Ibn Al-Haitham Jour. for Pure & Appl. Sci. 34 (1) 2021 
 

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Generalize Partial Metric Spaces 

Norhan I. Abdullah                                                  Laith K. Shaakir 

Department of Mathematics, College of Computer and Mathematics Sciences, University of Tikrit, 

Tikrit, Iraq 

   esamanorhan221@gmail.com                                       dr.laithkhaleel@tu.edu.iq 

   

 

 

 Abstract 

        The purpose of this research is to introduce a concept of general partial metric spaces as a 

generalization of partial metric space, give some results and properties and find relations 

between the general partial metric space, partial metric spaces and D-metric spaces. 

Keywords: partial metric space, D-metric space, general partial metric space.    

1 Introduction and preliminaries   

    Metric spaces are very important in mathematics introduced and studied by the French 

mathematicians. Many researchers tried to generalized the metric space to different types for 

example, b-metric space defined  by Stefan Czerwik [1], G-metric space defined  by Mustafa 

and Sims [2], 2-metric space defined  by Gahler [3], D*-metric space[4], partial metric space 

defined  by Mathews[5]  and D-metric space defined  by Dhage [11] 

First , give some definitions  and  properties of the partial metric spaces that can  be found in [5-

10] 

              A non-empty set 𝑌 is said to be partial metric space if there exists a function 𝑝: 𝑌2 →
[0, ∞) satisfies the following condition:  

p1 𝛼 = 𝛽 ↔  𝑝(𝛼 , 𝛼) =   𝑝(𝛼 , 𝛽) =  𝑝(𝛽 , 𝛽) 
p2 𝑝(𝛼 , 𝛼) ≤ 𝑝(𝛼 , 𝛽) 
p3 𝑝(𝛼, 𝛽) = 𝑝( 𝛽 , 𝛼) 

p4 𝑝(𝛼, 𝛽) ≤ 𝑝(𝛼 , 𝜇) + 𝑝 (𝜇 , 𝛽 ) – 𝑝(𝜇, 𝜇) 

∀ 𝛼 , 𝛽 and 𝜇 ∈  Y ,where 𝑝 is a partial metric on  𝑌 . 

 A basic example of a partial metric space is (𝑅+, 𝑝), where 𝑝(𝛼, 𝛽) = max{𝛼, 𝛽}  ∀𝛼, 𝛽 ∈ 𝑅+  

Ibn Al Haitham Journal for Pure and Applied Science 

Journal homepage: http://jih.uobaghdad.edu.iq/index.php/j/index 

 

Doi: 10.30526/34.1.2566 

Article history: Received 2, February, 2020, Accepted20, February, 2020, Published in January 2021  

 

mailto:esamanorhan221@gmail.com
mailto:dr.laithkhaleel@tu.edu.iq


  

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 Also, the partial metric space p on  𝑌  generates a 𝑇0 topology 𝜏𝑝on  𝑌 , which has as a base of 

family open balls{𝐵𝑝(𝛼, 𝜖) ∶   𝛼 ∈ 𝑌 , 𝜖 > 0}, where  𝐵𝑝(𝛼, ) = {𝛽 ∈ 𝑌: 𝑝(𝛼, 𝛽) < 𝑝(𝛼, 𝛼) +

 ϵ} ∀ 𝛼 ∈ Y  and 𝜖 > 0. (see [9])   

The sequence {𝛼n} in a partial metric space (𝑌, 𝑝) converge sequence if 
𝑙𝑖𝑚𝑛→∞ 𝑝(𝛼 , 𝛼𝑛) = 𝑝(𝛼, 𝛼)   
The sequence {𝛼𝑛} in a partial metric space (𝑌, 𝑝) is said to be Cauchy sequences if 
lim𝑛,𝑚 →∞ 𝑝(𝛼𝑛, 𝛼𝑚) exists (finite). 
The partial metric space (𝑌, 𝑝) is said to be complete if every cauchy sequence {𝛼𝑛 } convergent 
to a point 𝛼 in Y. 

 A mapping 𝐹: (𝑌, 𝑝)  →  (𝑌 `, 𝑝`) is said to be continuous at 𝛼0∈Y, if for every𝜖 > 0, there 
exists 𝛿 > 0  such that 

F(B𝑝(𝛼0, δ )) ⊆ B𝑝(F𝛼0, ϵ ). (See [10]) 

If p is a partial metric space, then the function 𝑝𝑠: 𝑌2 → 𝑅+ defined by
 
 

 𝑝𝑠(𝛼 , 𝛽) = 2𝑝(𝛼, 𝛽) – 𝑝(𝛼, 𝛼) – 𝑝(𝛽, 𝛽)  

is a metric on Y   (See [6]) 

Not that, the sequence {𝛼𝑛} is a Cauchy sequence in a partial metric space(𝑌, 𝑝) if and only if 
{𝛼n} is a Cauchy sequence in the metric space(𝑌, 𝑝

𝑠). (See [9], [10]) 

 A partial metric space (𝑌, 𝑝) is said complete if and only if the metric space (𝑌, 𝑝𝑠) is 
complete. Furthermore,limn→∞ p

𝑠(𝛼𝑛, 𝛼) = 0 if and only if 

 𝑝(𝛼, 𝛼) = 𝑙𝑖𝑚𝑛→∞𝑝(𝛼𝑛, 𝛼) = 𝑙𝑖𝑚𝑛 ,𝑚 →∞𝑝(𝛼𝑛, 𝛼𝑚). (𝑠𝑒𝑒[9] , [10]) 

Second, we recall definition of D-metric space that can be found in [11], [12], [13] 

        A non-empty set Y is said to be D-metric space if there exists a function 𝐷: 𝑌3 →

 [0, ∞) satisfies the following conditions: 

D1. 𝐷(𝛼, 𝛽, 𝛾) = 0 ⇔ 𝛼 = 𝛽 = 𝛾 
D2. 𝐷(𝛼, 𝛽, 𝛾) = 𝐷(𝛽, 𝛼, 𝛾) = 𝐷(𝛾, 𝛼, 𝛽) = 𝐷 (𝛽, 𝛾, 𝛼) =  …  (𝑆𝛽𝑚𝑚𝑒𝑡𝑟𝛽) 
D3. 𝐷(𝛼, 𝛽, 𝛾) ≤ 𝐷(𝜇 , 𝛽, 𝛾) + 𝐷(𝛼, 𝜇, 𝛾) + 𝐷(𝛼, 𝛽, 𝜇) 

∀α, β, γ 𝑎𝑛𝑑 𝜇 ∈ 𝑌, where D is D-metric on  𝑌 . 

Y if for given𝜖 > 0, A sequence {𝛼𝑛} in D-metric space (𝑌, 𝐷) is said converge to 𝛼∈ there 

exists a positive integer 𝔪0 such that𝐷(𝛼𝑛, 𝛼𝑚, 𝛼 ) < 𝜖  ∀ 𝑚 ≥  𝔪 0, 𝑛 ≥  𝔪0. [13]  

 A sequence {𝛼𝑛} in D-metric space (𝑌, 𝐷) is said Cauchy if for given  𝜖 > 0; there exists an 

positive integer 𝔪0 such that D(𝛼𝑛, 𝛼𝑚, 𝛼𝑙 ) < ϵ  ∀m ≥ 𝔪0, n ≥ 𝔪0, l ≥ 𝔪0.[13] 

 A sequence {𝛼𝑛} in D-metric space (𝑌, 𝐷) is said to be complete if every Cauchy sequence in 
Y converges to a point 𝛼 in Y.  

 A sequence {𝛼𝑛} in D-metric space (𝑌, 𝐷) converges strongly to an element 𝛼 in Y if 

(𝑖)𝐷(𝛼𝑛, 𝛼𝑚, 𝛼) → 0 𝑎𝑠 𝑛, 𝑚 → ∞. (𝑖𝑖) {𝐷(𝛽, 𝛽, 𝛼𝑛 )} converges to 𝐷(𝛽, 𝛽, 𝛼) ∀ 𝛽 ∈ 𝑌. [13] 

 A sequence {𝛼𝑛} in D-metric space (𝑌, 𝐷) is said to be very strongly converges to an element 
𝛼 in Y if  



  

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 (𝑖)𝐷(𝛼𝑛, 𝛼𝑚, 𝛼) → 0  𝑎𝑠 𝑛, 𝑚 → ∞. (𝑖𝑖){𝐷(𝛽, 𝑧 , 𝛼𝑛)} 𝑐𝑜𝑛𝑣𝑒𝑟𝑔𝑒𝑠 𝑡𝑜 𝐷(𝛽, 𝑧, 𝛼)  ∀𝛽, 𝑧 ∈
𝑌 . [13] 

2. Generalize Partial Metric Spaces 

 Definition 1 

              A non-empty set Y is said to be a general partial metric space if there exists a function 

𝐷𝑝: 𝑌
3 → [0, ∞) satisfy the following condition: 

Dp1.  𝐷𝑝(𝛼, 𝛼, 𝛼) = 𝐷𝑝( 𝛼, 𝛽, 𝛾) = 𝐷𝑝(  𝛽, 𝛽, 𝛽) = 𝐷𝑝(𝛾, 𝛾, 𝛾) ⇔ 𝛼 =  𝛽 = 𝛾  

Dp2.  𝐷𝑝(𝛼, 𝛼, 𝛼) ≤ 𝐷𝑝(𝛼, 𝛽, 𝛾) 

Dp3.  𝐷𝑝(𝛼, 𝛽, 𝛾) = 𝐷𝑝( 𝛼, 𝛾, 𝛽) = 𝐷𝑝(𝛽, 𝛼, 𝛾) = 𝐷𝑝( 𝛽, 𝛾, 𝛼) =  … (𝑆𝛽𝑚𝑚𝑒𝑡𝑟𝛽) 

Dp4.  𝐷𝑝(𝛼 , 𝛽, 𝛾) ≤ 𝐷𝑝(𝜇, 𝛽, 𝛾) + 𝐷𝑝(𝛼, 𝜇, 𝛾) + 𝐷𝑝( 𝛼, 𝛽, 𝜇) −  𝐷𝑝(𝜇, 𝜇, 𝜇) 

∀ 𝛼, 𝛽, 𝛾 𝑎𝑛𝑑 𝜇 ∈ 𝑌, where 𝐷𝑝 is general partial metric space on Y.  

