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Nadler’s fixed point theorem in νgeneralized metric spaces
Fixed Point Theory and Applications volume 2017, Article number: 18 (2017)
Abstract
We extend Nadler’s fixed point theorem to νgeneralized metric spaces. Through the proof of the above extension, we understand more deeply the mathematical structure of a νgeneralized metric space. In particular, we study the completeness of the space. We also improve Caristi’s and Subrahmanyam’s fixed point theorems in the space.
1 Introduction and preliminaries
In 1969, Nadler proved the following; the splendid fixed point theorem for setvalued contractions, which is one of generalizations of the Banach contraction principle [1, 2]. See also, e.g., [3–8].
Theorem 1
Theorem 5 in Nadler [9]
Let \((X, d)\) be a complete metric space and let T be a mapping from X into \(\operatorname{CB} (X)\), where \(\operatorname{CB} (X)\) is the set of all nonempty bounded closed subsets of X. Assume that there exists \(r \in[0, 1)\) such that
for all \(x, y \in X\), where δ is a function from \(\operatorname{CB}(X)^{2}\) into \([0,\infty)\) defined by
Then there exists \(z \in X\) such that \(z \in Tz\).
Remark
It is obvious that (1) is equivalent to the following:
where H is the Hausdorff metric, that is,
In 2000, Branciari introduced the following, very interesting concept.
Definition 2
Branciari [10]
Let X be a set, let d be a function from \(X \times X\) into \([0, \infty)\) and let \(\nu\in \mathbb {N}\). Then \((X,d)\) is said to be a νgeneralized metric space if the following hold:

(N1)
\(d(x,y) = 0\) iff \(x = y\) for any \(x, y \in X\).

(N2)
\(d(x,y) = d(y,x)\) for any \(x, y \in X\).

(N3)
\(d(x,y) \leq D(x,u_{1},u_{2},\ldots,u_{\nu},y)\) for any \(x, u_{1}, u_{2}, \ldots, u_{\nu}, y \in X\) such that \(x, u_{1}, u_{2}, \ldots, u_{\nu}, y\) are all different, where
$$D(x,u_{1},u_{2},\ldots,u_{\nu},y) = d(x,u_{1}) + d(u_{1},u_{2}) + \cdots+ d(u_{\nu},y). $$
We have studied the topological structure of this space. Indeed, recent studies tell that 1 and 3generalized metric spaces have the compatible topology and that all νgeneralized metric spaces have the strongly compatible topology. Also we have proved several fixed point theorems in this space. See, e.g., [11–25]. However, we have not generalized Theorem 1. Motivated by this fact, in this paper, we generalize Theorem 1. Another purpose of this paper is to understand more deeply the mathematical structure of this space. In particular, we study the completeness of this space. We also improve Caristi’s and Subrahmanyam’s fixed point theorems in this space.
Throughout this paper we denote by \(\mathbb {N}\) the set of all positive integers and by \(\mathbb {R}\) the set of all real numbers. For an arbitrary set A, we also denote by #A the cardinal number of A.
2 Completeness
In this section, we begin with definitions. Some of them are new.
Definition 3
Let \((X,d)\) be a νgeneralized metric space and let \(\{ x_{n} \}\) be a sequence in X. Let \(\kappa\in \mathbb {N}\).

(i)
\(\{ x_{n} \}\) is said to be Cauchy [10] if \(\lim_{n} \sup_{m > n} d(x_{n},x_{m}) = 0\) holds.

(ii)
\(\{ x_{n} \}\) is said to be κCauchy [11] if
$$\lim_{n \to\infty} \sup\bigl\{ d(x_{n}, x_{n + 1 + j \kappa}) : j = 0, 1, 2, \ldots\bigr\} = 0 $$holds.

(iii)
\(\{ x_{n} \}\) is said to be \((\sum,\neq)\)Cauchy if \(x_{n}\) \((n \in \mathbb {N})\) are all different and
$$\sum_{j=1}^{\infty}d(x_{j}, x_{j+1}) < \infty $$holds.

(iv)
\(\{ x_{n} \}\) is said to converge to x [10] if \(\lim_{n} d(x_{n}, x) = 0\) holds.

(v)
\(\{ x_{n} \}\) is said to converge only to x [11] if
$$\lim_{n \to\infty} d(x_{n}, x) = 0 \quad\text{and}\quad \limsup_{n \to\infty} d(x_{n}, y) > 0 $$hold for any \(y \in X \setminus\{ x \}\).

