝑋, +,⋅) if it satisfies the following 

needs: 



Communications on Applied Nonlinear Analysis 

ISSN: 1074-133X 

Vol 32 No. 9s (2025) 

 

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𝑝𝑗(𝑥) = 0 if only if 𝑥 = 0. 

Definition (3.1): The cancellative cone (𝑋, ℝ+,⋅)is defined as follows: 𝑥, 𝑦, 𝑧 ∈ 𝑋, 𝑧 + 𝑥 = 𝑧 +

𝑦  implies 𝑥 = 𝑦 for any 𝑥, 𝑦, 𝑧 ∈ 𝑋. 

One relatively direct way of interpreting any linear space X, R+, · as cone consists in the fact 

that the operation · has to be confined to R+ × X. As is well documented, every norm on a 

space X generates a metric on X. We continue this useful result by showing how quasi-metrics 

can naturally arise from quasi-norms on cancellative cones. In this work, we analyze the 

bicompletion of these structures for bijective isometries. This we prove to be a noninjective 

isometry between quasi-normed cones. We note that the standard Sorgenfrey line on R+ can 

be obtained through the extended quasi-metric created by a norm on R+ itself. 

Moreover, we extend the approach to complexity functions, giving a rather remarkable 

example of a space documented in different parts of Theoretical Computer Science (see 

Example 3.2 below). The notion of quasi-metric space is specially introduced and developed 

in [1].  

A quasi-metric X → ℝ + explains if the set X describes a nonnegative real number element in 

the set X. However, this definition must be true for all sets of elements taken from elements set 

x, y, and z, Elements set X: 

(a) 𝑑(𝑥, 𝑦) = 𝑑(𝑦, 𝑥) = 0 if and only if 𝑥 = 𝑦 and 

(b) 𝑑(𝑥, 𝑧) ≤ 𝑑(𝑥, 𝑦) + 𝑑(𝑦, 𝑧): 

We'll also talk about extended quasi-metric. Except for the fact that 𝑑(𝑥, 𝑦) = +∞.is permitted,       

they adhere to the first three axioms.A (n extended) quasi-metric space is a pair (𝑋, 𝑑) in which 

𝑋 is a (nonempty) set and d is a (n extended) quasi-metric on 𝑋. With the family of open d-

balls {𝐵𝑑(𝑥, 𝜌) ∶ 𝑥 ∈ 𝑋,   𝜌 > 0}, as its basis, every extended quasi-metric d on a set 𝑋 yields 

a  𝑇0 topology 𝒯(𝑑)on 𝑋; for all 𝑥 ∈ 𝑋 and 𝜌 > 0, 𝐵𝑑(𝑥, 𝜌) = {𝑦 ∈ 𝑋 ∶ 𝑑(𝑥, 𝑦) < 𝜌} 

𝑑𝑠(𝑥, 𝑦) = max{𝑑(𝑥, 𝑦), 𝑑(𝑦, 𝑥)}, defined on 𝑋 × 𝑋, is a (n extended) metric on 𝑋 if d is a (n 

extended)  quasi-metric on a set X.  

When 𝑑𝑠 is a complete extended metric on a set 𝑋, then d on 𝑋 is a bicomplete extended quasi-

metric. 

2. Producing Extended Quasi Metrics 

An extended quasi-metric d on a cone (𝑋, +,⋅)is considered constant if 𝑑(𝑥 + 𝑧, 𝑦 + 𝑧) =

𝑑(𝑥, 𝑦) and 𝑑(𝜌𝑥, 𝜌𝑦) = 𝜌𝑑(𝑥, 𝑦), as in [3].assuming 𝜌 ∈ℝ+ and 𝑥, 𝑦 𝑎𝑛𝑑 𝑧 ∈ 𝑋. 

Definition (2.1): A pair (𝑋, 𝑑) is considered to be an extended quasi-metric cone if 𝑋 is a cone 

and d is an invariant extended quasi-metric on 𝑋.  

Assume (𝑋, +,⋅) is a cone. For each 𝑥 ∈ 𝑋, the formula 𝑥 + 𝑋 = {𝑥 + 𝑦: 𝑦 ∈ 𝑋} is defined. 

Proposition (2.2): Consider that on the cancellative cone (𝑋, +,⋅). p is a quasi-norm. 𝑒𝑝𝑗
defined 

on 𝑋 × 𝑋 is an invariant extended quasi-metric on  𝑋 If 𝑥 ∈ 𝑋 and 𝑦 ∈ 𝑥 + 𝑋 with 𝑦 = 𝑥 + 𝑎, 



Communications on Applied Nonlinear Analysis 

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and 𝑒𝑝𝑗
(𝑥, 𝑦) = +∞ if 𝑥 ∈ 𝑋 and 𝑦 ∉ 𝑥 + 𝑋, respectively. Consequently, (𝑋, 𝑒𝑝𝑗

) is a 

stretched quasi-metric cone. Also, for every x in X, ρ in ℝ+, and every 𝜇 > −1, the translations 

are 𝑇 (𝑒𝑝𝑗
)--open, and 𝐵𝑒𝑝𝑗

(𝑥, 𝜇 + 1) = 𝛿𝑥 + {𝑦 ∈ 𝑋 ∶ 𝑝𝑗(𝑦) < 𝜌(𝜇 + 1)} 

Proof:  

