# On semi-vector spaces and semi-algebras

Giuliano G. La Guardia, Jocemar de Q. Chagas, Ervin K. Lenzi, Leonardo Pires
Giuliano G. La Guardia (gguardia@uepg.br), Jocemar de Q. Chagas
<EMAIL_ADDRESS>and Leonardo Pires<EMAIL_ADDRESS>are with
Department of Mathematics and Statistics, State University of Ponta Grossa
(UEPG), 84030-900, Ponta Grossa - PR, Brazil. Ervin K. Lenzi<EMAIL_ADDRESS>is with Department of Physics, State University of Ponta Grossa (UEPG),
84030-900, Ponta Grossa - PR, Brazil. Corresponding author: Giuliano G. La
Guardia (gguardia@uepg.br).

###### Abstract

It is well-known that the theories of semi-vector spaces and semi-algebras –
which were not much studied over time – are utilized/applied in Fuzzy Set
Theory in order to obtain extensions of the concept of fuzzy numbers as well
as to provide new mathematical tools to investigate properties and new results
on fuzzy systems. In this paper we investigate the theory of semi-vector
spaces over the semi-field of nonnegative real numbers ${\mathbb{R}}_{0}^{+}$.
We prove several results concerning semi-vector spaces and semi-linear
transformations. Moreover, we introduce in the literature the concept of
eigenvalues and eigenvectors of a semi-linear operator, describing in some
cases how to compute them. Topological properties of semi-vector spaces such
as completeness and separability are also investigated. New families of semi-
vector spaces derived from semi-metric, semi-norm, semi-inner product, among
others are exhibited. Additionally, some results on semi-algebras are
presented.

_keywords_ : semi-vector space; semi-algebras; semi-linear operators

##  1 Introduction

The concept of semi-vector space was introduced by Prakash and Sertel in [11].
Roughly speaking, semi-vector spaces are “vector spaces” where the scalars are
in a semi-field. Although the concept of semi-vector space was investigated
over time, there exist few works available in the literature dealing with such
spaces [13, 11, 12, 10, 4, 5, 8]. This fact occurs maybe due to the
limitations that such concept brings, i.e., the non-existence of (additive)
symmetric for some (for all) semi-vector. A textbook in such a topic of
research is the book by Kandasamy [6].

Although the seminal paper on semi-vector spaces is [11], the idea of such a
concept was implicit in [13], where Radstrom shown that a semi-vector space
over the semi-field of nonnegative real numbers can be extended to a real
vector space (see [13, Theorem 1-B.]). In [11], Prakash and Sertel
investigated the structure of topological semi-vector spaces. The authors were
concerned with the study of the existence of fixed points in compact convex
sets and also to generate min-max theorems in topological semi-vector spaces.
In [12], Prakash and Sertel investigated properties of the topological semi-
vector space consisting of nonempty compact subsets of a real Hausdorff
topological vector space. In [10], Pap investigated and formulated the concept
of integrals of functions having, as counter-domain, complete semi-vector
spaces. W. Gahler and S. Gahler [4] showed that a (ordered) semi-vector space
can be extended to a (ordered) vector space and a (ordered) semi-algebra can
be extended to a (ordered) algebra. Moreover, they provided an extension of
fuzzy numbers. Janyska et al. [5] developed such theory (of semi-vector space)
by proving useful results and defining the semi-tensor product of (semi-free)
semi-vector spaces. They were also interested to propose an algebraic model of
physical scales. Canarutto [1] explored the concept of semi-vector spaces to
express aspects and to exploit nonstandard mathematical notions of basics of
quantum particle physics on a curved Lorentzian background. Moreover, he dealt
with the case of electroweak interactions. Additionally, in [2], Canarutto
provided a suitable formulation of the fundamental mathematical concepts with
respect to quantum field theory. Such a paper presents a natural application
of the concept of semi-vector spaces and semi-algebras. Recently, Bedregal et
al. [8] investigated (ordered) semi-vector spaces over a weak semi-field $K$
(i.e., both $(K,+)$ and $(K,\bullet)$ are monoids) in the context of fuzzy
sets and applying the results in multi-criteria group decision-making.

In this paper we extend the theory of semi-vector spaces. The semi-field of
scalars considered here is the semi-field of nonnegative real numbers. We
prove several results in the context of semi-vector spaces and semi-linear
transformations. We introduce the concept of semi-eigenvalues and semi-
eigenvectors of an operator and of a matrix, showing how to compute it in
specific cases. We investigate topological properties such as completeness,
compactness and separability of semi-vector spaces. Additionally, we present
interesting new families of semi-vector spaces derived from semi-metric, semi-
norm, semi-inner product, metric-preserving functions among others.
Furthermore, we show some results concerning semi-algebras. Summarizing, we
provide new results on semi-vector spaces and semi-algebras, although such
theories are very difficult to be investigated due to the fact that vectors do
not even have (additive) symmetrical. These new results can be possibly
utilized in the theory of fuzzy sets in order to extend it or in the
generation of new results concerning such a theory.

The paper is organized as follows. In Section 2 we recall some concepts on
semi-vector spaces which will be utilized in this work. In Section 3 we
present and prove several results concerning semi-vector spaces and semi-
linear transformations. We introduce naturally the concepts of eigenvalue and
eigenvector of a semi-linear operator and of matrices. Additionally, we
exhibit and show interesting examples of semi-vector spaces derived from semi-
metric, semi-norms, metric-preserving functions among others. Results
concerning semi-algebras are also presented. In Section 4 we show
relationships between Fuzzy Set Theory and the theory of semi-vector spaces
and semi-algebras. Finally, a summary of this paper is presented in Section 5.

##  2 Preliminaries

In this section we recall important facts on semi-vector spaces necessary for
the development of this work. In order to define formally such concept, it is
necessary to define the concepts of semi-ring and semi-field.

###### Definition 2.1

A semi-ring $(S,+,\bullet)$ is a set $S$ endowed with two binary operations,
$+:S\times S\longrightarrow S$ (addition), $\bullet:S\times S\longrightarrow
S$ (multiplication) such that: $\operatorname{(1)}$ $(S,+)$ is a commutative
monoid; $\operatorname{(2)}$ $(S,\bullet)$ is a semigroup;
$\operatorname{(3)}$ the multiplication $\bullet$ is distributive with respect
to $+$: $\forall\ x,y,z\in S$, $(x+y)\bullet z=x\bullet z+y\bullet z$ and
$x\bullet(y+z)=x\bullet y+x\bullet z$.

We write $S$ instead of writing $(S,+,\bullet)$ if there is not possibility of
confusion. If the multiplication $\bullet$ is commutative then $S$ is a
commutative semi-ring. If there exists $1\in S$ such that, $\forall\ x\in S$
one has $1\bullet x=x\bullet 1=x$, then $S$ is a semi-ring with identity.

###### Definition 2.2

[6, Definition 3.1.1] A semi-field is an ordered triple $(K,+,\bullet)$ which
is a commutative semi-ring with unit satisfying the following conditions:
$\operatorname{(1)}$ $\forall\ x,y\in K$, if $x+y=0$ then $x=y=0$;
$\operatorname{(2)}$ if $x,y\in K$ and $x\bullet y=0$ then $x=0$ or $y=0$.

Before proceeding further, it is interesting to observe that in [4] the
authors considered the additive cancellation law in the definition of semi-
vector space. In [5], the authors did not assume the existence of the zero
(null) vector.

In this paper we consider the definition of a semi-vector space in the context
of that shown in [4], Sect.3.1.

###### Definition 2.3

A semi-vector space over a semi-field $K$ is a ordered triple $(V,$
$+,\cdot)$, where $V$ is a set endowed with the operations $+:V\times
V\longrightarrow V$ (vector addition) and $\cdot:K\times V\longrightarrow V$
(scalar multiplication) such that:

  * $\operatorname{(1)}$

$(V,+)$ is an abelian monoid equipped with the additive cancellation law:
$\forall\ u,v,w\in V$, if $u+v=u+w$ then $v=w$;

  * $\operatorname{(2)}$

$\forall$ $\alpha\in K$ and $\forall$ $u,v\in V$, $\alpha(u+v)=\alpha u+\beta
v$;

  * $\operatorname{(3)}$

$\forall$ $\alpha,\beta\in K$ and $\forall$ $v\in V$, $(\alpha+\beta)v=\alpha
v+\beta v$;

  * $\operatorname{(4)}$

$\forall$ $\alpha,\beta\in K$ and $\forall$ $v\in V$,
$(\alpha\beta)v=\alpha(\beta v)$;

  * $\operatorname{(5)}$

$\forall$ $v\in V$ and $1\in K$, $1v=v$.

Note that from Item $\operatorname{(1)}$ of Definition 2.3, all semi-vector
spaces considered in this paper are _regular_ , that it, the additive
cancellation law is satisfied. The zero (or null) vector of $V$, which is
unique, will be denoted by $0_{V}$. Let $v\in V$, $v\neq 0$. If there exists
$u\in V$ such that $v+u=0$ then $v$ is said to be _symmetrizable_. A semi-
vector space $V$ is said to be _simple_ if the unique symmetrizable element is
the zero vector $0_{V}$. In other words, $V$ is simple if it has none nonzero
symmetrizable elements.

###### Definition 2.4

[5, Definition 1.4] Let $V$ be a simple semi-vector space over
${\mathbb{R}}_{0}^{+}$. A subset $B\subset V$ is called a semi-basis of $V$ if
every $v\in V$, $v\neq 0$, can be written in a unique way as
$v=\displaystyle\sum_{i\in I_{v}}v^{(i)}b_{i}$, where
$v^{(i)}\in{\mathbb{R}}^{+}$, $b_{i}\in B$ and $I_{v}$ is a finite family of
indices uniquely determined by $v$. The finite subset $B_{v}\subset B$ defined
by $B_{v}:=\\{b_{i}\\}_{i\in I_{v}}$ is uniquely determined by $v$. If a semi-
vector space $V$ admits a semi-basis then it is said to be semi-free.

The concept of semi-dimension can be defined in analogous way to semi-free
semi-vector spaces due to the next result.

###### Corollary 2.1

[5, Corollary 1.7] Let $V$ be a semi-free semi-vector space. Then all semi-
bases of $V$ have the same cardinality.

Therefore, the semi-dimension of a semi-free semi-vector space is the
cardinality of a semi-basis (consequently, of all semi-bases) of $V$. We next
present some examples of semi-vector spaces.

###### Example 2.1

All real vector spaces are semi-vector spaces, but they are not simple.

###### Example 2.2

The set
${[{\mathbb{R}}_{0}^{+}]}^{n}=\underbrace{{\mathbb{R}}_{0}^{+}\times\ldots\times{\mathbb{R}}_{0}^{+}}_{n\operatorname{times}}$
endowed with the usual sum of coordinates and scalar multiplication is a semi-
vector space over ${\mathbb{R}}_{0}^{+}$.

###### Example 2.3

The set ${\mathcal{M}}_{n\times m}({\mathbb{R}}_{0}^{+})$ of matrices $n\times
m$ whose entries are nonnegative real numbers equipped with the sum of
matrices and multiplication of a matrix by a scalar (in
${\mathbb{R}}_{0}^{+}$, of course) is a semi-vector space over
${\mathbb{R}}_{0}^{+}$.

###### Example 2.4

The set ${\mathcal{P}}_{n}[x]$ of polynomials with coefficients from
${\mathbb{R}}_{0}^{+}$ and degree less than or equal to $n$, equipped with the
usual of polynomial sum and scalar multiplication, is a semi-vector space.

###### Definition 2.5

Let $(V,+,\cdot)$ be a semi-vector space over ${\mathbb{R}}_{0}^{+}$. We say
that a non-empty subset $W$ of $V$ is a semi-subspace of $V$ if $W$ is closed
under both addition and scalar multiplication of $V$, that is,

  * $\operatorname{(1)}$

$\forall\ w_{1},w_{2}\in W\Longrightarrow w_{1}+w_{2}\in W$;

  * $\operatorname{(2)}$

$\forall\ \lambda\in{\mathbb{R}}_{0}^{+}$ and $\forall\ w\in
W\Longrightarrow\lambda w\in W$.

