#  Remarks on some Limit Geometric Properties related to an Idempotent and
Non-Associative Algebraic Structure

Walter Briec University of Perpignan, LAMPS, 52 avenue Villeneuve, 66000
Perpignan, France<EMAIL_ADDRESS>
(September, 2020)

###### Abstract

This paper analyses the geometric properties of an idempotent and non-
associative algebraic structure which extends the Max-Times semi-ring. This
algebraic structure is useful for analyzing systems of Max-Times and Max-Plus
equations (using an appropriate notion of non-associative determinant). We
consider a related ultrametric distance and show that it implies, among other
things, an analogue of the Pythagorean relation. To that end, we propose a
suitable notion of right angle between two vectors and analyse a trigonometric
like concept related to the Tshebishev unit ball. Along this line, we study
the possible implications of these properties for the complex plane. We give
the algebraic definition of a line passing through two points which turns out
to be the Painlevé-Peano-Kuratowski limit on a sequence of generalized lines.
We show that this definition implies some particular geometric properties; in
particular two distinct parallel lines can have an infinite number of points
in common.

AMS: 06D50, 32F17

Keywords: Generalized mean, convexity, convex hull, ultrametric, semilattice,
$\mathbb{B}$-convexity.

## Introduction

It is known that the algebras associated with idempotent semi-rings are
associated with particular geometric representations. This article studies
some of these properties through the non-associative symmetrization of Max-
Times algebras. $\mathbb{B}$-convexity was defined in [11], as a suitable
Painlevé-Kuratowski limit of generalized convexities. The general idea is
related to some works by Ben-Tal [4] and Avriel [2]. When it is restricted to
the non-negative orthant $\mathbb{B}$-convexity reduces to Max-Times
convexity. Along this line, an algebraic formulation was established over the
Max-Times idempotent semi-ring $(\mathbb{R}_{+},\max,\times)$ making the
formal transformation $+\rightarrow\max$ over the nonnegative Euclidean
orthant. The Max-Times idempotent semi-ring is homeomorphic to the Max-Plus
idempotent semiring $(\mathbb{R}\cup\\{-\infty\\},\max,+)$ via a logarithmic
transformation and $\mathbb{B}$-convex sets can be derived from a large class
of idempotent algebraic structures introduced in [22] and derived from the
notion of dequantization (see also [21]). Hanh-Banach like separation
properties [13] as well as fixed point results have been established in [12].
Some alternative idempotent convex structures were also proposed in [1] and
[26]. The algebraic form of $\mathbb{B}$-convexity was previously limited to
the Max-Times semi-module $(\mathbb{R}_{+}^{n},\vee,\times)$. To circumvent
this difficulty an idempotent and non-associative binary operation has been
introduced in [7] where a special class of idempotent magma was considered. In
this paper an n-ary operation has been introduced and associated with a class
of suitable regularized semi-continuous operators. Note that in the case where
they involve only two points, the idempotent and non-associative algebraic
structures considered are also mentioned by Gaubert [15] and Viro [27] (who
consider the complex case).

The relaxation of associativity makes it possible to preserve both symmetry
and idempotence. It was shown that this notion of convexity is equivalently
characterized from the Kuratowski-Painlevé limit of the generalized convex
hull of two points as defined in [11]. Some separation results over
$\mathbb{R}^{n}$ were established in [8] and the limit structure of polytopes
was analysed in [9].

More recently, these structures have been studied from a more algebraic point
of view in [10] to calculate the limit solutions of a system of Max-Times
equations, in particular by introducing a special form of determinant. These
results also have geometric implications to find the equation of a limit
hyperplane passing trough $n$ points and can be extended to Max-Plus algebra.

This paper continues the previous investigations on this subject from a more
geometrical point of view, while introducing some suitable notions of
distance. Throughout the paper an idempotent analogue of the notion of
subtraction plays an important role.

First, we define a notion of space suitable for the algebraic structure which
is considered and generalize the notion of $\mathbb{B}$-space introduced in
[12].