If 𝑒𝑝𝑗
(𝑥, 𝑥) = 𝑝𝑗(0) = 0, then For any 𝑥 𝑖𝑛 𝑋. Allow 𝑒𝑝𝑗

(𝑥, 𝑦) = 𝑒𝑝𝑗
(𝑦, 𝑥) = 0at this 

point.Hence, 𝑦 = 𝑥 + 𝑎, 𝑥 = 𝑦 + 𝑏, occur when 𝑎, 𝑏 ∈ 𝑋. Given that 𝑎 + 𝑏 = 0 and 𝑋 is 

cancellative, we can deduce that b=-a.Thus, 𝑎 = 0  since 𝑝𝑗(𝑎) = 𝑝𝑗(−𝑎) = 0. Therefore, 𝑥 =

𝑦.  

Furthermore, we show that for each 𝑥, 𝑦, 𝑧 𝑖𝑛 𝑋, 𝑒𝑝𝑗
(𝑥, 𝑧)𝑒𝑝𝑗

(𝑥, 𝑦) + 𝑒𝑝𝑗
(𝑦, 𝑧),Think about just 

the case when 𝑦 ∈ 𝑥 + 𝑋 and 𝑧 ∈ 𝑦 + 𝑋. At that point, 𝑒𝑝𝑗
(𝑥, 𝑦) = 𝑝𝑗(𝑎) and 𝑒𝑝𝑗

(𝑦, 𝑧) =

𝑝𝑗(𝑏). have the properties 𝑦 = 𝑥 + 𝑎, 𝑧 = 𝑦 + 𝑏 for each 𝑎, 𝑏 ∈ 𝑋. Because of this, 𝑧 = 𝑥 +

𝑎 + 𝑏, and so 

𝑒𝑝𝑗
(𝑥, 𝑧) = 𝑝𝑗(𝑎 + 𝑏) ≤ 𝑝𝑗(𝑎) + 𝑝𝑗(𝑏) = 𝑒𝑝𝑗

(𝑥, 𝑦) + 𝑒𝑝𝑗
(𝑦, 𝑧). 

From this, we conclude that 𝑒𝑝𝑗
 is an extended quasi-metric on 𝑋. We then verified that 𝑒𝑝𝑗

 is 

invariant. Let 𝑥, 𝑦, and 𝑧 be part of 𝑋. If 𝑒𝑝𝑗
(𝑥 + 𝑧, 𝑦 + 𝑧) = +∞., then 𝑒𝑝𝑗

(𝑥, 𝑦) = +∞  as 𝑋 

is cancellative. Otherwise, assume that an is such that 𝑒𝑝𝑗
(𝑥 + 𝑧, 𝑦 + 𝑧) = 𝑝𝑗(𝑎). 

If +𝑧 = 𝑥 + 𝑧 + 𝑎  𝑡ℎ𝑒𝑛 𝑦 = 𝑥 + 𝑎, and 𝑒𝑝𝑗
(𝑥, 𝑦) = 𝑝𝑗(𝑎). 

The same as we derive that 𝑒𝑝𝑗
(𝜌𝑥, 𝜌𝑦) = 𝜌𝑒𝑝𝑗

(𝑥, 𝑦).) for all 𝑥, 𝑦 ∈ 𝑋 and 𝜌 ∈ ℝ+.  

Similarly, for any 𝑥, 𝑦 ∈ 𝑋 and 𝜌 ∈ ℝ+, we find that 𝑒𝑝𝑗
(𝜌𝑥, 𝜌𝑦) = 𝜌𝑒𝑝𝑗

(𝑥, 𝑦).  

Finally, recall that for each x in X and 𝜌 𝑖𝑛 ℝ+, 𝑒𝑝𝑗
(0, 𝑥) = 𝑝𝑗(𝑥) .Thus, for any 𝜇 > −1 ,we 

get rB_(e_(p_j ) ) (0,δ)=B_(e_(p_j ) ) (0,ρδ) and 𝐵𝑒𝑝𝑗
(0, 𝜇 + 1) = {𝑥 ∈ 𝑋 ∶ 𝑝𝑗(𝑥) < 𝛿} .It is 

obvious that for each 𝑥 𝑖𝑛 𝑋 and each 𝛿 > 0, 

𝜌𝐵𝑒𝑝𝑗
(𝑥, 𝛿) = 𝜌𝑥 + 𝐵𝑒𝑝𝑗

(0, 𝜌𝛿), 

𝒯 (𝑒𝑝𝑗
) −open is the translation for + and ⋅ as a result. 

Example (2.3): For every x in ℝ+, find a quasi-norm p such that 𝑝𝑗(𝑥) = 0, given the standard 

addition and product on ℝ+.The Alexandr off extended quasi-metric on ℝ+ is the extended 

quasi-metric in this instance, or 𝑒𝑝𝑗
(𝑥, 𝑦) = 0 if 𝑥 ≤ 𝑦 and 𝑒𝑝𝑗

(𝑥, 𝑦) = +∞ otherwise. 