The uniqueness of the zero vector implies that for each
$\lambda\in{\mathbb{R}}_{0}^{+}$ on has $\lambda 0_{V}=0_{V}$. Moreover, if
$v\in V$, it follows that $0v=0v+0v$; applying the regularity one obtains
$0v=0_{V}$. Therefore, from Item $\operatorname{(2)}$, every semi-subspace
contains the zero vector.

###### Example 2.5

Let ${\mathbb{Q}}_{0}^{+}$ denote the set of nonnegative rational numbers. The
semi-vector space ${\mathbb{Q}}_{0}^{+}$ considered as an
${\mathbb{Q}}_{0}^{+}$ space is a semi-subspace of ${\mathbb{R}}_{0}^{+}$
considered as an ${\mathbb{Q}}_{0}^{+}$ space.

###### Example 2.6

For each positive integer $i\leq n$, the subset
${\mathcal{P}}_{(i)}[x]\cup\\{0_{p}\\}$, where
${\mathcal{P}}_{(i)}[x]=\\{p(x);\partial(p(x))=i\\}$ and $0_{p}$ is the null
polynomial, is a semi-subspace of ${\mathcal{P}}_{n}[x]$, shown in Example
2.4.

###### Example 2.7

The set of diagonal matrices of order $n$ with entries in
${\mathbb{R}}_{0}^{+}$ is a semi-subspace of
${\mathcal{M}}_{n}({\mathbb{R}}_{0}^{+})$, where the latter is the semi-vector
space of square matrices with entries in ${\mathbb{R}}_{0}^{+}$ (according to
Example 2.3).

###### Definition 2.6

[5, Definition 1.22] Let $V$ and $W$ be two semi-vector spaces and
$T:V\longrightarrow W$ be a map. We say that $T$ is a semi-linear
transformation if: $\operatorname{(1)}$ $\forall\ v_{1},v_{2}\in V$,
$T(v_{1}+v_{2})=T(v_{1})+T(v_{2})$; $\operatorname{(2)}$
$\forall\lambda\in{\mathbb{R}}_{0}^{+}$ and $\forall\ v\in V$, $T(\lambda
v)=\lambda T(v)$.

If $U$ and $V$ are semi-vector spaces then the set
$\operatorname{Hom}(U,V)=\\{T:U\longrightarrow V;T\operatorname{is\ semi-
linear}\\}$ is also a semi-vector space.

##  3 The New Results

We start this section with important remarks.

###### Remark 3.1

  * $\operatorname{(1)}$

Throughout this section we always consider that the semi-field $K$ is the set
of nonnegative real numbers, i.e.,
$K={\mathbb{R}}_{0}^{+}={\mathbb{R}}^{+}\cup\\{0\\}$.

  * $\operatorname{(2)}$

In the whole section (except Subsection 3.2) we assume that the semi-vector
spaces $V$ are simple, i.e., the unique symmetrizable element is the zero
vector $0_{V}$.

  * $\operatorname{(3)}$

It is well-known that a semi-vector space $(V,+,\cdot)$ can be always extended
to a vector space according to the equivalence relation on $V\times V$ defined
by $(u_{1},v_{1})\sim(u_{2},v_{2})$ if and only if $u_{1}+v_{2}=v_{1}+u_{2}$
(see [13]; see also [4, Section 3.4]). However, our results are obtained
without utilizing such a natural embedding. In other words, if one want to
compute, for instance, the eigenvalues of a matrix defined over
${\mathbb{R}}_{0}^{+}$ we cannot solve the problem in the associated vector
spaces and then discard the negative ones. Put differently, all computations
performed here are restricted to nonnegative real numbers and also to the fact
that none vector (with exception of $0_{V}$) has (additive) symmetrical.
However, we will show that, even in this case, several results can be
obtained.

###### Proposition 3.1

Let $V$ be a semi-vector space over ${\mathbb{R}}_{0}^{+}$. Then the following
hold:

  * $\operatorname{(1)}$

let $v\in V$, $v\neq 0_{V}$, and $\lambda\in{\mathbb{R}}_{0}^{+}$; if $\lambda
v=0_{V}$ then $\lambda=0$;

  * $\operatorname{(2)}$

if $\alpha,\beta\in{\mathbb{R}}_{0}^{+}$, $v\in V$ and $v\neq 0_{V}$, then the
equality $\alpha v=\beta v$ implies that $\alpha=\beta$.

Proof: $\operatorname{(1)}$ If $\lambda\neq 0$ then there exists its
multiplicative inverse ${\lambda}^{-1}$, hence $1v={\lambda}^{-1}0_{V}=0_{V}$,
i.e., $v=0_{V}$, a contradiction.
$\operatorname{(2)}$ If $\alpha\neq\beta$, assume w.l.o.g. that
$\alpha>\beta$, i.e., there exists a positive real number $c$ such that
$\alpha=\beta+c$. Thus, $\alpha v=\beta v$ implies $\beta v+cv=\beta v$. From
the cancellation law we have $cv=0_{V}$, and from Item $\operatorname{(1)}$ it
follows that $c=0$, a contradiction. $\Box$

We next introduce in the literature the concept of eigenvalue and eigenvector
of a semi-linear operator.

###### Definition 3.1

Let $V$ be a semi-vector space and $T:V\longrightarrow V$ be a semi-linear
operator. If there exist a non-zero vector $v\in V$ and a nonnegative real
number $\lambda$ such that $T(v)=\lambda v$, then $\lambda$ is an eigenvalue
of $T$ and $v$ is an eigenvector of $T$ associated with $\lambda$.

As it is natural, the zero vector joined to the set of the eigenvectors
associated with a given eigenvalue has a semi-subspace structure.

###### Proposition 3.2

Let $V$ be a semi-vector space over ${\mathbb{R}}_{0}^{+}$ and
$T:V\longrightarrow V$ be a semi-linear operator. Then the set
$V_{\lambda}=\\{v\in V;T(v)=\lambda v\\}\cup\\{0_{V}\\}$ is a semi-subspace of
$V$.

Proof: From hypotheses, $V_{\lambda}$ is non-empty. Let $u,v\in V_{\lambda}$,
i.e., $T(u)=\lambda u$ and $T(v)=\lambda v$. Hence,
$T(u+v)=T(u)+T(v)=\lambda(u+v)$, i.e., $u+v\in V_{\lambda}$. Further, if
$\alpha\in{\mathbb{R}}_{0}^{+}$ and $u\in V$, it follows that $T(\alpha
u)=\alpha T(u)=\lambda(\alpha u)$, that is, $\alpha u\in V_{\lambda}$.
Therefore, $V_{\lambda}$ is a semi-subspace of $V$. $\Box$

The next natural step would be to introduce the characteristic polynomial of a
matrix, according to the standard Linear Algebra. However, how to compute
$\det(A-\lambda I)$ if $-\lambda$ can be a negative real number? Based on this
fact we must be careful to compute the eigenvectors of a matrix. In fact, the
main tools to be utilized in computing eigenvalues/eigenvectors of a square
matrix whose entries are nonnegative real numbers is the additive cancellation
law in ${\mathbb{R}}_{0}^{+}$ and also the fact that positive real numbers
have multiplicative inverse. However, in much cases, such a tools are not
sufficient to solve the problem. Let us see some cases when it is possible to
compute eigenvalues/eigenvectors of a matrix.

###### Example 3.1

Let us see how to obtain (if there exists) an eigenvalue/eigenvector of a
diagonal matrix $A\in{\mathcal{M}}_{2}({\mathbb{R}}_{0}^{+})$,

$\displaystyle A=\left[\begin{array}[]{cc}a&0\\\ 0&b\\\ \end{array}\right],$

where $a\neq b$ not both zeros.

Let us assume first that $a,b>0$. Solving the equation $Av=\lambda v$, that
is,

$\displaystyle\left[\begin{array}[]{cc}a&0\\\ 0&b\\\
\end{array}\right]\left[\begin{array}[]{c}x\\\ y\\\
\end{array}\right]=\left[\begin{array}[]{c}\lambda x\\\ \lambda y\\\
\end{array}\right],$

we obtain $\lambda=a$ with associated eigenvector $x(1,0)$ and $\lambda=b$
with associated eigenvector $y(0,1)$.

If $a\neq 0$ and $b=0$, then $\lambda=a$ with eigenvectors $x(1,0)$.

If $a=0$ and $b\neq 0$, then $\lambda=b$ with eigenvectors $y(0,1)$.

###### Example 3.2

Let $A\in{\mathcal{M}}_{2}({\mathbb{R}}_{0}^{+})$ be a matrix of the form

$\displaystyle A=\left[\begin{array}[]{cc}a&b\\\ 0&a\\\ \end{array}\right],$

where $a\neq b$ are positive real numbers. Let us solve the matrix equation:

$\displaystyle\left[\begin{array}[]{cc}a&b\\\ 0&a\\\
\end{array}\right]\left[\begin{array}[]{c}x\\\ y\\\
\end{array}\right]=\left[\begin{array}[]{c}\lambda x\\\ \lambda y\\\
\end{array}\right].$

If $y\neq 0$, $\lambda=a$; hence $by=0$, which implies $b=0$, a contradiction.
If $y=0$, $x\neq 0$; hence $\lambda=a$ with eigenvectors $(x,0)$.

If $V$ and $W$ are semi-free semi-vector spaces then it is possible to define
the matrix of a semi-linear transformation $T:V\longrightarrow W$ as in the
usual case (vector spaces).

###### Definition 3.2

Let $T:V\longrightarrow W$ be a semi-liner transformation between semi-free
semi-vector spaces with semi-basis $B_{1}$ and $B_{2}$, respectively. Then the
matrix $[T]_{B_{1}}^{B_{2}}$ is the matrix of the transformation $T$.

###### Theorem 3.1

Let $V$ be a semi-free semi-vector space over ${\mathbb{R}}_{0}^{+}$ and let
$T:V\longrightarrow V$ be a semi-linear operator. Then $T$ admits a semi-basis
$B=\\{v_{1},v_{2},\ldots,v_{n}\\}$ such that ${[T]}_{B}^{B}$ is diagonal if
and only if $B$ consists of eigenvectors of $T$.

Proof: The proof is analogous to the case of vector spaces. Let
$B=\\{v_{1},v_{2},\ldots,$ $v_{n}\\}$ be a semi-basis of $V$ whose elements
are eigenvectors of $T$. We then have:

$\displaystyle T(v_{1})={\lambda}_{1}v_{1}+0v_{2}+\ldots+0v_{n},$
$\displaystyle T(v_{2})=0v_{1}+{\lambda}_{2}v_{2}+\ldots+0v_{n},$
$\displaystyle\vdots$ $\displaystyle
T(v_{n})=0v_{1}+0v_{2}+\ldots+{\lambda}_{n}v_{n},$

which implies that $[T]_{B}^{B}$ is of the form

$\displaystyle[T]_{B}^{B}=\left[\begin{array}[]{ccccc}{\lambda}_{1}&0&0&\ldots&0\\\
0&{\lambda}_{2}&0&\ldots&0\\\ \vdots&\vdots&\vdots&\ldots&\vdots\\\
0&0&0&\ldots&{\lambda}_{n}\\\ \end{array}\right].$

On the other hand, let $B^{*}=\\{w_{1},w_{2},\ldots,w_{n}\\}$ be a semi-basis
of $V$ such that $[T]_{B^{*}}^{B^{*}}$ is diagonal:

$\displaystyle[T]_{B^{*}}^{B^{*}}=\left[\begin{array}[]{ccccc}{\alpha}_{1}&0&0&\ldots&0\\\
0&{\alpha}_{2}&0&\ldots&0\\\ \vdots&\vdots&\vdots&\ldots&\vdots\\\
0&0&0&\ldots&{\alpha}_{n}\\\ \end{array}\right];$

thus,

$\displaystyle
T(w_{1})={\alpha}_{1}w_{1}+0w_{2}+\ldots+0w_{n}={\alpha}_{1}w_{1},$
$\displaystyle
T(w_{2})=0w_{1}+{\alpha}_{2}w_{2}+\ldots+0w_{n}={\alpha}_{2}w_{2},$
$\displaystyle\vdots$ $\displaystyle
T(w_{n})=0w_{1}+0w_{2}+\ldots+{\alpha}_{n}w_{n}={\alpha}_{2}w_{n}.$

This means that $w_{i}$ are eigenvectors of $T$ with corresponding eigenvalues
${\alpha}_{i}$, for all $i=1,2,\ldots,n$. $\Box$

###### Definition 3.3

Let $T:V\longrightarrow W$ be a semi-linear transformation. The set
$\operatorname{Ker}(T)=\\{v\in V;T(v)=0\\}$ is called kernel of $T$.

