# Pseudovaluations on WFI Algebras

Authors: Young Bae Jun; Min Su Kang; Eun Hwan Roh
Journal: International Journal of Mathematics and Mathematical Sciences
 (2012)

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## Abstract

Using Buşneag's model, the notion of pseudovaluations (valuations) on a WFI algebra is introduced, and a pseudometric is induced by a pseudovaluation on WFI algebras. Given a valuation with additional condition, we show that the binary operation in WFI algebras is uniformly continuous.


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## Body

## 1. Introduction

In 1990, Wu [1] introduced the notion of fuzzy implication algebras (FI algebra, for short) and investigated several properties. In [2], Li and Zheng introduced the notion of distributive (regular, and commutative, resp.) FI algebras and investigated the relations between such FI algebras and MV algebras. In [3], Jun discussed several aspects of WFI algebras. He introduced the notion of associative (normal and medial, resp.) WFI algebras and investigated several properties. He gave conditions for a WFI algebra to be associative/medial, provided characterizations of associative/medial WFI algebras, and showed that every associative WFI algebra is a group in which every element is an involution. He also verified that the class of all medial WFI algebras is a variety. Jun et al. [4] introduced the concept of ideals of WFI algebras, and gave relations between a filter and an ideal. Moreover, they provided characterizations of an ideal, and established an extension property for an ideal. Buşneag [5] defined pseudovaluation on a Hilbert algebra and proved that every pseudovaluation induces a pseudometric on a Hilbert algebra. Also, Buşneag [6] provided several theorems on extensions of pseudovaluations. Buşneag [7] introduced the notions of pseudovaluations (valuations) on residuated lattices, and proved some theorems of extension for these (using the model of Hilbert algebras ([6])).In this paper, using Buşneag’s model, we introduce the notion of pseudovaluations (valuations) on WFI algebras, and we induce a pseudometric by using a pseudovaluation on WFI algebras. Given a valuation with additional condition, we show that the binary operation in WFI algebras is uniformly continuous.


## 2. Preliminaries

LetK(τ) be the class of all algebras of type τ=(2,0). By a WFIalgebra, we mean an algebra (X;⊖,θ)∈K(τ) in which the following axioms hold:(a1)
(∀x∈X)(x⊖x=θ),(a2)
(∀x,y∈X)(x⊖y=y⊖x=θ⇒x=y),(a3)
(∀x,y,z∈X)(x⊖(y⊖z)=y⊖(x⊖z)),(a4)
(∀x,y,z∈X)((x⊖y)⊖((y⊖z)⊖(x⊖z))=θ).For the convenience of notation, we will write[x,y1,y2,…,yn] for(2.1)(⋯((x⊖y1)⊖y2)⊖⋯)⊖yn.
We define [x,y]0=x, and for n>0,[x,y]n=[x,y,y,…,y], where y occurs n-times.Proposition 2.1 (see [3]).
In a WFI algebraX, the following are true:(b1)
x⊖[x,y]2=θ,(b2)
θ⊖x=θ⇒x=θ,(b3)
θ⊖x=x,(b4)
x⊖y=θ⇒(y⊖z)⊖(x⊖z)=θ,(z⊖x)⊖(z⊖y)=θ,(b5)
(x⊖y)⊖θ=(x⊖θ)⊖(y⊖θ),(b6)
[x,y]3=x⊖y.We define a relation “⪯’’ on X by x⪯y if and only if x⊖y=θ. It is easy to verify that a WFI algebra is a partially ordered set with respect to ⪯. A nonempty subset S of a WFI algebra X is called a subalgebra of X if x⊖y∈S whenever x,y∈S. A nonempty subset F of a WFI algebra X is called a filter of X if it satisfies:(c1)
θ∈F,(c2)
(∀x∈F)(∀y∈X)(x⊖y∈F⇒y∈F).A filterF of a WFI algebra X is said to be closed (see [3]) if F is also a subalgebra of X. A nonempty subset I of a WFI algebra X is called an ideal of X (see [4]) if it satisfies the condition (c1) and(c3)
(∀x,y∈X)(∀z∈I)((x⊖z)⊖y∈I⇒x⊖y∈I).Proposition 2.2 (see [3]).
LetF be a filter of a WFI algebra X. Then F is closed if and only if x⊖θ∈F for all x∈F.Proposition 2.3 (see [3]).
In a finite WFI algebra, every filter is closed.Note that every ideal of a WFI algebra is a closed filter (see [4, Theorem  4.3]). For a WFI algebra X, the set
(2.2)S(X):={x∈X∣x⪯θ}
is called the simulative part of X.


