# Proving Fixed-Point Theorems Employing Fuzzy \((\sigma,\mathcal{Z})\)-Contractive-Type Mappings
## 3 Main Results
: Pick out an arbitrary point \(\alpha_{0}\) in \(\mathcal{K}\) such that \(\sigma(\alpha_{0},T\alpha_{0},t)\leq 1\), for each \(t>0\), and consider a Picard sequence \(\{\alpha_{n}\}\) in \(\mathcal{K}\), that is,  
\[\alpha_{n+1}=T\alpha_{n},\text{ for all }n\in\mathbb{N}_{0}.\]  
In case \(\alpha_{n_{0}}=\alpha_{n_{0}+1}\), for some \(n_{0}\in\mathbb{N}_{0}\), then the fixed point of the mapping \(T\) is nothing but \(\alpha_{n_{0}}\). Assume that \(\alpha_{n+1}\
eq\alpha_{n}\), for each \(n\in\mathbb{N}_{0}\). As \(T\) is \(\sigma\)-admissible, we have:  
\[\sigma(\alpha_{0},\alpha_{1},t)=\sigma(\alpha_{0},T\alpha_{0},t)\leq 1\ \implies\ \sigma(\alpha_{1},\alpha_{2},t)=\sigma(T\alpha_{0},T\alpha_{1},t)\leq 1.\]  
The induction on \(n\), gives rise to:  
\[T(x)=1-x,\text{ for all }x\in\mathcal{K}.\]0  
Moreover, if for some \(m>n\), \(\alpha_{n}=\alpha_{m}\), then the contractive condition (2) and Equation (3) imply that:  
\[T(x)=1-x,\text{ for all }x\in\mathcal{K}.\]1 \[T(x)=1-x,\text{ for all }x\in\mathcal{K}.\]2 \[T(x)=1-x,\text{ for all }x\in\mathcal{K}.\]3  
hence,  
\[T(x)=1-x,\text{ for all }x\in\mathcal{K}.\]4  
Continuing in this way, one can show that  
\[T(x)=1-x,\text{ for all }x\in\mathcal{K}.\]5  
Since \(\alpha_{n}=\alpha_{m}\) for some \(m>n\), we have \(\alpha_{n+1}=\alpha_{m+1}\). This together with the above relation leads to a contradiction. Therefore, \(\alpha_{n}\
eq\alpha_{m}\) for each \(m>n\). In view of the condition \((d)\), there exists \(k_{0}\in\mathbb{N}\) such that \(\sigma(\alpha_{n},\alpha_{m},t)\leq 1,\ \forall\ m,n\in\mathbb{N}\) with  
\[T(x)=1-x,\text{ for all }x\in\mathcal{K}.\]6  
Applying the contractive condition (2) and making use of the above inequality, we obtain  
\[T(x)=1-x,\text{ for all }x\in\mathcal{K}.\]7 \[T(x)=1-x,\text{ for all }x\in\mathcal{K}.\]8 \[T(x)=1-x,\text{ for all }x\in\mathcal{K}.\]9  
and hence,  
\[\frac{t}{t+|x-y|}=\mathcal{M}(x,y,t)=\mathcal{M}(Tx,Ty,t) \geq\xi(\mathcal{M}(Tx,Ty,t),\mathcal{M}(x,y,t))\]0  
In the above inequality, taking the infimum over \(m(>n)\) and letting \(a_{n}(t)=\inf_{m>n}\mathcal{M}(\alpha_{n},\,\alpha_{m},t)\) we obtain that \(a_{n}(t)\leq a_{n+1}(t)\), for each \(t>0\), and hence, \(\{a_{n}(t)\}\) is a nondecreasing and bounded. Therefore, there exists \(a(t)\) such that \(\lim_{n\to\infty}a_{n}(t)=a(t)\). Our claim is to justify that \(a(t)=1\), for each \(t>0\). On the contrary, we assume that \(a(s)>1\), for some \(s>0\). From the fact that the quadruple \((\mathcal{K},\mathcal{M},T,\xi)\) owns the property \((S)\), we obtain  
\[\frac{t}{t+|x-y|}=\mathcal{M}(x,y,t)=\mathcal{M}(Tx,Ty,t) \geq\xi(\mathcal{M}(Tx,Ty,t),\mathcal{M}(x,y,t))\]1  
Equation (4) gives rise to  
\[\frac{t}{t+|x-y|}=\mathcal{M}(x,y,t)=\mathcal{M}(Tx,Ty,t) \geq\xi(\mathcal{M}(Tx,Ty,t),\mathcal{M}(x,y,t))\]2 \[\frac{t}{t+|x-y|}=\mathcal{M}(x,y,t)=\mathcal{M}(Tx,Ty,t) \geq\xi(\mathcal{M}(Tx,Ty,t),\mathcal{M}(x,y,t))\]3  
Taking \(n\to\infty\) in the above relation and using Equation (5), we obtain  
\[\frac{t}{t+|x-y|}=\mathcal{M}(x,y,t)=\mathcal{M}(Tx,Ty,t) \geq\xi(\mathcal{M}(Tx,Ty,t),\mathcal{M}(x,y,t))\]4  
which is a contradiction to the assumption \((a(s)>1\) for some \(s>0)\). This contradiction concludes that, for each \(t>0\), \(\lim_{n,m\to\infty}\mathcal{M}(\alpha_{n},\alpha_{m},t)=1\), that is, \(\{\alpha_{n}\}\) is a Cauchy sequence. Due to the \(\mathcal{M}\)-completeness of the fuzzy metric space \((\mathcal{K},\mathcal{M},*)\), there is \(\gamma\in\mathcal{K}\) such that  
\[\frac{t}{t+|x-y|}=\mathcal{M}(x,y,t)=\mathcal{M}(Tx,Ty,t) \geq\xi(\mathcal{M}(Tx,Ty,t),\mathcal{M}(x,y,t))\]5  
for all \(t>0\). The continuity of the mapping \(T\) implies that \(\lim_{n\to\infty}\mathcal{M}(T\alpha_{n},T\gamma,t)=1\), for each \(t>0\), and hence,  
\[\frac{t}{t+|x-y|}=\mathcal{M}(x,y,t)=\mathcal{M}(Tx,Ty,t) \geq\xi(\mathcal{M}(Tx,Ty,t),\mathcal{M}(x,y,t))\]6  
for all \(t>0\). Therefore, \(T\gamma=\gamma\), due to the uniqueness of the limit. In order to support the above-obtained result, we provide an example.