###### Proposition 3.3

Let $T:V\longrightarrow W$ be a semi-linear transformation. Then the following
hold:

  * $\operatorname{(1)}$

$\operatorname{Ker}(T)$ is a semi-subspace of $V$;

  * $\operatorname{(2)}$

if $T$ is injective then $\operatorname{Ker}(T)=\\{0_{V}\\}$;

  * $\operatorname{(3)}$

if $V$ has semi-dimension $1$ then $\operatorname{Ker}(T)=\\{0_{V}\\}$ implies
that $T$ is injective.

Proof: $\operatorname{(1)}$ We have $T(0_{V})=T(0_{V})+T(0_{V})$. Since $W$ is
regular, it follows that $T(0_{V})=0_{W}$, which implies
$\operatorname{Ker}(T)\neq\emptyset$. If $u,v\in\operatorname{Ker}(T)$ and
$\lambda\in{\mathbb{R}}_{0}^{+}$, then $u+v\in\operatorname{Ker}(T)$ and
$\lambda v\in\operatorname{Ker}(T)$, which implies that
$\operatorname{Ker}(T)$ is a semi-subspace of $V$.
$\operatorname{(2)}$ Since $T(0_{V})=0_{W}$, it follows that
$\\{0_{V}\\}\subseteq\operatorname{Ker}(T)$. On the other hand, let
$u\in\operatorname{Ker}(T)$, that is, $T(u)=0_{W}$. Since $T$ is injective,
one has $u=0_{V}$. Hence, $\operatorname{Ker}(T)=\\{0_{V}\\}$.
$\operatorname{(3)}$ Let $B=\\{v_{0}\\}$ be a semi-basis of $V$. Assume that
$T(u)=T(v)$, where $u,v\in V$ are such that $u=\alpha v_{0}$ and $v=\beta
v_{0}$. Hence, $\alpha T(v_{0})=\beta T(v_{0})$. Since
$\operatorname{Ker}(T)=\\{0_{V}\\}$ and $v_{0}\neq 0$, it follows that
$T(v_{0})\neq 0$. From Item $\operatorname{(2)}$ of Proposition 3.1, one has
$\alpha=\beta$, i.e., $u=v$. $\Box$

###### Definition 3.4

Let $T:V\longrightarrow W$ be a semi-linear transformation. The image of $T$
is the set of all vectors $w\in W$ such that there exists $v\in V$ with
$T(v)=w$, that is, $\operatorname{Im}(T)=\\{w\in W;\exists\ v\in
V\operatorname{with}T(v)=w\\}$.

###### Proposition 3.4

Let $T:V\longrightarrow W$ be a semi-linear transformation. Then the image of
$T$ is a semi-subspace of $W$.

Proof: The set $\operatorname{Im}(T)$ is non-empty because $T(0_{V})=0_{W}$.
It is easy to see that if $w_{1},w_{2}\in\operatorname{Im}(T)$ and
$\lambda\in{\mathbb{R}}_{0}^{+}$, then $w_{1}+w_{2}\in\operatorname{Im}(T)$
and $\lambda w_{1}\in\operatorname{Im}(T)$. $\Box$

###### Theorem 3.2

Let $V$ be a $n$-dimensional semi-free semi-vector space over
${\mathbb{R}}_{0}^{+}$. Then $V$ is isomorphic to
$({\mathbb{R}}_{0}^{+})^{n}$.

Proof: Let $B=\\{v_{1},v_{2},\ldots,v_{n}\\}$ be a semi-basis of $V$ and
consider the canonical semi-basis $e_{i}=(0,0,\ldots,$
$0,\underbrace{1}_{i},0,\ldots,0)$ of $({\mathbb{R}}_{0}^{+})^{n}$, where
$i=1,2,\ldots,n$. Define the map
$T:V\longrightarrow({\mathbb{R}}_{0}^{+})^{n}$ as follows: for each
$v=\displaystyle\sum_{i=1}^{n}a_{i}v_{i}\in V$, put
$T(v)=\displaystyle\sum_{i=1}^{n}a_{i}e_{i}$. It is easy to see that $T$ is
bijective semi-linear transformation, i.e., $V$ is isomorphic to
$({\mathbb{R}}_{0}^{+})^{n}$, as required. $\Box$

###  3.1 Complete Semi-Vector Spaces

We here define and study complete semi-vector spaces, i.e., semi-vector spaces
whose norm (inner product) induces a metric under which the space is complete.

###### Definition 3.5

Let $V$ be a semi-vector space over ${\mathbb{R}}_{0}^{+}$. If there exists a
norm $\|\ \|:V\longrightarrow{\mathbb{R}}_{0}^{+}$ on $V$ we say that $V$ is a
normed semi-vector space (or normed semi-space, for short). If the norm
defines a metric on $V$ under which $V$ is complete then $V$ is said to be
Banach semi-vector space.

###### Definition 3.6

Let $V$ be a semi-vector space over ${\mathbb{R}}_{0}^{+}$. If there exists an
inner product $\langle\ ,\ \rangle:V\times
V\longrightarrow{\mathbb{R}}_{0}^{+}$ on $V$ then $V$ is an inner product
semi-vector space (or inner product semi-space). If the inner product defines
a metric on $V$ under which $V$ is complete then $V$ is said to be Hilbert
semi-vector space.

The well-known norms on ${\mathbb{R}}^{n}$ are also norms on
$[{\mathbb{R}}_{0}^{+}]^{n}$, as we show in the next propositions.

###### Proposition 3.5

Let $V=[{\mathbb{R}}_{0}^{+}]^{n}$ be the Euclidean semi-vector space (over
${\mathbb{R}}_{0}^{+}$) of semi-dimension $n$ . Define the function $\|\
\|:V\longrightarrow{\mathbb{R}}_{0}^{+}$ as follows: if
$x=(x_{1},x_{2},\ldots,$ $x_{n})\in V$, put
$\|x\|=\sqrt{x_{1}^{2}+x_{2}^{2}+\ldots+x_{n}^{2}}$. Then $\|\ \|$ is a norm
on $V$, called the Euclidean norm on $V$.

Proof: It is clear that $\|x\|=0$ if and only if $x=0$ and for all
$\alpha\in{\mathbb{R}}_{0}^{+}$ and $x\in V$, $\|\alpha x\|=|\alpha|\|x\|$. To
show the triangle inequality it is sufficient to apply the Cauchy-Schwarz
inequality in ${\mathbb{R}}_{0}^{+}$: if $x=(x_{1},x_{2},\ldots,x_{n})$ and
$y=(y_{1},y_{2},\ldots,y_{n})$ are semi-vectors in $V$ then
$\displaystyle\sum_{i=1}^{n}x_{i}y_{i}\leq{\left(\displaystyle\sum_{i=1}^{n}x_{i}^{2}\right)}^{1/2}\cdot{\left(\displaystyle\sum_{i=1}^{n}y_{i}^{2}\right)}^{1/2}$.
$\Box$

In the next result we show that the Euclidean norm on
$[{\mathbb{R}}_{0}^{+}]^{n}$ generates the Euclidean metric on it.

###### Proposition 3.6

Let $x=(x_{1},x_{2},\ldots,x_{n})$, $y=(y_{1},y_{2},\ldots,y_{n})$ be semi-
vectors in $V=[{\mathbb{R}}_{0}^{+}]^{n}$. Define the function $d:V\times
V\longrightarrow{\mathbb{R}}_{0}^{+}$ as follows: for every fixed $i$, if
$x_{i}=y_{i}$ put $c_{i}=0$; if $x_{i}\neq y_{i}$, put
${\varphi}_{i}={\psi}_{i}+c_{i}$, where ${\varphi}_{i}=\max\\{x_{i},y_{i}\\}$
and ${\psi}_{i}=\min\\{x_{i},y_{i}\\}$ (in this case, $c_{i}>0$); then
consider $d(x,y)=\sqrt{c_{1}^{2}+\ldots+c_{n}^{2}}$. The function $d$ is a
metric on $V$.

###### Remark 3.2

Note that in Proposition 3.6 we could have defined $c_{i}$ simply by the
nonnegative real number satisfying
$\max\\{x_{i},y_{i}\\}=\min\\{x_{i},y_{i}\\}+c_{i}$. However, we prefer to
separate the cases when $c_{i}=0$ and $c_{i}>0$ in order to improve the
readability of this paper.

Proof: It is easy to see that $d(x,y)=0$ if and only if $x=y$ and
$d(x,y)=d(y,x)$.

We will next prove the triangle inequality. To do this, let
$x=(x_{1},x_{2},\ldots,x_{n})$, $y=(y_{1},y_{2},\ldots,y_{n})$ and
$z=(z_{1},z_{2},\ldots,z_{n})$ be semi-vectors in
$V=[{\mathbb{R}}_{0}^{+}]^{n}$. We look first at a fixed $i$. If
$x_{i}=y_{i}=z_{i}$ or if two of them are equal then $d(x_{i},z_{i})\leq
d(x_{i},y_{i})+d(y_{i},z_{i})$. Let us then assume that $x_{i}$, $y_{i}$ and
$z_{i}$ are pairwise distinct. We have to analyze the six cases:
$\operatorname{(1)}$ $x_{i}<y_{i}<z_{i}$; $\operatorname{(2)}$
$x_{i}<z_{i}<y_{i}$; $\operatorname{(3)}$ $y_{i}<x_{i}<z_{i}$;
$\operatorname{(4)}$ $y_{i}<z_{i}<x_{i}$; $\operatorname{(5)}$
$z_{i}<x_{i}<y_{i}$; $\operatorname{(6)}$ $z_{i}<y_{i}<x_{i}$. In order to
verify the triangle inequality we will see what occurs in the worst cases.
More precisely, we assume that for all $i=1,2,\ldots,n$ we have
$x_{i}<y_{i}<z_{i}$ or, equivalently, $z_{i}<y_{i}<x_{i}$. Since both cases
are analogous we only verify the (first) case $x_{i}<y_{i}<z_{i}$, for all
$i$. In such cases there exist positive real numbers $a_{i}$, $b_{i}$, for all
$i=1,2,\ldots,n$, such that $y_{i}=x_{i}+a_{i}$ and $z_{i}=y_{i}+b_{i}$, which
implies $z_{i}=x_{i}+a_{i}+b_{i}$. We need to show that $d(x,z)\leq
d(x,y)+d(y,z)$, i.e.,
${\left(\displaystyle\sum_{i=1}^{n}(a_{i}+b_{i})^{2}\right)}^{1/2}\leq{\left(\displaystyle\sum_{i=1}^{n}a_{i}^{2}\right)}^{1/2}+{\left(\displaystyle\sum_{i=1}^{n}b_{i}^{2}\right)}^{1/2}$.
The last inequality is equivalent to the inequality
$\displaystyle\sum_{i=1}^{n}(a_{i}+b_{i})^{2}\leq\displaystyle\sum_{i=1}^{n}a_{i}^{2}+\displaystyle\sum_{i=1}^{n}b_{i}^{2}+2{\left(\displaystyle\sum_{i=1}^{n}a_{i}^{2}\right)}^{1/2}\cdot{\left(\displaystyle\sum_{i=1}^{n}b_{i}^{2}\right)}^{1/2}$.
Again, the last inequality is equivalent to
$\displaystyle\sum_{i=1}^{n}a_{i}b_{i}\leq{\left(\displaystyle\sum_{i=1}^{n}a_{i}^{2}\right)}^{1/2}\cdot{\left(\displaystyle\sum_{i=1}^{n}b_{i}^{2}\right)}^{1/2}$,
which is the Cauchy-Schwarz inequality in ${\mathbb{R}}_{0}^{+}$. Therefore,
$d$ satisfies the triangle inequality, hence it is a metric on $V$. $\Box$

###### Remark 3.3

Note that Proposition 3.6 means that the Euclidean norm on
$[{\mathbb{R}}_{0}^{+}]^{n}$ (see Proposition 3.5) generates the Euclidean
metric on $[{\mathbb{R}}_{0}^{+}]^{n}$. This result is analogous to the fact
that every norm defined on vector spaces generates a metric on it. Further, a
semi-vector space $V$ is Banach (see Definition 3.5) if the norm generates a
metric under which every Cauchy sequence in $V$ converges to an element of
$V$.