So, given an example of a general partial metric space we obtain  

Example 2  

           Let   𝑌 =  [0, ∞) and define a function 𝐷𝑝 on  𝑌
3 by 𝐷𝑝(𝛼, 𝛽, 𝛾)  =  𝑚𝑎𝛼{ 𝛼, 𝛽, 𝛾} 

Then 𝐷𝑝 is a general partial metric space on Y. 

Solution: 

           1) since 𝑚𝑎𝛼{𝛼, 𝛽, 𝛾} =  𝑚𝑎𝛼{𝛼, 𝛼, 𝛼} =  𝑚𝑎𝛼{𝛽, 𝛽, 𝛽} =  𝑚𝑎𝛼 {𝛾, 𝛾, 𝛾} if  and only if  
𝛼 =   𝛽 =  𝛾 then 𝐷𝑝(𝛼, 𝛽, 𝛾) =  𝐷𝑝(𝛼, 𝛼, 𝛼)  =  𝐷𝑝(𝛽, 𝛽, 𝛽)  =  𝐷𝑝(𝛾, 𝛾, 𝛾) if  and only if 

 𝛼 =   𝛽 =  𝛾 

2) 𝐷𝑝(𝛼, 𝛼, 𝛼)  =  𝑚𝑎𝛼{𝛼, 𝛼, 𝛼} = 𝛼 ≤  𝑚𝑎𝛼{𝛼, 𝛽, 𝛾} = 𝐷𝑝(𝛼, 𝛽, 𝛾) . 

3) Trivial 

4) Since 𝑚𝑎𝛼{𝛼, 𝛽, 𝛾}  ≤  𝑚𝑎𝛼{𝜇, 𝛽, 𝛾} +  𝑚𝑎𝛼{𝛼, 𝜇, 𝛾} +  𝑚𝑎𝛼{𝛼, 𝛽, 𝜇}–  𝑚𝑎𝛼{𝜇, 𝜇, 𝜇} 

Then 𝐷𝑝(𝛼, 𝛽, 𝛾) + 𝐷𝑝(𝜇, 𝜇, 𝜇)  ≤ 𝐷𝑝(𝜇, 𝛽, 𝛾) + 𝐷𝑝(𝛼, 𝜇, 𝛾) + 𝐷𝑝(𝛼, 𝛽, 𝜇)                          ⎕ 

Definition 3  

            Let (𝑌, 𝐷𝑝) be a general partial metric space, then 

(1) A sequence {𝛼𝑛} in (𝑌, 𝐷𝑝) converges to a point 𝛼 ∈ 𝑌if  

 limn ,m →∞ D𝑝(𝛼𝑛, 𝛼𝑚, 𝛼) = D𝑝(𝛼, 𝛼, 𝛼) 

(2) A sequence {𝛼𝑛} in (𝑌, 𝐷𝑝) is Cauchy sequence if  

limn ,m ,l →∞ D𝑝(𝛼𝑛, 𝛼𝑚, 𝛼𝑙 ) exists (finite) 

(3) A general partial metric space (𝑌, 𝐷𝑝) is said to be complete if every Cauchy sequence is 

converge to a point 𝛼 in Y. 

(4) A sequence {𝛼𝑛} in (𝑌, 𝐷𝑝) is a strongly converge to 𝛼 if  



  

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i. 𝐷𝑝(𝛼𝑛, 𝛼𝑚, 𝛼)  → 𝐷𝑝(𝛼 , 𝛼 , 𝛼) 𝑎𝑠 𝑛 , 𝑚 → ∞  

ii. 𝐷𝑝(𝛽, 𝛽, 𝛼𝑛) → 𝐷𝑝(𝛽 , 𝛽 , 𝛼) 𝑎𝑠 𝑛 → ∞  ∀𝛽 ∈ 𝑌   

(5) A sequence {𝛼𝑛} in (𝑌, 𝐷𝑝) is a very strongly converge to 𝛼 if  

i. 𝐷𝑝(𝛼𝑛, 𝛼𝑚, 𝛼) →  𝐷𝑝(𝛼 , 𝛼 , 𝛼)  𝑎𝑠 𝑛 , 𝑚 → ∞  

ii. 𝐷𝑝(𝛽 , 𝛾, 𝛼𝑛) → 𝐷𝑝(𝛽 , 𝛾, 𝛼) 𝑎𝑠 𝑛 → ∞  ∀ 𝛽 , 𝛾 ∈ 𝑌  

(6) For 𝛼 ∈ 𝑌 and 𝜖 > 0, the open ball of the general partial metric space with center 𝛼 and 

radius 𝜖 is  𝐵𝐷𝑝 (𝛼, 𝜖) = { 𝛽 ∈ 𝑌 ∶ 𝐷𝑝(𝛼 , 𝛽 , 𝛽) < 𝐷𝑝(𝛼 , 𝛼 , 𝛼) + 𝜖}. 

(7) A mapping 𝐹: ( 𝑌 , 𝐷𝑝)  →  ( 𝑌 ′, 𝐷𝑝′) is said to be continuous at 𝛼 if for each open ball 

𝐵𝐷𝑝 (𝐹(𝛼), 𝜖′) in (𝑌
′, 𝐷𝑝′) there exists a ball  𝐵𝑝(𝛼 , 𝜖) in (𝑌, 𝐷𝑝) such that 

 𝐹(𝐵𝐷𝑝 (𝛼 , 𝜖))  ⊆ 𝐵𝐷𝑝 (𝐹(𝛼), 𝜖′). 

Not that, if {𝛼𝑛} is a converge sequence in(𝑌, 𝐷𝑝) then the converge point is not unique. 

Example4  

            Let  𝑌 = [0, ∞) with 𝐷𝑝(𝛼, 𝛽, 𝛾)  =  𝑚𝑎𝛼{𝛼, 𝛽, 𝛾} then (𝑌, 𝐷𝑝) is a general partial 

metric observe that if the sequence {1 +  
1

𝑛2
 }, 𝛼 ≥ 1 then 𝐷𝑝(𝛼𝑛, 𝛼𝑚, 𝛼) = 𝑙𝑖𝑚𝑛,𝑚→∞𝑚𝑎𝛼{1 +

 
1

𝑛2
 , 1 +  

1

𝑚2
 , 𝛼} = 𝛼 = 𝐷𝑝(𝛼 , 𝛼 , 𝛼). 

Hence, every  𝛼 ∈ [1, ∞) is a convergent point for the sequence {1 +
1

𝑛2
}.  

Thus, the converge point is not unique. 

Theorem 5  

             Every converge sequence in (𝑌, 𝐷𝑝) is a Cauchy sequence. 

 Proof: 

            Let (𝑌, 𝐷𝑝) be a general partial metric space and {𝛼𝑛} is a converge sequence to 𝛼 and 

𝜖 >  0. 

Since {𝛼𝑛} is converge to 𝛼 then there exists 𝑘 ∈ 𝑁 such that 

 |𝐷𝑝(𝛼𝑛, 𝛼𝑚 , 𝛼) –  𝐷𝑝(𝛼 , 𝛼 , 𝛼) | < 𝜖 ∀ 𝑛, 𝑚 > 𝑘 

So that  

𝐷𝑝(𝛼𝑛, 𝛼𝑚, 𝛼𝑙 ) ≤ 𝐷𝑝(𝛼, 𝛼𝑚, 𝛼𝑙 ) + 𝐷𝑝(𝛼𝑛, 𝛼, 𝛼𝑙 ) +  𝐷𝑝(𝛼𝑛, 𝛼𝑚, 𝛼) −  𝐷𝑝(𝛼, 𝛼, 𝛼)   

≤  2𝐷𝑝(𝛼, 𝛼, 𝛼)  + 𝜖           ∀ 𝑛, 𝑚, 𝑙 > 𝐾 

Hence, limn ,m ,l→∞ D𝑝(𝛼𝑛, 𝛼𝑚, 𝛼𝑙 )  exists, thus {𝛼𝑛} is a Cauchy sequence.                            ⎕ 

Remark 6  

       It is clear from the definition that every strong converge sequence is a converge but the 

opposite is not true, as we see in example (4) that a sequence {𝛼𝑛}  =  {1 +  
1

𝑛2
} is a converge to 

2 but not strongly a converge to 2, to see this 



  

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Take 1 <  𝛽 <  2, 𝑙𝑖𝑚𝑛→∞𝐷𝑝(𝛽 , 𝛽 , 𝛼𝑛) =  𝛽 ≠  2 = 𝐷𝑝(𝛽 , 𝛽 , 2) 

Thus, 𝛼𝑛 =  {1 +
1

𝑛2
} is not strongly converging to 2. 

Theorem 7 

       If {𝛼𝑛} is a strongly converge, then the converge point is unique. 