Example (2.4): For each x in ℝ+, 𝑝𝑗(𝑥) = 𝑥, therefore let 𝑝𝑗 ∶ ℝ+ → ℝ+ . 𝑝𝑗 is clearly a norm 

on ℝ+, and if 𝑥 ≤ 𝑦, then 𝑒𝑝𝑗
(𝑥, 𝑦) = +∞ and 𝑦 − 𝑥. The Sorgenfrey topology on ℝ+ is thus 

the topology generated by 𝑒𝑝𝑗
 , since 𝑒𝑝𝑗

 is the Sorgenfrey extended quasi-metric on ℝ+. 



Communications on Applied Nonlinear Analysis 

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Remark (2.5):Because 𝑝𝑗 is a (quasi) norm on a linear space (𝑋, +,⋅), the (extended) 

quasimetric 𝑝𝑗of  Proposition 2.2 is the classical (quasi)metric on 𝑋 that 𝑝𝑗  generates. 

𝑒𝑝𝑗
(𝑥, 𝑦) = 𝑝𝑗(𝑦 − 𝑥), 

for all 𝑥, 𝑦 belonging to 𝑋.Since there is a good solution to the bicompletion problem in the 

context of quasi-metric spaces ([5]), we will focus on quasi-norms defined on cancellative 

cones. However, some instances of spaces that naturally emerge from modeling particular 

processes in Theoretical Computer Science can be viewed as extended cancellative quasi-

metric cones (see Example 3.2) below). Consequently, we propose the following notion. 

Definition (2.6): Quasi-normed cones are pairs (𝑋, 𝑝𝑗) in which 𝑋 is a cancellative cone and 

𝑝𝑗 is a quasi-norm on 𝑋.  

3. The Semi-Normal Cone's Bicompletion 

Remember that 𝑓𝑗 ∶ 𝑋 → 𝑌 is a linear function from a cone  (𝑋, +,⋅) to a cone (𝑌,⊕,⊗). such 

that 𝑓𝑗(𝛼 ⋅ 𝑥 + 𝛽 ⋅ 𝑦) = 𝛼 ⊗ 𝑓𝑗(𝑥) ⊕ 𝛽 ⊗ 𝑓𝑗(𝑦). 

Definition (3.1):The quasi-normed cones (𝑋, 𝑝𝑗) and (𝑌, 𝑞𝑗). are isometric to a linear function 

𝑓𝑗 ∶ 𝑋 → 𝑌. It guarantees  that for every 𝑥 𝑖𝑛 𝑋, 𝑞𝑗 (𝑓𝑗(𝑥)) = 𝑝𝑗(𝑥). The following illustration 

shows that, unlike the quasi-metric case, there are non-injective isometries between quasi-

normed cones. 

Example (3.2): Taking cue from [6]'s applications for program and algorithmic complexity 

analysis, [4] introduces and investigates the idea of the so-called dual complexity space, which 

consists of the pair(𝒞∗, 𝑑𝒞∗), where 

 

𝒞∗ = {𝑓𝑗 ∈ (ℝ+)𝜔 ∶ ∑ (∑ 2−𝑛𝑓𝑗(𝑛)

∞

𝑛=0

< +∞)

𝑗

}, 

and 𝑑𝒞∗ is the quasi-metric on 𝒞∗ given by  

𝑑𝒞∗(𝑓𝑗 , 𝑔𝑗) = ∑ ∑ 2−𝑛

∞

𝑛=0

[(𝑔𝑗(𝑛) − 𝑓𝑗(𝑛)) ⋁0]

𝑗

 . 

There are several 𝑑𝒞∗properties discussed in [4]. Note specifically that 𝑇1 topology is not 

caused by 𝑑𝒞∗ . With the neutral element 𝑓0𝑗
∈ 𝒞∗ given by 𝑓0𝑗

(𝑛) = 0  for all 𝑛 ∈ 𝜔, and ⋅ 

being the operation specified by (𝜆 ⋅ 𝑓𝑗)(𝑛) = 𝜆𝑓𝑗(𝑛) for all 𝑛 ∈ 𝜔, (𝒞∗, +,⋅) is clearly a 

cancellative cone.  

Let's say that  

∑ 𝑝𝑗(𝑓𝑗)

𝑗

= ∑ (∑ 2−𝑛

∞

𝑛=0

𝑓𝑗(𝑛)) 

𝑗

  



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Suppose that 𝑝𝑗 ∶ 𝒞∗ → ℝ+.The fact that p_j is a quasi-norm on 𝒞∗ is well accepted. Next, the 

induced extended quasi-metric 𝑝𝑗 on 𝒞∗ is given by  

∑ 𝑒𝑝𝑗
(𝑓𝑗 , 𝑔𝑗)

∗

𝑗

= ∑ (∑ 2−𝑛 (𝑔𝑗(𝑛) − 𝑓𝑗(𝑛))

∞

𝑛=0

) 

𝑗

  

if 𝑓𝑗 ≤ 𝑔𝑗, and 𝑒𝑝𝑗
(𝑓𝑗 , 𝑔𝑗) = +∞ otherwise. 

Let 𝑋 = {𝑓𝑗 ∈ 𝒞∗ ∶ 𝑓𝑗(0) > 0} ∪ {𝑓𝑗0
}. It is frequently observed that 𝑋 is a subclone of 𝒞∗ 

𝑞(𝑓𝑗) = 𝑓𝑗(0)  is the definition of 𝑓𝑗 ∶ 𝑋 → ℝ+. There is no doubt that 𝑞𝑗 is a quasi-norm on X. 

