# Using Dual Double Fuzzy Semi-Metric to Study the Convergence
## 3 Double Fuzzy Semi-Metric
: To prove part (i), the canonical condition says that:  
\[\zeta(x,z;u,w,t+s)\geq\zeta(x,y;u,v,t)*\zeta(z,y;w,v,s)\]7  
The left-continuity of t-norm \(*\) at 1 also says that:  
\[\zeta(x,z;u,w,t+s)\geq\zeta(x,y;u,v,t)*\zeta(z,y;w,v,s)\]8  
To prove part (ii), the rational condition says that:  
\[\zeta(x,z;u,w,t+s)\geq\zeta(x,y;u,v,t)*\zeta(z,y;w,v,s)\]9  
The right-continuity of t-norm \(*\) at 0 also says that:  
\[\zeta(x,z;u,w,t+s)\geq\zeta(y,x;v,u,t)*\zeta(y,z;v,w,s)\]0  
This completes the proof. **Example 3**. : _Continued from Example 1, it is not hard to check that \(M\) satisfies both the canonical and rational conditions. Suppose that we take t-norm as \(a*b=a\cdot b\). Then Proposition 5 says that:_  
\[\zeta(x,z;u,w,t+s)\geq\zeta(y,x;v,u,t)*\zeta(y,z;v,w,s)\]1  
**4. Convergence Based on the Infimum**  
From Definition 1, we see that the double fuzzy semi-metric \(\zeta\) is a mapping from \(X^{4}\times[0,\infty)\) into \([0,1]\). Here, we shall consider its dual sense by considering the mapping from \((0,1]\times X^{4}\) into \([0,\infty)\). The formal definition is given below. **Definition 3**. : _Let \((X,M)\) be a fuzzy semi-metric space along with a t-norm \(*\). We also assume that \(M\) satisfies the canonical condition, and that the t-norm \(*\) is left-continuous at \(1\) with respect to the first or second argument. Given any fixed \(x,y,u,v\in X\) and any fixed \(\lambda\in(0,1]\), we consider the following set:_  
\[\zeta(x,z;u,w,t+s)\geq\zeta(y,x;v,u,t)*\zeta(y,z;v,w,s)\]2  
_which is used to define a mapping \(\Psi^{\downarrow}(\lambda,\cdot,\cdot;\cdot,\cdot):X^{4}\to[0,+\infty)\) by:_  
\[\zeta(x,z;u,w,t+s)\geq\zeta(y,x;v,u,t)*\zeta(y,z;v,w,s)\]3  
_In this case, the mapping \(\Psi^{\downarrow}\) from \((0,1]\times X^{4}\) into \([0,\infty)\) is called the infimum type of dual double fuzzy semi-metric._  
**Example 4**. : _Continued from Example 2, we have:_  
\[\zeta(x,z;u,w,t+s)\geq\zeta(y,x;v,u,t)*\zeta(y,z;v,w,s)\]4  
_where:_  
\[\zeta(x,z;u,w,t+s)\geq\zeta(y,x;v,u,t)*\zeta(y,z;v,w,s)\]5_We also have:_  
\[\zeta(x,z;u,w,t+s)\geq\zeta(y,x;v,u,t)*\zeta(y,z;v,w,s)\]6  
The potential application of dual double fuzzy semi-metric will be used to study the fixed point theorems in fuzzy semi-metric space. However, we first need to claim that the set \(\Pi^{\downarrow}(\lambda,x,y;u,v)\) is nonempty. Suppose that \(\Pi^{\downarrow}(\lambda,x,y;u,v)=\emptyset\). The definition says that \(\zeta(x,y;u,v,t)<1-\lambda\) for all \(t>0\); that is:  
\[\zeta(x,z;u,w,t+s)\geq\zeta(y,x;v,u,t)*\zeta(y,z;v,w,s)\]7  
which contradicts Equation (8). Therefore, Definition 3 is well-defined and \(\Pi^{\downarrow}(\lambda,x,y;u,v)\
eq\emptyset\). **Remark 2**. : _The following observations will be useful for further discussion._  
* _For any_ \(\lambda\in(0,1]\)_, we have:_ \[\zeta(x,z;u,w,t+s)\geq\zeta(y,x;v,u,t)*\zeta(y,z;v,w,s)\]8 _and:_ \[\zeta(x,z;u,w,t+s)\geq\zeta(y,x;v,u,t)*\zeta(y,z;v,w,s)\]9 \[\zeta(x,z;u,w,t+s)\geq\zeta(y,x;v,u,t)*\zeta(z,y;w,v,s)\]0 (10)
* _Given any fixed_ \(x,y,u,v\in X\)_, if_ \(\lambda_{1}>\lambda_{2}\)_, then:_ \[\zeta(x,z;u,w,t+s)\geq\zeta(y,x;v,u,t)*\zeta(z,y;w,v,s)\]1 (11)  
**Proposition 6**. : _Let \((X,M)\) be a fuzzy semi-metric space along with a \(t\)-norm \(*\). We also assume that \(M\) satisfies the canonical condition, and that the \(t\)-norm \(*\) is left-continuous at \(1\) with respect to the first or second argument. Given any fixed \(x,y,u,v\in X\), suppose that the following conditions are satisfied:_  
\[\zeta(x,z;u,w,t+s)\geq\zeta(y,x;v,u,t)*\zeta(z,y;w,v,s)\]2  
_and:_  
\[\zeta(x,z;u,w,t+s)\geq\zeta(y,x;v,u,t)*\zeta(z,y;w,v,s)\]3  
_Then we have:_  
\[\zeta(x,z;u,w,t+s)\geq\zeta(y,x;v,u,t)*\zeta(z,y;w,v,s)\]4  
_Moreover, the following limit exists:_  
\[\zeta(x,z;u,w,t+s)\geq\zeta(y,x;v,u,t)*\zeta(z,y;w,v,s)\]5  
**Proof. ** The assumption \(\Pi^{\downarrow}(0+,x,y;u,v)\
eq\emptyset\) says that we can consider \(t\in\Pi^{\downarrow}(0+,x,y;u,v)\), i.e., \(\zeta(x,y;u,v,t)\geq 1-\lambda\) for all \(\lambda\in(0,1]\).