Second, we consider a sequence of generalized metrics that are constructed
from a suitable homomorphic transformation of the vector space structure. It
is shown that their limit properties yield an ultra-metric distance that is
intimately related to the algebraic structure considered in the paper. Along
this line, a suitable notion of orthogonality is derived from the limit inner
product defined in [7]. It follows that some geometric properties of a special
class or right-angled triples are established. More precisely, a Pythagorean
relation is established in the context of this geometry. Paralleling the
properties of the usual inner product a pseudo-cosine function is constructed
from the limit inner-product. The pseudo-cosine function is derived from the
limit scalar product while the pseudo-sine function is derived from the two-
dimensional determinant formula established in [10]. Along this line, we
investigate the possible implications of these properties for analysis in the
complex plane.

Finally, the article considers a notion of line relative to the non-
associative algebraic structure considered in the article. It its shown that
the Painlevé-Kuratowski of a sequence of generalized lines has an algebraic
description extending an earlier result established in [7] where the result
was established for a segment line. We deduce a condition guaranteeing the
parallelism in limit between two limit lines which is analogous (from a
certain point of view) to that characterizing two parallel lines in a vector
space. We conclude the paper by focusing on the two dimensional case in which
a special idempotent geometry is analysed. From [10], it is shown how to
compute the equation of a two-dimensional line. The final result shows a
geometry of the plane in which through two distinct points passes a single
straight line and in which a straight line segment can be extended to a half-
line. However, two distinct and parallel lines can have an infinity of common
points.

The paper unfolds as follows. Section 1 presents the framework of the non-
associative idempotent algebraic structure defined in [7] and some related
notions of convexity. Section 2 defines some notions of field and space
associated with the considered algebraic structure. Examples are given to
illustrate these concepts. Section 3 derives an ultra-metric distance defined
as the limit of a sequence of generalized metrics. Along this line a
triangular equality is established for a special class of right-angled
triples. Section 4 analyses the properties of the inner product established in
[8] and introduces a related notion of pseudo-cosine. Section 5 gives the
algebraic description of a line and analyses it from an axiomatic viewpoint.
We conclude by analysing the geometric implications of the considered
algebraic structure.

##  1 Idempotent and Non-Associative Algebraic and Convex Structure

###  1.1 Isomorphism of vector-Space Structures in Limit

For all $p\in\mathbb{N}$, let us consider a bijection
$\varphi_{p}:\mathbb{R}\longrightarrow\mathbb{R}$ defined by:

$\varphi_{p}:x\longrightarrow x^{2p+1}$ (1)

and $\phi_{p}(x_{1},...,x_{n})=(\varphi_{p}(x_{1}),...,\varphi_{p}(x_{n}))$.
This is closely related to the approach proposed by Ben-Tal [4] and Avriel
[2]. One can induce a field structure on $\mathbb{R}$ for which $\varphi_{p}$
becomes a field isomorphism. Given this change of notation via $\varphi_{p}$
and $\phi_{p}$, we can define a $\mathbb{R}$-vector space structure on
$\mathbb{R}^{n}$ by:
$\lambda\stackrel{{\scriptstyle\varphi_{p}}}{{\cdot}}x=\phi_{p}^{-1}\big{(}\varphi_{p}(\lambda)\cdot\phi_{p}(x)\big{)}=\lambda.x$
and
$x\stackrel{{\scriptstyle\varphi_{p}}}{{+}}y=\phi_{p}^{-1}\big{(}\phi_{p}(x)+\phi_{p}(y)\big{)}$.
We call these two operations the indexed scalar product and the indexed sum
(indexed by $\varphi_{p}$).

The $\varphi_{p}$-sum denoted $\stackrel{{\scriptstyle\varphi_{p}}}{{\sum}}$
of $(x_{1},...,x_{m})\in\mathbb{R}^{n\times m}$ is defined by111For all
positive natural numbers $n$, $[n]=\\{1,...,n\\}$.

