In this section, we employ a binary relation \(\mathcal{S}\) to present some coincidence point results under Rcontractivity conditions. If ϱ denotes an Rfunction, we will consider the following three classes of contractivity conditions.
 (C_{1}):

\(\varrho ( d ( Tx,Ty ) ,d ( gx,gy ) ) >0 \) for all \(x,y\in X\) such that \(Tx\,\mathcal{S}^{\ast}\,Ty\) and \(gx\, \mathcal{S}^{\ast}\, gy\).
 (C_{2}):

\(\varrho ( d ( Tx,Ty ) ,d ( gx,gy ) ) >0 \) for each \(x,y\in X\) verifying \(gx\, \mathcal{S}^{\ast}\, gy\).
 (C_{3}):

\(\varrho ( d ( Tx,Ty ) ,d ( gx,gy ) ) >0 \) for each \(x,y\in X\) satisfying \(x\neq y\) and \(gx\,\mathcal{S}\,gy\).
Obviously, (C_{3}) ⇒ (C_{2}) ⇒ (C_{1}) because the only difference between them is the kind of points for which the inequality holds. The contractivity condition can also be useful to prove some properties of the binary relation.
Proposition 32
Suppose that
\(\varrho:D\times D\rightarrow\mathbb{R}\)
is a function for which (C_{3}) holds, and assume that
\(\varrho ( t,0 ) \leq0\)
for all
\(t>0\). Then
\(\mathcal{S}\)
is
\(( T,g ) \)compatible.
Proof
Let \(x,y\in X\) be such that \(gx=gy\) and \(gx\,\mathcal{S}\,gy\). By contradiction, assume that \(Tx\neq Ty\). Then, necessarily, \(x\neq y\). Let \(t_{0}=d ( Tx,Ty ) \in ( 0,\infty ) \cap\operatorname{ran}(d)\subseteq ( 0,\infty ) \cap D\). By (C_{3}), \(\varrho ( d ( Tx,Ty ) ,d ( gx,gy ) ) =\varrho ( t_{0},0 ) >0\), which contradicts \(\varrho ( t_{0},0 ) \leq0\). Thus \(Tx=Ty\). □
4.1 Coincidence point theorems under \(\mathcal{S}\)continuity
The first main result of the present manuscript is the following one, in which we use the weaker contractivity condition.
Theorem 33
Let
\(\mathcal{S}\)
be a binary relation on a metric space
\(( X,d ) \)
and let
\(T,g:X\rightarrow X\)
be two
\(\mathcal {S}\)continuous mappings such that
\(T(X)\)
is precomplete. Suppose that conditions (A) and (B) holds.
 (A):

There is on
X
a
\((T,g,\mathcal{S})\)PicardJungck sequence.
 (B):

There is an
Rfunction
\(\varrho\in R_{D}\)
such that
\(\operatorname{ran}(d)\subseteq D\)
and
$$ \varrho \bigl( d ( Tx,Ty ) ,d ( gx,gy ) \bigr) >0 \quad \textit{for all }x,y\in X\textit{ such that }Tx\, \mathcal{S}^{\ast}\, Ty\textit{ and }gx\, \mathcal{S}^{\ast}\, gy. $$
(4)
In addition to this, suppose, at least, one of the following assumptions holds.

(a)
The pair
\((T,g)\)
is
\(( O,\mathcal{S} ) \)compatible; or

(b)
T
and
g
are commuting.
Then
T
and
g
have a coincidence point. In fact, if
\(\{x_{n}\}\)
is any
\((T,g,\mathcal{S})\)PicardJungck sequence, either
\(\{gx_{n}\}\)
contains a coincidence point of
T
and
g, or
\(\{gx_{n}\}\)
converges to a coincidence point of
T
and
g.
We must point out the weakness of all hypotheses in the previous theorem because the following conditions, which we can find in many previous results in this field, are stronger.

If T and g are continuous, then T and g are \(\mathcal {S}\)continuous.

If there is a subset A such that \(T(X)\subseteq A\subseteq X\) and A is precomplete or complete, then \(T(X)\) is precomplete. In particular, if one of \(T(X)\), or \(g(X)\), or X is complete, then \(T(X)\) is precomplete.

