Our main result is the following improvement of Theorem 2, that is, the main result from [1]. Note that compactness of the space is not assumed, nor is the continuity of T. Also, the subsets \(A_{i}\) of X need not be closed. Moreover, the proof is much shorter than the proof of the relevant theorem in [1].
Theorem 4
Let
\((X,d)\)
be a complete metric space and
\(T:X\rightarrow X\)
be a mapping. Suppose that
\(X=\bigcup_{i=1}^{p}A_{i}\)
is a cyclic representation of
X
with respect to
T. If, for some
\(\phi\in \mathcal{F}\), relation (2.2) holds for any
\(x\in A_{i}\)
and
\(y\in A_{i+1}\), \(i=1,2,\ldots,p\), where
\(A_{p+1}=A_{1}\), then
T
has a unique fixed point
\(z\in X\)
and
\(z\in\bigcap_{i=1}^{p}A_{i}\). Moreover, each Picard sequence
\(x_{n}=T^{n}x_{0}\), \(x_{0}\in X\)
converges to
z.
Proof
Take arbitrary \(x_{0}\in X\). It belongs to some of the subsets \(A_{i}\), \(i\in\{1,\ldots,p\}\), say \(x_{0}\in A_{1}\). Then the Picard sequence \(x_{n}=Tx_{n1}\), \(n\in\mathbb{N}\), is divided into the following p subsequences, each belonging to some \(A_{i}\), \(i\in\{1,\ldots,p\}\):
$$ \{x_{np}\}\subset A_{1},\qquad \{x_{np+1}\} \subset A_{2}, \qquad\ldots,\qquad \{x_{np+p1}\}\subset A_{p}. $$
Suppose that \(x_{n+1}\ne x_{n}\), for each n (otherwise there is nothing to prove). It easily follows that \(x_{n}\ne x_{m}\) for \(n\ne m\). Moreover, it can be proved in a standard way (e.g., as in the proof of [1], Theorem 2.1) that
$$d(x_{n},x_{n+1})\downarrow0\quad\mbox{as }n\to\infty. $$
We will prove that \(\{x_{n}\}\) is a Cauchy sequence.
Suppose that this is not true. Then, using Lemma 1, we conclude that there exist an \(\varepsilon>0\) and two sequences \(\{ n( k) \} \) and \(\{ m( k) \} \) of positive integers, with \(n(k)>m(k)>k\), such that the sequences (2.3) tend to \(\varepsilon^{+}\) when \(k\rightarrow\infty\). Applying (2.2) with \(x=x_{m(k)j(k)}\), \(y=x_{n(k)}\), we get
$$d(x_{m(k)j(k)+1},x_{n(k)+1})\leq d(x_{m(k)j(k)},x_{n(k)}) \phi \bigl(d(x_{m(k)j(k)},x_{n(k)}) \bigr), $$
i.e.,
$$0< \phi \bigl(d(x_{m(k)j(k)},x_{n(k)}) \bigr) \leq d(x_{m(k)j(k)},x_{n(k)})d(x_{m(k)j(k)+1},x_{n(k)+1}), $$
since \(x_{m(k)j(k)}\ne x_{n(k)}\) (which follows from \(n(k)>m(k)\)). Passing to the limit as \(k\to\infty\), we get
$$\phi \bigl(d(x_{m(k)j(k)},x_{n(k)}) \bigr)\to0, \quad\mbox{as }k\to \infty. $$
Since \(\varepsilon\leq d(x_{m(k)j(k)},x_{n(k)})\) for k sufficiently large, we obtain \(0<\phi(\varepsilon)\leq\phi(d(x_{m(k)j(k)},x_{n(k)}))\), a contradiction. Hence, \(\{x_{n}\}\) is a Cauchy sequence.
