# Proving Fixed-Point Theorems Employing Fuzzy \((\sigma,\mathcal{Z})\)-Contractive-Type Mappings
## 3 Main Results
Hence, \(T\) is not a fuzzy \(\mathcal{Z}\)-contractive mapping._  
One of the advantages of \(\sigma\)-admissible mappings is that the continuity of the mapping is no longer required for the existence of a fixed point provided that the space under consideration satisfies a suitable condition (namely \((e^{\prime})\) given in the next theorem). Precisely, we state and prove the following theorem:  
**Theorem 4**. : _Let \((\mathcal{K},\mathcal{M},*)\) be an \(\mathcal{M}\)-complete FMS and \(\sigma:\mathcal{K}\times\mathcal{K}\times(0,\infty)\rightarrow[0,\infty)\). Assume that \(T:\mathcal{K}\rightarrow\mathcal{K}\) is a fuzzy \((\sigma,\mathcal{Z})\)-contractive mapping satisfy the following assumptions:_  
\((a)\): \(T\) _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_ \(m>n\geq k_{0}\)_,_ \(t>0\)_;_ \((e^{\prime})\): _If_ \(\{\alpha_{n}\}\) _is a sequence in_ \(\mathcal{K}\) _such that_ \(\lim_{n\rightarrow\infty}\alpha_{n}=\alpha\in\mathcal{K}\) _and_ \(\sigma(\alpha_{n},\alpha_{n+1},t)\leq 1\)_, for each_ \(n\in\mathbb{N}\) _and_ \(t>0\)_, then_ \(\sigma(\alpha_{n},\alpha,t)\leq 1\)_._ _Then,_ \(T\) _admits a fixed point._  
Proof. : The frame of the proof is the same as that in the previous theorem (Theorem 3). Therefore, for a Cauchy sequence \(\{\alpha_{n}\}\) in a complete FMS \((\mathcal{K},\mathcal{M},*)\), there exists \(\gamma\in\mathcal{K}\) such that \[\lim_{n\to\infty}(\alpha_{n},\gamma,t)=1,\ \forall\ t>0. \tag{6}\]  
Furthermore, we have \(\sigma(\alpha_{n},\alpha_{n+1},t)\leq 1\), for each \(n\in\mathbb{N}\) and \(t>0\), and hence, as a consequence of the condition \((e^{\prime})\), we obtain  
\[\sigma(\alpha_{n},\gamma,t)\leq 1, \tag{7}\]  
for each \(n\in\mathbb{N}\) and \(t>0\). Now, we have to show that \(T\) admits a fixed point (say \(\gamma\)). On the contrary, assume that \(T\gamma\
eq\gamma\), for all \(n\in\mathbb{N}\). Without loss of generality, one can assume that \(\alpha_{n}\
eq\gamma\) and \(\alpha_{n}\
eq T\gamma\). Then, there is \(s>0\) such that  
\[\mathcal{M}(\alpha_{n},\gamma,s)<1,\ \ \mathcal{M}(\alpha_{n},T\gamma,s)<1\ \text{ and }\ \mathcal{M}(T\alpha_{n},T\gamma,s)<1,\ \forall\ n\in\mathbb{N}. \tag{8}\]  
Using (2), (7) and (8), we obtain  
\[\mathcal{M}(\alpha_{n+1},T\gamma,s) \geq\sigma(\alpha_{n},\gamma,s)\mathcal{M}(T\alpha_{n},T\gamma,s)\] \[\geq\xi(\mathcal{M}(T\alpha_{n},T\gamma,s),\mathcal{M}(\alpha_{n},\gamma,s))\] \[\geq\xi(\mathcal{M}(T\alpha,T\beta,t),\mathcal{M}(\alpha,\beta,t))\]0  
Taking \(n\to\infty\) and making use of (6), we obtain \(\mathcal{M}(\gamma,T\gamma,s)\geq 1\), a contradiction. Therefore, for all \(t>0\), \(\mathcal{M}(\gamma,T\gamma,t)=1\), that is \(\gamma\) is the fixed point of \(T\). Next, we support Theorem 4 by an example in which the mapping \(T\) is not continuous. Moreover, we show the applicability of Theorem 4 over Theorems 1 and 3. **Example 6**. : _Let \(\mathcal{K}=[0,\infty)\), the set of all nonnegative real numbers, \(*\) be a minimum \(t\)-norm, and \(\mathcal{M}\) be a fuzzy set on \(\mathcal{K}^{2}\times(0,\infty)\) given by \(\mathcal{M}(x,y,t)=e^{-\frac{|x-y|}{t}}\), for all \(x,y\in\mathcal{K}\) and all \(t>0\). Then, \((\mathcal{K},\mathcal{M},*)\) is an \(\mathcal{M}\)-complete fuzzy metric space. Consider the mapping \(T:\mathcal{K}\to\mathcal{K}\) defined by_  
\[T(x)=\left\{ \begin{array}{ll}\frac{x}{2},&\text{ if }x\in[0,1],\\ x+1,&\text{ if }x\in(1,\infty).\end{array} \right.\]  
_It is obvious that \(T\) is not continuous at \(x=1\), and hence, Theorem 3 cannot be applied to this example.