By Proposition 26, hypothesis (A) is guaranteed under the following conditions.
 (A′):

\(\mathcal{S}\) is \(( T,g ) \)transitive (or gtransitive), \(TX\subseteq g(X)\), T is \(( g,\mathcal{S} ) \)nondecreasing and there exists a point \(x_{0}\in X\) such that \(gx_{0}\,\mathcal{S}\,Tx_{0}\).
 (A^{′′}):

\(\mathcal{S}\) is \(( T,g ) \)transitive (or gtransitive) and \(T(X)\subseteq_{\mathcal {S}}g(X)\).
Proof
By hypothesis (A), we can find a \((T,g,\mathcal{S})\)PicardJungck sequence \(\{x_{n}\}_{n\geq0}\) on X, that is, a sequence verifying \(gx_{n+1}=Tx_{n}\) and \(gx_{n}\,\mathcal{S}\,gx_{m}\) for all \(n< m\), \(n,m\in\mathbb{N}\). If there exists some \(n_{0}\in\mathbb{N}\) such that \(gx_{n_{0}+1}=gx_{n_{0}}\), then \(x_{n_{0}}\) is a coincidence point of T and g. Assume that \(gx_{n}\neq gx_{n+1}\) for all \(n\in\mathbb{N}\). Hence \(gx_{n}\, \mathcal{S}^{\ast}\, gx_{n+1}\) and \(Tx_{n}\, \mathcal{S}^{\ast }\, Tx_{n+1}\) for all \(n\in\mathbb{N}\). Let \(\{a_{n}\}\subset ( 0,\infty ) \) be the sequence defined by \(a_{n}=d(gx_{n},gx_{n+1})>0\) for all \(n\in\mathbb{N}\). Using the contractivity condition (4),
$$\begin{aligned} \varrho ( a_{n+1},a_{n} ) &=\varrho \bigl( d(gx_{n+1},gx_{n+2}),d(gx_{n},gx_{n+1}) \bigr) \\ &=\varrho \bigl( d(Tx_{n},Tx_{n+1}),d(gx_{n},gx_{n+1}) \bigr) >0 \end{aligned}$$
for all \(n\in\mathbb{N}\). Applying (\(\varrho_{1}\)) we deduce that \(\{d(gx_{n},gx_{n+1})\}=\{a_{n}\}\rightarrow0\), that is, \(\{ gx_{n}\}\) is an asymptotically regular sequence.
Next we prove that \(\{gx_{n}\}\) is Cauchy reasoning by contradiction. If it is not Cauchy, then there are \(\varepsilon_{0}>0\) and two subsequences \(\{gx_{n(k)}\}\) and \(\{gx_{m(k)}\}\) of \(\{gx_{n}\}\) such that
$$\begin{aligned}& k\leq n(k)< m(k),\quad d(gx_{n(k)},gx_{m(k)1})\leq\varepsilon _{0}< d(gx_{n(k)},gx_{m(k)})\quad \text{for all }k\in \mathbb{N}, \\& \lim_{k\rightarrow\infty}d(gx_{n(k)},gx_{m(k)})=\lim _{k\rightarrow \infty }d(gx_{n(k)1},gx_{m(k)1})= \varepsilon_{0}. \end{aligned}$$
As the sequences \(\{d(gx_{n(k)},gx_{m(k)})\}\) and \(\{d(gx_{n(k)1},gx_{m(k)1})\}\) converge to \(\varepsilon_{0}>0\), there exists \(k_{0}\in\mathbb{N}\) such that
$$d(gx_{n(k)},gx_{m(k)})>0\quad \text{and}\quad d(gx_{n(k)1},gx_{m(k)1})>0 $$
for all \(k\geq k_{0}\). In particular, \(d(Tx_{n(k)1},Tx_{m(k)1})=d(gx_{n(k)},gx_{m(k)})>0\) for all \(k\geq k_{0}\), which implies that
$$Tx_{n(k)1}\, \mathcal{S}^{\ast}\, Tx_{m(k)1}\quad \text{and}\quad gx_{n(k)1}\, \mathcal{S}^{\ast}\, gx_{m(k)1} \quad \text{for all }k\geq k_{0}. $$
Let \(L=\varepsilon_{0}>0\), \(\{a_{k}=d(gx_{n(k)},gx_{m(k)})\}_{k\geq k_{0}}\rightarrow L\) and \(\{b_{k}=d(gx_{n(k)1},gx_{m(k)1})\}_{k\geq k_{0}}\rightarrow L\). Since \(L=\varepsilon_{0}< d(gx_{n(k)},gx_{m(k)})=a_{k}\) and
$$\begin{aligned} \varrho ( a_{k},b_{k} ) & =\varrho \bigl( d(gx_{n(k)},gx_{m(k)}),d(gx_{n(k)1},gx_{m(k)1}) \bigr) \\ & =\varrho \bigl( d(Tx_{n(k)1},Tx_{m(k)1}),d(gx_{n(k)1},gx_{m(k)1}) \bigr) >0 \end{aligned}$$
for all \(k\geq k_{0}\), condition (\(\varrho_{2}\)) guarantees that \(\varepsilon_{0}=L=0\), which is a contradiction. As a result, the sequence \(\{gx_{n}\}\) is Cauchy. Since \(\{gx_{n+1}=Tx_{n}\}\subseteq T(X)\) and \(T(X)\) is precomplete, there exists \(z\in X\) such that \(\{gx_{n}\} \rightarrow z\). Furthermore, as T and g are \(\mathcal{S}\)continuous and \(gx_{n}\,\mathcal{S}\,gx_{m}\) for all \(n< m\), then \(\{Tgx_{n}\}\rightarrow Tz\) and \(\{ggx_{n}\}\rightarrow gz\). Next, we consider two cases.
Case (a). Assume the
\(( O,\mathcal{S} ) \)compatibility of the pair
\((T,g)\). Taking into account that \(gx_{n}\,\mathcal{S}\,gx_{m}\) for all \(n< m\) and \(\{Tx_{n}=gx_{n+1}\} \rightarrow z\), the \(( O,S ) \)compatibility of \(( T,g ) \) yields
$$\lim_{n\rightarrow\infty}d(gTx_{n},Tgx_{n})=0. $$
Notice that
$$d ( Tz,gz ) =d \Bigl( \lim_{n\rightarrow\infty}Tgx_{n}, \lim_{n\rightarrow\infty}ggx_{n+1} \Bigr) =\lim_{n\rightarrow\infty } d ( Tgx_{n},gTx_{n} ) =0, $$
so \(Tz=gz\) and z is a coincidence point of T and g.
Case (b). Assume that
T
and
g
are commuting. Then Case (a) is applicable because commutativity implies \((O,\mathcal{S} ) \)compatibility. □
The following consequence is obtained by using the binary relation \(x\,\mathcal{S}\,y\) for all \(x,y\in X\).
Corollary 34
Let
\(( X,d ) \)
be a metric space and let
\(T,g:X\rightarrow X\)
be two continuous mappings such that
\(T(X)\)
is precomplete. Suppose that conditions (A) and (B) holds.
 (A):