Let 𝐹(𝑓𝑗)(0) = 𝑓𝑗(0) and 𝐹(𝑓𝑗)(𝑛) = 0 define 𝐹 ∶ 𝑋 → 𝒞∗ for each 𝑓𝑗 ∈ 𝑋and 𝑛 ∈ ℕ.  

F is obviously linear from (𝑋, +,⋅) to (𝒞∗, +,⋅). Moreover, for any 𝑓𝑗𝑖𝑛 𝑋, 

∑ 𝑝𝑗 ((𝐹(𝑓𝑗))

𝑗

= ∑ (∑ 2−𝑛𝐹 (𝑓𝑗(𝑛))

∞

𝑛=0

) 

𝑗

 = ∑ 𝑓𝑗(0)

𝑗 𝑗

= ∑ 𝑞𝑗(𝑓𝑗)

𝑗

 

F, thus, is an isometry between (𝑋, 𝑞𝑗  ) and (𝒞∗, 𝑝𝑗). 

But 𝐹 is not injective if 𝑓𝑗 , 𝑔𝑗 ∈ 𝑋 satisfy 𝑓𝑗(0) = 𝑔𝑗(0) and 𝑓𝑗(1) ≠ 𝑔𝑗(1).This is because we 

get 𝐹(𝑓𝑗) = 𝐹(𝑔𝑗).F, thus, is an isometry between (𝑋, 𝑞𝑗  ) and (𝒞∗, 𝑝𝑗). However, if 𝑓𝑗 , 𝑔𝑗 ∈

𝑋 complete 𝑓𝑗(0) = 𝑔𝑗(0) and 𝑓𝑗(1) ≠ 𝑔𝑗(1), then F is not injective. The reason for this is 

that we obtain 𝐹(𝑓𝑗) = 𝐹(𝑔𝑗). 

Definition (3.3): It is contended that two quasi-normed cones (𝑋, 𝑓𝑗) and (𝑌, 𝑞𝑗), are isometric 

if there is a bijective isometry 𝑓𝑗 ∶ 𝑋 → 𝑌 between them. 

Proposition (3.4): The quasi-metric spaces (𝑋, 𝑝𝑗) and (𝑌, 𝑞𝑗) are also isometric to 𝑓𝑗 if a 

(bijective) isometry 𝑓𝑗  isometric to(𝑋, 𝑒𝑝𝑗
) and (𝑌, 𝑒𝑞𝑗

)quasi-normed cones. 

Proof: Let 𝑥, 𝑦 are in 𝑋. 𝑒𝑞𝑗
(𝑓𝑗(𝑥), 𝑓𝑗(𝑦)) = 𝑒𝑝𝑗

(𝑥, 𝑦) = +∞. is equal to e_(q_j). whether 

𝑓𝑗(𝑦) ∈ 𝑓𝑗(𝑥) + 𝑌. For some 𝑧 𝑖𝑛 𝑌, 𝑒𝑞𝑗
(𝑓𝑗(𝑥), 𝑓𝑗(𝑦)) = 𝑞𝑗(𝑧)if 𝑓𝑗(𝑦) = 𝑓𝑗(𝑥) + 𝑧 . Only one 

𝑎 ∈ 𝑋 can be covered by 𝑓𝑗(𝑎) = 𝑧 since 𝑓 is bijective. Because of this 

𝑓𝑗(𝑦) = 𝑓𝑗(𝑥) + 𝑓𝑗(𝑎) = 𝑓𝑗(𝑥 + 𝑎). 

Since 𝑦 = 𝑥 + 𝑎, 𝑒𝑝𝑗
(𝑥, 𝑦) = 𝑝𝑗(𝑎). follows. (𝑋, 𝑒𝑝𝑗

) and (𝑌, 𝑒𝑞𝑗
) are isometrically 

represented by 𝑓, we find. 

Definition (3.5): An extended quasi-metric 𝑒𝑝𝑗
 is bicomplete on 𝑋, and a quasi-normed cone 

(𝑋, 𝑝𝑗)  is bicomplete if it is. 



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Definition (3.6): Consider the quasi-normed cone (𝑋, 𝑝𝑗) .If (𝑋, 𝑝𝑗) .is isometric to a dense 

subspace of (𝑌, 𝑞𝑗)  in the extended metric space (𝑦, (𝑒𝑞𝑗
)

𝑠

), then (𝑋, 𝑝𝑗) is a bicompletion in 

terms of a bicomplete quasi-normed cone (𝑌, 𝑝𝑗). 

One bicompletion of each quasi-normed cone (𝑋, 𝑝𝑗)  is (�̃�, 𝑝𝑗) which is isometric to any 

bicompletion of (𝑋, 𝑝𝑗) .This is what we shall demonstrate. 

