# Proving Fixed-Point Theorems Employing Fuzzy \((\sigma,\mathcal{Z})\)-Contractive-Type Mappings
## 3 Main Results  
Throughout this article, \((\mathcal{K},\mathcal{M},*)\) is a fuzzy metric space in the [PERSON] and Veeramani sense. First of all, we start by introducing the notion of fuzzy \((\sigma,\mathcal{Z})\)-contractive mappings, which include many existing and familiar concepts as special cases. Definition 9: Let \(T\) be a self-mapping of an FMS \((\mathcal{K},\mathcal{M},*)\). We say that \(T\) is a fuzzy \((\sigma,\mathcal{Z})\)-contractive with respect to \(\xi\in\mathcal{Z}\) if there is \(\sigma:\mathcal{K}\times\mathcal{K}\times(0,\infty)\to(0,\infty)\) such that  
\[\sigma(\alpha,\beta,t)\mathcal{M}(T\alpha,T\beta,t)\geq\xi(\mathcal{M}(T \alpha,T\beta,t),\mathcal{M}(\alpha,\beta,t)), \tag{2}\]  
for all \(\alpha,\beta\in\mathcal{K},t>0\) with \(T\alpha\
eq T\beta\). Remark 2: By adopting the functions \(\xi\) and \(\sigma\) suitably in Definition 9, we deduce some well-known contractions as demonstrated below (for all \(\alpha,\beta\in\mathcal{K}\) and \(t>0\)):  
1. If \(\sigma(\alpha,\beta,t)=1\), for each \(\alpha,\beta\in\mathcal{K}\) and \(t>0\), then Definition 9 reduces to Definition 4. 2. Taking \(\xi(l,s)=\psi(s)\), for each \(l,s\in(0,1]\) and \(\psi\in\Psi\) in Definition 9, we deduce Definition 7. It is worth mentioning here that every fuzzy \(\mathcal{Z}\)-contractive is a fuzzy \((\sigma,\mathcal{Z})\)-contractive mapping, but the reverse is not in general true, as demonstrated by the following example:  
Example 4: Let \(\mathcal{K}=[0,1]\) and \(d(x,y)=|x-y|\), for all \(x,y\in\mathcal{K}\). Let \(\mathcal{M}\) be a fuzzy set on \(\mathcal{K}^{2}\times(0,\infty)\) given by \(\mathcal{M}(x,y,t)=\frac{t}{i+d(x,y)}\), \(t>0\). Define a mapping \(T:\mathcal{K}\to\mathcal{K}\) by  
\[T(x)=1-x,\text{ for all }x\in\mathcal{K}.\]  
\(T\) is not a fuzzy \(\mathcal{Z}\)-contractive mapping. On the contrary, we assume that \(T\) is a fuzzy \(\mathcal{Z}\)-contractive with respect to some \(\xi\in\mathcal{Z}\). Take \(x,y\in\mathcal{K}\) such that \(Tx\
eq Ty\). Since \(\mathcal{M}(x,y,t)=\mathcal{M}(Tx,Ty,t)=\frac{t}{t+|x-y|}\in(0,1)\), using Remark 1, we have  
\[\frac{t}{t+|x-y|}=\mathcal{M}(x,y,t)=\mathcal{M}(Tx,Ty,t) \geq\xi(\mathcal{M}(Tx,Ty,t),\mathcal{M}(x,y,t))\] \[>\mathcal{M}(x,y,t)=\frac{t}{t+|x-y|},\]  
for all \(t>0\), which is a contradiction. Hence, \(T\) is not a fuzzy \(\mathcal{Z}\)-contractive mapping. To show that \(T\) is a fuzzy \((\sigma,\mathcal{Z})\)-contractive mapping, we need to 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  
\[\xi(l,s)=s+l\,\text{ and }\,\sigma(x,y,t)=2.\]  
It is clear that \(\xi\in\mathcal{Z}\). Then, for all \(x,y\in\mathcal{K},t>0\), we have  
\[\frac{2t}{t+|x-y|} =\sigma(x,y,t)\mathcal{M}(Tx,Ty,t)\] \[\geq\xi(\mathcal{M}(Tx,Ty,t),\mathcal{M}(x,y,t))\] \[=\mathcal{M}(x,y,t)+\mathcal{M}(Tx,Ty,t)=\frac{2t}{t+|x-y|},\]  
which shows that \(T\) is a fuzzy \((\sigma,\mathcal{Z})\)-contractive mapping. Now, we are able to formulate our first main result as follows:  
Theorem 3.2: _Let \((\mathcal{K},\mathcal{M},*)\) be an \(\mathcal{M}\)-complete FMS and \(\sigma:\mathcal{K}\times\mathcal{K}\times(0,\infty)\to(0,\infty)\). Assume that \(T:\mathcal{K}\to\mathcal{K}\) is a fuzzy \((\sigma,\mathcal{Z})\)-contractive mapping and the following properties hold:_  
1. \(T\) _is_ \(\sigma\)_-admissible;_\((b)\): _The quadruple_ \((\mathcal{K},\mathcal{M},T,\xi)\) _owns the property_ \((S)\)_;_ \((c)\): _There exists_ \(\alpha_{0}\in\mathcal{K}\) _with_ \(\sigma(\alpha_{0},T\alpha_{0},t)\leq 1\)_, for each_ \(t>0\)_;_ \((d)\): _For each sequence_ \(\{\alpha_{n}\}\) _of_ \(\mathcal{K}\) _with the property that_ \(\sigma(\alpha_{n},\alpha_{n+1},t)\leq 1\)_, for each_ \(t>0\)_, there exists_ \(k_{0}\in\mathbb{N}\) _such that_ \(\sigma(\alpha_{n},\alpha_{m},t)\leq 1\)_, for each_ \(m,n\in\mathbb{N}\) _with_ \(m>n\geq k_{0}\)_,_ \(t>0\)_;_ \((e)\): \(T\) _is continuous._ _Then,_ \(T\) _admits a fixed point._  
Proof.