There exists on
X
a
\((T,g)\)PicardJungck sequence.
 (B):

There exists an
Rfunction
\(\varrho\in R_{D}\)
such that
\(\operatorname{ran}(d)\subseteq D\)
and
$$\varrho \bigl( d ( Tx,Ty ) ,d ( gx,gy ) \bigr) >0 \quad \textit{for all }x,y\in X\textit{ such that }Tx\neq Ty\textit{ and }gx\neq gy. $$
Additionally, assume that the pair
\((T,g)\)
is
Ocompatible or
T
and
g
are commuting.
Then
T
and
g
have, at least, a coincidence point. In fact, if
\(\{ x_{n}\}\)
is any
\((T,g)\)PicardJungck sequence, either
\(\{gx_{n}\}\)
contains a coincidence point of
T
and
g, or
\(\{gx_{n}\}\)
converges to a coincidence point of
T
and
g.
In the following result, we denote by ≺ a transitive binary relation (for instance, a preorder or a partial order), which is not necessarily reflexive. In this case, we replace hypothesis (A) by (A′) by virtue of Proposition 26 and the new condition (C).
Corollary 35
Let
\(( X,d ) \)
be a metric space endowed with a binary relation ≺ and let
\(T,g:X\rightarrow X\)
be two ≺continuous mappings such that
\(T(X)\)
is precomplete. Suppose that conditions (A) and (B) holds.
 (A′):