  Proof  

       Let {𝛼𝑛} be a strongly converge to w, 𝔃 

{𝐷𝑝(𝛽 , 𝛽 , 𝛼𝑛)}is real sequence converge to 𝐷𝑝(𝛽, 𝛽, 𝑤) , 𝐷𝑝(𝛽 , 𝛽, 𝓏) since the converge point 
is unique             

∴ 𝐷𝑝(𝛽, 𝛽, 𝑤)  =  𝐷𝑝(𝛽, 𝛽, 𝓏)   ∀ 𝛽   

 Take 𝛽 =  𝓏 then 𝐷𝑝(𝓏 , 𝓏 , 𝑤)  =  𝐷𝑝(𝓏 , 𝓏 , 𝓏)  … 1 

 Take 𝛽 =  𝑤 then 𝐷𝑝(𝑤, 𝑤, 𝑤)  =  𝐷𝑝(𝑤 , 𝑤 , 𝓏)  …  2 

  by defying  𝐷𝑝(𝓏 , 𝓏 , 𝓏)  ≤  𝐷𝑝(𝛽 , 𝛽 , 𝓏)    ∀ 𝛽 

 If 𝛽 =  𝑤 then 𝐷𝑝(𝓏 , 𝓏 , 𝓏)  ≤ 𝐷𝑝(𝑤 , 𝑤 , 𝓏)  =  𝐷𝑝(𝑤 , 𝑤 , 𝑤) 

 ∴  𝐷𝑝(𝓏 , 𝓏 , 𝓏) ≤  𝐷𝑝(𝑤 , 𝑤 , 𝑤)  …  3 

 Also, 𝐷𝑝(𝑤, 𝑤, 𝑤)  ≤  𝐷𝑝(𝛽, 𝛽, 𝑤)   ∀ 𝛽  

  If 𝛽 =  𝓏 then 𝐷𝑝(𝑤 , 𝑤 , 𝑤)  ≤  𝐷𝑝(𝓏 , 𝓏 , 𝑤)  =  𝐷𝑝(𝓏 , 𝓏 , 𝓏) 

 ∴ 𝐷𝑝(𝑤 , 𝑤 , 𝑤)  ≤  𝐷𝑝(𝓏 , 𝓏 , 𝓏)   …   4 

  𝑏𝛽 3, 4; 𝑤𝑒 ℎ𝑎𝑣𝑒  𝐷𝑝(𝑤 , 𝑤 , 𝑤)  = 𝐷𝑝(𝓏 , 𝓏 , 𝓏) 

∴ 𝐷𝑝(𝑤 , 𝑤 , 𝓏)  = 𝐷𝑝(𝑤 , 𝑤 , 𝑤)  = 𝐷𝑝(𝓏 , 𝓏 , 𝓏), 𝑡ℎ𝑢𝑠  𝑤 =  𝓏.                                   ⎕ 

Example 8 

Let  𝑌 =  [0, ∞) and 𝐷𝑝(𝛼, 𝛽 , 𝛾) = 𝑚𝑎𝛼{𝛼, 𝛽, 𝛾} then ( 𝑌 , 𝐷𝑝) is general partial metric space 

as we see in example 2, if 𝛼n= 2
1

𝑛, then the sequence {𝛼𝑛} is a very strongly converge to1. 

Indeed 𝐷𝑝(𝛼𝑛, 𝛼𝑚, 𝛼𝑙 ) =  𝑚𝑎𝛼 {2
1

𝑛, 2
1

𝑚, 1} → 1 =  𝐷𝑝(1, 1, 1) 𝑎𝑠 𝑛, 𝑚 → ∞     

 𝐴𝑙𝑠𝑜 𝐷𝑝(𝛽, 𝛾, 𝛼𝑛 )  =  𝑚𝑎𝛼{𝛽, 𝛾, 2
1

𝑛}  ⟶  𝑚𝑎𝛼{𝛽, 𝛾, 1} 𝑎𝑠 𝑛 → ∞   ∀ 𝛽, 𝛾 ∈   𝑌  

Remark 9  

          It is clear from definition that every very strong converge sequence is a strongly 

converge, so that If {𝛼𝑛} is very strongly converge then the converge point is unique.  

Theorem 10 

        Let (𝑌, 𝐷𝑝) be general partial metric space, if 𝐷𝑝(𝛼, 𝛽 , 𝛾)  =  0 then  𝛼 =   𝛽 =  𝛾  

 Proof  



  

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          We have 𝐷𝑝(𝛼 , 𝛼 , 𝛼 ) ≤ 𝐷𝑝(𝛼 , 𝛽 , 𝛾) = 0  , 𝐷𝑝(𝛽 , 𝛽 , 𝛽) ≤ 𝐷𝑝(𝛼 , 𝛽 , 𝛾) = 0  𝑎𝑛𝑑  

 𝐷𝑝(𝛾 , 𝛾 , 𝛾)  ≤  𝐷𝑝( 𝛼 , 𝛽 , 𝛾) = 0 

  ⇒ 𝐷𝑝( 𝛼 , 𝛼 , 𝛼) =  𝐷𝑝(𝛽, 𝛽, 𝛽) =  𝐷𝑝(𝛾, 𝛾 , 𝛾) = 𝐷𝑝(𝛼, 𝛽, 𝛾) = 0  

 Therefore 𝛼 =  𝛽 = 𝛾.                                                                                                         ⎕   

Remark 11  

             If 𝛼 = 𝛽 = 𝛾, then 𝐷𝑝(𝛼 , 𝛽, 𝛾) may not be zero. 

3. Relations between D-metric, partial metric and general partial metric spaces 

Theorem 12  

   Let ( 𝑌, 𝐷𝑝) be a general partial metric space, then the functions 𝐷
𝑔: 𝑌3 →  [0, ∞) given by 

𝐷 𝑔( 𝛼 , 𝛽 , 𝛾) = 𝐷𝑝(𝛼 , 𝛼 , 𝛽)  +  𝐷𝑝(𝛼 , 𝛼 , 𝛾)  +  𝐷𝑝(𝛽 , 𝛽 , 𝛼)  +  𝐷𝑝(𝛽 , 𝛽 , 𝛾)  +

 𝐷𝑝(𝛾 , 𝛾, 𝛼)  + 𝐷𝑝(𝛾 , 𝛾 , 𝛽) –  2𝐷𝑝( 𝛼 , 𝛼 , 𝛼) –  2𝐷𝑝( 𝛽 , 𝛽 , 𝛽) –  2𝐷𝑝(𝛾 , 𝛾 , 𝛾).              (1.1) 

Is a D-metric space on 𝑌  

 Proof   

         1)  Since 𝐷𝑝(𝛼 , 𝛼 , 𝛽)–  𝐷𝑝(𝛼 , 𝛼 , 𝛼) ≥ 0, 𝐷𝑝(𝛼 , 𝛼 , 𝛾)–  𝐷𝑝(𝛼 , 𝛼, 𝛼) ≥ 0 ,  

𝐷𝑝( 𝛽 , 𝛽 , 𝛼)–  𝐷𝑝( 𝛽 , 𝛽 , 𝛽) ≥ 0 , 

 𝐷𝑝( 𝛽 , 𝛽 , 𝛾)–  𝐷𝑝( 𝛽 , 𝛽 , 𝛽) ≥ 0 , 

 𝐷𝑝(𝛾 , 𝛾 , 𝛼) –  𝐷𝑝(𝛾 , 𝛾 , 𝛾) ≥ 0 𝑎𝑛𝑑 𝐷𝑝(𝛾 , 𝛾 , 𝛽) –  𝐷𝑝(𝛾 , 𝛾 , 𝛾) ≥ 0  𝑠𝑜 𝐷
𝑔( 𝛼 , 𝛽 , 𝛾) ≥ 0. 

 2) 𝐼𝑓 𝐷 𝑔(𝛼 , 𝛽 , 𝛾) =  0  

𝑡ℎ𝑒𝑛 𝐷𝑝( 𝛼 , 𝛼 , 𝛽)  +  𝐷𝑝( 𝛼 , 𝛼 , 𝛾)  +  𝐷𝑝( 𝛽 ,   𝛽 ,   𝛼)  +  𝐷𝑝( 𝛽, 𝛽 , 𝛾)  +  𝐷𝑝(𝛾 , 𝛾 , 𝛼 )   

+  𝐷𝑝(𝛾 , 𝛾  𝛽) –  2𝐷𝑝( 𝛼 , 𝛼 , 𝛼) –  2𝐷𝑝( 𝛽 , 𝛽 , 𝛽) –  2𝐷𝑝(𝛾 , 𝛾  𝛾)  =  0 

 𝑇𝑎𝑘𝑒 𝐷𝑝( 𝛼, 𝛼, 𝛽) + 𝐷𝑝( 𝛼, 𝛼, 𝛾) − 2𝐷𝑝( 𝛼, 𝛼, 𝛼) = 0  

 ⇒  2𝐷𝑝( 𝛼, 𝛼, 𝛼) = 𝐷𝑝( 𝛼, 𝛼, 𝛽) + 𝐷𝑝( 𝛼, 𝛼, 𝛾) … 1 

  𝐷𝑝( 𝛼, 𝛼, 𝛾) + 𝐷𝑝( 𝛽, 𝛽, 𝛼) − 2𝐷𝑝( 𝛼, 𝛼, 𝛼) = 0  

 ⇒  2𝐷𝑝( 𝛼, 𝛼, 𝛼) = 𝐷𝑝( 𝛼, 𝛼, 𝛾) + 𝐷𝑝( 𝛽, 𝛽, 𝛼) … 2 

 𝐹𝑟𝑜𝑚 1&2, 𝑤𝑒 𝑔𝑒𝑡 𝐷𝑝( 𝛼, 𝛼, 𝛽) + 𝐷𝑝( 𝛼, 𝛼, 𝛾) = 𝐷𝑝( 𝛼, 𝛼, 𝛾) + 𝐷𝑝( 𝛽, 𝛽, 𝛼) 

 𝑡ℎ𝑒𝑛 𝐷𝑝( 𝛼, 𝛼, 𝛽) = 𝐷𝑝( 𝛽, 𝛽, 𝛼) … 3 

 𝑆𝑖𝑛𝑐𝑒 𝐷𝑝( 𝛼, 𝛼, 𝛼) = 𝐷𝑝( 𝛼, 𝛼, 𝛽) + 𝐷𝑝( 𝛼, 𝛽, 𝛽) = 𝐷𝑝( 𝛼, 𝛼, 𝛽) + 𝐷𝑝( 𝛼, 𝛼, 𝛽) 