(vi)
\(\{ x_{n} \}\) is said to converge exclusively to x [25] if
$$\lim_{n \to\infty} d(x_{n}, x) = 0 \quad\text{and}\quad \liminf_{n \to\infty} d(x_{n}, y) > 0 $$hold for any \(y \in X \setminus\{ x \}\).

(vii)
\(\{ x_{n} \}\) is said to converge to x in the strong sense [25] if \(\{ x_{n} \}\) is Cauchy and \(\{ x_{n} \}\) converges to x.
Remark
We know the following.

\(\{ x_{n} \}\) is Cauchy iff \(\{ x_{n} \}\) is 1Cauchy.

If \(\{ x_{n} \}\) is Cauchy, then \(\{ x_{n} \}\) is 2Cauchy; see Proposition 6(i) in [11].

\(\text{(vii)} \Longrightarrow\text{(vi)} \Longrightarrow\text{(v)} \Longrightarrow\text{(iv)}\) holds; see Proposition 2.3(ii) in [25].
Definition 4
Let \((X,d)\) be a νgeneralized metric space. Let \(\kappa\in \mathbb {N}\).

X is said to be complete [10] if every Cauchy sequence converges.

X is κcomplete [11] if every κCauchy sequence converges.

X is \((\sum,\neq)\)complete if every \((\sum,\neq)\)Cauchy sequence converges.
Remark
We know the following.

X is complete iff X is 1complete.

If X is 2complete, then X is complete; see Proposition 6(ii) in [11].
We next study \((\sum,\neq)\)completeness.
Lemma 5
Proposition 7 in [11]
Let \((X,d)\) be a νgeneralized metric space where ν is odd. Let \(\{ x_{n} \}\) be a νCauchy sequence such that \(x_{n}\) are all different. Then \(\{ x_{n} \}\) is Cauchy.
Lemma 6
Proposition 8 in [11]
Let \((X,d)\) be a νgeneralized metric space where ν is even. Let \(\{ x_{n} \}\) be a νCauchy sequence such that \(x_{n}\) are all different. Then \(\{ x_{n} \}\) is 2Cauchy.
Lemma 7
Lemma 9 in [11]
Let \((X,d)\) be a νgeneralized metric space. Then every \((\sum,\neq)\)Cauchy sequence is νCauchy.
Lemma 8
Let \((X,d)\) be a νgeneralized metric space and let \(\kappa\in \mathbb {N}\). Let \(\{ x_{n} \}\) be a κCauchy sequence converging to some \(z \in X\). Assume that \(x_{n}\) are all different. Then \(\{ x_{n} \}\) is Cauchy.
Remark
We need the difference of \(x_{n}\). See Example 28(v) below.
Proof
Fix \(\varepsilon> 0\). Then from the assumption, there exists some \(\mu\in \mathbb {N}\) satisfying
for any \(n \in \mathbb {N}\) with \(n \geq\mu\). Fix \(m, n \in \mathbb {N}\) with \(\mu\leq n < m\). Then we have
Thus, we obtain the desired result. □
Lemma 9
Let \((X,d)\) be a νgeneralized metric space and let \(\{ x_{n} \}\) be a \((\sum,\neq)\)Cauchy sequence in X. Then the following hold:

(i)
If ν is odd, then \(\{ x_{n} \}\) is Cauchy.

(ii)
\(\{ x_{n} \}\) is 2Cauchy.

(iii)
If \(\{ x_{n} \}\) converges, then \(\{ x_{n} \}\) is Cauchy, that is, \(\{ x_{n} \}\) converges in the strong sense.
Proof
(i) follows from Lemmas 5 and 7. Similarly, (ii) follows from (i), Lemmas 6 and 7. (iii) follows from (ii) and Lemma 8. □
Lemma 10
Let \((X,d)\) be a νgeneralized metric space satisfying either of the following:

ν is odd and X is complete.