\(TX\subseteq g(X)\), T
is
\(( g,\prec ) \)nondecreasing and there exists a point
\(x_{0}\in X\)
such that
\(gx_{0}\prec Tx_{0}\).
 (B):

There exists an
Rfunction
\(\varrho\in R_{D}\)
such that
\(\operatorname{ran}(d)\subseteq D\)
and
$$\varrho \bigl( d ( Tx,Ty ) ,d ( gx,gy ) \bigr) >0\quad \textit{for all }x,y\in X \textit{ such that }Tx\prec Ty\textit{ and }gx\prec gy. $$
 (C):

The binary relation ≺ is transitive (or
gtransitive, or
\(( T,g ) \)transitive).
Additionally, assume that the pair
\((T,g)\)
is
\(( O,\mathcal {S} ) \)compatible or
T
and
g
are commuting.
Then
T
and
g
have, at least, a coincidence point. In fact, if
\(\{ x_{n}\}\)
is any
\((T,g,\prec)\)PicardJungck sequence, either
\(\{gx_{n}\}\)
contains a coincidence point of
T
and
g, or
\(\{gx_{n}\}\)
converges to a coincidence point of
T
and
g.
If \(g=I_{X}\) is the identity mapping on X, then we derive the following result.
Corollary 36
Let
\(\mathcal{S}\)
be a binary relation on a metric space
\(( X,d ) \)
and let
\(T:X\rightarrow X\)
be an
\(\mathcal{S}\)continuous mapping such that
\(T(X)\)
is precomplete. Suppose that conditions (A) and (B) holds.
 (A):

There exists on
X
a Picard sequence
\(\{x_{n+1}=Tx_{n}\}\)
such that
\(x_{n}\,\mathcal{S}\,x_{m}\)
for all
\(n,m\in \mathbb{N}\)
such that
\(n< m\).
 (B):

There exists an
Rfunction
\(\varrho\in R_{D}\)
such that
\(\operatorname{ran}(d)\subseteq D\)
and
$$\varrho \bigl( d ( Tx,Ty ) ,d ( x,y ) \bigr) >0 \quad \textit{for all }x,y\in X \textit{ such that }x\, \mathcal{S}^{\ast}\, y\textit{ and }Tx\, \mathcal{S}^{\ast}\, Ty. $$
Then
T
has, at least, a fixed point.
4.2 Coincidence point theorems under \(\mathcal{S}\)regularity and condition (C_{2})
In the following result, the contractivity condition is stronger because we do not assume that \(Tx\, \mathcal{S}^{\ast}\, Ty\) and \(gx\, \mathcal{S}^{\ast}\, gy\). However, the following result is applicable even if T and g are not \(\mathcal{S}\)continuous.
Theorem 37
Let
\(\mathcal{S}\)
be a binary relation on a metric space
\(( X,d ) \)
such that
X (or
\(g(X)\)) is
\((d,\mathcal{S})\)regular, and let
\(T,g:X\rightarrow X\)
be two mappings. Suppose that conditions (A) and (B) hold.
 (A):

There exists on
X
a
\((T,g,\mathcal{S})\)PicardJungck sequence.
 (B):

There exists an
Rfunction
\(\varrho\in R_{D}\)
such that (\(\varrho_{3}\)) holds, \(\operatorname{ran}(d)\subseteq D\)
and
$$ \varrho \bigl( d ( Tx,Ty ) ,d ( gx,gy ) \bigr) >0 \quad \textit{for all }x,y\in X\textit{ such that }gx\, \mathcal{S}^{\ast }\, gy. $$
(5)
In addition to this, suppose, at least, one of the following assumptions holds.