= 2𝐷𝑝( 𝛼, 𝛼, 𝛽) 

  𝐷𝑝(𝛼, 𝛼, 𝛼)  = 𝐷𝑝( 𝛼, 𝛼, 𝛽) … 4 

 𝑁𝑜𝑤, 𝑡𝑎𝑘𝑒 𝐷𝑝( 𝛽, 𝛽, 𝛾) + 𝐷𝑝( 𝛽, 𝛽, 𝛼) − 2𝐷𝑝( 𝛽, 𝛽, 𝛽) = 0  



  

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 ⇒ 2𝐷𝑝( 𝛽, 𝛽, 𝛽) = 𝐷𝑝( 𝛽, 𝛽, 𝛾) + 𝐷𝑝( 𝛽, 𝛽, 𝛼) … 5 

 𝐷𝑝( 𝛼, 𝛼, 𝛽) + 𝐷𝑝( 𝛽, 𝛽, 𝛾) − 2𝐷𝑝( 𝛽, 𝛽, 𝛽) = 0  

 ⇒ 2𝐷𝑝( 𝛽, 𝛽, 𝛽) = 𝐷𝑝( 𝛼, 𝛼, 𝛽) + 𝐷𝑝( 𝛽, 𝛽, 𝛾) … 6 

 𝐹𝑟𝑜𝑚 5&6, 𝑤𝑒 𝑔𝑒𝑡 𝐷𝑝( 𝛽, 𝛽, 𝛼) + 𝐷𝑝( 𝛽, 𝛽, 𝛾) = 𝐷𝑝( 𝛼, 𝛼, 𝛽) + 𝐷𝑝( 𝛽, 𝛽, 𝛾) 

  𝑡ℎ𝑒𝑛 𝐷𝑝(𝛽, 𝛽, 𝛼) = 𝐷𝑝( 𝛼, 𝛼, 𝛽) … 7 

 𝑆𝑖𝑛𝑐𝑒 2𝐷𝑝( 𝛽, 𝛽, 𝛽) = 𝐷𝑝( 𝛽, 𝛽, 𝛼) + 𝐷𝑝( 𝛼, 𝛼, 𝛽) = 𝐷𝑝( 𝛼, 𝛼, 𝛽) + 𝐷𝑝( 𝛼, 𝛼, 𝛽) 

=  2𝐷𝑝(𝛼, 𝛼, 𝛽) 

 ⇒  𝐷𝑝( 𝛽, 𝛽, 𝛽) = 𝐷𝑝( 𝛼, 𝛼, 𝛽) … 8 

 𝐹𝑟𝑜𝑚 4&8, 𝑤𝑒 𝑔𝑒𝑡𝐷𝑝( 𝛼, 𝛼, 𝛼) = 𝐷𝑝( 𝛼, 𝛼, 𝛽) = 𝐷𝑝( 𝛽, 𝛽, 𝛽) 

  𝑠𝑜 𝑏𝑦 𝑑𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛  𝛼 = 𝛽 … 9 

  𝑎𝑛𝑑 𝑡𝑎𝑘𝑒 𝐷𝑝(𝛾, 𝛾, 𝛼) + 𝐷𝑝(𝛾, 𝛾, 𝛽) − 2𝐷𝑝(𝛾, 𝛾, 𝛾) = 0  

 ⇒ 2𝐷𝑝(𝛾, 𝛾, 𝛾) = 𝐷𝑝(𝛾, 𝛾, 𝛼)  + 𝐷𝑝(𝛾, 𝛾, 𝛽) 𝑖𝑓  𝛽 =  𝛼 

 𝑇ℎ𝑒𝑛 2𝐷𝑝(𝛾, 𝛾, 𝛾)  = 𝐷𝑝(𝛾, 𝛾, 𝛼)  + 𝐷𝑝(𝛾, 𝛾, 𝛼)  =  2𝐷𝑝(𝛾, 𝛾, 𝛼) 

 ⇒  𝐷𝑝(𝛾, 𝛾, 𝛾)  = 𝐷𝑝(𝛾, 𝛾, 𝛼) … 10 

  𝑎𝑛𝑑 𝐷𝑝( 𝛼, 𝛼, 𝛽) + 𝐷𝑝( 𝛼, 𝛾, 𝛾) − 2𝐷𝑝( 𝛼, 𝛼, 𝛼) = 0 

  ⇒  2𝐷𝑝( 𝛼, 𝛼, 𝛼) = 𝐷𝑝( 𝛼, 𝛼, 𝛽) + 𝐷𝑝( 𝛼, 𝛾, 𝛾) 𝑖𝑓  𝛽 =  𝛼 

 𝑇ℎ𝑒𝑛, 2𝐷𝑝( 𝛼, 𝛼, 𝛼) = 𝐷𝑝( 𝛼, 𝛼, 𝛼) + 𝐷𝑝( 𝛼, 𝛾, 𝛾) 

  ⇒  𝐷𝑝( 𝛼, 𝛼, 𝛼) = 𝐷𝑝(𝛼, 𝛾, 𝛾) … 11 

 𝐹𝑟𝑜𝑚 10&11, 𝑤𝑒 𝑔𝑒𝑡𝐷𝑝(𝛾, 𝛾, 𝛾) = 𝐷𝑝(𝛾, 𝛾, 𝛼)  = 𝐷𝑝(𝛼, 𝛼, 𝛼) 

  𝑠𝑜 𝑏𝑦 𝑑𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛  𝛼 = 𝛾 … 12 

 𝑇ℎ𝑒𝑛 𝑏𝑦 9&12 𝑤𝑒 𝑔𝑒𝑡 𝛼 = 𝛽 = 𝛾.   

 3) 𝑇𝑟𝑖𝑣𝑖𝑎𝑙  

4) 𝑏𝑦 𝑑𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛 𝑠𝑖𝑛𝑐𝑒 𝐷𝑝(𝜇 , 𝜇 , 𝛽)  +  𝐷𝑝(𝜇 , 𝜇 , 𝛾)  − 2𝐷𝑝(𝜇 , 𝜇 , 𝜇) ≥ 0 

𝐷𝑝(𝛽, 𝛽 , 𝜇) + 𝐷𝑝(𝛽, 𝛽, 𝜇) – 2𝐷𝑝( 𝛽 , 𝛽 , 𝛽) ≥ 0 

   𝐷𝑝(𝛾 , 𝛾, 𝜇)  + 𝐷𝑝(𝛾 , 𝛾, 𝜇)  − 2𝐷𝑝(𝛾 , 𝛾 , 𝛾) ≥ 0 

 𝐷𝑝( 𝛼 , 𝛼 , 𝜇) + 𝐷𝑝( 𝛼 , 𝛼 , 𝜇)– 2𝐷𝑝( 𝛼 , 𝛼 , 𝛼) ≥ 0 , 

  𝐷𝑝(𝜇 , 𝜇 , 𝛼) + 𝐷𝑝(𝜇 , 𝜇 , 𝛾)– 2𝐷𝑝(𝜇 , 𝜇 , 𝜇) ≥ 0 

 𝐷𝑝(𝜇 , 𝜇 , 𝛼) +  𝐷𝑝(𝜇 , 𝜇 , 𝛽) –  2𝐷𝑝(𝜇 , 𝜇 , 𝜇)  ≥ 0 



  

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When combined, it is more than  and equal to zero and when adding these values 

 𝐷𝑝( 𝛼 , 𝛼, 𝛽), 𝐷𝑝( 𝛼 , 𝛼 , 𝛾), 𝐷𝑝( 𝛽 , 𝛽 , 𝛼), 𝐷𝑝( 𝛽 , 𝛽 , 𝛾), 𝐷𝑝(𝛾 , 𝛾, 𝛼), 𝐷𝑝(𝛾 , 𝛾, 𝛽), −2𝐷𝑝( 𝛼 , 𝛼 , 𝛼),  

−2𝐷𝑝( 𝛽, 𝛽, 𝛽), −2𝐷𝑝(𝛾 , 𝛾 , 𝛾) 𝑡𝑜 𝑡𝑤𝑜 𝑝𝑎𝑟𝑡𝑖𝑒𝑠, 𝑤𝑒 𝑔𝑒𝑡 

𝐷𝑝( 𝛼 , 𝛼 , 𝛽) +  𝐷𝑝( 𝛼 , 𝛼 , 𝛾) + 𝐷𝑝( 𝛽 , 𝛽 , 𝛼) + 𝐷𝑝( 𝛽 , 𝛽 , 𝛾) +  𝐷𝑝(𝛾 , 𝛾 , 𝛼)  

+  𝐷𝑝(𝛾 , 𝛾 , 𝛽)–  2𝐷𝑝( 𝛼 , 𝛼 , 𝛼) –  2𝐷𝑝( 𝛽 , 𝛽 , 𝛽) –  2𝐷𝑝(𝛾 , 𝛾 , 𝛾)  

≤ 𝐷𝑝( 𝛼 , 𝛼 , 𝛽) + 𝐷𝑝( 𝛼 , 𝛼 , 𝛾) + 𝐷𝑝( 𝛽 , 𝛽 , 𝛼) + 𝐷𝑝( 𝛽 , 𝛽 , 𝛾)  

+ 𝐷𝑝(𝛾 , 𝛾 , 𝛼)                                                                                                                         

+ 𝐷𝑝𝛾 , 𝛾 , 𝛽)– 2𝐷𝑝(𝛼 , 𝛼 , 𝛼) –  2𝐷𝑝( 𝛽 , 𝛽 , 𝛽) − 2𝐷𝑝(𝛾 , 𝛾 , 𝛾)  + 𝐷𝑝(𝜇 , 𝜇 , 𝛽) + 𝐷𝑝(𝜇 , 𝜇 , 𝛾)