X is 2complete.
Then X is \((\sum,\neq)\)complete.
Proof
Let \(\{ x_{n} \}\) be a \((\sum,\neq)\)Cauchy sequence. Then from the assumption and Lemma 9(i) and (ii), \(\{ x_{n} \}\) converges. □
Lemma 11
Let \((X,d)\) be a νgeneralized metric space and let \(\{ x_{n} \}\) be a Cauchy sequence in X converging to some \(z \in X\). Let \(\{ y_{n} \}\) be a sequence in X satisfying \(\lim_{n} d(x_{n}, y_{n}) = 0\). Then \(\{ y_{n} \}\) also converges to z.
Proof
We consider the following two cases:

(i)
\(\# \{ x_{n} : n \in \mathbb {N}\} < \infty\),

(ii)
\(\# \{ x_{n} : n \in \mathbb {N}\} = \infty\).
In the first case, there exists \(\mu\in \mathbb {N}\) satisfying \(x_{n} = z\) for any \(n \geq\mu\). Therefore \(\lim_{n} d(z, y_{n}) = \lim_{n} d(x_{n}, y_{n}) = 0\) holds. In the second case, we fix \(\varepsilon> 0\). Then from the assumption, there exists some \(\mu\in \mathbb {N}\) satisfying
for any \(n \geq\mu\). Fix \(n \in \mathbb {N}\) with \(n \geq\mu\). We further consider the following two cases:

(ii1)
\(x_{n} = z\) or \(y_{n} = z\) or \(x_{n} = y_{n}\),

(ii2)
\(x_{n} \neq z\), \(y_{n} \neq z\), \(x_{n} \neq y_{n}\).
In the case of (ii1), \(d(y_{n}, z) < \varepsilon\) obviously holds. In the case of (ii2), we choose \(n_{1}, \ldots, n_{\nu1} \in \mathbb {N}\) such that \(n_{j} \geq\mu\) holds and \(x_{n}, y_{n}, z, x_{n_{1}}, \ldots, x_{n_{\nu1}}\) are all different. Then we have
Thus, we obtain the desired result. □
Lemma 12
Let \((X,d)\) be a νgeneralized metric space and let \(\{ x_{n} \}\) be a Cauchy sequence in X satisfying \(\liminf_{n} d(x_{n}, z) = 0\) for some \(z \in X\). Then \(\{ x_{n} \}\) converges to z.
Proof
There exists a subsequence \(\{ f(n) \}\) of the sequence \(\{ n \}\) in \(\mathbb {N}\) such that \(\{ x_{f(n)} \}\) converges to z. We note that \(\{ x_{f(n)} \}\) is Cauchy and that \(\lim_{n} d(x_{f(n)}, x_{n}) = 0\) holds. So by Lemma 11, we obtain the desired result. □
Lemma 13
Let \((X,d)\) be a \((\sum,\neq)\)complete, νgeneralized metric space. Then X is complete.
Proof
Let \(\{ x_{n} \}\) be a Cauchy sequence in X. We consider the following two cases:

\(\# \{ x_{n} : n \in \mathbb {N}\} < \infty\),

\(\# \{ x_{n} : n \in \mathbb {N}\} = \infty\).
In the first case, we can prove that \(\{ x_{n} \}\) converges as in the proof of Lemma 11. In the second case, we can choose a subsequence \(\{ f(n) \}\) of \(\{ n \}\) such that \(x_{f(n)} \) are all different and
holds for any \(n \in \mathbb {N}\). We have
Since X is \((\sum,\neq)\)complete, \(\{ x_{f(n)} \}\) converges to some \(z \in X\). By Lemma 12, \(\{ x_{n} \}\) itself converges to z. We have shown that X is complete. □
Definition 14
see Example 1.1 in [15]
Let \((X,d)\) be a νgeneralized metric space. X is said to be Hausdorff if \(\lim_{n} d(x_{n},x) = \lim_{n} d(x_{n},y) = 0\) implies \(x = y\).
Lemma 15
Let \((X,d)\) be a 2complete, νgeneralized metric space. Then X is Hausdorff.
Proof
Arguing by contradiction, we assume that X is not Hausdorff, that is, there exists a sequence \(\{ x_{n} \}\) in X converging to some u and v, where \(u \neq v\) holds. Define a sequence \(\{ y_{n} \}\) in X by
We note that \(\{ y_{n} \}\) is as follows:
It is obvious that \(\{ y_{n} \}\) is 2Cauchy. Since X is 2complete, \(\{ y_{n} \}\) converges to some z. However, we have
which implies a contradiction. Therefore X is Hausdorff. □
Lemma 16
Let \((X,d)\) be a \((\sum,\neq)\)complete, Hausdorff, νgeneralized metric space. Then X is 2complete.
Proof
Let \(\{ x_{n} \}\) be a 2Cauchy sequence. Define two subsets \(A_{1}\) and \(A_{2}\) of X by
We consider the following two cases:

\(\# A_{1} < \infty\) or \(\# A_{2} < \infty\),

\(\# A_{1} = \infty\) and \(\# A_{2} = \infty\).
In the first case, without loss of generality, we may assume \(\# A_{2} < \infty\). Define a subset \(B_{2}\) of \(A_{2}\) by
We note \(0 < \# B_{2} < \# A_{2} < \infty\). Since \(\{ x_{n} \}\) is 2Cauchy,
holds for any \(x, y \in B_{2}\). Since X is Hausdorff, we obtain \(x = y\). Therefore we have shown \(\# B_{2} = 1\). We let \(x \in X\) satisfy \(B_{2} = \{ x \}\). Then we obtain
Therefore \(\{ x_{n} \}\) converges to x. In the second case, we can choose a subsequence \(\{ f(n) \}\) of \(\{ n \}\) such that \(x_{f(n)} \) are all different, \(f(2 n  1)\) is odd, \(f(2 n)\) is even and
holds for any \(n \in \mathbb {N}\). We have
Since X is \((\sum,\neq)\)complete, \(\{ x_{f(n)} \}\) converges to some \(z \in X\). Since \(\{ x_{f(n)} \}\) is still 2Cauchy, \(\{ x_{f(n)} \}\) is Cauchy by Lemma 8. Noting that \(f(n+1)  n\) is odd for any \(n \in \mathbb {N}\), we have \(\lim_{n} d(x_{f(n+1)}, x_{n}) = 0\). By Lemma 11, we obtain \(\lim_{n} d(x_{n}, z) = 0\). We have shown that X is 2complete. □
Proposition 17
Let \((X,d)\) be a νgeneralized metric space where ν is odd. Then the following are equivalent:

X is complete.

X is \((\sum,\neq)\)complete.
Proof
The conclusion follows from Lemmas 10 and 13. □
Proposition 18
Let \((X,d)\) be a νgeneralized metric space. Then the following are equivalent:

X is 2complete.

X is \((\sum,\neq)\)complete and Hausdorff.
Proof
The conclusion follows from Lemmas 10, 15 and 16. □
Proposition 19
Let \((X,d)\) be a Hausdorff, νgeneralized metric space where ν is odd. Then the following are equivalent:

X is complete.

X is \((\sum,\neq)\)complete.

X is 2complete.
Proof
3 Fixed point theorems
In this section, we first generalize Theorem 1.
Theorem 20
Let \((X,d)\) be a \((\sum,\neq)\)complete, νgeneralized metric space. Let T be a setvalued mapping on X satisfying the following:

For any \(x \in X\), Tx is a nonempty subset of X.

If a sequence \(\{ y_{n} \}\) in Tx converges to y, then \(y \in Tx\) holds.

There exists \(r \in[0,1)\) satisfying \(\delta(Tx, Ty) \leq r d(x, y) \) for all \(x,y \in X\), where δ is defined by (2).
Then there exists \(z \in X\) satisfying \(z \in Tz\).
Proof
Replace the value of r by \(r := (1+r)/2 \in(0,1)\). We note \(r > 0\) and the following:

For any \(x, y \in X\) and \(u \in Tx\) with \(x \neq y\), there exists \(v \in Ty\) satisfying \(d(u,v) < r d(x,y) \).
Define a function f from X into \([0,\infty)\) by
Arguing by contradiction, we assume \(f(x) > 0\) for any \(x \in X\). Fix \(u_{1} \in X\) and choose \(u_{2} \in T u_{1}\) satisfying \(d(u_{1}, u_{2}) < (1/r) f(u_{1})\). Since \(f(u_{2}) < r d(u_{1}, u_{2}) \), we can choose \(u_{3} \in T u_{2}\) satisfying
Then we have
Continuing this argument, we can define a sequence \(\{ u_{n} \}\) in X satisfying
for any \(n \in \mathbb {N}\). Since \(\{ f(u_{n}) \}\) is strictly decreasing, \(u_{n} \) (\(n \in \mathbb {N}\)) are all different. We also have
Since X is \((\sum,\neq)\)complete, \(\{ u_{n} \}\) converges to some \(y \in X\). We note that \(\{ u_{n} \}\) is Cauchy by Lemma 9(iii). From the assumption, we can choose a sequence \(\{ v_{n} \}\) in Ty satisfying
for any \(n \in \mathbb {N}\). Then \(\lim_{n} d(u_{n+1}, v_{n}) = 0\) holds. By Lemma 11, we have \(\lim_{n} d(v_{n}, y) = 0\). Hence \(f(y) = 0\) holds, which implies a contradiction. Therefore we have shown that there exists \(z \in X\) satisfying \(f(z) = 0\). From the assumption, \(z \in Tz\) holds. □
As a direct consequence of Theorem 20, we obtain the following.
Corollary 21
Branciari [10]
Let \((X,d)\) be a complete, νgeneralized metric space and let T be a contraction on X, that is, there exists \(r \in[0, 1)\) such that
for any \(x,y \in X\). Then T has a fixed point.
Remark
We improve Caristi’s fixed point theorem; see [26, 27].
Definition 22
Let \((X,d)\) be a νgeneralized metric space.