Since the space X is complete, there is \(z\in X=\bigcup_{i=1}^{p}A_{i}\) such that \(x_{n}\to z\) as \(n\to\infty\). This means that \(z\in A_{m}\) for some \(m\in\{1,\ldots,p\}\) and, hence, \(Tz\in A_{m+1}\). Consider the subsequence \(\{x_{np+m}\}\subset A_{m+1}\) of \(\{x_{n}\}\). Applying (2.2), we get
$$\begin{aligned} d(x_{np+m+1},Tz)&=d(Tx_{np+m},Tz) \\ &\leq d(x_{np+m},z)\phi \bigl(d(x_{np+m},z) \bigr) \\ &\leq d(x_{np+m},z)\to0,\quad\mbox{as }n\to\infty. \end{aligned}$$
It follows that \(x_{np+m+1}\to Tz\) as \(n\to\infty\) and, by the uniqueness of the limit, \(Tz=z\). By the cyclic property (2.1) of T, it follows that \(z\in\bigcap_{i=1}^{p}A_{i}\).
The uniqueness of the fixed point follows from the contractive condition (2.2), since, again by (2.1), each fixed point of T belongs to \(\bigcap_{i=1}^{p}A_{i}\). □
Taking \(\phi(t)=(1k)t\), \(t\in[0,+\infty)\), \(k\in[0,1)\) in Theorem 4, we obtain the following version of a Banachtype cyclic contraction result.
Corollary 1
Let
\((X,d)\)
be a complete metric space and
\(T:X\rightarrow X\)
be a mapping. Suppose that
\(X=\bigcup_{i=1}^{p}A_{i}\)
is a cyclic representation of
X
with respect to
T. If, for some
\(k\in[0,1)\), the inequality
\(d(Tx,Ty)\leq kd(x,y)\)
holds for any
\(x\in A_{i}\)
and
\(y\in A_{i+1}\), \(i=1,2,\ldots,p\), where
\(A_{p+1}=A_{1}\), then
T
has a unique fixed point
\(z\in X\)
and
\(z\in\bigcap_{i=1}^{p}A_{i}\). Moreover, each Picard sequence
\(x_{n}=T^{n}x_{0}\), \(x_{0}\in X\), converges to
z.
Example 1
Let \(X=\mathbb{R}\) be endowed with the standard metric. Take \(A_{1}=(\infty,2)\), \(A_{2}=(2,+\infty)\), and define \(T:X\to X\) and \(\phi\in\mathcal{F}\) by
$$Tx= \textstyle\begin{cases} \frac{1}{2}x,&x< 2,\\ \frac{1}{3}x,&x\ge2, \end{cases}\displaystyle \qquad \phi(t)= \textstyle\begin{cases} \frac{1}{2}t,&t\in[0,2),\\ \frac{1}{3}t,&t\in[2,+\infty). \end{cases} $$
Then all the assumptions of Theorem 4 are fulfilled; let us check the condition (2.2). Suppose that \(x\in A_{1}\) and \(y\in A_{2}\) (the other case can be treated symmetrically). Consider the following possibilities:
(1) \(x\in(\infty,2]\), (2) \(x\in(2,0)\), (3) \(x\in[0,2)\), as well as: (a) \(y\in(2,0)\), (b) \(y\in[0,2)\), (c) \(y\in[2,+\infty)\).
Case 1(a). \(Tx=\frac{1}{3}x<\frac{1}{2}x\), \(Ty=\frac{1}{2}y\), and \(d(Tx,Ty)<\frac{1}{2}(x(y))=\frac{1}{2}(yx)=\frac{1}{2}d(x,y)\).
Case 1(b). \(Tx=\frac{1}{3}x<\frac{1}{2}x\), \(Ty=\frac{1}{2}y\), \(d(Tx,Ty)<\frac{1}{2}(x(y))=\frac{1}{2}(yx)=\frac{1}{2}d(x,y)\).
Case 1(c). \(Tx=\frac{1}{3}x\), \(Ty=\frac{1}{3}y\), \(d(Tx,Ty)=\frac{1}{3}(x(y))=\frac{1}{3}(yx)<\frac{1}{2}d(x,y)\).