+ 𝐷𝑝(𝛽 , 𝛽 , 𝜇) + 𝐷𝑝( 𝛽 , 𝛽 , 𝜇) + 𝐷𝑝(𝛾 , 𝛾 , 𝜇)  + 𝐷𝑝(𝛾 , 𝛾 , 𝜇)  + 𝐷𝑝( 𝛼 , 𝛼 , 𝜇)  + 𝐷𝑝(𝛼 , 𝛼 , 𝜇)

+ 𝐷𝑝(𝜇 , 𝜇 , 𝛼)  + 𝐷𝑝(𝜇 , 𝜇, 𝛾)  + 𝐷𝑝(𝜇 , 𝜇 , 𝛼)  +  𝐷𝑝(𝜇 , 𝜇, 𝛽) 

 –  2𝐷𝑝(𝜇 , 𝜇 , 𝜇) –  2𝐷𝑝( 𝛽 , 𝛽 , 𝛽) –  2𝐷𝑝(𝛾 , 𝛾 , 𝛾) –  2𝐷𝑝( 𝛼 , 𝛼 , 𝛼) –  2𝐷𝑝(𝜇 , 𝜇 , 𝜇) –  2𝐷𝑝(𝜇 , 𝜇, 𝜇) 

= [𝐷𝑝(𝜇 , 𝜇 , 𝛽) + 𝐷𝑝(𝜇 , 𝜇 , 𝛾) + 𝐷𝑝( 𝛽 , 𝛽, 𝜇) + 𝐷𝑝( 𝛽 , 𝛽 , 𝛾)          

+ 𝐷𝑝(𝛾 , 𝛾, 𝜇)                                                                                                                        

+ 𝐷𝑝(𝛾 , 𝛾 , 𝛽)–  2𝐷𝑝(𝜇 , 𝜇 , 𝜇)–  2𝐷𝑝( 𝛽 , 𝛽 , 𝛽)–  2𝐷𝑝(𝛾 , 𝛾, 𝛾)] +  [𝐷𝑝( 𝛼 , 𝛼, 𝜇)

+ 𝐷𝑝( 𝛼 , 𝛼 , 𝛾) + 𝐷𝑝(𝜇 , 𝜇, 𝛼)  +  𝐷𝑝(𝜇, 𝜇 , 𝛾) +  𝐷𝑝(𝛾 , 𝛾 , 𝛼)  

+  𝐷𝑝(𝛾 , 𝛾, 𝜇) –  2𝐷𝑝( 𝛼 , 𝛼, 𝛼) –  2𝐷𝑝(𝜇, 𝜇 , 𝜇) –  2𝐷𝑝(𝛾 , 𝛾 , 𝛾)]  

+  [𝐷𝑝( 𝛼 , 𝛼, 𝛽) + 𝐷𝑝( 𝛼 , 𝛼 , 𝜇)  +  𝐷𝑝( 𝛽 , 𝛽 , 𝛼)  +  𝐷𝑝(𝛽 , 𝛽 , 𝜇)  

+  𝐷𝑝(𝜇 , 𝜇 , 𝛼)  +  𝐷𝑝(𝜇 , 𝜇 , 𝛽) – 2𝐷𝑝( 𝛼 , 𝛼, 𝛼) –  2𝐷𝑝(𝛽 , 𝛽, 𝛽) – 2𝐷𝑝(𝜇 , 𝜇 , 𝜇)] 

⇒ 𝐷 𝑔( 𝛼 , 𝛽, 𝛾)  ≤  𝐷 𝑔 (𝜇 , 𝛽 , 𝛾)  + 𝐷 𝑔( 𝛼 , 𝜇, 𝛾)  + 𝐷 𝑔( 𝛼 , 𝛽 , 𝜇)                                   ⎕ 

Corollary 13 

Let (𝑌, 𝐷𝑝) be a general partial metric space, the function 𝐷
𝑔 : 𝑌3 ⟶  [0, ∞) given by  

𝐷 𝑔(𝛼, 𝛽, 𝛾) = 𝐷𝑝(𝛼, 𝛽, 𝛾) + 𝐷𝑝(𝛼, 𝛼, 𝛽) + 𝐷𝑝(𝛼, 𝛼, 𝛾) + 𝐷𝑝(𝛽, 𝛽, 𝛼) + 𝐷𝑝( 𝛽, 𝛽, 𝛾)

+ 𝐷𝑝(𝛾, 𝛾, 𝛼) + 𝐷𝑝(𝛾, 𝛾, 𝛽) − 2𝐷𝑝( 𝛼, 𝛼, 𝛼) − 2𝐷𝑝( 𝛽, 𝛽, 𝛽) − 3𝐷𝑝(𝛾, 𝛾, 𝛾) 

𝑖𝑠 𝐷 −  𝑚𝑒𝑡𝑟𝑖𝑐 𝑠𝑝𝑎𝑐𝑒.                                                                                                     (1.2) 

 Proof: the prove is similar to theorem 12 

Remark 14  

      It is clear from definition that every D-metric space (𝑌, 𝐷) is a general partial metric 
space(𝑌, 𝐷𝑝), but the converse is not true. as we saw in example 2, by definition of D-metric 

space if 𝛼 = 𝛽 = 𝛾 then 𝐷(𝛼, 𝛽, 𝛾) = 0 
Suppose 𝛼 = 5 =  𝛽 = 𝛾 then  𝐷𝑝(5, 5, 5) = 𝑚𝑎𝛼{5, 5, 5} = 5 ≠ 0  

Lemma 15 

           Let (𝑌, 𝐷𝑝) be a general partial metric space if {𝐷𝑝(𝛼𝑚 , 𝛼𝑚 , 𝛼𝑚)}  →  𝛼  𝑎𝑠 𝑚 → ∞  

and {𝐷 𝑔(𝛼𝑛, 𝛼𝑚, 𝛼𝑙 )} is a Cauchy sequence as 𝑛 , 𝑚 , 𝑙 → ∞, then 𝐷𝑝(𝛼𝑛, 𝛼𝑚, 𝛼𝑙 )  →   𝛼  as 

𝑛, 𝑚, 𝑙 ⟶ ∞  where 𝐷 𝑔define in corollary 13. 

 Proof  

          Since 𝐷𝑝(𝛼𝑛 , 𝛼𝑚, 𝛼𝑙 ) →  𝛼  𝑎𝑠 𝑚 ⟶ ∞, then from every𝜖 > 0  there exists   𝑛0 ∈ 𝑁 such 

that   



  

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|𝐷𝑝(𝛼𝑚, 𝛼𝑚, 𝛼𝑚) − 𝛼| < 
𝜖

2
    ∀ 𝑚 >  𝑛0, and 𝐷

𝑔 (𝛼𝑛, 𝛼𝑚, 𝛼𝑙 )  <  
𝜀

2
    ∀ 𝑛 , 𝑚, 𝑙 >  𝑛0 

𝜖

2
 > 𝐷 𝑔(𝛼𝑛, 𝛼𝑚, 𝛼𝑙 ) 

=  𝐷𝑝(𝛼𝑛, 𝛼𝑚, 𝛼𝑙 )  +  𝐷𝑝(𝛼𝑛, 𝛼𝑛 , 𝛼𝑚)  +  𝐷𝑝(𝛼𝑛, 𝛼𝑛, 𝛼𝑙 )  +  𝐷𝑝(𝛼𝑚, 𝛼𝑚, 𝛼𝑛)  

+  𝐷𝑝(𝛼𝑚, 𝛼𝑚, 𝛼𝑙 )  +  𝐷𝑝(𝛼𝑙 , 𝛼𝑙 , 𝛼𝑛)  

+  𝐷𝑝(𝛼𝑙 , 𝛼𝑙 , 𝛼𝑚) –  2𝐷𝑝(𝛼𝑛, 𝛼𝑛, 𝛼𝑛 ) –  2𝐷𝑝(𝛼𝑚, 𝛼𝑚, 𝛼𝑚) –  3𝐷𝑝(𝛼𝑙 , 𝛼𝑙 , 𝛼𝑙 ). 

 ⇒  𝐷𝑝(𝛼𝑛, 𝛼𝑚, 𝛼𝑙 ) –  𝐷𝑝(𝛼𝑚, 𝛼𝑚, 𝛼𝑚)  <  
𝜖

2
 

 So that |𝐷𝑝(𝛼n, 𝛼m, 𝛼l) − 𝛼|=|𝐷𝑝(𝛼𝑛, 𝛼𝑚, 𝛼𝑙 ) − 𝐷𝑝(𝛼𝑚, 𝛼𝑚, 𝛼𝑚) + 𝐷𝑝(𝛼𝑚, 𝛼𝑚, 𝛼𝑚) −  𝛼| ≤

|𝐷𝑝(𝛼𝑛, 𝛼𝑚, 𝛼𝑙 ) − 𝐷𝑝(𝛼𝑚, 𝛼𝑚, 𝛼𝑚)| + |𝐷𝑝(𝛼𝑚, 𝛼𝑚, 𝛼𝑚) −  𝛼| < 
𝜖

2
 + 

𝜖

2
 <  𝜖  

 Hence 𝐷𝑝(𝛼𝑛, 𝛼𝑚, 𝛼𝑙 )  →  𝛼  𝑎𝑠 𝑛, 𝑚, 𝑙 ⟶ ∞                                                                        ⎕ 

Theorem 16 

           Let (𝑌, 𝐷𝑝) be a general partial metric space, then  

𝑖) A sequence {𝛼𝑛} is a Cauchy sequence in a general partial metric space (𝑌, 𝐷𝑝) if and only if 

{𝛼𝑛} is a Cauchy sequence in (𝑌, 𝐷
𝑔). 