A function f from X into \((\infty,+\infty]\) is proper if \(\{ x \in X : f(x) \in \mathbb {R}\}\) is nonempty.

A function f from X into \((\infty,+\infty]\) is said to be sequentially lower semicontinuous if \(f(x) \leq\liminf_{n} f(x_{n})\) holds whenever \(\{ x_{n} \}\) converges to x.

A mapping T on X is said to be sequentially continuous if \(\{ T x_{n} \}\) converges to Tx whenever \(\{ x_{n} \}\) converges to x.
Theorem 23
Theorem 2 in [28], Theorem 14 in [11]
Let \((X,d)\) be a \((\sum,\neq)\)complete, νgeneralized metric space and let T be a mapping on X. Let f be a proper, sequentially lower semicontinuous function from X into \((\infty,+\infty]\) bounded from below. Assume that
for all \(x \in X\). Then T has a fixed point.
Proof
We use Lemma 9(iii) in this paper instead of Lemma 12 in [11]. Then we can prove the conclusion as in the proof of Theorem 14 in [11]. □
Remark
We can weaken the assumption on the continuity of f as follows:

\(f(x) \leq\liminf_{n} f(x_{n})\) holds whenever \(\{ x_{n} \}\) converges to x in the strong sense.
We next improve Subrahmanyam’s fixed point theorem; see [29–32].
Theorem 24
Theorem 13 in [11]
Let \((X,d)\) be a \((\sum,\neq)\)complete, νgeneralized metric space and let T be a sequentially continuous mapping on X. Assume that there exists \(r \in[0,1)\) satisfying
for all \(x \in X\). Then for any \(x \in X\), \(\{ T^{n} x \}\) converges to a fixed point of T in the strong sense.
Proof
We use Lemma 9(iii) in this paper instead of Lemma 12 in [11]. Then we can prove the conclusion as in the proof of Theorem 13 in [11]. □
Remark
We can weaken the assumption on the continuity of T as follows:

\(\{ T x_{n} \}\) converges to Tx whenever \(\{ x_{n} \}\) converges to x in the strong sense.
4 Counterexamples
In this section, we give counterexamples on some results in Sections 2 and 3. The following example is a counterexample on Proposition 17 and Theorem 20.
Example 25
see Example 1 in [14]
Put \(X = \mathbb {N}\) and define a function d from \(X \times X\) into \([0,\infty)\) by
Define a setvalued mapping T by
Then the following hold:

(i)
\((X,d)\) is a 2generalized metric space.

(ii)
X is complete.

(iii)
\(\sum_{j=1}^{\infty}d(j,j+1) = 1 < \infty\) holds, however, \(\{ j \}\) is not Cauchy.

(iv)
X is not \((\sum,\neq)\)complete. Hence X is not 2complete.