## 3. WFI Algebras with Pseudovaluations

In what follows, letX denote a WFI algebra unless otherwise specified.Definition 3.1.
A mappingf:X→ℝ is called a pesudovaluation on X if it satisfies the following two conditions:(i)
f(θ)=0,(ii)
(∀x,y∈X)(f(x⊖y)+f(x)≥f(y)).A pseudovaluationf on X satisfying the following condition:
(3.1)(∀x∈X)(x≠θ⟹f(x)≠0)
is called a valuation on X.Obviously, a mapping(3.2)f:X⟶R,x⟼0
is a pseudovaluation on X, which is called the trivial pseudovaluation.Example 3.2.
Letf:X→ℝ be a mapping defined by
(3.3)f(x)={0ifx=θ,kifx∈X∖{θ},
where k is a positive real number. Then, f is a pseudovaluation on X. Moreover, it is a valuation on X.Example 3.3.
Letℤ be the set of integers. Then, (ℤ;⊖,θ) is a WFI algebra, where θ=0 and x⊖y=y-x for all x,y∈ℤ (see [8]). Let f:ℤ→ℝ be a mapping defined by
(3.4)f(x)={0ifx=θ,ax+botherwise,
for all x∈ℤ, where a and b are real numbers with a≠0 and b≥0. Then, f is a pseudovaluation on ℤ.Example 3.4.
LetX={θ,a,b} be a set with the following Cayley table:
(3.5)
Then, (X;⊖,θ) is a WFI algebra (see [3]). Define a mapping f:X→ℝ by f(θ)=0,f(a)=2 and f(b)=9. Then, f is a pseudovaluation on X. Also, it is a valuation on X.Proposition 3.5.
Every pseudovaluationf on X satisfies the following conditions:(1)
(∀x,y∈X)(x⪯y⇒f(x)≥f(y)),(2)
(∀x,y,z∈X)(f(x⊖z)≤f(x⊖y)+f(y⊖z)),(3)
(∀x,y∈X)(f(x⊖y)+f(y⊖x)≥0).Proof.
(1) Letx,y∈X be such that x⪯y. Then, x⊖y=θ, and so
(3.6)f(y)≤f(x⊖y)+f(x)=f(θ)+f(x)=0+f(x)=f(x).
(2) Using (a4), we havex⊖y⪯(y⊖z)⊖(x⊖z) for all x,y,z∈X. It follows from (1) and Definition 3.1(ii) that
(3.7)f(x⊖y)≥f((y⊖z)⊖(x⊖z))≥f(x⊖z)-f(y⊖z),
so that f(x⊖z)≤f(x⊖y)+f(y⊖z) for all x,y,z∈X.
(3) Letx,y∈X. Using Definition 3.1(ii), we have f(x⊖y)+f(x)≥f(y) and f(y⊖x)+f(y)≥f(x); that is, f(x⊖y)≥f(y)-f(x) and f(y⊖x)≥f(x)-f(y). It follows that f(x⊖y)+f(y⊖x)≥0.Corollary 3.6.
Letf:X→ℝ be a pseudovaluation on X. Then, f(x)≥0 for all x∈𝒮(X).Proof.
Sincex⪯θ for all x∈𝒮(X), we have f(x)≥f(θ)=0 by Proposition 3.5(1) and Definition 3.1(i).The following example shows that the converse of Corollary3.6 may not be true.Example 3.7.
LetX be a WFI algebra which is considered in Example 3.4. Let g:X→ℝ be a mapping defined by
(3.8)g=(θab0-32).
Then, 𝒮(X)={θ,b},g(θ)=0 and g(b)=2≥0. But g is not a pseudovaluation on X, since
(3.9)g(a⊖θ)+g(a)=g(θ)+g(a)=-3≱0=g(θ).Letf:X→ℝ be a pseudovaluation on X. If x1⊖x=θ, that is, x1⪯x, for all x,x1∈X, then f(x)≤f(x1) by Proposition 3.5(1). If x2⊖(x1⊖x)=θ for all x,x1,x2∈X, then x2⪯x1⊖x, and so, f(x2)≥f(x1⊖x)≥f(x)-f(x1) by Proposition 3.5(1) and Definition 3.1(ii). Hence, f(x)≤f(x1)+f(x2). Now, if x3⊖(x2⊖(x1⊖x))=θ for all x,x1,x2,x3∈X, then x3⪯x2⊖(x1⊖x). It follows from Proposition 3.5(1) and Definition 3.1(ii) that(3.10)f(x3)≥f(x2⊖(x1⊖x))≥f(x1⊖x)-f(x2)≥f(x)-f(x1)-f(x2),
