# Using Dual Double Fuzzy Semi-Metric to Study the Convergence
## 3 Double Fuzzy Semi-Metric  
Let \((X,M)\) be a fuzzy semi-metric space along with a t-norm \(*\). Given any four elements \(x,y,u,v\in\ X\), recall that the value \(M(x,y,t)\) means the membership degree of the distance that is less than \(t\) between \(x\) and \(y\), and the value \(M(u,v,t)\) means the membership degree of the distance that is less than \(t\) between \(u\) and \(v\). In this case, we can define a value:  
\[\zeta(x,y;u,v,t)=\min\left\{M(x,y,t),M(u,v,t)\right\},\]  
which means the membership degree of the distance that is simultaneously less than \(t\) between \(x\) and \(y\) and between \(u\) and \(v\). In general, instead of considering the min function, we shall use the t-norm. The formal definition is given below. **Definition 1**. : _Let \((X,M)\) be a fuzzy semi-metric space along with a t-norm \(*\). We define the mapping \(\zeta:X^{4}\times[0,+\infty)\rightarrow[0,1]\) by:_  
\[\zeta(x,y;u,v,t)=M(x,y,t)*M(u,v,t).\]  
_Then \(\zeta\) is called a double fuzzy semi-metric._  
**Example 2**. : _Continued from Example 1, we consider:_  
\[M(x,y,t)=\left\{ \begin{array}{ll}\dfrac{t}{t+d(x,y)}&\text{if }t>0\\ 1&\text{if }t=0\text{ and }x=y\\ 0&\text{if }t=0\text{ and }x\
eq y.\end{array} \right.\]  
_If we take t-norm as \(a*b=a\cdot b\), then the double fuzzy semi-metric can be obtained as:_  
\[\zeta(x,y;u,v,t) =M(x,y,t)\cdot M(u,v,t)\] \[=\left\{ \begin{array}{ll}\dfrac{t}{t+d(x,y)}\cdot\dfrac{t}{t+d (u,v)}&\text{if }t>0\\ 1&\text{if }t=0\text{ and }x=y\text{ and }u=v\\ 0&\text{if }t=0\text{ and }x\
eq y\text{ or }u\
eq v.\end{array} \right.\]  
The potential application for considering the double fuzzy semi-metric is to study the new type of fixed point theorems in fuzzy semi-metric space. **Proposition 2**. : _(Triangle Inequalities for Dual Fuzzy Semi-Metric) Let \((X,M)\) be a fuzzy semi-metric space along with a t-norm \(*\). Given any \(x,y,z,u,v,w\in X\), we have the following properties:_1. _Suppose that_ \(M\) _satisfies the_ \(\bowtie\)_-triangle inequality. Then we have the inequality:_ \[\zeta(x,z;u,w,t+s)\geq\zeta(x,y;u,v,t)*\zeta(y,z;v,w,s)\] _for_ \(s,t>0\)_._
2. _Suppose that_ \(M\) _satisfies the_ \(\triangleright\)_-triangle inequality. Then we have the inequality:_ \[\zeta(x,z;u,w,t+s)\geq\zeta(x,y;u,v,t)*\zeta(z,y;w,v,s)\] _for_ \(s,t>0\)_._
3. _Suppose that_ \(M\) _satisfies the_ \(\triangleleft\)_-triangle inequality. Then we have the inequality:_ \[\zeta(x,z;u,w,t+s)\geq\zeta(y,x;v,u,t)*\zeta(y,z;v,w,s)\] _for_ \(s,t>0\)_._
4. _Suppose that_ \(M\) _satisfies the_ \(\diamond\)_-triangle inequality. Then we have the inequality:_ \[\zeta(x,z;u,w,t+s)\geq\zeta(y,x;v,u,t)*\zeta(z,y;w,v,s)\] _for_ \(s,t>0\)_._  
Proof. : It suffices to prove part (i); we have:  
\[\zeta(x,z;u,w,t+s) =M(x,z,t+s)*M(u,w,t+s)\] \[\geq(M(x,y,t)*M(y,z,s))*M(u,w,t+s)\] \[\quad\text{(using the}\bowtie\text{-triangle}\text{-triangle}\text { inequality and the increasing property of t-norm})\] \[\geq(M(x,y,t)*M(y,z,s))*(M(u,v,t)*M(v,w,s))\] \[\quad\text{(using the}\bowtie\text{-triangle}\text{ and the increasing property of t-norm})\] \[=(M(x,y,t)*M(u,v,t))*(M(y,z,s)*M(v,w,s))\] \[\zeta(x,z;u,w,t+s)\geq\zeta(x,y;u,v,t)*\zeta(z,y;w,v,s)\]0 \[\zeta(x,z;u,w,t+s)\geq\zeta(x,y;u,v,t)*\zeta(z,y;w,v,s)\]1  
This completes the proof. **Definition 2**. : _Let \((X,M)\) be a fuzzy semi-metric space along with a t-norm \(*\), and let \(\zeta\) be a double fuzzy semi-metric given by:_  
\[\zeta(x,z;u,w,t+s)\geq\zeta(x,y;u,v,t)*\zeta(z,y;w,v,s)\]2  
_Given any fixed_ \(x,y,u,v\in X\)_, we define the following concepts of monotonicity:_  
* _The mapping_ \(\zeta(x,y;u,v,\cdot)\) _is said to be nondecreasing if and only if_ \(\zeta(x,y;u,v,t_{1})\geq\zeta(x,y;u,v,t_{2})\) _for_ \(t_{1}>t_{2}\)_. The mapping_ \(\zeta(x,y;u,v,\cdot)\) _is said to be increasing if and only if_ \(\zeta(x,y;u,v,t_{1})>\zeta\ (x,y;u,v,t_{2})\) _for_ \(t_{1}<t_{2}\)_._
* _The mapping_ \(\zeta(x,y;u,v,\cdot)\) _is said to be symmetrically nondecreasing if and only if_ \(\zeta(x,y;u,v,t_{1})\geq\zeta(y,x;v,u,t_{2})\) _for_ \(t_{1}>t_{2}\)_.