# Using Dual Double Fuzzy Semi-Metric to Study the Convergence
## 3 Double Fuzzy Semi-Metric
Then we have the following properties:_  
* \(x_{n}\stackrel{{ M^{\triangleright}}}{{\longrightarrow}}x\) _and_ \(y_{n}\stackrel{{ M^{\triangleright}}}{{\longrightarrow}}y\) _as_ \(n\to\infty\) _if and only if_ \(\Psi^{\downarrow}(\lambda,x_{n},x;y_{n},y)\to 0\) _as_ \(n\to\infty\) _for all_ \(\lambda\in(0,1)\)_._
* \(x_{n}\stackrel{{ M^{\triangleright}}}{{\longrightarrow}}x\) _and_ \(y_{n}\stackrel{{ M^{\triangleright}}}{{\longrightarrow}}y\) _as_ \(n\to\infty\) _if and only if_ \(\Psi^{\downarrow}(\lambda,x_{n},x;y,y_{n})\to 0\) _as_ \(n\to\infty\) _for all_ \(\lambda\in(0,1)\)_._
* \(x_{n}\stackrel{{ M^{\triangle}}}{{\longrightarrow}}x\) _and_ \(y_{n}\stackrel{{ M^{\triangleright}}}{{\longrightarrow}}y\) _as_ \(n\to\infty\) _if and only if_ \(\Psi^{\downarrow}(\lambda,x,x_{n};y_{n},y)\to 0\) _as_ \(n\to\infty\) _for all_ \(\lambda\in(0,1)\)_._
* \(x_{n}\stackrel{{ M^{\triangle}}}{{\longrightarrow}}x\) _and_ \(y_{n}\stackrel{{ M^{\triangleright}}}{{\longrightarrow}}y\) _as_ \(n\to\infty\) _if and only if_ \(\Psi^{\downarrow}(\lambda,x,x_{n};y,y_{n})\to 0\) _as_ \(n\to\infty\) _for all_ \(\lambda\in(0,1)\)_._  
Proof. : For any fixed \(\lambda\in(0,1)\), using Lemma 1, it follows that there exists \(\lambda_{0}\in(0,1)\), satisfying:  
\[\Psi^{\downarrow}(\mu,x_{1},x_{p};y_{p},y_{1}) \leq\Psi^{\downarrow}(\lambda,x_{1},x_{2};y_{2},y_{1})+\Psi^{ \downarrow}(\lambda,x_{2},x_{3};y_{3},y_{2})+\Psi^{\downarrow}(\lambda,x_{4}, x_{3};y_{3},y_{4})\]6  
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 \(n\geq n_{t,\delta}^{(2)}\). Therefore, given any \(\epsilon\in(0,1)\), there exists \(n_{\epsilon}\in\mathbb{N}\), satisfying:  
\[\Psi^{\downarrow}(\mu,x_{1},x_{p};y_{p},y_{1}) \leq\Psi^{\downarrow}(\lambda,x_{1},x_{2};y_{2},y_{1})+\Psi^{ \downarrow}(\lambda,x_{2},x_{3};y_{3},y_{2})+\Psi^{\downarrow}(\lambda,x_{4}, x_{3};y_{3},y_{4})\]7  
for \(n\geq n_{\epsilon}\). We also have:  
\[\Psi^{\downarrow}(\mu,x_{1},x_{p};y_{p},y_{1}) \leq\Psi^{\downarrow}(\lambda,x_{1},x_{2};y_{2},y_{1})+\Psi^{ \downarrow}(\lambda,x_{2},x_{3};y_{3},y_{2})+\Psi^{\downarrow}(\lambda,x_{4}, x_{3};y_{3},y_{4})\]8  
for \(n\geq n_{\epsilon}\). The increasing property of t-norm says that:  
\[\Psi^{\downarrow}(\mu,x_{1},x_{p};y_{p},y_{1}) \leq\Psi^{\downarrow}(\lambda,x_{1},x_{2};y_{2},y_{1})+\Psi^{ \downarrow}(\lambda,x_{2},x_{3};y_{3},y_{2})+\Psi^{\downarrow}(\lambda,x_{4}, x_{3};y_{3},y_{4})\]9  
The definition of \(\Psi^{\downarrow}\) says that:  
\[\quad+\Psi^{\downarrow}(\lambda,x_{4},x_{5};y_{5},y_{4})+\Psi^{ \downarrow}(\lambda,x_{6},x_{5};y_{5},y_{6})+\Psi^{\downarrow}(\lambda,x_{6}, x_{7};y_{7},y_{6})\]0  
for \(n\geq n_{\epsilon}\). This shows that \(\Psi^{\downarrow}(\lambda,x_{n},x;y_{n},y)\to 0\) as \(n\to\infty\). Conversely, assume that \(\Psi^{\downarrow}(\lambda,x_{n},x;y_{n},y)\to 0\) as \(n\to\infty\) for all \(\lambda\in(0,1)\). Now, given any \(\delta>0\) and \(\lambda\in(0,1]\), there exists \(n_{\delta,\lambda}\in\mathbb{N}\), satisfying \(|\Psi^{\downarrow}(\lambda,x_{n},x;y_{n},y)|<\delta\) for all \(n\geq n_{\delta,\lambda}\).