International Journal of Analysis and Applications

Volume 18, Number 1 (2020), 99-103

URL: https://doi.org/10.28924/2291-8639

DOI: 10.28924/2291-8639-18-2020-99

ON FIXED POINT THEOREM IN NON-ARCHIMEDEAN FUZZY NORMED SPACES

M.E. EGWE∗

Department of Mathematics, University of Ibadan, Ibadan, Nigeria

∗Corresponding author: murphy.egwe@ui.edu.ng

Abstract. Let (X, N) be a non-archimedean fuzzy normed space and (X, ‖.‖), a non-archimedean normed

space where X is a linear space over a linearly ordered non-archimedean field K with a non-archimedean

valuation. We give a proof of the fixed point theorem in non-archimedean Fuzzy normed space.

1. Introduction

Definition 1.1 [9]: A valuation is a map | · | from a field K into a non-negative reals such that

(i) |a| = 0 if and only if a = 0

(ii) |ab| = |a||b|

(iii) |a + b| ≤ |a| + |b| for all a,b ∈ K (triangle inequality).

When a field K carries an absolute value | · |, it is called a valued field (K, | · |). Examples of the pair (K, | · |)

is called a valued field. Examples of valuations are provided by the usual absolute values of R and C. In the

definition above, if the triangle inequality is replaced by a strong triangle inequality, i.e, |a+b| ≤ max(|a|, |b|)

for all a,b ∈ K, the map | | is then called a non-archimedean or ultrametric valuation.

Theorem 1.2 [1]: Let be a Complete space, 0 < λ < 1, and f : X → X be a map such that ‖f(x)−f(y)‖≤

‖x−y‖ for all x,y ∈ X. Then there exists a unique point x◦ such that f(x◦) = x◦.

This fixed point in several cases have been obtained in non-archimedean normed and metric spaces (see [2],

[4] [5], [6]). In this paper, we shall prove a version given in [4] for non-archimedean fuzzy normed spaces.

Received 2019-10-08; accepted 2019-11-14; published 2020-01-02.

2010 Mathematics Subject Classification. 46S10, 46S40, 47H10.

Key words and phrases. Fixed point, Non-archimedean, Fuzzy normed space, Spherically complete.

c©2020 Authors retain the copyrights

of their papers, and all open access articles are distributed under the terms of the Creative Commons Attribution License.

99

https://doi.org/10.28924/2291-8639
https://doi.org/10.28924/2291-8639-18-2020-99


Int. J. Anal. Appl. 18 (1) (2020) 100

2. Main Result

Definition 2.1 [4], [11]: Let (X,‖.‖), a non-archimedean normed space. A function N : X × R → [0, 1]

is called a nonarhimedean norm on X if for all x,y ∈ X and all s,t ∈ R,

(i) N(x,t) = 0 for t ≤ 0,

(ii) N(x,t) = 1 if and only if x = 0 for all t > 0,

(iii) N(λx,t) = N(x,
t

|λ|
) for λ 6= 0

(iv) N(x + y, max{s,t}) ≥ min{N(x,s),N(y,t)}

(v) N(x,∗) is nondecreasing function of R and lim
t→∞

N(x,t) = 1

Definition 2.2: Let (X,N) be a non-archimedean fuzzy normed space. A closed ball in (X,N) with centre

a is the set of points B(x,t) := {N(x−a,t) ≤ r,} where t ∈ R+.

Definition 2.3: A sequence of n-closed balls B(x1, t) ⊆ B(x2, t) ⊆ ··· ⊂ B(xn, t) is called a sequence of

closed balls ordered by inclusion.

Definition 2.4: Let (X,N) be a non-archimedean fuzzy normed space. Let {B(xi, t)}n be a sequence of

closed balls ordered by inclusion. Then, (X,N) is said to be spherically complete if the sequence of closed

balls {B(x,t)}n satisfies the finite intersection property in (X,N). i.e.,
n⋂

i=1

{B(xi, t)} 6= 0.

Theorem 2.5 [1] Let V be a complete normed linear space, 0 < α < 1, and ϕ : V → V such that

‖ϕ(u) −ϕ(ν)‖≤ α‖u−ν‖ for all u,ν ∈ V. Then, there exists a fixed point, u◦ ∈ V such that ϕ(u◦) = u◦.

A version of this theorem on non-archimedean normed space was proved in [6] as seen in the next result.

Proposition 2.6 [6]: Suppose that X and Y are non-archimedean normed over a non-archimedean field

K with |p| 6= 1 for some p ∈ N. Assume that X or Y is spherically complete. If f : X → Y is a surjective

isometry, then for each x ∈ X, there exists a unique y ∈ X such that f(x) + f(y) = f
(
x + y

p

)
.

We now state and prove a version of this result for the non-archimedean fuzzy normed spaces

Proposition 2.7: Let (X,N) and (Y,N) be non-archimedean fuzzy normed spaces over a non-archimedean

field K with |p| > 1 for some p ∈ N. Assume that X or Y is spherically complete. If f : (X,N) → (Y,N) is

a surjective isometry, then for each u ∈ X, there exists a unique ν ∈ X such that f(u) + f(ν) = f
(
u + ν

p

)
.

