# Proving Fixed-Point Theorems Employing Fuzzy \((\sigma,\mathcal{Z})\)-Contractive-Type Mappings
## 3 Main Results
Define two essential functions \(\xi:(0,1]\times(0,1]\to\mathbb{R}\) and \(\sigma:\mathcal{K}\times\mathcal{K}\times(0,\infty)\to(0,\infty)\) by_  
\[\xi(l,s)=\left\{ \begin{array}{ll}l,&\text{ if }l>s,\\ \sqrt{s}&\text{ if }s\geq l,\end{array} \right.\quad\text{ and }\quad\sigma(x,y,t)=\left\{ \begin{array}{ll}1,&x,y\in[0,1],\\ e^{\frac{|\frac{x}{2}-y-1|}{t}},&x\in[0,1]\ \&\ y\in(1,\infty),\\ e^{\frac{|\frac{y}{2}-x-1|}{t}},&y\in[0,1]\ \&\ x\in(1,\infty),\\ e^{\frac{|x-y|}{t}},&x,y\in(1,\infty),\end{array} \right..\]  
_Let \(x,y\in\mathcal{K}\) such that \(\sigma(x,y,t)\leq 1\). Then, either \(x,y\in[0,1]\) or \(x=y\in(1,\infty)\). In case \(x,y\in[0,1]\), by the definition of \(T\), we have \(Tx,Ty\in[0,1]\), and hence, \(\sigma(Tx,Ty)=1\). In the other case, if \(x=y\in(1,\infty)\), then again, by the definition of \(T\), we have \(Tx=Ty\in(1,\infty)\), and hence, \(\sigma(Tx,Ty)=1\). Therefore, \(T\) is a \(\sigma\)-admissible mapping. Furthermore, \(1\in\mathcal{K}\) and \(\sigma(1,T1,t)=\sigma(1,\frac{1}{2},t)=1\). Further, let \(\{\alpha_{n}\}\) be a sequence in \(\mathcal{K}\) such that \(\lim_{n\to\infty}\alpha_{n}=x\) with \(k_{0}=1\) and \(\sigma(\alpha_{n},\alpha_{n+1},t)\leq 1\), for all \(n\in\mathbb{N}\). From the definition of \(\alpha\), it follows that \(\alpha_{n}\in[0,1]\), for all \(n\in\mathbb{N}\), if we assume that \(x\in(1,\infty)\), then we assume_  
\[\mathcal{M}(\alpha_{n},x,t)=e^{-\frac{|x_{n}-x|}{t}}<1,\ \text{ for all }t>0,\]_which is a contradiction of the assumption that \(\lim_{n\to\infty}\alpha_{n}=x\). Thus, we have \(x\in[0,1]\). Therefore, \(\alpha(\alpha_{n},x,t)\leq 1\) and \(\alpha(\alpha_{n},\alpha_{m},t)\leq 1\) for all \(m,n\in\mathbb{N}\) and \(t>0\)._  
_Finally, we show that \(T\) is a fuzzy \((\alpha,\mathcal{Z})\)-contractive mapping. To do so, for all \(x,y\in x\) with \(Tx\
eq Ty\), we consider the following four cases._  
**Case I:**: _If \(x,y\in[0,1]\), then (as \(\alpha(x,y,t)=1\) and \(\mathcal{M}(Tx,Ty,t)>\mathcal{M}(x,y,t)\)), and we have_  
\[e^{\frac{-|x-y|}{2}} =\alpha(x,y,t)\mathcal{M}(Tx,Ty,t)\] \[\geq\xi(\mathcal{M}(Tx,Ty,t),\mathcal{M}(x,y,t))\] \[=\mathcal{M}(Tx,Ty,t)=e^{\frac{(-|x-y|}{2})}.\]
**Case II:**: _If \(x\in[0,1]\) and \(y\in(1,\infty)\), then (as \(\alpha(x,y,t)=e^{\frac{(|\frac{x}{2}-y-1|)}{t}}\)), and we distinguish two subcases._  
_Subcase I: If \(\mathcal{M}(Tx,Ty,t)<\mathcal{M}(x,y,t)\), then we have_  
\[1=e^{\frac{|\frac{x}{2}-y-1|}{t}}\cdot e^{\frac{-|\frac{x}{2}-y- 1|}{t}} =\alpha(x,y,t)\mathcal{M}(Tx,Ty,t)\] \[\geq\xi(\mathcal{M}(Tx,Ty,t),\mathcal{M}(x,y,t))\] \[=\mathcal{M}(Tx,Ty,t)=e^{\frac{-|\frac{x}{2}-y-1|}{t}}.\]  
_Subcase II: If \(\mathcal{M}(Tx,Ty,t)\geq\mathcal{M}(x,y,t)\), then we have_  
\[1=e^{\frac{|\frac{x}{2}-y-1|}{t}}\cdot e^{\frac{-|\frac{x}{2}-y- 1|}{t}} =\alpha(x,y,t)\mathcal{M}(Tx,Ty,t)\] \[\geq\xi(\mathcal{M}(Tx,Ty,t),\mathcal{M}(x,y,t))\] \[=\sqrt{\mathcal{M}(x,y,t)}=e^{\frac{-|\frac{x}{2}-y-1|}{2}}.\]
**Case III:**: _This case is similar to that in_ **Case II**_._
**Case IV:**: _If \(x,y\in(1,\infty)\), then (as \(\alpha(x,y,t)=e^{\frac{(|x-y|)}{t}}\) and \(\mathcal{M}(Tx,Ty,t)=\mathcal{M}(x,y,t)\)), and we have_  
\[e^{\frac{-|x-y|}{2}} =\alpha(x,y,t)\mathcal{M}(Tx,Ty,t)\]0 \[e^{\frac{-|x-y|}{2}} =\alpha(x,y,t)\mathcal{M}(Tx,Ty,t)\]1 \[e^{\frac{-|x-y|}{2}} =\alpha(x,y,t)\mathcal{M}(Tx,Ty,t)\]2  
_Hence, in all cases, \(T\) is a fuzzy \((\alpha,\mathcal{Z})\)-contractive mapping. Therefore, all the hypotheses of Theorem 4 are satisfied. Hence, \(T\) has a fixed point (namely \(x=0\))._  
_However, \(T\) is not a fuzzy \(\mathcal{Z}\)-contractive mapping.