$\stackrel{{\scriptstyle\varphi_{p}}}{{\sum_{i\in[m]}}}x_{j}=\phi_{p}^{-1}\Big{(}\sum_{j\in[m]}\phi_{p}(x_{j})\Big{)}.$
(2)

For simplicity, throughout the paper we denote for all $x,y\in\mathbb{R}^{n}$:

$x\stackrel{{\scriptstyle
p}}{{+}}y=x\stackrel{{\scriptstyle\varphi_{p}}}{{+}}y.$ (3)

Recall that Kuratowski-Painlevé lower limit of the sequence of sets
$\\{A_{n}\\}_{n\in\mathbb{N}}$, denoted
$\operatorname*{\mathrm{Li}}_{n\to\infty}A_{n}$, is the set of points $x$ for
which there exists a sequence $\\{x^{(n)}\\}_{n\in\mathbb{N}}$ of points such
that $x^{(n)}\in A_{n}$ for all $n$ and $x=\lim_{n\to\infty}x^{(n)}$. The
Kuratowski-Painlevé upper limit of the sequence of sets
$\\{A_{n}\\}_{n\in\mathbb{N}}$, denoted
$\operatorname*{\mathrm{Ls}}_{n\to\infty}A_{n}$, is the set of points $x$ for
which there exists a subsequence $\\{x_{n_{k}}\\}_{k\in\mathbb{N}}$ of points
such that $x^{(n_{k})}\in A_{n_{k}}$ for all $k$ and
$x=\lim_{k\to\infty}x^{(n_{k})}$. A sequence $\\{A_{n}\\}_{n\in\mathbb{N}}$ of
subsets of $\mathbb{R}^{n}$ is said to converge, in the Kuratowski-Painlevé
sense, to a set $A$ if
$\operatorname*{\mathrm{Ls}}_{n\to\infty}A_{n}=A=\operatorname*{\mathrm{Li}}_{n\to\infty}A_{n}$,
in which case we write $A=\operatorname*{\mathrm{Lim}}_{n\to\infty}A_{n}$.

In [7] it was shown that for all $x,y\in\mathbb{R}$ we have:

$\lim_{p\longrightarrow+\infty}x\stackrel{{\scriptstyle
p}}{{+}}y=\left\\{\begin{matrix}x\ &\hbox{ if }&|x|&>&|y|\\\
\frac{1}{2}(x+y)&\hbox{ if }&|x|&=&|y|\\\ y&\hbox{ if
}&|x|&<&|y|.\end{matrix}\right.$ (4)

Along this line one can introduce the binary operation $\boxplus$ defined for
all $x,y\in\mathbb{R}$ by:

$x\boxplus y=\lim_{p\longrightarrow+\infty}x\stackrel{{\scriptstyle
p}}{{+}}y.$ (5)

Though the operation $\boxplus$ does not satisfy associativity, it can be
extended by constructing a non-associative algebraic structure which returns
to a given $n$-tuple a real value. For all $x\in\mathbb{R}^{n}$ and all
subsets $I$ of $[n]$, let us consider the map
$\xi_{I}[x]:\mathbb{R}\longrightarrow\mathbb{Z}$ defined for all
$\alpha\in\mathbb{R}$ by

$\xi_{I}[x](\alpha)=\operatorname*{\mathrm{Card}}\\{i\in
I:x_{i}=\alpha\\}-\operatorname*{\mathrm{Card}}\\{i\in I:x_{i}=-\alpha\\}.$
(6)

This map measures the symmetry of the occurrences of a given value $\alpha$ in
the coordinates of a vector $x$.

For all $x\in\mathbb{R}^{n}$ let $\mathcal{J}_{I}(x)$ be a subset of $I$
defined by

$\mathcal{J}_{I}(x)=\Big{\\{}j\in
I:\xi_{I}[x](x_{j})\not=0\Big{\\}}=I\backslash\big{\\{}i\in
I:\xi_{I}[x](x_{i})=0\big{\\}}.$ (7)

$\mathcal{J}_{I}(x)$ is called the residual index set  of $x$. It is obtained
by dropping from $I$ all the $i$’s such that
$\operatorname*{\mathrm{Card}}\\{j\in
I:x_{j}=x_{i}\\}=\operatorname*{\mathrm{Card}}\\{j\in I:x_{j}=-x_{i}\\}$.