###### Proposition 3.7

Let $V=[{\mathbb{R}}_{0}^{+}]^{n}$ and define the function $\langle\ ,\
\rangle:V\times V\longrightarrow{\mathbb{R}}_{0}^{+}$ as follows: if
$u=(x_{1},x_{2},\ldots,x_{n})$ and $v=(y_{1},y_{2},\ldots,y_{n})$ are semi-
vectors in $V$, put $\langle
u,v\rangle=\displaystyle\sum_{i=1}^{n}x_{i}y_{i}$. Then $\langle\ ,\ \rangle$
is an inner product on $V$, called dot product.

Proof: The proof is immediate. $\Box$

###### Proposition 3.8

The dot product on $V=[{\mathbb{R}}_{0}^{+}]^{n}$ generates the Euclidean norm
on $V$.

Proof: If $x=(x_{1},x_{2},\ldots,x_{n})\in V$, define the norm of $x$ by
$\|x\|=\sqrt{\langle x,x\rangle}$. Note that the norm is exactly the Euclidean
norm given in Proposition 3.5. $\Box$

###### Remark 3.4

We observe that if an inner product on a semi-vector space $V$ generates a
norm $\|\ \|$ and such a norm generates a metric $d$ on $V$, then $V$ is a
Hilbert space (according to Definition 3.6) if every Cauchy sequence in $V$
converges w.r.t. $d$ to an element of $V$.

###### Proposition 3.9

Let $V=[{\mathbb{R}}_{0}^{+}]^{n}$ and define the function ${\|\
\|}_{1}:V\longrightarrow{\mathbb{R}}_{0}^{+}$ as follows: if
$x=(x_{1},x_{2},\ldots,$ $x_{n})\in V$,
${\|x\|}_{1}=\displaystyle\sum_{i=1}^{n}x_{i}$. Then ${\|x\|}_{1}$ is a norm
on $V$.

Proof: The proof is direct. $\Box$

###### Proposition 3.10

Let $x=(x_{1},x_{2},\ldots,x_{n})$, $y=(y_{1},y_{2},\ldots,y_{n})$ be semi-
vectors in $V=[{\mathbb{R}}_{0}^{+}]^{n}$. Define the function $d_{1}:V\times
V\longrightarrow{\mathbb{R}}_{0}^{+}$ in the following way. For every fixed
$i$, if $x_{i}=y_{i}$, put $c_{i}=0$; if $x_{i}\neq y_{i}$, put
${\varphi}_{i}={\psi}_{i}+c_{i}$, where ${\varphi}_{i}=\max\\{x_{i},y_{i}\\}$
and ${\psi}_{i}=\min\\{x_{i},y_{i}\\}$. Let us consider that
$d_{1}(x,y)=\displaystyle\sum_{i=1}^{n}c_{i}$. Then the function $d_{1}$ is a
metric on $V$ derived from the norm ${\|\ \|}_{1}$ shown in Proposition 3.9.

Proof: We only prove the triangle inequality. To avoid stress of notation, we
consider the same that was considered in the proof of Proposition 3.6. We then
fix $i$ and only investigate the worst case $x_{i}<y_{i}<z_{i}$. In this case,
there exist positive real numbers $a_{i}$, $b_{i}$ for all $i=1,2,\ldots,n$,
such that $y_{i}=x_{i}+a_{i}$ and $z_{i}=y_{i}+b_{i}$, which implies
$z_{i}=x_{i}+a_{i}+b_{i}$. Then, for all $i$, $d_{1}(x_{i},z_{i})\leq
d_{1}(x_{i},y_{i})+d_{1}(y_{i},z_{i})$; hence,
$d_{1}(x,z)=\displaystyle\sum_{i=1}^{n}d_{1}(x_{i},z_{i})=\displaystyle\sum_{i=1}^{n}(a_{i}+b_{i})=\displaystyle\sum_{i=1}^{n}a_{i}+\displaystyle\sum_{i=1}^{n}b_{i}=\displaystyle\sum_{i=1}^{n}d_{1}(x_{i},y_{i})+\displaystyle\sum_{i=1}^{n}d_{1}(y_{i},z_{i})=d_{1}(x,y)+d_{1}(y,z)$.
Therefore, $d_{1}$ is a metric on $V$. $\Box$

###### Proposition 3.11

Let $V=[{\mathbb{R}}_{0}^{+}]^{n}$ be the Euclidean semi-vector space of semi-
dimension $n$. Define the function ${\|\
\|}_{2}:V\longrightarrow{\mathbb{R}}_{0}^{+}$ as follows: if
$x=(x_{1},x_{2},\ldots,$ $x_{n})\in V$, take
${\|x\|}_{2}=\displaystyle\max_{i}\\{x_{i}\\}$. Then ${\|x\|}_{2}$ is a norm
on $V$.

###### Proposition 3.12

Keeping the notation of Proposition 3.6, define the function $d_{2}:V\times
V\longrightarrow{\mathbb{R}}_{0}^{+}$ such that
$d_{2}(x,y)=\max_{i}\\{c_{i}\\}$. Then $d_{2}$ is a metric on $V$. Moreover,
$d_{2}$ is obtained from the norm ${\|\ \|}_{2}$ exhibited in Proposition
3.11.

###### Proposition 3.13

The norms $\|\ \|$, ${\|\ \|}_{1}$ and ${\|\ \|}_{2}$ shown in Propositions
3.5, 3.9 and 3.11 are equivalent.

Proof: It is immediate to see that ${\|\ \|}_{2}\leq\|\ \|\leq{\|\ \|}_{1}\leq
n{\|\ \|}_{2}$. $\Box$

In a natural way we can define the norm of a bounded semi-linear
transformation.

###### Definition 3.7

Let $V$ and $W$ be two normed semi-vector spaces and let $T:V\longrightarrow
W$ be a semi-linear transformation. We say that $T$ is bounded if there exists
a real number $c>0$ such that $\|T(v)\|\leq c\|v\|$.

If $T:V\longrightarrow W$ is bounded and $v\neq 0$ we can consider the
quotient $\frac{\|T(v)\|}{\|v\|}$. Since such a quotient is upper bounded by
$c$, the supremum $\displaystyle\sup_{v\in V,v\neq 0}\frac{\|T(v)\|}{\|v\|}$
exists and it is at most $c$. We then define

$\|T\|=\displaystyle\sup_{v\in V,v\neq 0}\frac{\|T(v)\|}{\|v\|}.$

###### Proposition 3.14

Let $T:V\longrightarrow W$ be a bounded semi-linear transformation. Then the
following hold:

  * $\operatorname{(1)}$

$T$ sends bounded sets in bounded sets;

  * $\operatorname{(2)}$

$\|T\|$ is a norm, called norm of $T$;

  * $\operatorname{(3)}$

$\|T\|$ can be written in the form $\|T\|=\displaystyle\sup_{v\in
V,\|v\|=1}\|T(v)\|$.

Proof: Items $\operatorname{(1)}$ and $\operatorname{(2)}$ are immediate. The
proof of Item $\operatorname{(3)}$ is analogous to the standard proof but we
present it here to guarantee that our mathematical tools are sufficient to
perform it. Let $v\neq 0$ be a semi-vector with norm $\|v\|=a\neq 0$ and set
$u=(1/a)v$. Thus, $\|u\|=1$ and since $T$ is semi-linear one has

$\|T\|=\displaystyle\sup_{v\in V,v\neq
0}\frac{1}{a}\|T(v)\|=\displaystyle\sup_{v\in V,v\neq
0}\|T((1/a)v)\|=\displaystyle\sup_{u\in V,\|u\|=1}\|T(u)\|=$

$=\displaystyle\sup_{v\in V,\|v\|=1}\|T(v)\|$. $\Box$

####  3.1.1 The Semi-Spaces ${l}_{+}^{\infty}$, ${l}_{+}^{p}$ and
${\operatorname{C}}_{+}[a,b]$

In this subsection we investigate topological aspects of some semi-vector
spaces over ${\mathbb{R}}_{0}^{+}$ such as completeness and separability. We
investigate the sequence spaces ${l}_{+}^{\infty}$, ${l}_{+}^{p}$,
${\operatorname{C}}_{+}[a,b]$, which will be defined in the sequence.

We first study the space ${l}_{+}^{\infty}$, the set of all bounded sequences
of nonnegative real numbers. Before studying such a space we must define a
metric on it, since the metric in $l^{\infty}$ which is defined as
$d(x,y)=\displaystyle\sup_{i\in{\mathbb{N}}}|{x}_{i}-{y}_{i}|$, where
$x=({x}_{i})$ and $y=({y}_{i})$ are sequences in $l^{\infty}$, has no meaning
to us, because there is no sense in considering $-{y}_{i}$ if ${y}_{i}>0$.
Based on this fact, we circumvent this problem by utilizing the total order of
${\mathbb{R}}$ according to Proposition 3.6. Let $x=({\mu}_{i})$ and
$y=({\nu}_{i})$ be sequences in $l_{+}^{\infty}$. We then fix $i$, and define
$c_{i}$ as was done in Proposition 3.6: if ${\mu}_{i}={\nu}_{i}$ then we put
$c_{i}=0$; if ${\mu}_{i}\neq{\nu}_{i}$, let
${\gamma}_{i}=\max\\{{\mu}_{i},{\nu}_{i}\\}$ and
${\psi}_{i}=\min\\{{\mu}_{i},{\nu}_{i}\\}$; then there exists a positive real
number $c_{i}$ such that ${\gamma}_{i}={\psi}_{i}+c_{i}$ and, in place of
$|{\mu}_{i}-{\nu}_{i}|$, we put $c_{i}$. Thus, our metric becomes

$\displaystyle d(x,y)=\displaystyle\sup_{i\in{\mathbb{N}}}\\{c_{i}\\}.$ (7)

It is clear that $d(x,y)$ shown in Eq. (7) defines a metric. However, we must
show that the tools that we have are sufficient to proof this fact, once we
are working on ${\mathbb{R}}_{0}^{+}$.

###### Proposition 3.15

The function $d$ shown in Eq. (7) is a metric on ${l}_{+}^{\infty}$.

Proof: It is clear that $d(x,y)\geq 0$ and $d(x,y)=0\Longleftrightarrow x=y$.
Let $x=({\mu}_{i})$ and $y=({\nu}_{i})$ be two sequences in $l_{+}^{\infty}$.
Then, for every fixed $i\in{\mathbb{N}}$, if $c_{i}=d({\mu}_{i},{\nu}_{i})=0$
then ${\mu}_{i}={\nu}_{i}$, i.e.,
$d({\mu}_{i},{\nu}_{i})=d({\nu}_{i},{\mu}_{i})$. If $c_{i}>0$ then
$c_{i}=d({\mu}_{i},{\nu}_{i})$ is computed by ${\gamma}_{i}={\psi}_{i}+c_{i}$,
where ${\gamma}_{i}=\max\\{{\mu}_{i},{\nu}_{i}\\}$ and
${\psi}_{i}=\min\\{{\mu}_{i},{\nu}_{i}\\}$. Hence,
$d({\nu}_{i},{\mu}_{i})=c_{i}^{*}$ is computed by
${\gamma}_{i}^{*}={\psi}_{i}^{*}+c_{i}^{*}$, where
${\gamma}_{i}^{*}=\max\\{{\nu}_{i},{\mu}_{i}\\}$ and
${\psi}_{i}^{*}=\min\\{{\nu}_{i},{\mu}_{i}\\}$, which implies
$d({\mu}_{i},{\nu}_{i})=d({\nu}_{i},{\mu}_{i})$. Taking the supremum over all
$i$’s we have
$d(x,y)=\displaystyle\sup_{i\in{\mathbb{N}}}\\{c_{i}\\}=\displaystyle\sup_{i\in{\mathbb{N}}}\\{c_{i}^{*}\\}=d(y,x)$.