(v)
T satisfies the assumption of Theorem 20. However, T does not have a fixed point.
Proof
We have proved (i)(iii) in [14]. (iv) follows from (iii) and Lemma 10. Let us prove (v). Fix \(x, y \in X\) with \(x < y\). Since \(Ty \subset Tx\), we have
In the case where \(yx\) is odd, we have
In the other case, where \(yx\) is even, we have
It is clear that T satisfies the other assumption of Theorem 20. It is obvious that T does not have a fixed point. We have shown (v). □
Lemma 26
Proposition 4.1 in [20]
Let \((X,d)\) be a νgeneralized metric space and let \(\lambda\in \mathbb {N}\) such that λ is divisible by ν. Then \((X,d)\) is a λgeneralized metric space.
The following is a slight generalization of Lemma 4.1 in [19].
Lemma 27
Let \(\nu\in \mathbb {N}\). Let X be a nonempty set and let A and B be two subsets of X with \(A \cap B = \varnothing\). Assume that A consists of at most \((\nu1)/2\) elements in the case where ν is odd. Let \((Y,\rho)\) be a metric space and let S be a mapping from \(A \cup B\) into Y such that \(S(A) \cap S(B) = \varnothing\) holds and there exists some positive real number M satisfying
for all \(x \in A\) and \(y \in B\). Define a function d from \(X \times X\) into \([0, \infty)\) by
Then \((X,d)\) is a νgeneralized metric space.
Remark
The proof below employs the methods in the proofs of Lemma 4 in [18] and Lemmas 4.2 and 4.3 in [20].
Proof
It is obvious that (N1) and (N2) hold. In order to show (N3), we consider the following three cases:

(a)
\(\nu= 2\).

(b)
ν is odd.

(c)
ν is even.
In the case of (a), we let \(x, y, u, v \in X\) be all different. Put
In the case where \(t \geq M\), (N3) holds because \(d(x,y) \leq M\). In the other case, where \(t < M\), without loss of generality, we may assume \(x \in A\). Then we have \(v \in A\) and \(u, y \in B\) from the definition of d. Hence we obtain
In the case of (b), we let \(x_{1}, \ldots, x_{\nu+2} \in X\) be all different. Then we have
In the case of (c), from (a) and Lemma 26, we obtain the desired result. □
The following example is a counterexample on Lemma 8 and Proposition 18. Also this example tells that Theorem 20 is a true generalization of Theorem 1.
Example 28
see Example 1.1 in [15]
Put \(A = \{ 0, 2, 3 \}\), \(B =\{ 2^{n} : n \in \mathbb {N}\}\) and \(X = A \cup B\). Define a function d from \(X \times X\) into \([0,\infty)\) by
Define sequences \(\{ x_{n} \}\) and \(\{ y_{n} \}\) in X by
and
Define a setvalued mapping T on X by
Then the following hold:

(i)
\((X,d)\) is a νgeneralized metric space for \(\nu\in \mathbb {N}\setminus\{ 1, 3, 5 \}\).

(ii)
\(\{ 2^{n} \}\) converges to 0, 2 and 3. Therefore X is not Hausdorff.

(iii)
X is \((\sum,\neq)\)complete.

(iv)
\(\{ x_{n} \}\) is 2Cauchy, however, it does not converge. Therefore X is not 2complete.

(v)
\(\{ y_{n} \}\) is 2Cauchy and it converges to 0. However, \(\{ y_{n} \}\) is not Cauchy.

(vi)
All the assumptions of Theorem 20 are satisfied.

(vii)
There does not exist a metric q on X satisfying (1) with \(d := q\).
Proof
(i) follows from Lemma 27. (ii) obviously holds. There does not exist a \((\sum,\neq)\)Cauchy sequence in X. So (iii) holds. We note that \(\{ x_{n} \}\) is as follows:
It is obvious that \(\{ x_{n} \}\) is 2Cauchy. However, \(\{ x_{n} \}\) does not converge. We have shown (iv). (v) obviously holds. We can easily prove
for all \(x, y \in X\). So, (vi) holds. Let us prove (vii). Arguing by contradiction, we assume that there exist a metric q on X and \(r \in[0,1)\) satisfying \(\delta(Tx, Ty) \leq r q(x, y)\) for all \(x, y \in X\), where δ is defined by (2) with \(d := q\). Let \(x \in A\) and \(y \in B\) be arbitrary. We have
and hence
Therefore
holds. We also have
and hence
Combining this and (3),
holds. So we obtain \(2 = 3\), which implies a contradiction. □
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The author is supported in part by JSPS KAKENHI Grant Number 16K05207 from Japan Society for the Promotion of Science.
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Suzuki, T. Nadler’s fixed point theorem in νgeneralized metric spaces. Fixed Point Theory Appl 2017, 18 (2017). https://doi.org/10.1186/s1366301706112
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DOI: https://doi.org/10.1186/s1366301706112