For all positive natural numbers $n$ and for all subsets $I$ of $[n]$, let
$\digamma_{I}:\mathbb{R}^{n}\longrightarrow\mathbb{R}$ be the map defined for
all $x\in\mathbb{R}^{n}$ by

$\digamma_{I}(x)=\left\\{\begin{matrix}\max_{i\in\mathcal{J}_{I}(x)}x_{i}&\text{
if }&\xi_{I}[x]\big{(}\max_{i\in\mathcal{J}_{I}(x)}|x_{i}|\big{)}>0\\\
\min_{i\in\mathcal{J}_{I}(x)}x_{i}&\hbox{ if
}&\xi_{I}[x]\big{(}\max_{i\in\mathcal{J}_{I}(x)}|x_{i}|\big{)}<0\\\ 0&\text{
if
}&\xi_{I}[x]\big{(}\max_{i\in\mathcal{J}_{I}(x)}|x_{i}|\big{)}=0.\end{matrix}\right.$
(8)

where $\xi_{I}[x]$ is the map defined in (6) and $\mathcal{J}_{I}(x)$ is the
residual index set of $x$. The operation that takes an $n$-tuple
$(x_{1},....,x_{n})$ of $\mathbb{R}^{n}$ and returns a single real element
$\digamma_{I}(x_{1},...,x_{n})$ is called a $n$-ary extension of the binary
operation $\boxplus$ for all natural numbers $n\geq 1$ and all
$x\in\mathbb{R}^{n}$, if $I$ is a nonempty subset of $[n]$. From [7], for all
$n$-tuple $x=(x_{1},...,x_{n})$, one can define the operation:

$\operatorname*{\text{\Huge${}_{\boxplus}$}}_{i\in
I}x_{i}=\lim_{p\longrightarrow\infty}\stackrel{{\scriptstyle\varphi_{p}}}{{\sum_{i\in
I}}}x_{i}=\digamma_{I}(x).$ (9)

Clearly, this operation encompasses as a special case the binary operation
mentioned in [15] and defined in equation (4). For all
$(x_{1},x_{2})\in\mathbb{R}^{2}$:

$\operatorname*{\text{\Huge${}_{\boxplus}$}}_{i\in\\{1,2\\}}x_{i}=x_{1}\boxplus
x_{2}.$

For example, if $x=(-3,-2,3,3,1,-3)$, we have
$F_{[6]}(-3,-2,3,3,1,-3)=F_{[2]}(-2,1)=-2=\operatorname*{\text{\Huge${}_{\boxplus}$}}_{i\in[6]}x_{i}$.
There are some basic properties that can be inherited from the above algebraic
structure: $(i)$ For all $\alpha\in\mathbb{R}$, one has:
$\alpha\Big{(}\operatorname*{\text{\Huge${}_{\boxplus}$}}_{i\in
I}x_{i}\Big{)}=\operatorname*{\text{\Huge${}_{\boxplus}$}}_{i\in I}(\alpha
x_{i})$; $(ii)$ Suppose that $x\in\epsilon\mathbb{R}_{+}^{n}$ where $\epsilon$
is $+1$ or $-1$. Then $\operatorname*{\text{\Huge${}_{\boxplus}$}}_{i\in
I}x_{i}=\epsilon\max_{i\in I}\\{\epsilon x_{i}\\}$; $(iii)$ We have
$|\operatorname*{\text{\Huge${}_{\boxplus}$}}_{i\in
I}x_{i}|\leq\operatorname*{\text{\Huge${}_{\boxplus}$}}_{i\in I}|x_{i}|$;
$(iv)$ For all $x\in\mathbb{R}^{n}$:

$\Big{[}x_{i}\boxplus\big{(}\operatorname*{\text{\Huge${}_{\boxplus}$}}_{j\in
I\backslash\\{i\\}}x_{j}\big{)}\Big{]}\in\Big{\\{}0,\operatorname*{\text{\Huge${}_{\boxplus}$}}_{j\in
I}x_{j}\Big{\\}}\quad\text{ and
}\quad\operatorname*{\text{\Huge${}_{\boxplus}$}}_{i\in
I}x_{i}=\operatorname*{\text{\Huge${}_{\boxplus}$}}_{i\in
I}\Big{[}x_{i}\boxplus\big{(}\operatorname*{\text{\Huge${}_{\boxplus}$}}_{j\in
I\backslash\\{i\\}}x_{j}\big{)}\Big{]}.$ (10)