To show the triangle inequality, let $x=({\mu}_{i})$, $y=({\nu}_{i})$ and
$z=({\eta}_{i})$ be sequences in $l_{+}^{\infty}$. For every fixed $i$, we
will prove that $d({\mu}_{i},{\eta}_{i})\leq
d({\mu}_{i},{\nu}_{i})+d({\nu}_{i},{\eta}_{i})$. If
${\nu}_{i}={\mu}_{i}={\eta}_{i}$, the result is trivial. If two of them are
equal, the result is also trivial. Assume that ${\mu}_{i}$, ${\nu}_{i}$ and
${\eta}_{i}$ are pairwise distinct. As in the proof of Proposition 3.6, we
must investigate the six cases:
$\operatorname{(1)}$ ${\mu}_{i}<{\nu}_{i}<{\eta}_{i}$; $\operatorname{(2)}$
${\mu}_{i}<{\eta}_{i}<{\nu}_{i}$; $\operatorname{(3)}$
${\nu}_{i}<{\mu}_{i}<{\eta}_{i}$; $\operatorname{(4)}$
${\nu}_{i}<{\eta}_{i}<{\mu}_{i}$; $\operatorname{(5)}$
${\eta}_{i}<{\mu}_{i}<{\nu}_{i}$; $\operatorname{(6)}$
${\eta}_{i}<{\nu}_{i}<{\mu}_{i}$. We only show $\operatorname{(1)}$ and
$\operatorname{(2)}$.

To show $\operatorname{(1)}$, note that there exist positive real numbers
$c_{i}$ and $c_{i}^{{}^{\prime}}$ such that ${\nu}_{i}={\mu}_{i}+c_{i}$ and
${\eta}_{i}={\nu}_{i}+c_{i}^{{}^{\prime}}$, which implies
$\eta_{i}=\mu_{i}+c_{i}+c_{i}^{{}^{\prime}}$. Hence,
$d({\mu}_{i},{\eta}_{i})=c_{i}+c_{i}^{{}^{\prime}}=d({\mu}_{i},{\nu}_{i})+d({\nu}_{i},{\eta}_{i})$.

Let us show $\operatorname{(2)}$. There exist positive real numbers $b_{i}$
and $b_{i}^{{}^{\prime}}$ such that ${\eta}_{i}={\mu}_{i}+b_{i}$ and
${\nu}_{i}={\eta}_{i}+b_{i}^{{}^{\prime}}$, so
${\nu}_{i}={\mu}_{i}+b_{i}+b_{i}^{{}^{\prime}}$. Therefore,
$d({\mu}_{i},{\eta}_{i})=b_{i}<d({\mu}_{i},{\nu}_{i})+d({\nu}_{i},{\eta}_{i})=b_{i}+2b_{i}^{{}^{\prime}}$.

Taking the supremum over all $i$’s we have
$\displaystyle\sup_{i\in{\mathbb{N}}}\\{d({\mu}_{i},{\eta}_{i})\\}\leq\displaystyle\sup_{i\in{\mathbb{N}}}\\{d({\mu}_{i},{\nu}_{i})\\}+\displaystyle\sup_{i\in{\mathbb{N}}}\\{d({\nu}_{i},{\eta}_{i})\\}$,
i.e., $d(x,z)\leq d(x,y)+d(y,z)$. Therefore, $d$ is a metric on
${l}_{+}^{\infty}$. $\Box$

###### Definition 3.8

The metric space ${l}_{+}^{\infty}$ is the set of all bounded sequences of
nonnegative real numbers equipped with the metric
$d(x,y)=\displaystyle\sup_{i\in{\mathbb{N}}}\\{c_{i}\\}$ given previously.

We prove that ${l}_{+}^{\infty}$ equipped with the previous metric is
complete.

###### Theorem 3.3

The space ${l}_{+}^{\infty}$ with the metric
$d(x,y)=\displaystyle\sup_{i\in{\mathbb{N}}}\\{c_{i}\\}$ shown above is
complete.

Proof: The proof follows the same line as the standard proof of completeness
of ${l}^{\infty}$; however it is necessary to adapt it to the metric (written
above) in terms of nonnegative real numbers. Let $(x_{n})$ be a Cauchy
sequence in ${l}_{+}^{\infty}$, where
$x_{i}=({\eta}_{1}^{(i)},{\eta}_{2}^{(i)},\ldots)$. We must show that
$(x_{n})$ converges to an element of ${l}_{+}^{\infty}$. As $(x_{n})$ is
Cauchy, given $\epsilon>0$, there exists a positive integer $K$ such that, for
all $n,m>K$,

$d(x_{n},x_{m})=\displaystyle\sup_{j\in{\mathbb{N}}}\\{c_{j}^{(n,m)}\\}<\epsilon,$

where $c_{j}^{(n,m)}$ is a nonnegative real number such that, if
${\eta}_{j}^{(n)}={\eta}_{j}^{(m)}$ then $c_{j}^{(n,m)}=0$, and if
${\eta}_{j}^{(n)}\neq{\eta}_{j}^{(m)}$ then $c_{j}^{(n,m)}$ is given by
$\max\\{{\eta}_{j}^{(n)},{\eta}_{j}^{(m)}\\}=\min\\{{\eta}_{j}^{(n)},{\eta}_{j}^{(m)}\\}+c_{j}^{(n,m)}$.
This implies that for each fixed $j$ one has

$\displaystyle c_{j}^{(n,m)}<\epsilon,$ (8)

where $n,m>K$. Thus, for each fixed $j$, it follows that
$({\eta}_{j}^{(1)},{\eta}_{j}^{(2)},\ldots)$ is a Cauchy sequence in
${\mathbb{R}}_{0}^{+}$. Since ${\mathbb{R}}_{0}^{+}$ is a complete metric
space, the sequence $({\eta}_{j}^{(1)},{\eta}_{j}^{(2)},\ldots)$ converges to
an element ${\eta}_{j}$ in ${\mathbb{R}}_{0}^{+}$. Hence, for each $j$, we
form the sequence $x$ whose coordinates are the limits ${\eta}_{j}$, i.e.,
$x=({\eta}_{1},{\eta}_{2},{\eta}_{3},\ldots)$. We must show that
$x\in{l}_{+}^{\infty}$ and $x_{n}\longrightarrow x$.

To show that $x$ is a bounded sequence, let us consider the number
$c_{j}^{(n,\infty)}$ defined as follows: if ${\eta}_{j}={\eta}_{j}^{(n)}$ then
$c_{j}^{(n,\infty)}=0$, and if ${\eta}_{j}\neq{\eta}_{j}^{(n)}$, define
$c_{j}^{(n,\infty)}$ be the positive real number satisfying
$\max\\{{\eta}_{j},{\eta}_{j}^{(n)}\\}=\min\\{{\eta}_{j},{\eta}_{j}^{(n)}\\}+c_{j}^{(n,\infty)}$.
From the inequality $(\ref{distCauchy1})$ one has

$\displaystyle c_{j}^{(n,\infty)}\leq\epsilon.$ (9)

Because ${\eta}_{j}\leq{\eta}_{j}^{(n)}+c_{j}^{(n,\infty)}$ and since
${\eta}_{j}^{(n)}\in l_{+}^{\infty}$, it follows that ${\eta}_{j}$ is a
bounded sequence for every $j$. Hence,
$x=({\eta}_{1},{\eta}_{2},{\eta}_{3},\ldots)\in{l}_{+}^{\infty}$. From
$(\ref{distCauchy2})$ we have

$\displaystyle\sup_{j\in{\mathbb{N}}}\\{c_{j}^{(n,\infty)}\\}\leq\epsilon,$

which implies that $x_{n}\longrightarrow x$. Therefore, $l_{+}^{\infty}$ is
complete. $\Box$

Although $l_{+}^{\infty}$ is a complete metric space, it is not separable.

###### Theorem 3.4

The space ${l}_{+}^{\infty}$ with the metric
$d(x,y)=\displaystyle\sup_{i\in{\mathbb{N}}}\\{c_{i}\\}$ is not separable.

Proof: The proof is the same as shown in [7, 1.3-9], so it is omitted. $\Box$

Let us define the space analogous to the space $l^{p}$.

###### Definition 3.9

Let $p\geq 1$ be a fixed real number. The set ${l}_{+}^{p}$ consists of all
sequences $x=({\eta}_{1},{\eta}_{2},{\eta}_{3},\ldots)$ of nonnegative real
numbers such that $\displaystyle\sum_{i=1}^{\infty}({\eta}_{i})^{p}<\infty$,
whose metric is defined by
$d(x,y)={\left[\displaystyle\sum_{i=1}^{\infty}{[c_{i}]}^{p}\right]}^{1/p}$,
where $y=({\mu}_{1},{\mu}_{2},{\mu}_{3},\ldots)$ and $c_{i}$ is defined as
follows: $c_{i}=0$ if ${\mu}_{i}={\eta}_{i}$, and if ${\mu}_{i}>{\eta}_{i}$
(respect. ${\eta}_{i}>{\mu}_{i}$) then $c_{i}>0$ is such that
${\mu}_{i}={\eta}_{i}+c_{i}$.

###### Theorem 3.5

The space ${l}_{+}^{p}$ with the metric
$d(x,y)={\left[\displaystyle\sum_{i=1}^{\infty}{[c_{i}]}^{p}\right]}^{1/p}$
exhibited above is complete.

Proof: Recall that given two sequences $({\mu}_{i})$ and $({\eta}_{i})$ in
${l}_{+}^{p}$ the Minkowski inequality for sums reads as

$\displaystyle{\left[\displaystyle\sum_{i=1}^{\infty}{|{\mu}_{i}+{\eta}_{i}|}^{p}\right]}^{1/p}\leq{\left[\displaystyle\sum_{j=1}^{\infty}{|{\mu}_{j}|}^{p}\right]}^{1/p}+{\left[\displaystyle\sum_{k=1}^{\infty}{|{\eta}_{k}|}^{p}\right]}^{1/p}.$

Applying the Minkowski inequality as per [7, 1.5-4] with some adaptations, it
follows that $d(x,y)$ is, in fact, a metric. In order to prove the
completeness of ${l}_{+}^{p}$, we proceed similarly as in the proof of Theorem
3.3 with some adaptations. The main adaptation is performed according to the
proof of completeness of $l^{p}$ in [7, 1.5-4] replacing the last equality
$x=x_{m}+(x-x_{m})\in l^{p}$ (after Eq. (5)) by two equalities in order to
avoid negative real numbers.

  1. $\operatorname{(1)}$

If the $i$-th coordinate $x^{(i)}-x_{m}^{(i)}$ of the sequence $x-x_{m}$ is
positive, then define $c_{m}^{(i)}=x^{(i)}-x_{m}^{(i)}$ and write
$x^{(i)}=x_{m}^{(i)}+c_{m}^{(i)}$. From Minkowski inequality, it follows that
the sequence $(x^{(i)})_{i}$ is in $l_{+}^{p}$.

  2. $\operatorname{(2)}$

If $x^{(j)}-x_{m}^{(j)}$ is negative, then define
$c_{m}^{(j)}=x_{m}^{(j)}-x^{(j)}$ and write $x_{m}^{(j)}=x^{(j)}+c_{m}^{(j)}$.
Since $x_{m}\in l_{+}^{p}$, from the comparison criterion for positive series
it follows that the sequence $(x^{(j)})_{j}$ is also in $l_{+}^{p}$.

$\Box$

###### Theorem 3.6

The space ${l}_{+}^{p}$ is separable.

Proof: The proof follows the same line of [7, 1.3-10]. $\Box$

###### Definition 3.10

Let $I=[a,b]$ be a closed interval in ${\mathbb{R}}_{0}^{+}$, where $a\geq 0$
and $a<b$. Then ${\operatorname{C}}_{+}[a,b]$ is the set of all continuous
nonnegative real valued functions on $I=[a,b]$, whose metric is defined by
$d(f(t),g(t))=\displaystyle\max_{t\in I}\\{c(t)\\}$, where $c(t)$ is given by
$\max\\{f(t),g(t)\\}=\min\\{f(t),g(t)\\}+c(t)$.

###### Theorem 3.7

The metric space $({\operatorname{C}}_{+}[a,b],d)$, where $d$ is given in
Definition 3.10, is complete.

