In this section we extend some of the previous results using a weaker notion than Finvariant set. Throughout this section, let X be a nonempty set, let F:X\times X\to X be a mapping, and let M be a subset of {X}^{4}.
3.1 Fclosed sets and a related fixed point theorem
We extend the notion of Finvariant set as follows.
Definition 12 We say that M is an Fclosed subset of {X}^{4} if, for all x,y,u,v\in X,
(x,y,u,v)\in M\phantom{\rule{1em}{0ex}}\u27f9\phantom{\rule{1em}{0ex}}(F(x,y),F(y,x),F(u,v),F(v,u))\in M.
Obviously, every Finvariant set is an Fclosed set. In particular, ∅ and {X}^{4} are Fclosed sets.
Example 13 Let X=\{0,1\}, and let M=\{(0,0,0,0),(1,0,0,0)\}\subseteq {X}^{4}. If we consider the mapping F:X\times X\to X given by F(x,y)=0 for all x,y\in X, then M is Fclosed, but it is not Finvariant.
In Lemma 18 we will show some nontrivial examples of Fclosed sets. The following result presents a characterization of Fclosed sets.
Lemma 14 Let X be a nonempty set, let F:X\times X\to X be a mapping, and let M be a subset of {X}^{4}. Define:
\begin{array}{r}(x,y){\u2291}_{M}(u,v)\phantom{\rule{1em}{0ex}}\u27fa\phantom{\rule{1em}{0ex}}[(x,y)=(u,v)\mathit{\text{or}}(u,v,x,y)\in M],\phantom{\rule{1em}{0ex}}\mathit{\text{and}}\\ {T}_{F}:{X}^{2}\to {X}^{2},\phantom{\rule{2em}{0ex}}{T}_{F}(x,y)=(F(x,y),F(y,x))\phantom{\rule{1em}{0ex}}\mathit{\text{for all}}(x,y)\in {X}^{2}.\end{array}
(2)
Then the following properties hold.

(1)
{\u2291}_{M} is reflexive whatever M.

(2)
M satisfies the transitive property if and only if {\u2291}_{M} is a preorder on {X}^{2}.

(3)
M is Fclosed if and only if the mapping {T}_{F} is {\u2291}_{M}nondecreasing.