(𝑋, 𝑒𝑝𝑗
) is the symbol for the extended quasi-metric space created by (𝑋, 𝑝𝑗).In the extended 

metric space (𝑋, (𝑒𝑝𝑗
)

𝑠

) , �̃� ̃ is the sum of all Cauchy sequences. Consider that for each  

𝜇 > −1, there exists 𝑛0 ∈N such that (𝑒𝑝𝑗
)

𝑠
(𝑥𝑛, 𝑥𝑚) < 𝜇 + 1,and for any 𝑚, 𝑛 ≥ 𝑛0, 

𝑥 𝑚 ∈ 𝑥𝑛 + 𝑋   if  ̃ 𝑥 ∶= (𝑥𝑛)𝑛∈ℕ ∈ �̃� 

R is a relation of �̃� that has the following definition: For each 𝑥 ∶= (𝑥𝑛)𝑛∈ℕ and 𝑦 ∶= (𝑦𝑛)𝑛∈𝑁 

in �̂�, set 𝑅𝑦 ⟺ lim
𝑛→∞

(𝑒𝑝𝑗
)

𝑠
(𝑥𝑛, 𝑦𝑛) = 0 . R is an equivalency relation on X ̂ in this situation. 

Show the quotient of �̂� 𝑅⁄  by �̂�.For any 𝑥 in �̂� therefore,  

�̂� = {[𝑥] ∶ 𝑥 ∈ �̂�},where[𝑥] = {𝑦 ∈ �̂� ∶ 𝑥𝑅𝑦}. 

Insert [𝑥] + [𝑦] = [𝑥 + 𝑦] and 𝑎 ⋅ [𝑥] = [𝑎𝑥] for each 𝑥 ∶= (𝑥𝑛)𝑛∈ℕ 𝑎𝑛𝑑  𝑦 ∶= (𝑦𝑛)𝑛∈ℕ in �̂� 

and every 𝑎 ∈ ℝ+, where +𝑦 = (𝑥𝑛 + 𝑦𝑛)𝑛∈ℕ, and 𝑎𝑥 = (𝑎𝑥𝑛)𝑛∈ℕ. These processes are 

simple to comprehend due to their precise specification. This is the result that comes next. 

Lemma (3.7): Take the case of a quasi-normed cone (𝑋, 𝑝𝑗). A cancellative cone is therefore 

(�̃�, +,⋅) 

Proof:  

(�̃�, +) is a cancellative Abelian semigroup with neutral element [0] ∈ �̃� ̃, as we can easily 

conclude from (𝑋, +,⋅) being a cancellative cone. Furthermore, the previously defined 

operation ⋅, which was defined as ℝ+ × �̃�  to �̃�  , satisfies for any [𝑥], [𝑦] ∈ �̃�and 𝑟, 𝑠 ∈ ℝ+: 

(𝑎) 𝑟 ⋅ (𝑠[𝑥]) = (𝑟𝑠) ⋅ [𝑥]; 

(𝑏) 𝑟 ⋅ ([𝑥] + [𝑦]) = (𝑟 ⋅ [𝑥]) + (𝑟 ⋅ [𝑦]); 

(𝑐) (𝑟 + 𝑠) ⋅ [𝑥] = (𝑟 ⋅ [𝑥]) + (𝑠 ⋅ [𝑥]); 

(𝑑)1 ⋅ [𝑥] = [𝑥]. 

The proof of the following result on a general monoid context can be found in [99]. 

Lemma (3..8): Assume a quasi-normed cone (𝑋, 𝑝𝑗) and that 𝑥 ∶= (𝑥𝑛)𝑛∈ℕ ∈ �̂� then: 

(𝑎) lim
𝑛→∞

𝑝𝑗(𝑥𝑛) exists and is finite. 

(𝑏) lim
𝑛→∞

𝑝𝑗(𝑥𝑛) = lim
𝑛→∞

𝑝𝑗(𝑦𝑛) for all 𝑦 ∈ [𝑥]. 

We may define a function 𝑝�̃� ∶ �̃� → ℝ+ given the earlier lemma, which is defined as 



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𝑝�̃�([𝑥]) =  lim
𝑛→∞

𝑝𝑗(𝑥𝑛) for all 𝑥 ∈ �̂�. 

We will show how (�̃�, 𝑝�̃�) is a bicomplete quasi-normed cone in the following Lemma 3:9. 

Lemma (3.9): Consider the following two cones (𝑋, +,⋅) and (𝑌,⊕,⊗) If 𝑓𝑗 ∶ 𝐴 → 𝑌 is a linear 

function and A is a sub cone of 𝑋, then 𝑓𝑗(𝐴) is a sub cone of 𝑌. 

Lemma (3.10): A quasi-normed cone (𝑋, 𝑝𝑗) is considered. Then the following statements are 

exact: 

(a) The quasi-normed cone (�̃�, 𝑝𝑗)bicomplete. 

(b) (𝑋, 𝑝𝑗) is isometric to a dense subspace of (�̃�, 𝑝𝑗) in the metric space (�̃�, (𝑒�̃�𝑗
)

𝑠

). 

isometric to a dense subspace of the metric space  

Proof:  

(a) The cancellative condition of (X ̃,p ̃_j) is obtained from Lemma (3.7). 