𝑖𝑖) A general partial metric space (𝑌, 𝐷𝑝) is complete if (𝑌, 𝐷
𝑔) is complete. 

Where 𝐷 𝑔define in  corollary 13 

 Proof 𝒊  

           First, we must prove that each Cauchy sequence in (𝑌, 𝐷𝑝) is Cauchy in (𝑌, 𝐷
𝑔). 

Then, there exists 𝛼 ∈ 𝑅 such that, ∀𝜖 > 0 there is 𝑛0 ∈ 𝑁 with 

 |𝐷𝑝(𝛼𝑛, 𝛼𝑚, 𝛼𝑙 ) −  𝛼| <
𝜖

14
  ∀ n, m, l ≥ n0. Hence,   

 |Dg(𝛼n, 𝛼m, 𝛼l)|= |𝐷𝑝(𝛼𝑛, 𝛼𝑚, 𝛼𝑙 ) + 𝐷𝑝(𝛼𝑛, 𝛼𝑛 , 𝛼𝑚) + 𝐷𝑝(𝛼𝑛, 𝛼𝑛, 𝛼𝑙 ) + 𝐷𝑝(𝛼𝑚, 𝛼𝑚, 𝛼𝑛) +

𝐷𝑝(𝛼𝑚, 𝛼𝑚, 𝛼𝑙 ) + 𝐷𝑝(𝛼𝑙 , 𝛼𝑙 , 𝛼𝑛) + 𝐷 𝑝(𝛼𝑙 , 𝛼𝑙 , 𝛼𝑚 ) − 2𝐷𝑝(𝛼𝑛, 𝛼𝑛, 𝛼𝑛 ) − 2𝐷𝑝(𝛼𝑚, 𝛼𝑚, 𝛼𝑚) −

3𝐷𝑝(𝛼𝑙 , 𝛼𝑙 , 𝛼𝑙 )| ≤ |𝐷𝑝(𝛼𝑛, 𝛼𝑚, 𝛼𝑙 ) −  𝛼| + |𝐷𝑝(𝛼𝑛, 𝛼𝑛 , 𝛼𝑚) −  𝛼| + |𝐷𝑝(𝛼𝑛, 𝛼𝑛 , 𝛼𝑙 ) −  𝛼| +

|𝐷 𝑝(𝛼𝑚, 𝛼𝑚, 𝛼𝑛 ) −  𝛼| + |𝐷𝑝(𝛼𝑚, 𝛼𝑚, 𝛼𝑙 ) −  𝛼| + |𝐷𝑝(𝛼𝑙 , 𝛼𝑙 , 𝛼𝑛) −  𝛼| + |𝐷𝑝(𝛼𝑙 , 𝛼𝑙 , 𝛼𝑚) −

 𝛼|-2|𝐷𝑝(𝛼𝑛, 𝛼𝑛, 𝛼𝑛) −  𝛼 | − 2|𝐷𝑝(𝛼𝑚, 𝛼𝑚, 𝛼𝑚) −  𝛼| − 3|𝐷𝑝(𝛼𝑙 , 𝛼𝑙 , 𝛼𝑙 ) −  𝛼| 
ε

14
 <  ∀ n, m, l 

≥ n0. 

Hence, {𝛼𝑛} is a Cauchy sequence in(𝑌, 𝐷
𝑔). 

Conversely, now we must prove {𝛼𝑛} is Cauchy sequence in (𝑌, 𝐷𝑝) 

Since {𝛼𝑛} is a Cauchy sequence in (𝑌, 𝐷
𝑔) 𝑠𝑜 ∀𝜖 > 0, ∃ 𝑛0 ∈  𝑁 such that  

𝐷 𝑔(𝛼𝑛, 𝛼𝑚, 𝛼𝑙 ) <
𝜖

2
  ∀𝑛 , 𝑚, 𝑙 >  𝑛0  

𝜖

2
>𝐷 𝑔(𝛼𝑛, 𝛼𝑚, 𝛼𝑙 )                                                                        

=  𝐷𝑝(𝛼𝑛, 𝛼𝑚, 𝛼𝑙 )  +  𝐷𝑝(𝛼𝑛, 𝛼𝑛 , 𝛼𝑚)  +  𝐷𝑝(𝛼𝑛, 𝛼𝑛, 𝛼𝑙 )  +  𝐷𝑝(𝛼𝑚, 𝛼𝑚, 𝛼𝑛)  

+  𝐷𝑝(𝛼𝑚, 𝛼𝑚, 𝛼𝑙 )  +  𝐷𝑝(𝛼𝑙 , 𝛼𝑙 , 𝛼𝑛)  

+  𝐷𝑝(𝛼𝑙 , 𝛼𝑙 , 𝛼𝑚) –  2𝐷𝑝(𝛼𝑛, 𝛼𝑛, 𝛼𝑛 ) –  2𝐷𝑝(𝛼𝑚, 𝛼𝑚, 𝛼𝑚) –  3𝐷𝑝(𝛼𝑙 , 𝛼𝑙 , 𝛼𝑙 ) 



  

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 ⇒  𝐷𝑝(𝛼𝑛, 𝛼𝑚, 𝛼𝑙 ) –  𝐷𝑝(𝛼𝑛, 𝛼𝑛, 𝛼𝑛 )  ≤  𝐷
𝑔(𝛼𝑛, 𝛼𝑚, 𝛼𝑙 )  <

𝜖

2
 

 By compensation 𝐷𝑝(𝛼𝑛, 𝛼𝑛, 𝛼𝑛) to two parties, we have  

 𝐷𝑝(𝛼𝑛, 𝛼𝑚, 𝛼𝑙 )  ≤  𝐷
𝑔(𝛼𝑛, 𝛼𝑚, 𝛼𝑙 )  +  𝐷𝑝(𝛼𝑛, 𝛼𝑛 , 𝛼𝑛)  <

𝜖

2
 + 𝐷𝑝(𝛼𝑛, 𝛼𝑛, 𝛼𝑛) 

  𝐴𝑛𝑑 𝑠𝑖𝑛𝑐𝑒 𝐷𝑝(𝛼𝑚, 𝛼𝑚, 𝛼𝑚)  ≤  𝐷𝑝(𝛼𝑛, 𝛼𝑚, 𝛼𝑙 )  𝑠𝑜  

 𝐷𝑝(𝛼𝑚, 𝛼𝑚, 𝛼𝑚)  ≤  𝐷𝑝(𝛼𝑛, 𝛼𝑚, 𝛼𝑙 )  ≤  𝐷
𝑔(𝛼𝑛, 𝛼𝑚, 𝛼𝑙 )  +  𝐷𝑝(𝛼𝑛, 𝛼𝑛, 𝛼𝑛 )  

<
𝜖

2
+ 𝐷𝑝(𝛼𝑛, 𝛼𝑛 , 𝛼𝑛)     

 ⇒  𝐷𝑝(𝛼𝑚, 𝛼𝑚, 𝛼𝑚)  ≤
𝜖

2
+ 𝐷𝑝(𝛼𝑛, 𝛼𝑛, 𝛼𝑛 )     ∀ 𝑛 , 𝑚 >  𝑛0  

 𝐿𝑒𝑡 𝛼𝑛 =  𝐷𝑝(𝛼𝑛, 𝛼𝑛 , 𝛼𝑛)  ∈  𝑅 𝑠𝑢𝑐ℎ 𝑡ℎ𝑎𝑡 | 𝛼 𝑚–   𝛼𝑛 | <  
𝜖

2
 

∴{ α𝑚} is a Cauchy sequence,  ∴ { 𝛼𝑛}  →   𝛼 , ∴ 𝐷𝑝(𝛼𝑚, 𝛼𝑚, 𝛼𝑚) →   𝛼 ∈  𝑅 

Then by lemma 15, 𝐷𝑝(𝛼𝑚, 𝛼𝑚, 𝛼𝑚) is Cauchy sequence in ( 𝑌, 𝐷𝑝).                        ⎕ 

 ii : 

 If {𝛼𝑛} is a Cauchy sequence in (𝑌, 𝐷𝑝), then it is a Cauchy sequence in (𝑌, 𝐷
𝑔) and since D-

metric (𝑌, 𝐷 𝑔) is complete then there exists 𝛼 ∈   𝑌  such that  

limn ,m →∞ D
𝑔(𝛼𝑛, 𝛼𝑚, 𝛼)  =  0, hence 

 𝑙𝑖𝑚𝑛 ,𝑚→∞[𝐷𝑝(𝛼𝑛, 𝛼𝑚, 𝛼)  +  𝐷𝑝(𝛼𝑛, 𝛼𝑛 , 𝛼𝑚)  +  𝐷𝑝(𝛼𝑛, 𝛼𝑛, 𝛼)  +  𝐷𝑝(𝛼𝑚, 𝛼𝑚, 𝛼𝑛)  +

 𝐷𝑝(𝛼𝑚, 𝛼𝑚, 𝛼)  +  𝐷𝑝(𝛼 , 𝛼 , 𝛼𝑛 )  +

 𝐷𝑝(𝛼, 𝛼 , 𝛼𝑚) –  2𝐷𝑝(𝛼𝑛, 𝛼𝑛, 𝛼𝑛 ) –  2𝐷𝑝(𝛼𝑚, 𝛼𝑚, 𝛼𝑚) –  3𝐷𝑝(𝛼 , 𝛼 , 𝛼)]  =  0   

There for 𝑙𝑖𝑚𝑛 𝑚 →∞[𝐷𝑝(𝛼𝑛, 𝛼𝑚, 𝛼) –  𝐷𝑝(𝛼, 𝛼, 𝛼)]  =  0 

⇒𝑙𝑖𝑚𝑛  𝑚→∞ 𝐷𝑝(𝛼𝑛, 𝛼𝑚, 𝛼)  = 𝐷𝑝(𝛼 , 𝛼, 𝛼) hence (𝑌, 𝐷𝑝) is converge   

Thus (𝑌, 𝐷𝑝) is complete. 