The algebraic structure $(\mathbb{R},\boxplus,\cdot)$ can be extended to
$\mathbb{R}^{n}$. Suppose that $x,y\in\mathbb{R}^{n}$, and let us denote
$x\boxplus y=(x_{1}\boxplus y_{1},...,x_{n}\boxplus y_{n}).$ Moreover, let us
consider $m$ vectors $x_{1},...,x_{m}\in\mathbb{R}^{n}$, and define

$\displaystyle\operatorname*{\text{\Huge${}_{\boxplus}$}}_{j\in[m]}x_{j}$
$\displaystyle=\Big{(}\operatorname*{\text{\Huge${}_{\boxplus}$}}_{j\in[m]}x_{j,1},...,\operatorname*{\text{\Huge${}_{\boxplus}$}}_{j\in[m]}x_{j,n}\Big{)}.$
(11)

The $n$-ary operation
$(x_{1},...,x_{n})\rightarrow\operatorname*{\text{\Huge${}_{\boxplus}$}}_{i\in[n]}x_{i}$
is not associative. To simplify the notations of the paper, for all
$z\in\\{z_{i_{1},...,i_{m}}:i_{k}\in I_{k},k\in[m]\\}$, where
$I_{1},...,I_{m}$ are $m$ index subsets of $\mathbb{N}$, we use the notation:

$\operatorname*{\text{\Huge${}_{\boxplus}$}}_{\begin{subarray}{c}i_{k}\in
I_{k}\\\
k\in[m]\end{subarray}}z_{i_{1},...,i_{m}}=\operatorname*{\text{\Huge${}_{\boxplus}$}}_{\begin{subarray}{c}(i_{1},...,i_{m})\in\prod_{k\in[m]}I_{k}\end{subarray}}z_{i_{1},...,i_{m}}.$
(12)

In the remainder, we will adopt the following notational convention. For all
$x\in\mathbb{R}^{n}$:

$\operatorname*{\text{\Huge${}_{\boxplus}$}}_{i\in[n]}x_{i}=x_{1}\boxplus\cdots\boxplus
x_{n}=\digamma_{[n]}(x).$ (13)

In the sequel, for all $x,y\in\mathbb{R}^{n}$ we will often use the following
notation to define an idempotent analogue of the subtraction operation:

$x\boxminus y=x\boxplus(-y).$ (14)

###  1.2 Scalar Products and Determinants in Limit

This section presents the algebraic properties induced by an isomorphism
between a scalar field and a set and their implications on the scalar product.
If $\langle\cdot,\cdot\rangle$ is the canonical inner product over
$\mathbb{R}^{n}$, then there exists a symmetric bilinear form
$\langle\cdot,\cdot\rangle_{\varphi_{p}}:\mathbb{R}^{n}\times\mathbb{R}^{n}\longrightarrow\mathbb{R}$
defined by:

$\langle
x,y\rangle_{\varphi_{p}}=\varphi_{p}^{-1}\big{(}\langle\phi_{p}(x),\phi_{p}(y)\rangle\big{)}=\big{(}\sum_{i\in[n]}x_{i}^{2p+1}y_{i}^{2p+1}\big{)}^{\frac{1}{2p+1}}.$
(15)

Now, let us denote $\big{[}\langle
y,\cdot\rangle_{\varphi_{p}}\leq\lambda\big{]}=\left\\{x\in\mathbb{R}^{n}:\langle
y,x\rangle_{\varphi_{p}}\leq\lambda\right\\}$. For the sake of simplicity, let
us denote $\langle\cdot,\cdot\rangle_{p}$ this scalar product.

In what follows, we state some results obtained in [7, 8] by taking the limit.
We first introduce the operation
$\langle\cdot,\cdot\rangle_{\infty}:\mathbb{R}^{n}\times\mathbb{R}^{n}\longrightarrow\mathbb{R}$
defined for all $x,y\in\mathbb{R}^{n}$ by $\langle
x,y\rangle_{\infty}=\operatorname*{\text{\Huge${}_{\boxplus}$}}_{i\in[n]}x_{i}y_{i}$.
Let $\|\cdot\|_{\infty}$ be the Tchebychev norm defined by
$\|x\|_{\infty}=\max_{i\in[n]}|x_{i}|$. It is established in [7] that for all
$x,y\in\mathbb{R}^{n}$, we have: $(i)$ $\sqrt{\langle
x,x\rangle_{\infty}}=\|x\|_{\infty}$; $(ii)$ $|\langle
x,y\rangle_{\infty}|\leq\|x\|_{\infty}\|y\|_{\infty}$; $(iii)$ For all
$\alpha\in\mathbb{R}$, $\alpha\langle x,y\rangle_{\infty}=\langle\alpha
x,y\rangle_{\infty}=\langle x,\alpha y\rangle_{\infty}$. By definition, we
have for all $x,y\in\mathbb{R}^{n}$:

$\langle x,y\rangle_{\infty}=\lim_{p\longrightarrow\infty}\langle
x,y\rangle_{p}.$ (16)

A map $f:\mathbb{R}^{n}\longrightarrow\mathbb{R}$ defined as $f(x)=\langle
a,x\rangle_{\infty}$ for some $a\in\mathbb{R}^{n}$ is called a
$\mathbb{B}$-form. In the following, for all maps
$f:\mathbb{R}^{n}\longrightarrow\mathbb{R}$ and all real numbers $c$, the
notation $[\,f\leq c\,]$ stands for the set $f^{-1}(\,]-\infty,c]\,)$.
Similarly, $[\,f<c\,]$ stands for $f^{-1}(\,]-\infty,c[\,)$ and $[\,f\geq
c\,]=[\,-f\leq-c\,]$.

For all $u,v\in\mathbb{R}$, let us define the binary operation

$u\stackrel{{\scriptstyle-}}{{\smile}}v=\left\\{\begin{matrix}u&\hbox{ if
}&|u|&>&|v|\\\ \min\\{u,v\\}&\hbox{ if }&|u|&=&|v|\\\ v&\hbox{ if
}&|u|&<&|v|.\end{matrix}\right.$

An elementary calculus shows that $u\boxplus
v=\frac{1}{2}\Big{[}u\stackrel{{\scriptstyle-}}{{\smile}}v-\big{[}(-u)\stackrel{{\scriptstyle-}}{{\smile}}(-v)\big{]}\Big{]}$.

Similarly one can introduce a symmetrical binary operation defined for all
$u,v\in\mathbb{R}$, let us define the binary operation
$u\stackrel{{\scriptstyle+}}{{\smile}}v=-\big{[}(-u)\stackrel{{\scriptstyle-}}{{\smile}}(-v)\big{]}$.

Given $m$ elements $u_{1},...,u_{m}$ of $\mathbb{R}$, not all of which are
$0$, let $I_{+}$, respectively $I_{-}$, be the set of indices for which
$0<u_{i}$, respectively $u_{i}<0$. We can then write
$u_{1}\stackrel{{\scriptstyle-}}{{\smile}}\cdots\stackrel{{\scriptstyle-}}{{\smile}}u_{m}=(\stackrel{{\scriptstyle-}}{{\smile}}_{i\in
I_{+}}u_{i})\stackrel{{\scriptstyle-}}{{\smile}}(\stackrel{{\scriptstyle-}}{{\smile}}_{i\in
I_{-}}u_{i})=(\max_{i\in
I_{+}}u_{i})\stackrel{{\scriptstyle-}}{{\smile}}(\min_{i\in I_{-}}u_{i})$ from
which we have

$u_{1}\stackrel{{\scriptstyle-}}{{\smile}}\cdots\stackrel{{\scriptstyle-}}{{\smile}}u_{m}=\left\\{\begin{array}[]{lcc}\max_{i\in
I_{+}}u_{i}&\hbox{if}&I_{-}=\emptyset\hbox{ or }\max_{i\in
I_{-}}|u_{i}|<\max_{i\in I_{+}}u_{i}\\\ \min_{i\in
I_{-}}u_{i}&\hbox{if}&I_{-}=\emptyset\hbox{ or }\max_{i\in
I_{+}}u_{i}<\max_{i\in I_{-}}|u_{i}|\\\ \min_{i\in
I_{-}}u_{i}&\hbox{if}&\max_{i\in I_{-}}|u_{i}|=\max_{i\in
I_{+}}u_{i}.\end{array}\right.$ (17)

We define a lower ${\mathbb{B}}$-form on ${\mathbb{R}}^{n}_{+}$ as a map
$g:{\mathbb{R}}^{n}\to\mathbb{R}$ such that for all
$(x_{1},...,x_{n})\in{\mathbb{R}}^{n}_{+}$,