Proof: The proof follows the same lines as the standard one with some
modifications. Let $(f_{m})$ be a Cauchy sequence in
${\operatorname{C}}_{+}[a,b]$. Given $\epsilon>0$ there exists a positive
integer $N$ such that, for all $m,n>N$, it follows that

$\displaystyle d(f_{m},f_{n})=\displaystyle\max_{t\in
I}\\{c_{m,n}(t)\\}<\epsilon,$ (10)

where $\max\\{f_{m}(t),f_{n}(t)\\}=\min\\{f_{m}(t),f_{n}(t)\\}+c_{m,n}(t)$.
Thus, for any fixed $t_{0}\in I$ we have $c_{m,n}(t_{0})<\epsilon$, for all
$m,n>N$. This means that $(f_{1}(t_{0}),f_{2}(t_{0}),\ldots)$ is a Cauchy
sequence in ${\mathbb{R}}_{0}^{+}$, which converges to $f(t_{0})$ when
$m\longrightarrow\infty$ since ${\mathbb{R}}_{0}^{+}$ is complete. We then
define a function $f:[a,b]\longrightarrow{\mathbb{R}}_{0}^{+}$ such that for
each $t\in[a,b]$, we put $f(t)$. Taking $n\longrightarrow\infty$ in (10) we
obtain $\displaystyle\max_{t\in I}\\{c_{m}(t)\\}\leq\epsilon$ for all $m>N$,
where $\max\\{f_{m}(t),f(t)\\}=\min\\{f_{m}(t),f(t)\\}+c_{m}(t)$, which
implies $c_{m}(t)\leq\epsilon$ for all $t\in I$. This fact means that
$(f_{m}(t))$ converges to $f(t)$ uniformly on $I$, i.e.,
$f\in{\operatorname{C}}_{+}[a,b]$ because the functions $f_{m}$’s are
continuous on $I$. Therefore, ${\operatorname{C}}_{+}[a,b]$ is complete, as
desired. $\Box$

###  3.2 Interesting Semi-Vector Spaces

In this section we exhibit semi-vector spaces over $K={\mathbb{R}}_{0}^{+}$
derived from semi-metrics, semi-metric-preserving functions, semi-norms, semi-
inner products and sub-linear functionals.

###### Theorem 3.8

Let $X$ be a semi-metric space and ${\mathcal{M}}_{X}=\\{d:X\times
X\longrightarrow{\mathbb{R}};d$ $\operatorname{is\ a\ semi-metric\ on}X\\}$.
Then $({\mathcal{M}}_{X},+,\cdot)$ is a semi-vector space over
${\mathbb{R}}_{0}^{+}$, where $+$ and $\cdot$ are the addition and the scalar
multiplication (in ${\mathbb{R}}_{0}^{+}$) pointwise, respectively.

Proof: We first show that ${\mathcal{M}}_{X}$ is closed under addition. Let
$d_{1},d_{2}\in{\mathcal{M}}_{X}$ and set $d:=d_{1}+d_{2}$. It is clear that
$d$ is nonnegative real-valued function. Moreover, for all $x,y\in X$,
$d(x,y)=d(y,x)$. Let $x\in X$; $d(x,x)=d_{1}(x,x)+d_{2}(x,x)=0$. For all
$x,y,z\in X$,
$d(x,z)=d_{1}(x,z)+d_{2}(x,z)\leq[d_{1}(x,y)+d_{2}(x,y)]+[d_{1}(y,z)+d_{2}(y,z)]=d(x,y)+d(y,z)$.

Let us show that ${\mathcal{M}}_{X}$ is closed under scalar multiplication.
Let $d_{1}\in{\mathcal{M}}_{X}$ and define $d=\lambda d_{1}$, where
$\lambda\in{\mathbb{R}}_{0}^{+}$. It is clear that $d$ is real-valued
nonnegative and for all $x,y\in X$, $d(x,y)=d(y,x)$. Moreover, if $x\in X$,
$d(x,x)=0$. For all $x,y,z\in X$, $d(x,z)=\lambda
d_{1}(x,z)\leq\lambda[d_{1}(x,y)+d_{1}(y,z)]=d(x,y)+d(y,z)$. This means that
${\mathcal{M}}_{X}$ is closed under scalar multiplication.

It is easy to see that $({\mathcal{M}}_{X},+,\cdot)$ satisfies the other
conditions of Definition 2.3. $\Box$

Let $(X,d)$ be a metric space. In [3], Corazza investigated interesting
functions $f:{\mathbb{R}}_{0}^{+}\longrightarrow{\mathbb{R}}_{0}^{+}$ such
that the composite of $f$ with $d$, i.e., $X\times
X\xrightarrow{d}{{\mathbb{R}}_{0}^{+}}\xrightarrow{f}{{\mathbb{R}}_{0}^{+}}$
also generates a metric on $X$. Let us put this concept formally.

###### Definition 3.11

Let $f:{\mathbb{R}}_{0}^{+}\longrightarrow{\mathbb{R}}_{0}^{+}$ be a function.
We say that $f$ is metric-preserving if for all metric spaces $(X,d)$, the
composite $f\circ d$ is a metric.

To our purpose we will consider semi-metric preserving functions as follows.

###### Definition 3.12

Let $f:{\mathbb{R}}_{0}^{+}\longrightarrow{\mathbb{R}}_{0}^{+}$ be a function.
We say that $f$ is semi-metric-preserving if for all semi-metric spaces
$(X,d)$, the composite $f\circ d$ is a semi-metric.

We next show that the set of semi-metric preserving functions has a semi-
vector space structure.

###### Theorem 3.9

Let
${\mathcal{F}}_{pres}=\\{f:{\mathbb{R}}_{0}^{+}\longrightarrow{\mathbb{R}}_{0}^{+};f\operatorname{is\
semi-metric\ preserving}\\}$. Then $({\mathcal{F}}_{pres},+,\cdot)$ is a semi-
vector space over ${\mathbb{R}}_{0}^{+}$, where $+$ and $\cdot$ are the
addition and the scalar multiplication (in ${\mathbb{R}}_{0}^{+}$) pointwise,
respectively.

Proof: We begin by showing that ${\mathcal{F}}_{pres}$ is closed under
addition and scalar multiplication pointwise.

Let $f,g\in{\mathcal{F}}_{pres}$. Given a semi-metric space $(X,d)$, we must
prove that $(f+g)\circ d$ is also semi-metric preserving. We know that
$[(f+g)\circ d](x,y)\geq 0$ for all $x,y\in X$. Let $x\in X$; then
$[(f+g)\circ d](x,x)=f(d(x,x))+g(d(x,x))=0$. It is clear that $[(f+g)\circ
d](x,y)=[(f+g)\circ d](y,x)$. Let $x,y,z\in X$. One has: $[(f+g)\circ
d](x,y)=f(d(x,y))+g(d(x,y))\leq[f(d(x,z))+g(d(x,z))]+[f(d(z,y))+g(d(z,y))]=(f+g)(d(x,z))+(f+g)(d(z,y))=[(f+g)\circ
d](x,z)+[(f+g)\circ d](z,y)$.

Here, we show that for each $f\in{\mathcal{F}}_{pres}$ and
$\alpha\in{\mathbb{R}}_{0}^{+}$, it follows that $\alpha
f\in{\mathcal{F}}_{pres}$. We show only the triangular inequality since the
other conditions are immediate. Let us calculate: $[\alpha f\circ
d](x,y)=\alpha f(d(x,y))\leq\alpha f(d(x,z))+\alpha f(d(z,y))=[\alpha f\circ
d](x,z)+[\alpha f\circ d](z,y)$.

The null vector is the null function
$0_{f}:{\mathbb{R}}_{0}^{+}\longrightarrow{\mathbb{R}}_{0}^{+}$. The other
conditions are easy to verify. $\Box$

###### Theorem 3.10

Let $V$ be a semi-normed real vector space and ${\mathcal{N}}_{V}=\\{\|\
\|:V\longrightarrow{\mathbb{R}};\|\ \|$ $\operatorname{is\ a\ semi-norm\
on}V\\}$. Then $({\mathcal{N}}_{V},+,\cdot)$ is a semi-vector space over
${\mathbb{R}}_{0}^{+}$, where $+$ and $\cdot$ are addition and scalar
multiplication (in ${\mathbb{R}}_{0}^{+}$) pointwise, respectively.

Proof: From hypotheses, ${\mathcal{N}}_{V}$ is non-empty. Let ${\|\
\|}_{1},{\|\ \|}_{2}\in{\mathcal{N}}_{V}$ and set $\|\ \|:={\|\ \|}_{1}+{\|\
\|}_{2}$. For all $v\in V$, $\|v\|\geq 0$. If $v\in V$ and
$\alpha\in{\mathbb{R}}$ then $\|\alpha v\|=|\alpha|\|v\|$. For every $u,v\in
V$, it follows that
$\|u+v\|:={\|u+v\|}_{1}+{\|u+v\|}_{2}\leq({\|u\|}_{1}+{\|u\|}_{2})+({\|v\|}_{1}+{\|v\|}_{2})=\|u\|+\|v\|$.
Hence, ${\mathcal{N}}_{V}$ is closed under addition.

We next show that ${\mathcal{N}}_{V}$ is closed under scalar multiplication.
Let ${\|\ \|}_{1}\in{\mathcal{N}}_{V}$ and define $\|\ \|:=\lambda{\|\
\|}_{1}$, where $\lambda\in{\mathbb{R}}_{0}^{+}$. For all $v\in V$, $\|v\|\geq
0$. If $\alpha\in{\mathbb{R}}$ and $v\in V$, $\|\alpha
v\|=|\alpha|(\lambda{\|v\|}_{1})=|\alpha|\|v\|$. Let $u,v\in V$. Then
$\|u+v\|\leq\lambda{\|u\|}_{1}+\lambda{\|v\|}_{1}=\|u\|+\|v\|$. Therefore,
${\mathcal{N}}_{V}$ is closed under addition and scalar multiplication over
${\mathbb{R}}_{0}^{+}$.

The zero vector is the null function
$\textbf{0}:V\longrightarrow{\mathbb{R}}$. The other conditions of Definition
2.3 are straightforward. $\Box$

###### Remark 3.5

Note that ${\mathcal{N}}_{V}^{\diamond}=\\{\|\
\|:V\longrightarrow{\mathbb{R}};\|\ \|$ $\operatorname{is\ a\ norm\ on}V\\}$
is also closed under both function addition and scalar multiplication
pointwise.

###### Lemma 3.1

Let $T:V\longrightarrow W$ be a linear transformation.

  * $\operatorname{(1)}$

If $\|\ \|:W\longrightarrow{\mathbb{R}}$ is a semi-norm on $W$ then $\|\
\|\circ T:V\longrightarrow{\mathbb{R}}$ is a semi-norm on $V$.

  * $\operatorname{(2)}$

If $T$ is injective linear and $\|\ \|:W\longrightarrow{\mathbb{R}}$ is a norm
on $W$ then $\|\ \|\circ T$ is a norm on $V$.

Proof: We only show Item $\operatorname{(1)}$. It is clear that $[\|\ \|\circ
T](v)\geq 0$ for all $v\in V$. For all $\alpha\in{\mathbb{R}}$ and $v\in V$,
$[\|\ \|\circ T](\alpha v)=|\alpha|\|T(v)\|=|\alpha|[\|\ \|\circ T](v)$.
Moreover, $\forall\ v_{1},v_{2}\in V$, $[\|\ \|\circ T](v_{1}+v_{2})\leq[\|\
\|\circ T](v_{1})+[\|\ \|\circ T](v_{2})$. Therefore, $\|\ \|\circ T$ is a
semi-norm on $V$. $\Box$

###### Theorem 3.11

Let $V$ and $W$ be two semi-normed vector spaces and $T:V\longrightarrow W$ be
a linear transformation. Then

${\mathcal{N}}_{V_{T}}=\\{\|\ \|\circ T:V\longrightarrow{\mathbb{R}};\|\
\|\operatorname{is\ a\ semi-norm\ on}W\\}$

is a semi-subspace of $({\mathcal{N}}_{V},+,\cdot)$.