Corollary 17 

            we can get from the proof of theorem 16, (𝑖𝑖), 𝑙𝑖𝑚𝑛,𝑚 →∞[𝐷𝑝(𝛼𝑛, 𝛼𝑚, 𝛼) −

𝐷𝑝(𝛼, 𝛼, 𝛼)]=𝑙𝑖𝑚𝑛 ,𝑚 →∞)[𝐷𝑝(𝛼𝑛, 𝛼𝑚, 𝛼) –  𝐷𝑝(𝛼𝑚, 𝛼𝑚, 𝛼𝑚)]  = 

𝑙𝑖𝑚𝑛 ,𝑚 →∞[𝐷𝑝(𝛼𝑛, 𝛼𝑚𝛼) –  𝐷𝑝(𝛼𝑛, 𝛼𝑛, 𝛼𝑛)]  =  0, such that  𝑙𝑖𝑚𝑛 ,𝑚 →∞ 𝐷𝑝(𝛼𝑛, 𝛼𝑚, 𝛼) = 

𝑙𝑖𝑚𝑚 →∞) 𝐷𝑝(𝛼𝑚, 𝛼𝑚, 𝛼𝑚) = 𝑙𝑖𝑚𝑛→∞ 𝐷𝑝(𝛼𝑛, 𝛼𝑛, 𝛼𝑛) =  𝐷𝑝(𝛼 , 𝛼 , 𝛼).  

 Propositions 18  

             If {𝛼𝑛} is Cauchy sequence in (𝑌, 𝐷
𝑔), then 𝑙𝑖𝑚𝑛 ,𝑚 →∞ 𝐷𝑝(𝛼𝑛, 𝛼𝑛, 𝛼𝑚) 

=𝑙𝑖𝑚𝑛→∞ 𝐷𝑝(𝛼𝑛, 𝛼𝑛 , 𝛼𝑛).  

 Proof  

            Since {𝛼𝑛} is Cauchy sequence in (𝑌, 𝐷
𝑔) then 𝑙𝑖𝑚𝑛 ,𝑚,𝑙→∞ 𝐷

𝑔(𝛼𝑛, 𝛼𝑚, 𝛼𝑙 )  =  0 



  

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And 𝐷 𝑔(𝛼𝑛, 𝛼𝑚, 𝛼𝑙 ) =
𝐷𝑝(𝛼𝑛, 𝛼𝑛, 𝛼𝑚) +  𝐷𝑝(𝛼𝑛, 𝛼𝑛, 𝛼𝑙 ) + 𝐷𝑝(𝛼𝑚, 𝛼𝑚, 𝛼𝑛 ) + 𝐷𝑝(𝛼𝑚, 𝛼𝑚, 𝛼𝑙 ) + 𝐷𝑝(𝛼𝑙 , 𝛼𝑙 , 𝛼𝑛 ) +

𝐷𝑝(𝛼𝑙 , 𝛼𝑙 , 𝛼𝑚)– 𝐷𝑝(𝛼𝑛, 𝛼𝑛, 𝛼𝑛)– 2𝐷𝑝(𝛼𝑚, 𝛼𝑚, 𝛼𝑚 )– 2𝐷𝑝(𝛼𝑙 , 𝛼𝑙 , 𝛼𝑙 ). 

And 𝐷𝑝(𝛼𝑛, 𝛼𝑛 , 𝛼𝑚) − 𝐷𝑝(𝛼𝑛, 𝛼𝑛, 𝛼𝑛 ) ≤ 𝐷
𝑔(𝛼𝑛, 𝛼𝑚, 𝛼𝑙 ) then  

 𝑙𝑖𝑚𝑛 ,𝑚 →∞𝐷𝑝(𝛼𝑛, 𝛼𝑛 , 𝛼𝑚)– 𝐷𝑝(𝛼𝑛, 𝛼𝑛, 𝛼𝑛) →  0                                

Similarly 𝑙𝑖𝑚𝑛 ,𝑚 →∞𝐷𝑝 (𝛼𝑛, 𝛼𝑛, 𝛼𝑚)– 𝐷𝑝(𝛼𝑚, 𝛼𝑚 , 𝛼𝑚) →  0              

 Since 𝐷𝑝(𝛼𝑚, 𝛼𝑚, 𝛼𝑚) – 𝐷𝑝(𝛼𝑛, 𝛼𝑛, 𝛼𝑛 ) = 𝐷𝑝(𝛼𝑚, 𝛼𝑚, 𝛼𝑚) + 𝐷𝑝(𝛼𝑛, 𝛼𝑛, 𝛼𝑚) −

𝐷𝑝(𝛼𝑛, 𝛼𝑛, 𝛼𝑚) − 𝐷𝑝(𝛼𝑛, 𝛼𝑛, 𝛼𝑛) 𝑡ℎ𝑒𝑛 𝑙𝑖𝑚𝑛,𝑚⟶∞[𝐷𝑝(𝛼𝑚, 𝛼𝑚, 𝛼𝑚) − 𝐷𝑝(𝛼𝑛, 𝛼𝑛 , 𝛼𝑛)] =

𝑙𝑖𝑚𝑛,𝑚⟶∞[𝐷𝑝(𝛼𝑚, 𝛼𝑚, 𝛼𝑚) − 𝐷𝑝(𝛼𝑛, 𝛼𝑛, 𝛼𝑚)] + 𝑙𝑖𝑚𝑛,𝑚⟶∞[𝐷𝑝(𝛼𝑛, 𝛼𝑛, 𝛼𝑚) −

𝐷𝑝(𝛼𝑛, 𝛼𝑛, 𝛼𝑛 )]  ⟶ 0 

So that 𝑙𝑖𝑚𝑛,𝑚⟶∞[𝐷𝑝 (𝛼𝑚, 𝛼𝑚, 𝛼𝑚)  −  𝐷𝑝 (𝛼𝑛, 𝛼𝑛 , 𝛼𝑛)] ⟶ 0 

Let 𝛼𝑛 =  𝐷𝑝 (𝛼𝑛, 𝛼𝑛, 𝛼𝑛 ) 

 ∴ | 𝛼𝑚 –  𝛼𝑛|  → 0 𝑎𝑠 𝑛, 𝑚 ⟶ ∞ 

 Hence {𝛼𝑛} is a Cauchy sequence in R, therefor {𝐷𝑝(𝛼𝑛, 𝛼𝑛 , 𝛼𝑛)} converge to α.                 

 
Also , 𝑙𝑖𝑚𝑛⟶∞ 𝐷𝑝(𝛼𝑛 , 𝛼𝑛, 𝛼𝑚)  =

𝑙𝑖𝑚𝑛,𝑚⟶∞ [𝐷𝑝(𝛼𝑛, 𝛼𝑛 , 𝛼𝑚)  + 𝐷𝑝(𝛼𝑛, 𝛼𝑛, 𝛼𝑛) –  𝐷𝑝(𝛼𝑛, 𝛼𝑛, 𝛼𝑛)] 

Then [𝐷𝑝 (𝛼𝑛, 𝛼𝑛 , 𝛼𝑚)– 𝐷𝑝(𝛼𝑛, 𝛼𝑛, 𝛼𝑛 )] ⟶ 0 so 𝑙𝑖𝑚𝑛,𝑚⟶∞𝐷𝑝(𝛼𝑛, 𝛼𝑛 , 𝛼𝑚) = 𝛼 

Thus 𝑙𝑖𝑚𝑛,𝑚⟶∞𝐷𝑝 (𝛼𝑛, 𝛼𝑛 , 𝛼𝑚)  =  𝑙𝑖𝑚𝑛⟶∞𝐷𝑝(𝛼𝑛, 𝛼𝑛, 𝛼𝑛 ).                                            ⎕ 

Theorem 19 

            If (Y, 𝑝) be partial metric space then  

 𝐷𝑝(𝛼, 𝛽, 𝛾)  =   𝑝(𝛼, 𝛽)  +   𝑝(𝛼 , 𝛾)  +   𝑝(𝛽 , 𝛾) –   𝑝(𝛼 , 𝛼) –   𝛲(𝛽 , 𝛽) –   𝑝(𝛾 , 𝛾)      (1.3) 

is general partial metric space. 

 1)𝑆𝑖𝑛𝑐𝑒  𝑝(𝛼, 𝛽) –   𝑝( 𝛼, 𝛼) ≥ 0 , 𝑝(𝛼 , 𝛾) –   𝑝(𝛾 , 𝛾) ≥ 0 , 𝑝(𝛽 , 𝛾) –   𝑝( 𝛽, 𝛽) ≥ 0  

then 𝐷𝑝( 𝛼 , 𝛽, 𝛾) ≥ 0 

 2)𝑙𝑒𝑡 𝐷𝑝( 𝛼 , 𝛽, 𝛾)  =  𝐷𝑝( 𝛼 , 𝛼 , 𝛼)  =  𝐷𝑝( 𝛽, 𝛽, 𝛽)  =  𝐷𝑝(𝛾 , 𝛾 , 𝛾) 

  𝑠𝑖𝑛𝑐𝑒 𝐷𝑝(𝛼 , 𝛼 , 𝛼)  =  𝐷𝑝( 𝛽, 𝛽 , 𝛽)  =  𝐷𝑝(𝛾, 𝛾, 𝛾)  =  0 

⇒ 𝐷𝑝( 𝛼 , 𝛽 , 𝛾) = 0  

⇒ 𝑝( 𝛼, 𝛽)  +  𝑝( 𝛽 , 𝛾)  +  𝑝( 𝛼, 𝛾) –  𝑝(𝛼 , 𝛼) –  𝑝(𝛽, 𝛽) –   𝑝(𝛾 , 𝛾) =  0 

  ⇒  𝑝( 𝛼 , 𝛽)–  𝑝( 𝛼 , 𝛼) =  0 ⇒   𝑝( 𝛼, 𝛽) =  𝑝( 𝛼, 𝛼) … 1, 

 ⇒  𝑝( 𝛼 , 𝛾)–   𝑝(𝛾 , 𝛾) = 0 ⇒   𝑝( 𝛼, 𝛾) =  𝑝(𝛾, 𝛾) … 2,   



  