so that f(x)≤f(x1)+f(x2)+f(x3). Continuing this process, we have the following proposition.Proposition 3.8.
Letf:X→ℝ be a pseudovaluation on X. For any elements x,x1,x2,…,xn of X, if xn⊖(⋯⊖(x2⊖(x1⊖x))⋯)=θ, then f(x)≤∑k=1nf(xk).Theorem 3.9.
LetF be a filter of X, and let fF:X→ℝ be a mapping defined by
(3.11)fF(x)={0ifx∈F,kifx∉F,
where k is a positive real number. Then, fF is a pseudovaluation on X. In particular, fF is a valuation on X if and only if F={θ}.Proof.
Straightforward.We sayfF is a pseudovaluation induced by a filter F.Theorem 3.10.
If a mappingf:X→ℝ is a pseudovaluation on X, then the set
(3.12)Ff:={x∈X∣f(x)≤0}
is a filter of X.Proof.
Obviously,θ∈Ff. Let x,y∈X be such that x∈Ff and x⊖y∈Ff. Then, f(x)≤0 and f(x⊖y)≤0. It follows from Definition 3.1(ii) that f(y)≤f(x⊖y)+f(x)≤0 so that y∈Ff. Hence, Ff is a filter of X.We sayFf is a filter induced by a pseudovaluation f on X.Corollary 3.11.
If a mappingf:X→ℝ is a pseudovaluation on a finite WFI algebra X, then the set
(3.13)Ff:={x∈X∣f(x)≤0}
is a closed filter of X.Proof.
It follows from Proposition2.3 and Theorem 3.10.Remark 3.12.
A filter induced by a pseudovaluation onX may not be closed. Indeed, in Example 3.3, if we take a=1 and b=0, then f:ℤ→ℝ,x↦x, is a pseudovaluation on ℤ. Then, Ff={θ}∪{k∈ℤ∣k<θ} which is a filter but not a subalgebra of ℤ, since (-3)⊖(-1)=-1-(-3)=2∉Ff. Hence, Ff is not a closed filter of ℤ.Theorem 3.13.
For any pseudovaluationf:X→ℝ, if F is a filter of X, then FfF=F.Proof.
We haveFfF={x∈X∣fF(x)≤0}={x∈X∣x∈F}=F.The following example shows that the converse of Theorem3.10 may not be true; that is, there exist a WFI algebra X and a mapping f:X→ℝ such that(1)
f is not a pseudovaluation on X,(2)
Ff:={x∈X∣f(x)≤0} is a filter of X.Example 3.14.
LetX={θ,1,2,a,b} be a set with the following Cayley table:
(3.14)
Then (X;⊖,θ) is a WFI algebra. Let f:X→ℝ be a mapping defined by
(3.15)f=(θ12ab0-43-25).
Then, Ff={θ,1,a} is a filter of X. But f is not a pseudovaluation on X, since
(3.16)f(a⊖b)+f(a)=1≱5=f(b).Definition 3.15.
A pseudovaluation (or, valuation)f on X is said to be positive if f(x)≥0 for all x∈X.The pseudovaluationf on X which is given in Example 3.4 is positive.Theorem 3.16.
If a pseudovaluationf on X is positive, then the set
(3.17)Ff=:={x∈X∣f(x)=0}
is a filter of X.Proof.
Clearly,θ∈Ff=. Let x,y∈X be such that x∈Ff= and x⊖y∈Ff=. Then, f(x)=0 and f(x⊖y)=0. Since f is positive, it follows from Definition 3.1(ii) that
(3.18)0≤f(y)≤f(x⊖y)+f(x)=0,
so that f(y)=0, that is, y∈Ff=. Hence, Ff= is a filter of X.The following example shows that two distinct pseudovaluations induce the same filter.Example 3.17.
Consider a WFI algebraX={θ,1,2,a,b}which is given in Example 3.14. Let g and h be mappings from X to ℝ defined by