Proof: From hypotheses, it follows that ${\mathcal{N}}_{V_{T}}$ is non-empty.
From Item $\operatorname{(1)}$ of Lemma 3.1, it follows that $\|\ \|\circ T$
is a semi-norm on $V$. Let $f,g\in{\mathcal{N}}_{V_{T}}$, i.e., $f={\|\
\|}_{1}\circ T$ and $g={\|\ \|}_{2}\circ T$, where ${\|\ \|}_{1}$ and ${\|\
\|}_{2}$ are semi-norms on $W$. Then $f+g=[{\|\ \|}_{1}+{\|\ \|}_{2}]\circ
T\in{\mathcal{N}}_{V_{T}}$. For every nonnegative real number $\lambda$ and
$f\in{\mathcal{N}}_{V_{T}}$, $\lambda f=\lambda[\|\ \|\circ T]=(\lambda\|\
\|)\circ T\in{\mathcal{N}}_{V_{T}}$. $\Box$

###### Theorem 3.12

Let ${\mathcal{N}}$ be the class whose members are $\\{{\mathcal{N}}_{V}\\}$,
where the ${\mathcal{N}}_{V}$ are given in Theorem 3.10. Let
$\operatorname{Hom}({\mathcal{N}})$ be the class whose members are the sets

$\operatorname{hom}({\mathcal{N}}_{V},{\mathcal{N}}_{W})=\\{F_{T}:{\mathcal{N}}_{V}\longrightarrow{\mathcal{N}}_{W};F_{T}({\|\
\|}_{V})={\|\ \|}_{V}\circ T\\},$

where $T:W\longrightarrow V$ is a linear transformation and ${\|\ \|}_{V}$ is
a semi-norm on $V$. Then
$({\mathcal{N}},\operatorname{Hom}({\mathcal{N}}),Id,\circ)$ is a category.

Proof: The sets $\operatorname{hom}({\mathcal{N}}_{V},{\mathcal{N}}_{W})$ are
pairwise disjoint. For each ${\mathcal{N}}_{V}$, there exists
$Id_{({\mathcal{N}}_{V})}$ given by $Id_{({\mathcal{N}}_{V})}({\|\
\|}_{V})={\|\ \|}_{V}={\|\ \|}_{V}\circ Id_{(V)}$. It is clear that if
${F}_{T}:{\mathcal{N}}_{V}\longrightarrow{\mathcal{N}}_{W}$ then ${F}_{T}\circ
Id_{({\mathcal{N}}_{V})}={F}_{T}$ and
$Id_{({\mathcal{N}}_{W})}\circ{F}_{T}={F}_{T}$.

It is easy to see that for every $T:W\longrightarrow V$ linear transformation,
the map $F_{T}$ is semi-linear, i.e., $F_{T}({\|\ \|}_{V}^{(1)}+{\|\
\|}_{V}^{(2)})=F_{T}({\|\ \|}_{V}^{(1)})+F_{T}({\|\ \|}_{V}^{(2)})$ and
$F_{T}(\lambda{\|\ \|}_{V})=\lambda F_{T}({\|\ \|}_{V})$, for every ${\|\
\|}_{V},{\|\ \|}_{V}^{(1)},{\|\ \|}_{V}^{(2)}\in{\mathcal{N}}_{V}$ and
$\lambda\in{\mathbb{R}}_{0}^{+}$.

Let
${\mathcal{N}}_{U},{\mathcal{N}}_{V},{\mathcal{N}}_{W},{\mathcal{N}}_{X}\in{\mathcal{N}}$
and $F_{T_{1}}\in\operatorname{hom}({\mathcal{N}}_{U},{\mathcal{N}}_{V})$,
$F_{T_{2}}\in\operatorname{hom}({\mathcal{N}}_{V},{\mathcal{N}}_{W})$,
$F_{T_{3}}\in\operatorname{hom}({\mathcal{N}}_{W},{\mathcal{N}}_{X})$, i.e.,

${\mathcal{N}}_{U}\xrightarrow{F_{T_{1}}}{\mathcal{N}}_{V}\xrightarrow{F_{T_{2}}}{\mathcal{N}}_{W}\xrightarrow{F_{T_{3}}}{\mathcal{N}}_{X}.$

The linear transformations are of the forms

$X\xrightarrow{T_{3}}W\xrightarrow{T_{2}}V\xrightarrow{T_{1}}U\xrightarrow{{\|\
\|}_{U}}{\mathbb{R}}.$

The associativity $(F_{T_{3}}\circ F_{T_{2}})\circ
F_{T_{1}}=F_{T_{3}}\circ(F_{T_{2}}\circ F_{T_{1}})$ follows from the
associativity of composition of maps. Moreover, the map $F_{T_{3}}\circ
F_{T_{2}}\circ F_{T_{1}}\in\operatorname{Hom}({\mathcal{N}})$ because
$F_{T_{3}}\circ F_{T_{2}}\circ F_{T_{1}}=({\|\ \|}_{U})\circ(T_{1}\circ
T_{2}\circ T_{3})$ and $T_{1}\circ T_{2}\circ T_{3}$ is a linear
transformation. Therefore,
$({\mathcal{N}},\operatorname{Hom}({\mathcal{N}}),Id,\circ)$ is a category, as
required. $\Box$

###### Theorem 3.13

Let $V$ be a real vector space endowed with a semi-inner product and let
${\mathcal{P}}_{V}=\\{\langle\ ,\ \rangle:V\times
V\longrightarrow{\mathbb{R}};\langle\ ,\ \rangle$ $\operatorname{is\ a\ semi-
inner\ product\ on}V\\}$. Then $({\mathcal{P}}_{V},+,\cdot)$ is a semi-vector
space over ${\mathbb{R}}_{0}^{+}$, where $+$ and $\cdot$ are addition and
scalar multiplication (in ${\mathbb{R}}_{0}^{+}$) pointwise, respectively.

Proof: The proof is analogous to that of Theorems 3.8 and 3.10. $\Box$

###### Proposition 3.16

Let $V,W$ be two vector spaces and $T_{1},T_{2}:V\longrightarrow W$ be two
linear transformations. Let us consider the map $T_{1}\times T_{2}:V\times
V\longrightarrow W\times W$ given by $T_{1}\times
T_{2}(u,v)=(T_{1}(u),T_{2}(v))$. If $\langle\ ,\ \rangle$ is a semi-inner
product on $W$ then $\langle\ ,\ \rangle\circ T_{1}\times T_{2}$ is a semi-
inner product on $V$.

Proof: The proof is immediate, so it is omitted. $\Box$

Let $V$ be a real vector space. Recall that a sub-linear functional on $V$ is
a functional $t:V\longrightarrow{\mathbb{R}}$ which is sub-additive: $\forall\
u,v\in V$, $t(u+v)\leq t(u)+t(v)$; and positive-homogeneous: $\forall\
\alpha\in{\mathbb{R}}_{0}^{+}$ and $\forall\ v\in V$, $t(\alpha v)=\alpha
t(v)$.

###### Theorem 3.14

Let $V$ be a real vector space. Let us consider
${\mathcal{S}}_{V}=\\{S:V\longrightarrow{\mathbb{R}};$
$S\operatorname{is}\operatorname{sub-linear}\operatorname{on}V\\}$. Then
$({\mathcal{S}}_{V},+,\cdot)$ is a semi-vector space on
${\mathbb{R}}_{0}^{+}$, where $+$ and $\cdot$ are addition and scalar
multiplication (in ${\mathbb{R}}_{0}^{+}$) pointwise, respectively.

Proof: The proof follows the same line of that of Theorems 3.8 and 3.10 and
3.13. $\Box$

###  3.3 Semi-Algebras

We start this section by recalling the definition of semi-algebra and semi-
sub-algebra. For more details the reader can consult [4]. In [9], Olivier and
Serrato investigated relation semi-algebras, i.e., a semi-algebra being both a
Boolean algebra and an involutive semi-monoid, satisfying some conditions (see
page 2 in Ref. [9] for more details). Roy [14] studied the semi-algebras of
continuous and monotone functions on compact ordered spaces.

###### Definition 3.13

A semi-algebra $A$ over a semi-field $K$ (or a $K$-semi-algebra) is a semi-
vector space $A$ over $K$ endowed with a binary operation called
multiplication of semi-vectors $\bullet:A\times A\longrightarrow A$ such that,
$\forall\ u,v,w\in A$ and $\lambda\in K$:

  * $\operatorname{(1a)}$

$u\bullet(v+w)=(u\bullet v)+(u\bullet w)$ (left-distributivity);

  * $\operatorname{(1b)}$

$(u+v)\bullet w=(u\bullet w)+(v\bullet w)$ (right-distributivity);

  * $\operatorname{(2)}$

$\lambda(u\bullet v)=(\lambda u)\bullet v=u\bullet(\lambda v)$.

A semi-algebra $A$ is _associative_ if $(u\bullet v)\bullet
w=u\bullet(v\bullet w)$ for all $u,v,w\in A$; $A$ is said to be _commutative_
(or abelian) is the multiplication is commutative, that is, $\forall\ u,v\in
A$, $u\bullet v=v\bullet u$; $A$ is called a semi-algebra with identity if
there exists an element $1_{A}\in A$ such that $\forall\ u\in A$,
$1_{A}\bullet u=u\bullet 1_{A}=u$; the element $1_{A}$ is called identity of
$A$. The identity element of a semi-algebra $A$ is unique (if exists). If $A$
is a semi-free semi-vector space then the dimension of $A$ is its dimension
regarded as a semi-vector space. A semi-algebra is _simple_ if it is simple as
a semi-vector space.

###### Example 3.3

The set ${\mathbb{R}}_{0}^{+}$ is a commutative semi-algebra with identity
$e=1$.

###### Example 3.4

The set of square matrices of order $n$ whose entries are in
${\mathbb{R}}_{0}^{+}$, equipped with the sum of matrices, multiplication of a
matrix by a scalar (in ${\mathbb{R}}_{0}^{+}$, of course) and by
multiplication of matrices is an associative and non-commutative semi-algebra
with identity $e=I_{n}$ (the identity matrix of order $n$), over
${\mathbb{R}}_{0}^{+}$.

###### Example 3.5

The set ${\mathcal{P}}_{n}[x]$ of polynomials with coefficients from
${\mathbb{R}}_{0}^{+}$ and degree less than or equal to $n$, equipped with the
usual of polynomial sum and scalar multiplication is a semi-vector space.

###### Example 3.6

Let $V$ be a semi-vector space over a semi-field $K$. Then the set
${\mathcal{L}}(V,V)=\\{T:V\longrightarrow V;T\operatorname{is\ a\ semi-linear\
operator}\\}$ is a semi-vector space. If we define a vector multiplication as
the composite of semi-linear operators (which is also semi-linear) then we
have a semi-algebra over $K$.

###### Definition 3.14

Let $A$ be a semi-algebra over $K$. We say that a non-empty set $S\subseteq A$
is a semi-subalgebra if $S$ is closed under the operations of $A$, that is,

  * $\operatorname{(1)}$

$\forall\ u,v\in A$, $u+v\in A$;

  * $\operatorname{(2)}$

$\forall\ u,v\in A$, $u\bullet v\in A$;

  * $\operatorname{(3)}$

$\forall\ \lambda\in K$ and $\forall u\in A$, $\lambda u\in A$.

###### Definition 3.15

Let $A$ and $B$ two semi-algebras over $K$. We say that a map
$T:A\longrightarrow B$ is an $K$-semi-algebra homomorphism if, $\forall\
u,v\in A$ and $\lambda\in K$, the following conditions hold:

  * $\operatorname{(1)}$

$T(u+v)=T(u)+T(v)$;

  * $\operatorname{(2)}$

$T(u\bullet v)=T(u)\bullet T(v)$;

  * $\operatorname{(3)}$

$T(\lambda v)=\lambda T(v)$.

Definition 3.15 means that $T$ is both a semi-ring homomorphism and also semi-
linear (as semi-vector space).

###### Definition 3.16

Let $A$ and $B$ be two $K$-semi-algebras. A $K$-semi-algebra isomorphism
$T:A\longrightarrow B$ is a bijective $K$-semi-algebra homomorphism. If there
exists such an isomorphism, we say that $A$ is isomorphic to $B$, written
$A\cong B$.

The following results seems to be new, because semi-algebras over
${\mathbb{R}}_{0}^{+}$ are not much investigated in the literature.

###### Proposition 3.17

Assume that $A$ and $B$ are two $K$-semi-algebras, where
$K={\mathbb{R}}_{0}^{+}$ and $A$ has identity $1_{A}$. Let $T:A\longrightarrow
B$ be a $K$-semi-algebra homomorphism. Then the following properties hold:

  * $\operatorname{(1)}$

$T(0_{A})=0_{B}$;

  * $\operatorname{(2)}$

If $u\in A$ is invertible then its inverse is unique and $(u^{-1})^{-1}=u$;

  * $\operatorname{(3)}$

If $T$ is surjective then $T(1_{A})=1_{B}$, i.e., $B$ also has identity;
furthermore, $T(u^{-1})=[T(u)]^{-1}$;

  * $\operatorname{(4)}$

If $u,v\in A$ are invertible then $(u\bullet v)^{-1}=v^{-1}\bullet u^{-1}$;

  * $\operatorname{(5)}$

the composite of $K$-semi-algebra homomorphisms is also a $K$-semi-algebra
homomorphism;

  * $\operatorname{(6)}$

if $T$ is a $K$-semi-algebra isomorphism then also is $T^{-1}:B\longrightarrow
A$.