Let 𝑥 ∶= (𝑥𝑛)𝑛∈ℕ be an element of X ̂ such that – [𝑥] ∈ �̃� and 𝑝𝑗([𝑥]) = 𝑝𝑗(−[𝑥]) = 0. Since 

lim
𝑛→∞

𝑝𝑗(𝑥𝑛) = 0 = lim
𝑛→∞

𝑝𝑗(−𝑥𝑛) follows.  lim
𝑛→∞

(𝑒𝑝𝑗
)

𝑠
(0, 𝑥𝑛) = 0 because, finally, 

𝑒𝑝𝑗
(0, 𝑥𝑛) = 𝑝𝑗(𝑥𝑛)and 𝑒𝑝𝑗

(𝑥𝑛, 0) = 𝑝𝑗(−𝑥𝑛) 

  Consequently, [𝑥] = [0]. 

Lemma (3.7) yields the cancellative condition of (�̃�, 𝑝𝑗). 

Consider an element of �̂�, 𝑥 ∶= (𝑥𝑛)𝑛∈ℕ, such that – [𝑥] ∈ �̃� and 𝑝𝑗([𝑥]) = 𝑝𝑗(−[𝑥]) = 0. As 

a result, lim
𝑛→∞

𝑝𝑗(𝑥𝑛) = 0 = lim
𝑛→∞

𝑝𝑗(−𝑥𝑛). 

       lim
𝑛→∞

(𝑒𝑝𝑗
)

𝑠
(0, 𝑥𝑛) = 0 since 𝑒𝑝𝑗

(0, 𝑥𝑛) = 𝑝𝑗(𝑥𝑛) and 𝑒𝑝𝑗
(𝑥𝑛, 0) = 𝑝𝑗(−𝑥𝑛) at last. 

        As a result, [𝑥] = [0]. 

We have  

       𝑝𝑗(𝑎 ⋅ [𝑥]) = lim
𝑛→∞

𝑝𝑗(𝑎𝑥𝑛) = lim
𝑛→∞

𝑎𝑝𝑗(𝑥𝑛) = 𝑎 lim
𝑛→∞

𝑝𝑗(𝑥𝑛) = 𝑎𝑝𝑗([𝑥]). 

given 𝑥 ∶= (𝑥𝑛)𝑛∈ℕ ∈ �̂� and 𝑎 ∈ ℝ+. 

        Let 𝑥 ∶= (𝑥𝑛)𝑛∈ℕ and 𝑦 ∶= (𝑦𝑛)𝑛∈ℕ be two of �̂�.elements.  

𝑝𝑗(𝑥𝑛 + 𝑦𝑛) ≤ 𝑝𝑗(𝑥𝑛) + 𝑝𝑗(𝑦𝑛) is taken into consideration in order to explain the triangle 

inequality; therefore, 

lim
𝑛→∞

𝑝𝑗(𝑥𝑛 + 𝑦𝑛) ≤ lim
𝑛→∞

𝑝𝑗(𝑥𝑛) + lim
𝑛→∞

𝑝𝑗(𝑦𝑛). 

𝑝𝑗([𝑥] + [𝑦]) ≤ 𝑝𝑗([𝑥]) + 𝑝𝑗([𝑦]), 

for this. So 𝑝�̃� is a quasi-norm for �̃� 



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It well known that the bicompletion of the quasi-metric space (𝑋, 𝑒𝑝𝑗
) is a quasi-metric space 

(𝑋𝑏 , (𝑒𝑝𝑗
)

𝑏
), where 𝑋𝑏 = {[𝑥] ∶ 𝑥 is a Cauchy sequence in the metric space (𝑋, (𝑒𝑝𝑗

)
𝑠

)}, 

(𝑒𝑝𝑗
)

𝑏
([𝑥], [𝑦]) = lim

𝑛→∞
𝑒𝑝𝑗

(𝑥𝑛, 𝑦𝑛) for all [𝑥], [𝑦] ∈ 𝑋𝑏, and for each Cauchy sequence  

𝑥 ∶= (𝑥𝑛)𝑛∈ℕ  𝑖𝑛 (𝑋, (𝑒𝑝𝑗
)

𝑠

), 

[𝑥] = {𝑦 ∶= (𝑦𝑛) ∶ 𝑦 is a Cauchy sequence in (𝑋, (𝑒𝑝𝑗
)

𝑠

)  and lim
𝑛→∞

(𝑒𝑝𝑗
)

𝑠
(𝑥𝑛, 𝑦𝑛) = 0}. 

𝑋𝑏 = �̃� and (𝑒𝑝𝑗
)

𝑏

= 𝑒�̃�𝑗
, are the results. 

(b) The constant sequence 𝑥, 𝑥, … 𝑥, …. for every x in 𝑋 is defined by �̂�. provided that (𝑋𝑏 , (𝑒𝑝𝑗
)

𝑏
)  

is the bicompletion of (𝑋, 𝑒𝑝𝑗
), 𝑖(𝑋) is dense in (�̃�, (𝑒𝑝�̃�

)
𝑠

), where i is the one-to-one function 

from 𝑋 𝑡𝑜 �̃� provided by 𝑖(𝑥)=[ [�̂�] ] for every 𝑥 ∈ 𝑋. 

For each 𝑥 𝑖𝑛 𝑋, note that [�̂�] is the collection of all sequences in 𝑋 that converge to x in the 

metric space (𝑋, (𝑒𝑝𝑗
)

𝑠

).Since it is standard procedure to confirm that 𝑖 is a linear function, 

the preceding Lemma states that 𝑖(𝑋) is a semi linear subspace of �̃�. We may deduce that 

(𝑋, 𝑝𝑗) and (𝑖(𝑋), 𝑝𝑗|
𝑖(𝑋)

) are isometric quasi-normed cones because, for all 𝑥 ∈ 𝑋, 

𝑝𝑗(𝑖(𝑥)) = 𝑝𝑗([�̂�]) = 𝑝𝑗(𝑥). The proof is complete. 