Proof: First, we prove that (X,N) or (Y,N) is spherically complete. Suppose that (X,N) is spherically

complete and let {B(yi, t)}n be a sequence od closed balls in (Y,N) ordered by inclusion. Then by the

surjectivity of f, there is a sequence {B(xi, t)}n of closed balls in (X,N) ordered by inclusion with

B(x1, t) = f
−1(B(y1, t)) ⊆ B(x2, t) = f−1(B(y2, t)) ⊆ ···⊆ B(xn, t) = f−1(B(yn, t)).

Thus,
n⋂
f−1(B(yi, t)) 6= φ as φ 6=

n⋂
(B(xi, t)) = f

−1(B(yi, t))

because (X,N) is spherically complete. Thus, (Y,N) is spherically complete if (X,N) is spherically complete.



Int. J. Anal. Appl. 18 (1) (2020) 101

Conversely, let (Y,N) be spherically complete and {B(x,t)}n a sequence of closed balls in (X,N) ordered

by inclusion. Then there exists a sequence {f(B(xi, t))}n of closed balls in (Y,N) such that

f(B(x1, t)) ⊆ f(B(x2, t)) ⊆ ···f(B(xn, t)).

Then,

B(x1, t) = f(B(x1, t)) ⊆ f(B(x2, t)) = B(x2, t) ⊆ ···f(B(xn, t)) = B(xn, t)

and
n⋂
B(xi, t) 6= φ as φ 6=

n⋂
f(B(xi, t)) =

n⋂
(B(xi, t))

because (Y,N) is spherically complete. Thus, (X,N) is spherically complete if (Y,N) is. This implies that

(X,N) or (Y,N) is spherically complete.

Next, we show that there exists a unique ν ∈ X such that f(u) + f(ν) = f
(
u + ν

p

)
for each u ∈ X. To

do this,let u ∈ X, and consider the mapping ϕ : X → X : x 7→ px−u. Now,

N(ϕ(x) −ϕ(y), t) = N(px−u− (py −u), t)

= N(px−u−py + u), t)

= N((px−py), t)

= N(x−y,
t

|p|
)

< N(x−y,t).

Thus, there exists M > 1 such that M.N(ϕ(x) −ϕ(y), t) < N(x−y,t), i.e.,

N(ϕ(x) −ϕ(y), t) <
1

M
N(x−y,t).

Obviously, 0 <
1

M
< 1, and N(ϕ(x) −ϕ(y), t) ≤ N(x−y,t) which implies that ϕ is a contractive mapping.

Let ψ : Y → Y be an isometry defined by ψ(y) = f(u) + y. If h = ϕf−1ψf. Then,

N(ϕh(x) −h(y), t) = N((ϕf−1ψf)(x) − (ϕf−1ψf)(y), t)

= N(p(ϕf−1ψf)(x) −u− (p(ϕf−1ψf)(y)), t)

= N(p(ϕf−1ψf)(x) −u−p(ϕf−1ψf)(y) + u,t)

= N(p(ϕf−1ψf)(x) −p(ϕf−1ψf)(y), t)

= N((ϕf−1ψf)(x) − (ϕf−1ψf)(y),
t

|p|
)

< N((ϕf−1ψf)(x) − (ϕf−1ψf)(y), t)

= N((ψf)(x) − (ψf)(y), t)

= N(f(x) −f(y), t)

= N(x−y,t).

Similarly, there exists K∗ > 1 such that K∗.N(h(x) −h(y), t) ≤ N(x−y,t). This also implies that

N(h(x) −h(y), t) ≤
1

K∗
N(x−y,t).



Int. J. Anal. Appl. 18 (1) (2020) 102

Since
1

K∗
< 1 and N(h(x) − h(y), t) ≤ N(x − y,t), then, h is a contraction mapping. By the fixed point

theorem, h has a unique fixed point ν such that

(ψf−1ψf)(ν) = h(ν)

= ν.

But

ψ

(
u + ν

p

)
= p.

u + ν

p
−u

= u + ν −u

= ν.

Therefore, ψ(ν) = ψ(f(ν)) = ψ

(
u + ν

p

)
= f

(
u + ν

p

)
, as f, and ψ are injections. Since ψ(f(ν)) =

f(u) + f(ν) by definition, it follows that f(u) + f(ν) = f

(
u + ν

p

)
. �

Remark 2.8: It is necessary for |p| > 1 for

(i) if |p| > 1, then

N(ϕ(u) −ϕ(ν),p) = N(u−ν,t)

as

N(u−ν,
t

|p|
) = N(u−ν,t).

Also,

N(h(u) −h(ν), t) = N(u−ν,t)

as

N((f−1ψf)(u) − (f−1ψf)(ν),
t

|p|
) = N((f−1ψf)(u) − (f−1ψf)(ν), t).

So, ϕ and h are not contraction mappings.

(ii) if |p| = 0, then p = 0 by definition of valuation. This negates the assumption that p ∈ N.

Conflicts of Interest: The author(s) declare that there are no conflicts of interest regarding the publication

of this paper.

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	1. Introduction
	2. Main Result
	References