# Proving Fixed-Point Theorems Employing Fuzzy \((\sigma,\mathcal{Z})\)-Contractive-Type Mappings
## 3 Main Results
Precisely, we show that Theorem 3 can be used to cover this example while Theorem 1 is not applicable. **Example 5**. : _Consider \(\mathcal{K}=\{A_{1}=(0,0),A_{2}=(1,0),A_{3}=(1,2),A_{4}=(0,1),A_{5}=(1,3)\} \subseteq\mathbb{R}^{2}\). Define the fuzzy metric \(\mathcal{M}\) as_  
\[\frac{t}{t+|x-y|}=\mathcal{M}(x,y,t)=\mathcal{M}(Tx,Ty,t) \geq\xi(\mathcal{M}(Tx,Ty,t),\mathcal{M}(x,y,t))\]7  
_where \(d(\alpha,\beta)\) is the Euclidean metric on \(\mathbb{R}^{2}\). It is obvious that \((\mathcal{K},\mathcal{M},*)\) is an \(\mathcal{M}\)-complete FMS with respect to the product \(t\)-norm. Let \(T:\mathcal{K}\to\mathcal{K}\) be defined by_  
\[T(\alpha)=\left\{ \begin{array}{ll}A_{1},&\text{ if }\alpha\in\{A_{1},A_{3},A_{4},A _{5}\},\\ A_{5},&\text{ if }\alpha=A_{2}.\end{array} \right.\]  
_Furthermore, define \(\sigma:\mathcal{K}\times\mathcal{K}\times(0,\infty)\to(0,\infty)\) by_  
\[\sigma(\alpha,\beta,t)=\left\{ \begin{array}{ll}e^{\frac{t}{2}(2\sqrt{10}-3)},&\text{ if }\alpha=A_{2}\ \&\ \beta\in\{A_{1},A_{3},A_{4},A_{5}\}\\ &\text{ or }\alpha=\beta=A_{2}\\ 1&\text{ otherwise,}\end{array} \right.\]  
_and \(\xi:(0,1]\times(0,1]\to\mathbb{R}\) by_  
\[\xi(l,s)=\frac{s}{l},\text{ for all }l,s\in(0,1].\]_For all \(\alpha,\beta\in\mathcal{K}\), we have_  
\[\mathcal{M}(T\alpha,T\beta,t)=\left\{ \begin{array}{ll}e^{\frac{-2 }{t}(\sqrt{10})},&\text{ if }\alpha=A_{2}\ \&\ \beta\in\{A_{1},A_{3},A_{4},A_{5}\}\\ &\text{ or }\alpha\in\{A_{1},A_{3},A_{4},A_{5}\}\ \&\ \beta=A_{2}\\ 1&\text{ otherwise.}\end{array} \right.\]  
_Let \(\alpha,\beta\in\mathcal{K}\) such that \(\sigma(\alpha,\beta,t)\leq 1\). Then, \(\alpha,\beta\in\{A_{1},A_{3},A_{4},A_{5}\}\), and by the definition of \(T\), we conclude that \(T\alpha=T\beta\in\{A_{1},A_{3},A_{4},A_{5}\}\); hence, \(\sigma(T\alpha,T\beta)=1\). Therefore, the mapping \(T\) is \(\sigma\)-admissible. Furthermore, \(A_{4}\in\mathcal{K}\) and \(\sigma(A_{4},T\alpha_{4},t)=\alpha(A_{4},A_{1},t)=1\), for each \(t>0\). Further, let \(\{\alpha_{n}\}\subseteq\mathcal{K}\) such that \(\sigma(\alpha_{n},\alpha_{n+1},t)\leq 1\) with \(k_{0}=1\), for each \(n\in\mathbb{N}\). It follows that \(\alpha_{n}\in\{A_{1},A_{3},A_{4},A_{5}\}\), for each \(n\in\mathbb{N}\), and hence, \(\sigma(\alpha_{n},\alpha_{n},t)\leq 1\) for all \(m,n\in\mathbb{N}\) and \(t>0\). Furthermore, it is obvious that the quadruple \((\mathcal{K},\mathcal{M},T,\xi)\) has the property \((S)\)._  
_Finally, to show that \(T\) is a fuzzy \((\sigma,\mathcal{Z})\)-contractive mapping, we only need to consider the case \(\alpha=A_{2}\) and \(\beta\in\{A_{1},A_{3},A_{4},A_{5}\}\). In this case, \(\sigma(\alpha,\beta,t)=e^{\frac{t}{t}(2\sqrt{10}-1)}\), and hence,_  
\[e^{\frac{2}{t}(\sqrt{10}-3)}=e^{\frac{2}{t}(2\sqrt{10}-3)}\cdot e ^{\frac{-2}{t}(\sqrt{10})} =\sigma(\alpha,\beta,t)\mathcal{M}(T\alpha,T\beta,t)\] \[\geq\xi(\mathcal{M}(T\alpha,T\beta,t),\mathcal{M}(\alpha,\beta,t))\] \[=\frac{\mathcal{M}(\alpha,\beta,t)}{\mathcal{M}(T\alpha,T\beta,t )}=\frac{e^{\frac{-2\sqrt{10}}{t}}}{e^{\frac{-2}{t}(\sqrt{10}-3)}}=e^{\frac{ 2}{t}(\sqrt{10}-3)},\]  
_which shows that \(T\) is a fuzzy \((\sigma,\mathcal{Z})\)-contractive mapping. Therefore, all the hypotheses of Theorem 3 are satisfied. This ensures that the mapping \(T\) admits a fixed point (namely \(x=A_{1}\))._  
_However, \(T\) is not a fuzzy \(\mathcal{Z}\)-contractive mapping. On the contrary, we assume \(T\) is fuzzy \(\mathcal{Z}\)-contractive with respect to to some \(\xi\in\mathcal{Z}\). Take \(\alpha=A_{2}\) and \(\beta=A_{4}\). As \(\mathcal{M}(\alpha,\beta,t)=e^{\frac{-2\sqrt{2}}{t}}\in(0,1)\) and \(\mathcal{M}(T\alpha,T\beta,t)=e^{\frac{-2\sqrt{10}}{t}}\in(0,1)\), from the contractive condition and the definition of \(\xi\), we have_  
\[e^{\frac{-2\sqrt{10}}{t}}=\mathcal{M}(T\alpha,T\beta,t) \geq\xi(\mathcal{M}(T\alpha,T\beta,t),\mathcal{M}(\alpha,\beta,t))\] \[>\mathcal{M}(\alpha,\beta,t)=e^{\frac{-2\sqrt{2}}{t}},\]  
_for all \(t>0\), which is a contradiction.