Lemma (3.11): Take a quasi-normed cone (𝑋, 𝑝𝑗) , a bicomplete quasi-normed cone 

(𝑌, 𝑞𝑗),and a sub cone A of 𝑋 to 𝑌. If A is dense in (𝑋, (𝑒𝑝𝑗
)

𝑠

), then 𝑓 is a one-to-one isometry.  

From (𝑋, 𝑝𝑗) to (𝑌, 𝑞𝑗), 𝑓𝑗 then extends uniquely to a one-to-one isometry. 

Proof:A contains a sequence (𝑥𝑛)𝑛∈ℕsuch that, for all 𝑥 ∈ 𝑋\𝐴, lim
𝑛→∞

(𝑒𝑝𝑗
)

𝑠
(𝑥, 𝑥𝑛) = 0. In the 

metric space (𝑋, (𝑒𝑝𝑗
)

𝑠

) , the sequence (𝑥𝑛)𝑛∈ℕ (related to 𝑥 ∈ 𝑋\𝐴 is a Cauchy sequence, 

and for any 𝑚, 𝑛 ≥ 𝑛0, there exists 𝑛0 ∈ ℕ such that (𝑒𝑝𝑗
)

𝑠
(𝑥𝑛, 𝑥𝑚) < 𝜇 + 1.for any value of 

𝜇 > −1. Proposition (3.2.9) states that for any 𝑚, 𝑛 ≥ 𝑛0, (𝑒𝑝𝑗
)

𝑠

(𝑓(𝑥𝑛), 𝑓(𝑥𝑚) < 𝜇 + 1. The 

metric (𝑒𝑞𝑗
)

𝑠

(𝑓(𝑥𝑛), is therefore a Cauchy sequence in the metric space (𝑌, (𝑒𝑞𝑗
)

𝑠

), and 

converges to a point 𝑥∗ ∈ 𝑌. 

Define 𝑓𝑗
∗ ∶ 𝑋 → 𝑌 for each x in A, where 𝑓𝑗

∗(𝑥) = 𝑓𝑗(𝑥) , and for each 𝑥 𝑖𝑛 𝑋\𝐴 ,where 

  𝑓𝑗
∗(𝑥) = 𝑥∗*. It should be noted that the definition of 𝑓𝑗

∗
is not changed by the sequences 

(𝑥𝑛)𝑛∈ℕIn fact, if(𝑥𝑛)𝑛∈ℕ and (𝑦𝑛)𝑛∈ℕ) are sequences in A that converge to a point 𝑥 𝑖𝑛 𝑋\𝐴 

with respect to the metric (𝑒𝑞𝑗
)

𝑠

; additionally, if we designate by 𝑥∗ and 𝑦∗the limit points in 



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(𝑌, (𝑒𝑞𝑗
)

𝑠

) of lim
𝑛→∞

(𝑒𝑞𝑗
)

𝑠

𝑓𝑗(𝑥𝑛), 𝑓𝑗(𝑦𝑛) = 0,since lim
𝑛→∞

(𝑒𝑝𝑗
)

𝑠
(𝑥𝑛, 𝑦𝑛) = 0.Therefore, 𝑥∗ =

𝑦∗. 

We conclude that f extends uniquely to 𝑓∗based on Lemma (3.12) of [7], where 𝑓𝑗
∗
is a one-to-

one function such that 𝑞𝑗 (𝑓𝑗
∗(𝑥)) = 𝑝𝑗(𝑥), (𝑋, 𝑝𝑗) and (𝑌, 𝑞𝑗) are quasi-normed cones. We 

simply show that 𝑓𝑗
∗
 is linear on 𝑋 after that. Consider x and y to be in 𝑋.  

Only the case where 𝑥, 𝑦 ∈ 𝑋\𝐴, which is linear on A, is considered. Assume that the sequences 

(𝑥𝑛)𝑛∈ℕ and (𝑦𝑛)𝑛∈ℕ be in A converge to x and y, respectively, in the metric space (𝑋, (𝑒𝑝𝑗
)

𝑠

). 

Subsequently, with respect to (𝑒𝑞𝑗
)

𝑠

, 𝑓𝑗
∗
converges to (𝑓𝑗(𝑎𝑥𝑛 + 𝑏𝑦𝑛))

𝑛∈ℕ
. That's because, 

with considering (𝑒𝑞𝑗
)

𝑠

, (𝑎𝑥𝑛 + 𝑏𝑦𝑛)𝑛∈𝑁 converges to 𝑎𝑥 + 𝑏𝑦. With respect to (𝑒𝑞𝑗
)

𝑠

, the 

sequence (𝑎𝑓𝑗(𝑥𝑛) + 𝑏𝑓𝑗(𝑦𝑛))
𝑛∈ℕ

 converges to 𝑓𝑗
∗(𝑎𝑥 + 𝑏𝑦)since f is linear on A.On the 

other hand, (𝑓𝑗(𝑥𝑛))
𝑛∈ℕ

 converges to 𝑓𝑗
∗(𝑥) and (𝑓𝑗(𝑦𝑛))