(4)
If M is Finvariant, then the mapping {T}_{F} is {\u2291}_{M}nondecreasing.
Proof (1) is obvious. (2) Suppose that M satisfies the transitive property, and let (x,y){\u2291}_{M}(u,v) and (u,v){\u2291}_{M}(a,b). If (x,y)=(u,v) or (u,v)=(a,b), then it is apparent that (x,y){\u2291}_{M}(a,b). In other case, (u,v,x,y)\in M and (a,b,u,v)\in M. Since M satisfies the transitive property, then (a,b,x,y)\in M, which means that {\u2291}_{M} is a preorder on {X}^{2}. The converse is similar. (3) Suppose that M is Fclosed. Let (x,y),(u,v)\in {X}^{2} be such that (x,y){\u2291}_{M}(u,v). If (x,y)=(u,v), then it is clear that {T}_{F}(x,y)={T}_{F}(u,v). Now suppose that (u,v,x,y)\in M. Since M is Fclosed, we know that (F(u,v),F(v,u),F(x,y),F(y,x))\in M, that is, (F(x,y),F(y,x)){\u2291}_{M}(F(u,v),F(v,u)), which means that {T}_{F}(x,y){\u2291}_{M}{T}_{F}(u,v). Therefore, {T}_{F} is {\u2291}_{M}nondecreasing. The converse is similar. (4) It follows from the fact that M is also Fclosed. □
Notice that if d is a metric on X, then the mapping {D}_{d}:{X}^{2}\times {X}^{2}\to [0,\mathrm{\infty}), defined by
{D}_{d}((x,y),(u,v))=\frac{d(x,u)+d(y,v)}{2}\phantom{\rule{1em}{0ex}}\text{for all}(x,y),(u,v)\in {X}^{2},
is a metric on {X}^{2}. Furthermore, if F:X\times X\to X is a dcontinuous mapping, then {T}_{F}:{X}^{2}\to {X}^{2}, defined as in (2), is {D}_{d}continuous. Thus, the following result reduces a coupled fixed point theorem to a unidimensional case.
Theorem 15 Theorem 11 follows from Theorem 5.
Proof Let Y=X\times X={X}^{2}, provided with the metric {D}_{d} and the preorder {\u2291}_{M}. It is clear that (Y,{D}_{d}) is a complete metric space, and Lemma 14 assures us that {T}_{F} is {\u2291}_{M}nondecreasing. The condition (F({x}_{0},{y}_{0}),F({y}_{0},{x}_{0}),{x}_{0},{y}_{0})\in M means that the point {Z}_{0}=({x}_{0},{y}_{0}) verifies {Z}_{0}{\u2291}_{M}{T}_{F}({Z}_{0}). If F is dcontinuous, then {T}_{F} is {D}_{d}continuous. Taking into account that \{({x}_{m},{y}_{m})\} {D}_{d}converges to (x,y)\in Y if and only if \{{x}_{m}\} dconverges to x and \{{y}_{m}\} dconverges to y, it is clear that condition (b) in Theorem 11 implies that (Y,{D}_{d},{\u2291}_{M}) is nondecreasingregular. Finally, suppose that (x,y),(u,v)\in Y verify (x,y){\u2291}_{M}(u,v) and we are going to show that
{D}_{d}({T}_{F}(x,y),{T}_{F}(u,v))\le \phi \left({D}_{d}((x,y),(u,v))\right).
Indeed, if (x,y)=(u,v), there is nothing to prove. Suppose that (x,y)\ne (u,v). In this case, (u,v,x,y)\in M. By (1),
d(F(u,v),F(x,y))\le \phi \left(\frac{d(u,x)+d(v,y)}{2}\right)=\phi \left({D}_{d}((x,y),(u,v))\right).
Moreover, by condition (i) in Definition 8, we have
(u,v,x,y)\in M\phantom{\rule{1em}{0ex}}\u27f9\phantom{\rule{1em}{0ex}}(y,x,v,u)\in M,
and using (1) again, it follows that
d(F(y,x),F(v,u))\le \phi \left(\frac{d(y,v)+d(x,u)}{2}\right)=\phi \left({D}_{d}((x,y),(u,v))\right).
Combining the previous inequalities, we deduce that
\begin{array}{rl}{D}_{d}({T}_{F}(x,y),{T}_{F}(u,v))& ={D}_{d}((F(x,y),F(y,x)),(F(u,v),F(v,u)))\\ =\frac{d(F(x,y),F(u,v))+d(F(y,x),F(v,u))}{2}\le \phi \left({D}_{d}((x,y),(u,v))\right).\end{array}
Using Theorem 5, we conclude that {T}_{F} has a fixed point, that is, F has a coupled fixed point. □
3.2 Fixed point results without the mixed monotone property
In the previous result, M is Finvariant and satisfies the transitive property. Next we show that these conditions are not necessary in order to prove coupled fixed point theorems. Therefore, we can prove some results avoiding such property.
Theorem 16 Let (X,d) be a complete metric space, let F:X\times X\to X be a continuous mapping, and let M be a subset of {X}^{4}. Assume that:

(i)
M is Fclosed;

(ii)
there exists ({x}_{0},{y}_{0})\in {X}^{2} such that (F({x}_{0},{y}_{0}),F({y}_{0},{x}_{0}),{x}_{0},{y}_{0})\in M;