(c)
\((g(X),d)\)
is complete;

(d)
\((X,d)\)
is complete and
\(g(X)\)
is closed;

(e)
\((X,d)\)
is complete, the pair
\((T,g)\)
is
\(( O,\mathcal{S} ) \)compatible, \(\mathcal{S}\)
is
gclosed and
g
is injective on
\(g ( X ) \)
and
\(\mathcal{S}\)continuous.
Then
T
and
g
have, at least, a coincidence point.
Proof
Since (5) ⇒ (4), the proof of Theorem 33 can be followed, point by point, to prove that the sequence \(\{gx_{n}\}\), which satisfies \(gx_{n}\neq gx_{n+1}\) for all \(n\in\mathbb{N}\), is a Cauchy sequence. Next, we distinguish some cases.
Case (c). Assume that
\(g(X)\)
is complete. Since \(g ( X ) \) is complete, there exists \(u\in g ( X ) \) such that \(\{gx_{n}\}\rightarrow u\). As \(u\in g(X)\), then \(g^{1} ( u ) \) is not empty. We are going to show that any point \(z\in g^{1}(u)\) is a coincidence point of T and g. Indeed, let \(z\in X\) be an arbitrary point such that \(gz=u\). Since \(gx_{n}\,\mathcal{S}\,gx_{n+1}\) for all \(n\in\mathbb{N}\) and \(\{gx_{n}\}\rightarrow gz\), the \(( d,\mathcal{S} ) \)regularity of X (or \(g(X)\)) yields \(gx_{n}\,\mathcal{S}\,gz\) for all \(n\in\mathbb{N}\). If there exists some \(n_{0}\in\mathbb{N}\) such that \(gx_{n_{0}}=gz\), then \(gx_{n_{0}+1}\neq gz\). Therefore, the set \(\{ n\in \mathbb{N}:gx_{n}\neq gz \}\) is not finite. As a consequence, there exists a subsequence \(\{gx_{n(k)}\}\) of \(\{gx_{n}\}\) such that
$$d ( gx_{n(k)},gz ) >0 \quad \text{for all }k\in\mathbb{N}. $$
In particular, \(gx_{n(k)}\, \mathcal{S}^{\ast}\, gz\) for all \(k\in\mathbb{N}\). Applying the contractivity condition (5),
$$ \varrho \bigl( d ( Tx_{n(k)},Tz ) ,d ( gx_{n(k)},gz ) \bigr) >0 \quad \text{for all }k\in\mathbb{N}. $$
(6)
Let \(M=\{ k\in\mathbb{N}:Tx_{n(k)}=Tz \}\).
Subcase (c.1). Assume that
M
is finite. In this case, there exists \(k_{0}\in\mathbb{N}\) such that \(Tx_{n(k)}\neq Tz\) for all \(k\geq k_{0} \). Let \(\{a_{k}=d ( Tx_{n(k)},Tz ) \}_{k\geq k_{0}}\) and \(\{b_{k}=d ( gx_{n(k)},gz ) \}_{k\geq k_{0}}\). Then \(a_{k}>0\) and \(b_{k}>0\) for all \(k\geq k_{0}\). Moreover, \(\{b_{k}\}\rightarrow0\). As (6) means that \(\varrho ( a_{k},b_{k} ) >0\) for all \(k\geq k_{0}\), condition (\(\varrho_{3}\)) implies that \(\{a_{k}\}\rightarrow0\). It follows that \(\{gx_{n(k)+1}=Tx_{n(k)}\} \rightarrow Tz\). As this is a subsequence of \(\{gx_{n}\}\) and \(\{gx_{n}\} \rightarrow gz\), we conclude that \(Tz=gz\), that is, z is a coincidence point of T and g.
Subcase (c.2). Assume that
M
is not finite. In this case, there exists a subsequence \(\{Tx_{n^{\prime}(k)}\}\) of \(\{Tx_{n(k)}\}\) such that \(Tx_{n^{\prime}(k)}=Tz\) for all \(k\in\mathbb{N}\). Since \(gx_{n^{\prime }(k)+1}=Tx_{n^{\prime}(k)}=Tz\) for all \(k\in\mathbb{N}\), and \(\{gx_{n}\}\rightarrow gz\), we also conclude that \(Tz=gz\), that is, z is a coincidence point of T and g.
Case (d). Assume that
\(g(X)\)
is closed and
\((X,d)\)
is complete. In this case, we can apply item (c) because any closed subsets of complete spaces are also complete.