𝑛∈ℕ
 converges to 𝑓𝑗

∗(𝑦) with respect 

to (𝑒𝑞𝑗
)

𝑠

, as per the definition of 𝑓𝑗
∗. For the metric (𝑒𝑞𝑗

)
𝑠

, then((𝑎𝑓𝑗(𝑥𝑛) +

𝑏𝑓𝑗(𝑦𝑛)))
𝑛∈ℕ

converges to a 𝑎𝑓𝑗
∗(𝑥) + 𝑏𝑓𝑗

∗(𝑦).Consequently, 𝑓𝑗
∗(𝑎𝑥 + 𝑏𝑦) = 𝑎𝑓𝑗

∗(𝑥) +

𝑏𝑓𝑗
∗(𝑦). 

Lemma (3.2.12): A quasi-normed cone (𝑋, 𝑝𝑗)  has bicompletions of all kinds that are isometric 

to (�̃�, 𝑝𝑗). 

Proof: A bicompletion (𝑌, 𝑞𝑗) is assumed to exist for (𝑋, 𝑝𝑗). Suppose that i is the one-to-one 

isometry from (𝑋, 𝑝) to (�̃�, 𝑝𝑗) is defined in Lemma (3.2.14). Furthermore, because X is dense 

in the metric space (𝑌, (𝑒q𝑗
)

𝑠

), the prior Lemma leads to the conclusion that f has a unique 

one-to-one isometry extension 𝑓∗ to (𝑌, 𝑞𝑗). 

It remains to be shown that 𝑓𝑗
∗ ∶ 𝑌 → �̃�  is an onto mapping. Let x actually stand for any random 

�̃� point. For a sequence (𝑥𝑛)𝑛∈ℕ  in 𝑋, lim
𝑛→∞

𝑒�̃�𝑗

𝑠 (𝑥, 𝑓𝑗(𝑥𝑛)) = 0, given that 𝑓𝑗(𝑋) is dense in 

(�̃�, (𝑒�̃�𝑗
)

𝑠

). A Cauchy sequence is thus found in (�̃�, (𝑒�̃�𝑗
)

𝑠

), (𝑓𝑗(𝑥𝑛))
𝑛∈ℕ

. In (𝑌, (𝑒q𝑗
)

𝑠

), 

(𝑥𝑛)𝑛∈ℕ is a Cauchy sequence since 𝑓𝑗
∗
is an isometry. For every 𝑦 𝑖𝑛 𝑌,if lim

𝑛→∞
(𝑒q𝑗

)
𝑠

(𝑦, 𝑥𝑛) =

0, lim
𝑛→∞

(𝑒�̃�𝑗
)

𝑠

(𝑓𝑗
∗(𝑦), 𝑓𝑗

∗(𝑥𝑛)) = 0. For 𝑓𝑗
∗(𝑦) = 𝑥. This completes the proof. From the 

above stated lemmas, the following can be deduced immediately. 

Theorem (3.13): For any quasi-normed cone (𝑋, 𝑝𝑗), there is only one bicompletion (up to 

bijective isometry). 



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Corollary (3.2.19): (𝑋, 𝑒𝑝𝑗
) and (𝑌, 𝑒𝑞𝑗

) quasi-metric spaces that are isometric by 𝑓𝑗 if (𝑋, 𝑝𝑗) 

and (𝑌, 𝑞𝑗) are isometric quasi-normed cones by a (bijective) isometry 𝑓𝑗. 

Proof: Suppose that 𝑥, 𝑥 + 2(𝜇 + 1) ∈ 𝑋. 𝑓𝑗(𝑥 + 2(𝜇 + 1)) = 𝑒𝑝𝑗
(𝑥, 𝑥 + 2(𝜇 + 1)) = +∞.  if 

𝑓𝑗(𝑥 + 2(𝜇 + 1)) ∈ 𝑓𝑗(𝑥) + 𝑌. 

For 𝑧 ∈ 𝑌 such that 𝑓𝑗(𝑥 + 2(𝜇 + 1) = 𝑓𝑗(𝑥) + 𝑥 + 𝜇 + 1, otherwise   

𝑒𝑞𝑗
(𝑓𝑗(𝑥), 𝑓𝑗(𝑥 + 2(𝜇 + 1))) = 𝑞𝑗(𝑥 + 𝜇 + 1). 

There is a single 𝑎 ∈ 𝑋 with 𝑓𝑗(𝑎) = 𝑥 + 𝜇 + 1 since 𝑓𝑗 is bijective. Consequently, 

𝑓𝑗(𝑥 + 2(𝜇 + 1)) = 𝑓𝑗(𝑥) + 𝑓𝑗(𝑎) = 𝑓𝑗(𝑥 + 𝑎). Because 𝜇 =
𝑎

2
− 1, 𝑒𝑝𝑗

(𝑥, 𝑥 + 2(𝜇 + 1) =

𝑝𝑗(𝑎). The isometry of (𝑋, 𝑒𝑝𝑗
) and (𝑌, 𝑒𝑞𝑗

) by 𝑓𝑗,is what we deduce. 

 

References 

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Math.,77, Marcel Dekker ,New York. 

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