(iii)
there exists k\in [0,1) such that for all (x,y,u,v)\in M, we have
d(F(x,y),F(u,v))+d(F(y,x),F(v,u))\le k(d(x,u)+d(y,v)).
Then F has a coupled fixed point.
Proof Using ({x}_{0},{y}_{0})\in {X}^{2} and by recurrence, define {x}_{m+1}=F({x}_{m},{y}_{m}) and {y}_{m+1}=F({y}_{m},{x}_{m}) for all m\ge 0. We claim that ({x}_{m+1},{y}_{m+1},{x}_{m},{y}_{m})\in M for all m\ge 0. Indeed, ({x}_{1},{y}_{1},{x}_{0},{y}_{0})=(F({x}_{0},{y}_{0}),F({y}_{0},{x}_{0}),{x}_{0},{y}_{0})\in M. Assume that ({x}_{m+1},{y}_{m+1},{x}_{m},{y}_{m})\in M for some m\ge 0. Since M is Fclosed,
\begin{array}{r}({x}_{m+1},{y}_{m+1},{x}_{m},{y}_{m})\in M\\ \phantom{\rule{1em}{0ex}}\Rightarrow \phantom{\rule{1em}{0ex}}(F({x}_{m+1},{y}_{m+1}),F({y}_{m+1},{x}_{m+1}),F({x}_{m},{y}_{m}),F({y}_{m},{x}_{m}))\in M\\ \phantom{\rule{1em}{0ex}}\Rightarrow \phantom{\rule{1em}{0ex}}({x}_{m+2},{y}_{m+2},{x}_{m+1},{y}_{m+1})\in M.\end{array}
Applying the contractivity condition to ({x}_{m+1},{y}_{m+1},{x}_{m},{y}_{m})\in M, we deduce that, for all m\ge 0,
\begin{array}{r}d(F({x}_{m+1},{y}_{m+1}),F({x}_{m},{y}_{m}))+d(F({y}_{m+1},{x}_{m+1}),F({y}_{m},{x}_{m}))\\ \phantom{\rule{1em}{0ex}}\le k(d({x}_{m+1},{x}_{m})+d({y}_{m+1},{y}_{m})),\end{array}
that is,
d({x}_{m+2},{x}_{m+1})+d({y}_{m+2},{y}_{m+1})\le k(d({x}_{m+1},{x}_{m})+d({y}_{m+1},{y}_{m}))\phantom{\rule{1em}{0ex}}\text{for all}m\ge 0.
In other words,
\begin{array}{r}max(d({x}_{m+1},{x}_{m}),d({y}_{m+1},{y}_{m}))\le d({x}_{m+1},{x}_{m})+d({y}_{m+1},{y}_{m})\le {k}^{m}(d({x}_{1},{x}_{0})+d({y}_{1},{y}_{0}))\\ \phantom{\rule{1em}{0ex}}\text{for all}m\ge 0.\end{array}
This proves that \{{x}_{m}\} and \{{y}_{m}\} are Cauchy sequences in the complete metric space (X,d). Therefore, there are x,y\in X such that \{{x}_{m}\}\to x and \{{y}_{m}\}\to y. Since F is continuous, \{{x}_{m+1}\}=\{F({x}_{m},{y}_{m})\}\to F(x,y), so F(x,y)=x. Analogously, F(y,x)=y and (x,y) is a coupled fixed point of F. □
Example 17 Let X=[1,1] provided with the Euclidean metric. Let M={X}^{4}\setminus \{(0,0,0,1)\} and consider the mapping F:X\times X\to X given by F(x,y)=(xy)/4 for all x,y\in X. Then M is an Fclosed set but it is not an Finvariant set. Taking into account that, for all x,y,u,v\in X,
F(x,y)F(u,v)+F(y,x)F(v,u)=\frac{1}{2}(xu)+(vy)\le \frac{1}{2}(xu+yv),
and choosing ({x}_{0},{y}_{0})=(1,1), we conclude that all hypotheses of Theorem 16 are verified. Then F has a coupled fixed point, which is (0,0), but Theorem 11 cannot be applied because M is not an Finvariant set.
The previous theorem also holds using a weaker contractivity condition. To introduce it, we need some notation. Given (x,y),(u,v)\in {X}^{2}, we will use, for simplicity, the notation
({T}_{F}(x,y),{T}_{F}(u,v))=(F(x,y),F(y,x),F(u,v),F(v,u))\in {X}^{4}.
Given a point {Z}_{0}=({x}_{0},{y}_{0})\in {X}^{2}, let {T}_{F}^{0}({Z}_{0})={Z}_{0}, {T}_{F}^{1}({Z}_{0})={T}_{F}({Z}_{0}) and {T}_{F}^{m+1}({Z}_{0})={T}_{F}({T}_{F}^{m}({Z}_{0})) for all m\ge 0. We will denote
{M}_{F}({Z}_{0})=\{({T}_{F}^{m+1}({Z}_{0}),{T}_{F}^{m}({Z}_{0})):m\ge 0\}\subseteq {X}^{4}.
Lemma 18 Given {Z}_{0}\in {X}^{2}, the set {M}_{F}({Z}_{0}) is Fclosed. Indeed, if M\subseteq {X}^{4} is an Fclosed set verifying ({T}_{F}({Z}_{0}),{Z}_{0})\in M, then {M}_{F}({Z}_{0})\subseteq M.