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 𝑎𝑛𝑑 𝑝( 𝛽, 𝛾) –   𝑝( 𝛽 , 𝛽)  = 0 ⇒   𝑝( 𝛽, 𝛾) =  𝑝( 𝛽, 𝛽) … 3 

 𝐹𝑟𝑜𝑚 1  𝑝( 𝛼, 𝛼) =  𝑝( 𝛼, 𝛽)  𝑎𝑛𝑑 𝑏𝑦 𝑑𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛  𝑝( 𝛼, 𝛼) =  𝑝( 𝛼, 𝛽) ≤  𝑝( 𝛼, 𝛾) +  𝑝(𝛾,
𝛽) −  𝑝(𝛾, 𝛾) 

 𝑆𝑖𝑛𝑐𝑒  𝑝( 𝛼, 𝛾) =  𝑝(𝛾, 𝛾) & 𝑝( 𝛽, 𝛾) =  𝑝( 𝛽, 𝛽) 𝑤𝑒 𝑔𝑒𝑡  𝑝( 𝛼, 𝛼) =  𝑝( 𝛽, 𝛽) 

 𝐹𝑟𝑜𝑚 2   𝑝( 𝛽, 𝛽) =  𝑝( 𝛽, 𝛾)  𝑎𝑛𝑑 𝑏𝑦 𝑑𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛  𝑝( 𝛽, 𝛽) =  𝑝( 𝛽, 𝛾) ≤  𝑝( 𝛽, 𝛼) +
 𝑝( 𝛼, 𝛾) −  𝑝( 𝛼, 𝛼) 

  𝑏𝛽 1&2 𝑤𝑒 𝑔𝑒𝑡 𝑝(𝛽, 𝛽)  =  (𝛾, 𝛾) 

    𝐹𝑟𝑜𝑚 3 𝑝(𝛾, 𝛾) =  𝑝( 𝛼, 𝛾)𝑎𝑛𝑑 𝑏𝑦 𝑑𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛 𝑝(𝛾, 𝛾) =  𝑝( 𝛼, 𝛾) ≤  𝑝( 𝛼, 𝛽) +
 𝑝( 𝛽, 𝛾) − 𝑝( 𝛽, 𝛽)     

   𝑏𝑦 1&3 𝑤𝑒 𝑔𝑒𝑡 𝑝(𝛾, 𝛾)  =  𝑝 ( 𝛼, 𝛼) 

 𝐻𝑒𝑛𝑐𝑒 𝑝(𝛼 , 𝛼)  =  𝑝( 𝛽, 𝛽)  =  𝑝(𝛾, 𝛾)  

𝑇ℎ𝑢𝑠  𝑝( 𝛼, 𝛼) =  𝑝( 𝛼, 𝛽) =  𝑝( 𝛽, 𝛽) 𝑏𝑦 𝑑𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛  𝛼 =  𝛽, 𝑎𝑛𝑑  𝑝( 𝛽, 𝛽) =  𝑝( 𝛽, 𝛾)
=  𝑝(𝛾, 𝛾) 𝑏𝑦 𝑑𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛  𝛽 = 𝛾 𝑡ℎ𝑒𝑛 𝑤𝑒 𝑔𝑒𝑡  𝛼 =  𝛽 = 𝛾.  

 3) 𝑇𝑟𝑖𝑣𝑖𝑎𝑙  

4) 𝑠𝑖𝑛𝑐𝑒  𝑝(𝜇 , 𝛾) –   𝑝(𝛾 , 𝛾)  ≥ 0 , 𝑝(𝜇 , 𝛽) –   𝑝(𝜇 , 𝜇) ≥ 0 , 𝑝( 𝛼 , 𝜇) –   𝑝( 𝛼 , 𝛼) ≥ 0 ,
𝑝(𝜇 , 𝛾) –   𝑝(𝜇 , 𝜇 )  ≥ 0 , 𝑝( 𝛽 , 𝜇) –   𝑝( 𝛽 , 𝛽 )  ≥ 0 ,
𝑝( 𝛼 , 𝜇) –   𝑝(𝜇 , 𝜇) ≥ 0  

When combined, it is more than and equal to zero and when added these values   

𝑝(𝛼 , 𝛽) , 𝑝( 𝛽 , 𝛾) , 𝑝( 𝛼 , 𝛾) , − 𝑝( 𝛼 , 𝛼) , − 𝑝( 𝛽 , 𝛽) , − 𝑝(𝛾 , 𝛾) 𝑡𝑜 𝑏𝑜𝑡ℎ 𝑠𝑎𝑖𝑑 , 𝑤𝑒 𝑔𝑒𝑡  

  𝑝( 𝛼 , 𝛽)  +   𝑝( 𝛽 , 𝛾)  +   𝑝( 𝛼 , 𝛾) –   𝑝( 𝛼 , 𝛼) –   𝑝( 𝛽 , 𝛽) –   𝑝(𝛾 , 𝛾)  ≤  𝑝( 𝛼, 𝛽)  +

  𝑝( 𝛽 , 𝛾)  +   𝑝( 𝛼 , 𝛾) –   𝑝( 𝛼 , 𝛼) –   𝑝( 𝛽 , 𝛽) –   𝑝(𝛾 , 𝛾)  +   𝑝(𝜇 , 𝛾) –  𝑝(𝛾 , 𝛾)  +

  𝑝( 𝛼 , 𝜇) –   𝑝( 𝛼 , 𝛼)  +   𝑝( 𝛽 , 𝜇) –  𝑝( 𝛽 , 𝛽)  +  𝑝(𝜇 , 𝛽) –   𝑝(𝜇 , 𝜇)  +

  𝑝(𝜇 , 𝛾) –   𝑝(𝜇 , 𝜇)  +   𝑝( 𝛼 , 𝜇) –   𝑝(𝜇 , 𝜇). 

⇒ 𝐷𝑝( 𝛼 , 𝛽 , 𝛾)                                                                                                                                                  

≤ [ 𝑝(𝜇 , 𝛽)  +   𝑝(𝜇 , 𝛾)  +   𝑝(𝛾 , 𝛽) –  𝑝(𝜇 , 𝜇) –   𝑝( 𝛽 , 𝛽) –   𝑝(𝛾 , 𝛾)]  +  [ 𝑝( 𝛼 , 𝜇)  
+   𝑝( 𝛼 , 𝛾)  +   𝑝(𝜇 , 𝛾) –   𝑝( 𝛼 , 𝛼) –   𝑝(𝜇 , 𝜇) –   𝑝(𝛾 , 𝛾)]  +  [ 𝑝( 𝛼 , 𝛽)  +   𝑝( 𝛼 , 𝜇)  
+   𝑝( 𝛽 , 𝜇) –   𝑝( 𝛼 , 𝛼) –   𝑝( 𝛽 , 𝛽) –   𝑝(𝜇 , 𝜇)] 

𝐷𝑝(𝛼 , 𝛽 , 𝛾)  ≤ 𝐷𝑝(𝜇 , 𝛽, 𝛾)  +  𝐷𝑝( 𝛼 , 𝜇 , 𝛾)  +  𝐷𝑝( 𝛼 , 𝛽, 𝜇) –  𝐷𝑝(𝜇 , 𝜇 , 𝜇)            ⎕ 

 Proposition 20 

             Let (𝑌, 𝐷𝑝) be a general partial metric space and 

  𝐷𝑝( 𝛼 , 𝛽 , 𝛽)  ≤ 𝐷𝑝( 𝛼 , 𝛾 , 𝛾)  +  𝐷𝑝(𝛾 , 𝛽, 𝛽) –  𝐷𝑝(𝛾, 𝛾 , 𝛾)                                     (1.4)  

 holds then the function  𝑝: 𝑌2 →  [0 , ∞) which is defied by 𝑝(𝛼 , 𝛽)  =  𝐷𝑝( 𝛼 , 𝛽 , 𝛽) is a  

partial metric on  Y . 

 Proof   



  

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 1)𝑠𝑖𝑛𝑐𝑒 𝑝(𝛼, 𝛽) = 𝑝(𝛼, 𝛼) = 𝑝(𝛽, 𝛽) ⇔ 𝐷𝑝(𝛼, 𝛽, 𝛽) = 𝐷𝑝(𝛼, 𝛼, 𝛼) = 𝐷𝑝(𝛽, 𝛽, 𝛽) ⇔ 𝛼 = 𝛽.   

2) 𝑠𝑖𝑛𝑐𝑒 𝐷𝑝( 𝛼 , 𝛼, 𝛼)  ≤ 𝐷𝑝( 𝛼, 𝛽 , 𝛽) 𝑡ℎ𝑒𝑛  𝑝( 𝛼 , 𝛼) ≤  𝑝( 𝛼 , 𝛽)   ∀  𝛼, 𝛽 ∈ 𝛼     

 3) Trivial  

 4)  𝑝(𝛼 , 𝛽) = 𝐷𝑝( 𝛼 , 𝛽 , 𝛽) ≤  𝐷𝑝(𝛼 , 𝛾 , 𝛾)  +  𝐷𝑝(𝛾 , 𝛽, 𝛽) –  𝐷𝑝(𝛾 , 𝛾 , 𝛾)  𝑓𝑟𝑜𝑚 (1.3) 

                                 =   𝑝( 𝛼, 𝛾) +  𝑝(𝛾, 𝛽) −  𝑝(𝛾, 𝛾).                                                         ⎕ 

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