# Using Dual Double Fuzzy Semi-Metric to Study the Convergence
## 3 Double Fuzzy Semi-Metric
There exists_ \(\lambda\in(0,1)\) _such that the following inequalities are satisfied:_ * _If_ \(p\) _is even and_ \(\Psi^{\uparrow}(\mu,x_{1},x_{p};y_{1},y_{p})<+\infty\)_, then:_ \[\quad+\cdots+\Psi^{\downarrow}(\lambda,x_{p-1},x_{p};y_{p},y_{p-1 }),\]2 \[\quad+\cdots+\Psi^{\downarrow}(\lambda,x_{p-1},x_{p};y_{p},y_{p-1 }),\]3 \[\quad+\cdots+\Psi^{\downarrow}(\lambda,x_{p-1},x_{p};y_{p},y_{p-1 }),\]4 (48)  
To prove part (i), we just prove the first case, since the other cases can be similarly obtained. Suppose that \(M(x_{n},x,t)\to 1\) and \(M(y_{n},y,t)\to 1\) as \(n\to\infty\) for all \(t>0\). Then, given any \(t>0\) and \(\delta>0\), there exists \(n_{t,\delta}^{(1)},n_{t,\delta}^{(2)}\in\mathbb{N}\), satisfying \(|M(x_{n},x,t)-1|<\delta\) for \(n\geq n_{t,\delta}^{(1)}\) and \(|M(y_{n},y,t)-1|<\delta\) for \(n\geq n_{t,\delta}^{(2)}\). Given any \(\epsilon\in(0,1)\), there exists \(n_{\epsilon}\in\mathbb{N}\), satisfying  
\[\quad+\cdots+\Psi^{\downarrow}(\lambda,x_{p-1},x_{p};y_{p},y_{p-1 }),\]5  
for \(n\geq n_{\epsilon}\). We also have:  
\[\quad+\cdots+\Psi^{\downarrow}(\lambda,x_{p-1},x_{p};y_{p},y_{p-1 }),\]6  
for \(n\geq n_{\epsilon}\). The increasing property of t-norm says that:  
\[\quad+\cdots+\Psi^{\downarrow}(\lambda,x_{p-1},x_{p};y_{p},y_{p-1 }),\]7  
The first result of part (ii) of Proposition 17 says that:  
\[\quad+\cdots+\Psi^{\downarrow}(\lambda,x_{p-1},x_{p};y_{p},y_{p-1 }),\]8  
for \(n\geq n_{\epsilon}\). This shows that \(\Psi^{\dagger}(\lambda,x_{n},x;y_{n},y)\to 0\) as \(n\to\infty\). To prove the converse, suppose that \(\Psi^{\dagger}(\lambda,x_{n},x;y_{n},y)\to 0\) as \(n\to\infty\) for all \(\lambda\in(0,1)\). Given any \(\delta>0\) and \(\lambda\in(0,1]\), there exists \(n_{\delta,\lambda}\in\mathbb{N}\), satisfying \(|\Psi^{\dagger}(\lambda,x_{n},x;y_{n},y)|<\delta\) for all \(n\geq n_{\delta,\lambda}\). For any fixed \(t>0\) and given any \(\epsilon\in(0,1)\), there exists \(n_{\epsilon}\in\mathbb{N}\), satisfying:  
\[\quad+\cdots+\Psi^{\downarrow}(\lambda,x_{p-1},x_{p};y_{p},y_{p-1 }),\]9  
for \(n\geq n_{\epsilon}\), which implies:  
\[\Psi^{\downarrow}(\mu,x_{p},x_{1};y_{p},y_{1}) \leq\Psi^{\downarrow}(\lambda,x_{2},x_{1};y_{2},y_{1})+\Psi^{ \downarrow}(\lambda,x_{3},x_{2};y_{3},y_{2})+\Psi^{\downarrow}(\lambda,x_{3}, x_{4};y_{3},y_{4})\]0  
for \(n\geq n_{\epsilon}\) by the first result of part (i) of Proposition 16, i.e.,  
\[\Psi^{\downarrow}(\mu,x_{p},x_{1};y_{p},y_{1}) \leq\Psi^{\downarrow}(\lambda,x_{2},x_{1};y_{2},y_{1})+\Psi^{ \downarrow}(\lambda,x_{3},x_{2};y_{3},y_{2})+\Psi^{\downarrow}(\lambda,x_{3}, x_{4};y_{3},y_{4})\]1  
for \(n\geq n_{\epsilon}\). Lemma 2 says that:  
\[\Psi^{\downarrow}(\mu,x_{p},x_{1};y_{p},y_{1}) \leq\Psi^{\downarrow}(\lambda,x_{2},x_{1};y_{2},y_{1})+\Psi^{ \downarrow}(\lambda,x_{3},x_{2};y_{3},y_{2})+\Psi^{\downarrow}(\lambda,x_{3}, x_{4};y_{3},y_{4})\]2  
for \(n\geq n_{\epsilon}\). This shows that the sequences \(\{x_{n}\}_{n=1}^{\infty}\) and \(\{y_{n}\}_{n=1}^{\infty}\) in \(X\) converge to \(x\) and \(y\), respectively. To prove part (ii), the first result to the fourth result can be similarly obtained by the third result of part (ii) of Proposition 17. For proving the fifth result, the fact \(\Psi^{\dagger}(\lambda,x_{n},x;y_{n},y)\to 0\) implies the inequality (Equation (62)). The third result of part (i) of Proposition 16 says that \(\zeta(x,x_{n};y,y_{n},t)>1-\epsilon\), which implies \(M\left(x,x_{n},x,t\right)>1-\epsilon\) and \(M\left(y,y_{n},y,t\right)>1-\epsilon\). In other words, we have \(x_{n}\stackrel{{ M^{\epsilon}}}{{\longrightarrow}}x\) and \(y_{n}\stackrel{{ M^{\epsilon}}}{{\longrightarrow}}y\) as \(n\to\infty\). The remaining three results can be similarly obtained. This completes the proof. **Example 7**.