  * $\operatorname{(7)}$

the relation $A\sim B$ if and only if $A$ is isomorphic to $B$ is an
equivalence relation.

Proof: Note that Item $\operatorname{(1)}$ holds because the additive
cancelation law holds in the definition of semi-vector spaces (see
Definition2.3). We only show Item $\operatorname{(3)}$ since the remaining
items are direct. Let $v\in B$; then there exists $u\in A$ such that $T(u)=v$.
It then follows that $v\bullet T(1_{A})=T(u\bullet 1_{A})=v$ and
$T(1_{A})\bullet v=T(1_{A}\bullet u)=v$; which means that $T(1_{A})$ is the
identity of $B$, i.e., $T(1_{A})=1_{B}$.

We have: $T(u)\bullet T(u^{-1})=T(u\bullet u^{-1})=T(1_{A})=1_{B}$ and
$T(u^{-1})\bullet T(u)=T(u^{-1}\bullet u)=T(1_{A})=1_{B}$, which implies
$T(u^{-1})=[T(u)]^{-1}$. $\Box$

###### Proposition 3.18

If $A$ is a $K$-semi-algebra with identity $1_{A}$ then $A$ can be embedded in
${\mathcal{L}}(A,A)$, the semi-algebra of semi-linear operators on $A$.

Proof: For every fixed $v\in A$, define $v^{*}:A\longrightarrow A$ as
$v^{*}(x)=v\bullet x$. It is easy to see that $v^{*}$ is a semi-linear
operator on $A$. Define $h:A\longrightarrow{\mathcal{L}}(A,A)$ by
$h(v)=v^{*}$. We must show that $h$ is a injective $K$-semi-algebra
homomorphism where the product in ${\mathcal{L}}(A,A)$ is the composite of
maps from $A$ into $A$. Fixing $u,v\in A$, we have:
$[h(u+v)](x)=(u+v)^{*}(x)=(u+v)\bullet x=u\bullet x+v\bullet
x=u^{*}(x)+v^{*}(x)=[h(u)](x)+[h(v)](x)$, hence $h(u+v)=h(u)+h(v)$. For
$\lambda\in K$ and $v\in A$, it follows that $[h(\lambda v)](x)=(\lambda
v)^{*}(x)=(\lambda v)x=\lambda(vx)=[\lambda h(v)](x)$, i.e., $h(\lambda
v)=\lambda h(v)$. For fixed $u,v\in A$, $[h(u\bullet v)](x)=(u\bullet
v)^{*}(x)=(u\bullet v)\bullet x=u\bullet(v\bullet x)=u\bullet
v^{*}(x)=u^{*}(v^{*}(x))=[h(u)\circ h(v)](x)$, i.e., $h(u\bullet v)=h(u)\circ
h(v)$. Assume that $h(u)=h(v)$, that is, $u^{*}=v^{*}$; hence, for every $x\in
A$, $u^{*}(x)=v^{*}(x)$, i.e., $u\bullet x=v\bullet x$ . Taking in particular
$x=1_{A}$, it follows that $u=v$, which implies that $h$ is injective.
Therefore, $A$ is isomorphic to $h(A)$, where
$h(A)\subseteq{\mathcal{L}}(A,A)$. $\Box$

###### Definition 3.17

Let $A$ be a semi-vector space over a semi-field $K$. Then $A$ is said to be a
Lie semi-algebra if $A$ is equipped with a product $[\ ,\ ]:A\times
A\longrightarrow A$ such that the following conditions hold:

  * $\operatorname{(1)}$

$[\ ,\ ]$ is semi-bilinear, i.e., fixing the first (second) variable, $[\ ,\
]$ is semi-linear w.r.t. the second (first) one;

  * $\operatorname{(2)}$

$[\ ,\ ]$ is anti-symmetric, i.e., $[v,v]=0$ $\forall\ v\in A$;

  * $\operatorname{(3)}$

$[\ ,\ ]$ satisfies the Jacobi identity: $\forall\ u,v,w\in A$,
$[u,[v,w]]+[w,[u,v]]+[v,[w,u]]=0$

From Definition 3.17 we can see that a Lie semi-algebra can be non-associa-
tive, i.e., the product $[\ ,\ ]$ is not always associative.

Let us now consider the semi-algebra ${\mathcal{M}}_{n}({\mathbb{R}}_{0}^{+})$
of matrices of order $n$ with entries in ${\mathbb{R}}_{0}^{+}$ (see Example
3.4). We know that ${\mathcal{M}}_{n}({\mathbb{R}}_{0}^{+})$ is simple, i.e.,
with exception of the zero matrix (zero vector), no matrix has (additive)
symmetric. Therefore, the product of such matrices can be nonzero. However, in
the case of a Lie semi-algebra $A$, if $A$ is simple then the unique product
$[\ ,\ ]$ that can be defined over $A$ is the zero product, as it is shown in
the next result.

###### Proposition 3.19

If $A$ is a simple Lie semi-algebra over a semi-field $K$ then the semi-
algebra is abelian, i.e., $[u,v]=0$ for all $u,v\in A$.

Proof: Assume that $u,v\in A$ and $[u,v]\neq 0$. From Items
$\operatorname{(1)}$ and $\operatorname{(2)}$ of Definition 3.17, it follows
that $[u+v,u+v]=[u,u]+[u,v]+[v,u]+[v,v]=0$, i.e., $[u,v]+[v,u]=0$. This means
that $[u,v]$ has symmetric $[v,u]\neq 0$, a contradiction. $\Box$

###### Definition 3.18

Let $A$ be a Lie semi-algebra over a semi-field $K$. A Lie semi-subalgebra
$B\subseteq A$ is a semi-subspace of $A$ which is closed under $[u,v]$, i.e.,
for all $u,v\in B$, $[u,v]\in B$.

###### Corollary 3.1

All semi-subspaces of $A$ are semi-subalgebras of $A$.

Proof: Apply Proposition 3.19. $\Box$

##  4 Fuzzy Set Theory and Semi-Algebras

The theory of semi-vector spaces and semi-algebras is a natural generalization
of the corresponding theories of vector spaces and algebras. Since the scalars
are in semi-fields (weak semi-fields), some standard properties does not hold
in this new context. However, as we have shown in Section 3, even in case of
nonexistence of symmetrizable elements, several results are still true. An
application of the theory of semi-vector spaces is in the investigation on
Fuzzy Set Theory, which was introduced by Lotfali Askar-Zadeh [16]. In fact,
such a theory fits in the investigation/extension of results concerning fuzzy
sets and their corresponding theory. Let us see an example.

Let $L$ be a linearly ordered complete lattice with distinct smallest and
largest elements $0$ and $1$. Recall that a fuzzy number is a function
$x:{\mathbb{R}}\longrightarrow L$ on the field of real numbers satisfying the
following items (see [4, Sect. 1.1]): $\operatorname{(1)}$ for each $\alpha\in
L_{0}$ the set $x_{\alpha}=\\{\varphi\in{\mathbb{R}}|\alpha\leq x(\varphi)\\}$
is a closed interval $[x_{\alpha l},x_{\alpha r}]$, where $L_{0}=\\{\alpha\in
L|\alpha>0\\}$; $\operatorname{(2)}$
$\\{\varphi\in{\mathbb{R}}|0<x(\varphi)\\}$ is bounded.

We denote the set ${\mathbb{R}}_{L}$ to be the set of all fuzzy numbers;
${\mathbb{R}}_{L}$ can be equipped with a partial order in the following
manner: $x\leq y$ if and only if $x_{\alpha l}\leq y_{\alpha l}$ and
$x_{\alpha r}\leq y_{\alpha r}$ for all $\alpha\in L_{0}$. In this scenario,
Gahler et al. showed that the concepts of semi-algebras can be utilized to
extend the concept of fuzzy numbers, according to the following proposition:

###### Proposition 4.1

[4, Proposition 19] The set ${\mathbb{R}}_{L}$ is an ordered commutative semi-
algebra.

Thus, a direct utilization of the investigation of the structures of semi-
vector spaces and semi-algebras is the possibility to generate new interesting
results on the Fuzzy Set Theory.

Another work relating semi-vector spaces and Fuzzy Set Theory is the paper by
Bedregal et al. [8]. In order to study the aggregation functions (geometric
mean, weighted average, ordered weighted averaging, among others) w.r.t. an
admissible order (a total order $\preceq$ on $L_{n}([0,1])$ such that for all
$x,y\in L_{n}([0,1])$, $x\ {\leq}_{n}^{p}\ y\Longrightarrow x\preceq y$), the
authors worked with semi-vector spaces over a weak semi-field.

Let $L_{n}([0,1])=\\{(x_{1},x_{2},\ldots,x_{n})\in{[0,1]}^{n}|x_{1}\leq
x_{2}\leq\ldots\leq x_{n}\\}$ and $U=([0,1],\oplus,\cdot)$ be a weak semi-
field defined as follows: for all $x,y\in[0,1]$, $x\oplus y=\min\\{1,x+y\\}$
and $\cdot$ is the usual multiplication. The product order proposed by Shang
et al. [15] is given as follows: for all $x=\\{(x_{1},x_{2},\ldots,x_{n})$ and
$y=\\{(y_{1},y_{2},\ldots,y_{n})$ vectors in $L_{n}([0,1])$, define $x\
{\leq}_{n}^{p}\ y\Longleftrightarrow{\pi}_{i}(x)\leq{\pi}_{i}(x)$ for each
$i\in\\{1,2,\ldots,n\\}$, where ${\pi}_{i}:L_{n}([0,1])\longrightarrow[0,1]$
is the $i$-th projection ${\pi}_{i}(x_{1},x_{2},\ldots,x_{n})=x_{i}$. With
these concepts in mind, the authors showed two important results:

###### Theorem 4.1

(see [8, Theorem 1]) ${\mathcal{L}}_{n}([0,1])=(L_{n}([0,1],\dotplus,\odot)$
is a semi-vector space over $U$, where $r\odot v=(rx_{1},\ldots,rx_{n})$ and
$u\dotplus v=(x_{1}\oplus y_{1},\ldots,x_{n}\oplus y_{n})$. Moreover,
$({\mathcal{L}}_{n}([0,1]),{\leq}_{n}^{p})$ is an ordered semi-vector space
over $U$, where ${\leq}_{n}^{p}$ is the product order.

###### Proposition 4.2

(see [8, Propostion 2]) For any bijection
$f:\\{1,2,\ldots,n\\}\longrightarrow\\{1,2,\ldots,n\\}$, the pair
$({\mathcal{L}}_{n}([0,1]),{\preceq}_{f})$ is an ordered semi-vector space
over $U$, where ${\preceq}_{f}$, defined in [8, Example 1], is an admissible
order.

As a consequence of the investigation made, the authors propose an algorithm
to perform a multi-criteria and multi-expert decision making method.

Summarizing the ideas: the better the theory of semi-vector spaces is extended
and developed, the more applications and more results we will have in the
Fuzzy Set Theory. Therefore, it is important to understand deeply which are
the algebraic and geometry structures of semi-vector spaces, providing, in
this way, support for the development of the own theory as well as other
interesting theories as, for example, the Fuzzy Set Theory.

##  5 Summary

In this paper we have extended the theory of semi-vector spaces, where the
semi-field of scalars considered here is the nonnegative real numbers. We have
proved several results in the context of semi-vector spaces and semi-linear
transformations. We introduced the concept of eigenvalues and eigenvectors of
a semi-linear operator and of a matrix and shown how to compute it in specific
cases. Topological properties of semi-vector spaces such as completeness and
separability were also investigated. We have exhibited interesting new
families of semi-vector spaces derived from semi-metric, semi-norm, semi-inner
product, among others. Additionally, some results concerning semi-algebras
were presented. The results presented in this paper can be possibly utilized
in the development and/or investigation of new properties of fuzzy systems and
also in the study of correlated areas of research.

## Acknowledgment

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