(3.19)g=(θ12ab00435),h=(θ12ab00423).
Then, g and h are pseudovaluations on X, and Fg={θ,1}=Fh.For a mappingf:X→ℝ, define a mapping df:X×X→ℝ by df(x,y)=f(x⊖y)+f(y⊖x) for all (x,y)∈X×X. Note that df(x,y)≥0 for all (x,y)∈X×X.Theorem 3.18.
Iff:X→ℝ is a pseudovaluation on X, then df is a pseudometric on X, and so (X,df) is a pseudometric space.We saydf is called the pseudometric induced by pseudovaluationf.Proof.
Letx,y,z∈X. Then, df(x,y)=f(x⊖y)+f(y⊖x)≥0 by Proposition 3.5(3), and obviously, df(x,y)=df(y,x) and df(x,x)=0. Now,
(3.20)df(x,y)+df(y,z)=[f(x⊖y)+f(y⊖x)]+[f(y⊖z)+f(z⊖y)]=[f(x⊖y)+f(y⊖z)]+[f(z⊖y)+f(y⊖x)]≥f(x⊖z)+f(z⊖x)=df(x,z).
Therefore, (X,df) is a pseudometric space.Proposition 3.19.
Every pseudometricdf induced by pseudovaluation f satisfies the following inequalities:(1)
df(x,y)≥df(x⊖a,y⊖a),(2)
df(x,y)≥df(a⊖x,a⊖y),(3)
df(x⊖y,a⊖b)≤df(x⊖y,a⊖y)+df(a⊖y,a⊖b), for all x,y,a,b∈X.Proof.
(1) Letx,y,a∈X. Since (x⊖y)⊖((y⊖a)⊖(x⊖a))=θ and (y⊖x)⊖((x⊖a)⊖(y⊖a))=θ, it follows from Proposition 3.5(1) that f(x⊖y)≥f((y⊖a)⊖(x⊖a)) and f(y⊖x)≥f((x⊖a)⊖(y⊖a)) so that
(3.21)df(x,y)=f(x⊖y)+f(y⊖x)≥f((y⊖a)⊖(x⊖a))+f((x⊖a)⊖(y⊖a))=df(x⊖a,y⊖a).
(2) It is similar to the proof of (1).
(3) Using Proposition3.5(2), we have
(3.22)f((x⊖y)⊖(a⊖b))≤f((x⊖y)⊖(a⊖y))+f((a⊖y)⊖(a⊖b)),f((a⊖b)⊖(x⊖y))≤f((a⊖b)⊖(a⊖y))+f((a⊖y)⊖(x⊖y)),
for all x,y,a,b∈X. Hence,
(3.23)df(x⊖y,a⊖b)=f((x⊖y)⊖(a⊖b))+f((a⊖b)⊖(x⊖y))≤[f((x⊖y)⊖(a⊖y))+f((a⊖y)⊖(a⊖b))]+[f((a⊖b)⊖(a⊖y))+f((a⊖y)⊖(x⊖y))]=[f((x⊖y)⊖(a⊖y))+f((a⊖y)⊖(x⊖y))]+[f((a⊖b)⊖(a⊖y))+f((a⊖y)⊖(a⊖b))]=df(x⊖y,a⊖y)+df(a⊖y,a⊖b)
for all x,y,a,b∈X.Theorem 3.20.
Letf:X→ℝ be a pseudovaluation on X such that Ff={x∈X∣f(x)≤0} is a closed filter of X. If df is a metric on X, then f is a valuation on X.Proof.
Suppose thatf is not a valuation on X. Then, there exists x∈X such that x≠θ and f(x)=0. Thusθ,x∈Ff and so x⊖θ∈Ff, since Ff is a closed filter of X. It follows that f(x⊖θ)≤0 so that
(3.24)0=f(θ)≤f(x⊖θ)+f(x)=f(x⊖θ)≤0.
Hence, f(x⊖θ)=0, and thus df(x,θ)=f(x⊖θ)+f(θ⊖x)=f(x⊖θ)+f(x)=0. Thus, x=θ since df is a metric on X. This is a contradiction. Therefore, f is a valuation on X.Consider the pseudovaluationf on ℤ which is described in Example 3.3. If a=-1, then(3.25)f(x)={0ifx=θ,-x+botherwise,
for all x∈ℤ, and Ff={x∈ℤ∣b≤x}∪{θ} which is not a closed filter of ℤ. Since f is a pseudovaluation on ℤ, we know that (ℤ,df) is a pseudometric space by Theorem 3.18. If x≠y in ℤ, then(3.26)df(x,y)=f(x⊖y)+f(y⊖x)=f(y-x)+f(x-y)=-y+x+b-x+y+b=2b≠0.
Hence, (ℤ,df) is a metric space. But f(b)=0, and so, f is not a valuation on ℤ. This shows that Theorem 3.20 may not be true when Ff is not a closed filter of X.Theorem 3.21.
For a mappingf:X→ℝ, if df is a pseudometric on X, then (X×X,df*) is a pseudometric space, where
(3.27)df*((x,y),(a,b))=max⁡⁡{df(x,a),df(y,b)}