Case (e). Assume that
\((X,d)\)
is complete, the pair
\((T,g)\)
is
\(( O,\mathcal{S} ) \)compatible, \(\mathcal{S}\)
is
gclosed and
g
is injective on
\(g ( X ) \)
and
\(\mathcal{S}\)continuous. The completeness of X guarantees that there is \(u\in X\) satisfying \(\{gx_{n}\}\rightarrow u\). As g is \(\mathcal{S}\)continuous and \(gx_{n}\,\mathcal{S}\,gx_{m}\) for all \(n< m\), then \(\{ggx_{m}\}\rightarrow gu\). Moreover, the \(( O,\mathcal{S} ) \)compatibility of \((T,g)\) leads to
$$\lim_{n\rightarrow\infty}d(gu,Tgx_{n})=\lim_{n\rightarrow\infty}d(ggx_{n+1},Tgx_{n})= \lim_{n\rightarrow\infty}d(gTx_{n},Tgx_{n})=0. $$
Hence \(\{Tgx_{n}\}\rightarrow gu\). We are going to show that \(\{ Tgx_{n}\}\) has a subsequence converging to Tu, and this will also finish the proof (because, in such a case, \(Tu=gu\)). Let us consider \(M=\{ n\in\mathbb{N} :Tgx_{n}=Tu \}\). If M is not finite, then there is a partial subsequence \(\{gx_{n(k)}\}\) of \(\{gx_{n}\}\) such that \(Tgx_{n(k)}=Tu\) for all \(k\in\mathbb{N}\). As \(\{Tgx_{n}\}\rightarrow gu\), then \(Tu=gu\), and the proof is finished. On the contrary, assume that M is finite. In such a case, there exists \(n_{0}\in\mathbb{N}\) such that \(Tgx_{n}\neq Tu\) for all \(n\geq n_{0}\). For simplicity, we will assume that
$$Tgx_{n}\neq Tu \quad \text{for all }n\in\mathbb{N}. $$
Since X is \((d,\mathcal{S})\)regular, \(\{gx_{n}\}\rightarrow u\) and \(gx_{n}\,\mathcal{S}\,gx_{m}\) for all \(n< m\), then \(gx_{n}\,\mathcal{S}\,u\) for all \(n\in\mathbb{N}\). Moreover, as \(\mathcal{S}\) is gclosed, then \(ggx_{n}\,\mathcal{S}\,gu\) for all \(n\in\mathbb{N}\). We must consider two cases depending on \(M^{\prime}=\{ n\in\mathbb{N}:gx_{n}=u \}\).
Subcase (e.1). Assume that
\(M^{\prime}\)
is finite. In this case, there exists \(n_{0}\in\mathbb{N}\) such that \(gx_{n}\neq u\) for all \(n\geq n_{0}\). Then \(ggx_{n}\neq gu\) for all \(n\geq n_{0}\) because g is injective. Hence \(ggx_{n}\, \mathcal{S}^{\ast}\, gu\) for all \(n\geq n_{0}\). Let \(\{a_{n}\}\) and \(\{b_{n}\}\) be the sequences of positive real numbers given by
$$a_{n}=d ( Tgx_{n},Tu ) >0 \quad \text{and}\quad b_{n}=d ( ggx_{n},gu ) >0\quad \text{for all }n\geq n_{0}. $$
Clearly, \(\{b_{n}\}\rightarrow0\). Applying the contractivity condition (5),
$$\varrho ( a_{n},b_{n} ) =\varrho \bigl( d ( Tgx_{n},Tu ) ,d ( ggx_{n},gu ) \bigr) >0\quad \text{for all }n\geq n_{0}. $$
By (\(\varrho_{3}\)), we deduce that \(\{a_{n}\}\rightarrow0\), which implies that \(\{Tgx_{n}\}\rightarrow Tu\).
Subcase (e.2). Assume that
\(M^{\prime}\)
is not finite. In this case, there exists a subsequence \(\{gx_{n(k)}\}\) of \(\{gx_{n}\}\) such that \(gx_{n(k)}=u\) for all \(k\in\mathbb{N}\). Hence \(\{Tgx_{n(k)}\} \rightarrow Tu\). □
If \(x\,\mathcal{S}\,y\) for all \(x,y\in X\), we derive the following consequence.
Corollary 38
Given two maps
\(T,g:X\rightarrow X\)
in a metric space
\(( X,d ) \), assume that conditions (A) and (B) hold.
 (A):