Case 2(a). \(Tx=\frac{1}{2}x\), \(Ty=\frac{1}{2}y\), \(d(Tx,Ty)=\frac{1}{2}xy=\frac{1}{2}d(x,y)\).
Case 2(b). \(Tx=\frac{1}{2}x\), \(Ty=\frac{1}{2}y\), \(d(Tx,Ty)=\frac{1}{2}(xy)=\frac{1}{2}d(x,y)\).
The cases 2(c), 3(a), 3(b), and 3(c) are symmetric to the cases 1(b), 2(b), 2(a), and 1(a), respectively. On the other hand, \(\phi(t)\geq\frac{1}{2}t\) for all \(t\geq0\).
Hence, in all possible cases,
$$d(Tx,Ty)\leq\frac{1}{2}d(x,y)\leq d(x,y)\phi \bigl(d(x,y) \bigr), $$
and the contractive condition (2.2) of Theorem 4 is fulfilled. We conclude that T has a unique fixed point (which is \(z=0\)). Since:

1.
X is not compact;

2.
T is not continuous;

3.
\(A_{i}\), \(i\in\{1,2\}\), are not closed,
Theorems 1 and 2 cannot be used to reach this conclusion.
The following modification of Theorem 4 can be proved nearly in the same way.
Theorem 5
Let
X
be a nonempty set, and
\(A_{1}\), \(A_{2}\), …, \(A_{p}\)
be its nonempty subsets, with at least one of them being closed. Let
\(Y=\bigcup_{i=1}^{p}A_{i}\)
be a cyclic representation of
Y
with respect to a mapping
\(T:Y\to Y\). If, for some
\(\phi\in \mathcal{F}\), relation (2.2) holds for any
\(x\in A_{i}\)
and
\(y\in A_{i+1}\), \(i=1,2,\ldots,p\), where
\(A_{p+1}=A_{1}\), then
T
has a unique fixed point
\(z\in Y\)
and
\(z\in\bigcap_{i=1}^{p}A_{i}\). Moreover, each Picard sequence
\(x_{n}=T^{n}x_{0}\), \(x_{0}\in Y\)
converges to
z.
Proof
Suppose, without loss of generality, that the subset \(A_{1}\) is closed (and hence complete) in X. The only difference in the proof, compared with Theorem 4, is that, after proving that \(\{x_{n}\}\) is a Cauchy sequence (in Y), we conclude that there is \(z\in A_{1} ({}\subseteq Y)\) such that \(x_{n}\to z\), as \(n\to\infty\). See also the proof of [3], Theorem 2.1. □
As another consequence of Theorem 4, we obtain the following improvement of Theorem 3.
Corollary 2
Under the assumptions of Theorem
4, the fixed point problem for
T
is well posed, that is, if
\(\{y_{n}\} \)
is a sequence in
X
satisfying
\(d( y_{n},Ty_{n}) \rightarrow0\)
as
\(n\rightarrow\infty\), then
\(y_{n}\rightarrow z\)
as
\(n\rightarrow \infty\), where
z
is the unique fixed point of
T (whose existence is guaranteed by Theorem
4).
Proof
Since, according to Theorem 4, it follows that the unique fixed point z of T belongs to \(\bigcap_{i=1}^{p}A_{i}\), the rest of the proof is the same as in [1], p. 282. Namely, this proof uses neither compactness of X, nor continuity of T. □
Remark 3
As a kind of conclusion, we state once more that there are two kinds of cyclic fixed point results  those treating mappings \(T:Y\to Y\), where \(Y=\bigcup_{i=1}^{p}A_{i}\subset X\) (possibly with \(Y\ne X\)) and those where \(Y=X\). In results of the first kind, it is enough to assume that one of the sets \(A_{i}\) is closed, while in results of the second kind, no closedness assumption is needed. In both cases, if the considered mapping T is continuous, all the results reduce to the case when all \(A_{i}\)’s are closed.