for all (x,y),(a,b)∈X×X.Proof.
Supposedf is a pseudometric on X. For any (x,y),(a,b)∈X×X, we have
(3.28)df*((x,y),(x,y))=max⁡{df(x,x),df(y,y)}=0,df*((x,y),(a,b))=max⁡{df(x,a),df(y,b)}=max⁡{df(a,x),df(b,y)}=df*((a,b),(x,y)).
Now, let (x,y),(a,b),(u,v)∈X×X. Then,
(3.29)df*((x,y),(u,v))+df*((u,v),(a,b))=max⁡{df(x,u),df(y,v)}+max⁡{df(u,a),df(v,b)}≥max⁡{df(x,u)+df(u,a),df(y,v)+df(v,b)}≥max⁡{df(x,a),df(y,b)}=df*((x,y),(a,b)).
Therefore, (X×X,df*) is a pseudometric space.Corollary 3.22.
Iff:X→ℝ is a pseudovaluation on X, then (X×X,df*) is a pseudometric space.It is natural to ask that iff:X→ℝ is a valuation on X, then is (X,df) a metric space. But, we see that it is incorrect in the following example.Example 3.23.
For a WFI algebra(ℤ;⊖,θ), a mapping f:ℤ→ℝ defined by f(x)=(1/2)x for all x∈ℤ is a valuation on ℤ. Then, df is a pseudometric on ℤ. Note that df(-2,3)=f(-2⊖3)+f(3⊖(-2))=0, but -2≠3. Hence, (X,df) is not a metric space.Theorem 3.24.
Iff:X→ℝ is a positive valuation on X, then (X,df) is a metric space.Proof.
Suppose thatf is a positive valuation on X. Then, (X,df) is a pseudometric space by Theorem 3.18. Let x,y∈X be such that df(x,y)=0. Then, 0=df(x,y)=f(x⊖y)+f(y⊖x), and so f(x⊖y)=0 and f(y⊖x)=0, since f is positive. Also, since f is a valuation on X, it follows that x⊖y=θ and y⊖x=θ so from (a2) that x=y. Therefore, (X,df) is a metric space.Corollary 3.25.
Iff:X→ℝ is a valuation on X such that Ff={θ}, then (X,df) is a metric space.Theorem 3.26.
Iff:X→ℝ is a positive valuation on X, then (X×X,df*) is a metric space.Proof.
Note from Corollary3.22 that (X×X,df*) is a pseudometric space. Let (x,y),(a,b)∈X×X be such that df*((x,y),(a,b))=0. Then,
(3.30)0=df*((x,y),(a,b))=max⁡{df(x,a),df(y,b)},
and so df(x,a)=0=df(y,b), since df(x,y)≥0 for all (x,y)∈X×X. Hence,
(3.31)0=df(x,a)=f(x⊖a)+f(a⊖x),0=df(y,b)=f(y⊖b)+f(b⊖y).
Since f is positive, it follows that f(x⊖a)=0=f(a⊖x) and f(y⊖b)=0=f(b⊖y) so that x⊖a=θ=a⊖x and y⊖b=θ=b⊖y. Using (a2), we have a=x and b=y, and so (x,y)=(a,b). Therefore, (X×X,df*) is a metric space.Theorem 3.27.
Iff is a positive valuation on X, then the operation ⊖:X×X→X is uniformly continuous. (Suppose that (X,d) and (Y,ρ) are metric spaces and f:X→Y. We say that f is uniformly continuous provided that for every ɛ>0, there exists δ>0 such that for any points x1 and x2 in X, if d(x1,x2)<δ, then ρ(f(x1),f(x2))<ɛ.)Proof.
For anyɛ>0, if df*((x,y),(a,b))<ɛ/2, then df(x,a)<ɛ/2, and df(y,b)<ɛ/2. Using Proposition 3.19, we have
(3.32)df(x⊖y,a⊖b)≤df(x⊖y,a⊖y)+df(a⊖y,a⊖b)≤df(x,a)+df(y,b)<ɛ2+ɛ2=ɛ.
Therefore, the operation ⊖:X×X→X is uniformly continuous.Corollary 3.28.
Iff is a valuation on X such that Ff={θ}, then the operation ⊖:X×X→X is uniformly continuous.



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