There is on
X
a
\((T,g)\)PicardJungck sequence.
 (B):

There exists an
Rfunction
\(\varrho\in R_{D}\)
such that (\(\varrho_{3}\)) holds, \(\operatorname{ran}(d)\subseteq D\)
and
$$\varrho \bigl( d ( Tx,Ty ) ,d ( gx,gy ) \bigr) >0\quad \textit{for all }x,y\in X \textit{ such that }gx\neq gy. $$
In addition to this, suppose, at least, one of the following assumptions holds.

(c)
\((g(X),d)\)
is complete;

(d)
\((X,d)\)
is complete and
\(g(X)\)
is closed;

(e)
\((X,d)\)
is complete, g
is injective on
\(g ( X ) \)
and continuous, and the pair
\((T,g)\)
is
Ocompatible.
Then
T
and
g
have, at least, a coincidence point.
In the following result, we denote by ≺ a transitive binary relation (for instance, a preorder or a partial order), which is not necessarily reflexive.
Corollary 39
Let
\(( X,d ) \)
be a metric space endowed with a binary relation ≺ such that
X (or
\(g(X)\)) is
\((d,\prec)\)nondecreasingregular, and let
\(T,g:X\rightarrow X\)
be two mappings. Assume that conditions (A) and (B) hold.
 (A):

\(TX\subseteq g(X)\), T
is
\(( g,\prec ) \)nondecreasing and there exists a point
\(x_{0}\in X\)
such that
\(gx_{0}\prec Tx_{0}\).
 (B):

There exists an
Rfunction
\(\varrho\in R_{D}\)
such that (\(\varrho_{3}\)) holds, \(\operatorname{ran}(d)\subseteq D\), and
$$\varrho \bigl( d ( Tx,Ty ) ,d ( gx,gy ) \bigr) >0 \quad \textit{for all }x,y\in X\textit{ such that }gx\, \mathcal{S}^{\ast}\, gy. $$
 (C):

The binary relation ≺ is transitive (or
gtransitive, or
\(( T,g ) \)transitive).
In addition to this, suppose, at least, one of the following assumptions holds.

(c)
\((g(X),d)\)
is complete;

(d)
\((X,d)\)
is complete and
\(g(X)\)
is closed;

(e)
\((X,d)\)
is complete, the pair
\((T,g)\)
is
\(( O,\prec ) \)compatible, g
is ≺nondecreasing, and
g
is injective on
\(g ( X ) \)
and
\(\mathcal{S}\)continuous.
Then
T
and
g
have, at least, a coincidence point.
Furthermore, if \(g=I_{X}\) is the identity mapping on X, then we derive the following fixed point result.
Corollary 40
Let
\(( X,d ) \)
be a complete, \((d,\mathcal{S})\)regular metric space endowed with a binary relation
\(\mathcal{S}\)
and let
\(T:X\rightarrow X\)
be a mapping. Assume that conditions (A) and (B) hold.
 (A):

There is on
X
a Picard sequence
\(\{x_{n+1}=Tx_{n}\}\)
such that
\(x_{n}\,\mathcal{S}\,x_{m}\)
for all
\(n,m\in\mathbb{N}\)
such that
\(n< m\).
 (B):

There exists an
Rfunction
\(\varrho\in R_{D}\)
such that (\(\varrho_{3}\)) holds, \(\operatorname{ran}(d)\subseteq D\), and
$$\varrho \bigl( d ( Tx,Ty ) ,d ( x,y ) \bigr) >0 \quad \textit{for all }x,y\in X \textit{ such that }x\, \mathcal{S}^{\ast}\, y. $$
Then
T
has, at least, a fixed point.
4.3 Coincidence point theorems under \(\mathcal{S}\)regularity and condition (C_{3})
If we assume that the contractivity condition is more restrictive, we can avoid the injectivity of g in condition (e) in Theorem 37.
Theorem 41
Let
\(\mathcal{S}\)
be a binary relation on a metric space
\(( X,d ) \)
and let
\(T,g:X\rightarrow X\)
be two mappings. Assume that conditions (A) and (B) hold.
 (A):