In particular,
{M}_{F}({Z}_{0})=\bigcap \{M\subseteq {X}^{4}:M\text{is}F\text{closed and}({T}_{F}({Z}_{0}),{Z}_{0})\in M\}.
Proof Let {Z}_{0}=({x}_{0},{y}_{0})\in {X}^{2} and define {x}_{m+1}=F({x}_{m},{y}_{m}) and {y}_{m+1}=F({y}_{m},{x}_{m}) for all m\ge 0. Let us prove that {T}_{F}^{m}({Z}_{0})=({x}_{m},{y}_{m}) for all m\ge 0. If m=0, it is obvious. If m=1, then ({x}_{1},{y}_{1})=(F({x}_{0},{y}_{0}),F({y}_{0},{x}_{0}))={T}_{F}({x}_{0},{y}_{0})={T}_{F}({Z}_{0}). By recurrence,
({x}_{m+1},{y}_{m+1})=(F({x}_{m},{y}_{m}),F({y}_{m},{x}_{m}))={T}_{F}({x}_{m},{y}_{m})={T}_{F}({T}_{F}^{m}({Z}_{0}))={T}_{F}^{m+1}({Z}_{0}).
Next we claim that {M}_{F}({Z}_{0}) is Fclosed. Let (x,y,u,v)\in {M}_{F}({Z}_{0}). Then there exists m\ge 0 such that (x,y,u,v)=({T}_{F}^{m+1}({Z}_{0}),{T}_{F}^{m}({Z}_{0}))=({x}_{m+1},{y}_{m+1},{x}_{m},{y}_{m}). Therefore
\begin{array}{r}(F(x,y),F(y,x),F(u,v),F(v,u))\\ \phantom{\rule{1em}{0ex}}=(F({x}_{m+1},{y}_{m+1}),F({y}_{m+1},{x}_{m+1}),F({x}_{m},{y}_{m}),F({y}_{m},{x}_{m}))\\ \phantom{\rule{1em}{0ex}}=({x}_{m+2},{y}_{m+2},{x}_{m+1},{y}_{m+1})=({T}_{F}^{m+2}({Z}_{0}),{T}_{F}^{m+1}({Z}_{0}))\in {M}_{F}({Z}_{0}).\end{array}
Finally, let M\subseteq {X}^{4} be an Fclosed set verifying ({T}_{F}({Z}_{0}),{Z}_{0})\in M, and we are going to show that {M}_{F}({Z}_{0})\subseteq M. In particular, we will prove that ({T}_{F}^{m+1}({Z}_{0}),{T}_{F}^{m}({Z}_{0}))\in M for all m\ge 0 by the induction method. Indeed, if m=0, by hypothesis, ({T}_{F}({Z}_{0}),{Z}_{0})\in M. Suppose that ({T}_{F}^{m+1}({Z}_{0}),{T}_{F}^{m}({Z}_{0}))\in M for some m\ge 0. Since M is Fclosed,
\begin{array}{r}({x}_{m+1},{y}_{m+1},{x}_{m},{y}_{m})=({T}_{F}^{m+1}({Z}_{0}),{T}_{F}^{m}({Z}_{0}))\in M\\ \phantom{\rule{1em}{0ex}}\Rightarrow \phantom{\rule{1em}{0ex}}(F({x}_{m+1},{y}_{m+1}),F({y}_{m+1},{x}_{m+1}),F({x}_{m},{y}_{m}),F({y}_{m},{x}_{m}))\in M.\end{array}
Hence,
\begin{array}{rl}({T}_{F}^{m+2}({Z}_{0}),{T}_{F}^{m+1}({Z}_{0}))& =({x}_{m+2},{y}_{m+2},{x}_{m+1},{y}_{m+1})\\ =(F({x}_{m+1},{y}_{m+1}),F({y}_{m+1},{x}_{m+1}),F({x}_{m},{y}_{m}),F({y}_{m},{x}_{m}))\in M.\end{array}
This completes the induction. Thus, {M}_{F}({Z}_{0})\subseteq M. □
If we particularize Theorem 16 to the Fclosed set M={M}_{F}({Z}_{0}), we obtain the following result.
Corollary 19 Let (X,d) be a complete metric space, let F:X\times X\to X be a continuous mapping, and suppose that there exist {Z}_{0}\in {X}^{2} and k\in [0,1) such that
d(F(x,y),F(u,v))+d(F(y,x),F(v,u))\le k(d(x,u)+d(y,v))
(3)
for all (x,y,u,v)\in {M}_{F}({Z}_{0}). Then F has a coupled fixed point.
Notice that in the previous result, we have not necessarily a partial order on X nor a mapping verifying the mixed monotone property.
Theorem 20 Theorem 9 follows from Corollary 19.
Proof Following the proof given in [29], we can consider a constant
k=\frac{\alpha +\beta +\gamma +2\delta +\theta}{2(\alpha +\gamma +\beta +\theta )}\in [0,1)
and a sequence {\{({x}_{m+1},{y}_{m+1})=(F({x}_{m},{y}_{m}),F({y}_{m},{x}_{m}))\}}_{m\ge 0} such that
d({x}_{m+2},{x}_{m+1})+d({y}_{m+2},{y}_{m+1})\le k(d({x}_{m+1},{x}_{m})+d({y}_{m+1},{y}_{m}))\phantom{\rule{1em}{0ex}}\text{for all}m\ge 0,
which is exactly condition (3). Then F has a coupled fixed point. □