There is on
X
a
\((T,g,\mathcal{S})\)PicardJungck sequence.
 (B):

There exists an
Rfunction
\(\varrho\in R_{D}\)
such that (\(\varrho_{4}\)) holds, \(\operatorname{ran}(d)\subseteq D\), and
$$ \varrho \bigl( d ( Tx,Ty ) ,d ( gx,gy ) \bigr) >0 \quad \textit{for each }x,y\in X\textit{ such that }x\neq y\textit{ and }gx\,\mathcal{S}\,gy. $$
(7)
 (e′):

\((X,d)\)
is complete, \(\mathcal{S}\)
is
gclosed, \((T,g)\)
is
\(( O,\mathcal{S} ) \)compatible, g
is
\(\mathcal {S}\)continuous and
X
is
\((d,\mathcal{S})\)regular.
Then
T
and
g
have, at least, a coincidence point.
Proof
Since (7) ⇒ (4), the proof of Theorem 33 can be followed, point by point, to deduce that \(\{gx_{n}\}\) is Cauchy. The completeness of \((X,d)\) implies the existence of \(u\in X\) such that \(\{gx_{n}\}\rightarrow u\). Since \(gx_{n}\,\mathcal{S}\,gx_{m}\) for each \(n< m\) and g is \(\mathcal{S}\)continuous, then \(\{ggx_{m}\}\rightarrow gu\). Furthermore, from the \(( O,\mathcal{S} ) \)compatibility of the pair \((T,g)\),
$$\lim_{n\rightarrow\infty}d(gu,Tgx_{n})=\lim_{n\rightarrow\infty}d(ggx_{n+1},Tgx_{n})= \lim_{n\rightarrow\infty}d(gTx_{n},Tgx_{n})=0. $$
Hence \(\{Tgx_{n}\}\rightarrow gu\). Let us consider \(M=\{ n\in\mathbb{N}:Tgx_{n}=Tu \}\). If M is not finite, then there is a subsequence \(\{gx_{n(k)}\}\) of \(\{gx_{n}\}\) such that \(Tgx_{n(k)}=Tu\) for all \(k\in\mathbb{N}\). As \(\{Tgx_{n}\}\rightarrow gu\), then \(Tu=gu\), and the proof is finished. On the other case, if M is finite, there is \(n_{0}\in \mathbb{N}\) such that \(Tgx_{n}\neq Tu\) for each \(n\geq n_{0}\). For simplicity, we will assume that
$$Tgx_{n}\neq Tu \quad \text{for all }n\in\mathbb{N}. $$
The proof will be finished if we prove that \(\{Tgx_{n}\}\rightarrow Tu\). Indeed, the \((d,\mathcal{S})\)regularity of X yields \(\{gx_{n}\}\rightarrow u\) and \(gx_{n}\,\mathcal{S}\,gx_{m}\) for each \(n< m\), so \(gx_{n}\,\mathcal{S}\,u\) for all \(n\in\mathbb{N}\). In addition, as \(\mathcal{S}\) is gclosed, \(ggx_{n}\,\mathcal{S}\,gu\) for each \(n\in\mathbb{N}\). Let \(\{a_{n}\}\) and \(\{b_{n}\}\) be the sequences of nonnegative real numbers given by
$$a_{n}=d ( Tgx_{n},Tu )\quad \text{and}\quad b_{n}=d ( ggx_{n},gu ) \quad \text{for all }n\in\mathbb{N}. $$
Clearly, \(\{b_{n}\}\rightarrow0\). Applying the contractivity condition (7),
$$\varrho ( a_{n},b_{n} ) =\varrho \bigl( d ( Tgx_{n},Tu ) ,d ( ggx_{n},gu ) \bigr) >0 \quad \text{for all }n\in\mathbb{N}. $$
By (\(\varrho_{4}\)), we deduce that \(\{a_{n}\}\rightarrow0\), which implies that \(\{Tgx_{n}\}\rightarrow Tu\). □
The reader may particularize the previous results to the cases: (1) \(x\,\mathcal{S}\,y\) for all \(x,y\in X\); (2) ≺ is a transitive binary relation on X; (3) g is the identity mapping on X.