$g(x_{1},...,x_{n})={\langle
a,x\rangle}_{\infty}^{-}=a_{1}x_{1}\stackrel{{\scriptstyle-}}{{\smile}}\cdots\stackrel{{\scriptstyle-}}{{\smile}}a_{n}x_{n}.$
(18)

It was established in [7] that for all $c\in\mathbb{R}$,
$g^{-1}\left(\,]-\infty,c]\right)=\left\\{x\in\mathbb{R}^{n}:g(x)\leq
c\right\\}$ is closed. It follows that a ${\mathbb{B}}$-form is lower semi-
continuous. It was established in [13] that
$g^{-1}\left(\,]-\infty,c]\right)\cap\mathbb{R}_{+}^{n}$ is a
$\mathbb{B}$-halfspace, that is a $\mathbb{B}$-convex subset of
$\mathbb{R}_{+}^{n}$ whose the complement in $\mathbb{R}_{+}^{n}$ is also
$\mathbb{B}$-convex.

Similarly, one can define an upper $\mathbb{B}$-form as a map
$h:{\mathbb{R}}^{n}\to\mathbb{R}$ such that, for all
$(x_{1},...,x_{n})\in{\mathbb{R}}^{n}$,

$h(x_{1},...,x_{n})={\langle
a,x\rangle}_{\infty}^{+}=a_{1}x_{1}\stackrel{{\scriptstyle+}}{{\smile}}\cdots\stackrel{{\scriptstyle+}}{{\smile}}a_{n}x_{n}.$
(19)

For all $x\in\mathbb{R}^{n}$, we clearly, have the following identities

$\langle a,x\rangle_{\infty}^{+}=-\langle a,-x\rangle_{\infty}^{-}\text{ and
}\langle a,x\rangle_{\infty}^{-}=-\langle a,-x\rangle_{\infty}^{+}.$ (20)

The largest (smallest) lower (upper) semi-continuous minorant (majorant) of a
map $f$ is said to be the lower (upper) semi-continuous regularization of $f$.
In the next statements it is shown that the lower (upper) $\mathbb{B}$-forms
are the lower (upper) semi-continuous regularized of the $\mathbb{B}$-form.
These forms can be related to the construction of a ring involving a balance
relation symmetrizing the semi-ring $(\mathbb{R}_{+},\vee,\cdot)$ (see [10],
see also [17] and [24] in a Max-Plus context).

It was shown in [7] that if $g$ is a lower $\mathbb{B}$-form defined by
$g(x_{1},...,x_{n})=a_{1}x_{1}\stackrel{{\scriptstyle-}}{{\smile}}\cdots\stackrel{{\scriptstyle-}}{{\smile}}a_{n}x_{n},$
for some $a\in\mathbb{R}^{n}$, then $g$ is the lower semi-continuous
regularization of the map $x\mapsto\langle
a,x\rangle_{\infty}=\operatorname*{\text{\Huge${}_{\boxplus}$}}_{i\in[n]}a_{i}x_{i}$.
Similarly, if $h$ is an upper $\mathbb{B}$-form defined by
$h(x_{1},...,x_{n})=a_{1}x_{1}\stackrel{{\scriptstyle+}}{{\smile}}\cdots\stackrel{{\scriptstyle+}}{{\smile}}a_{n}x_{n},$
for some $a\in\mathbb{R}^{n}$, then $h$ is the upper semi-continuous
regularization of the map $x\mapsto\langle
a,x\rangle_{\infty}=\operatorname*{\text{\Huge${}_{\boxplus}$}}_{i\in[n]}a_{i}x_{i}$.

Recently, the following notion of determinant was introduced in [10]. The set
$\mathbb{R}$ equipped with the operation $\boxplus$ makes it possible to
define a suitable notion of determinant (see [9, 10]). We denote by
$\mathcal{M}_{n}(\mathbb{R})$ the set of matrices set to $\mathbb{R}$. We
introduce over $\mathcal{M}_{n}(\mathbb{R})$ the map
$\Phi_{p}:\mathcal{M}_{n}(\mathbb{R})\longrightarrow\mathcal{M}_{n}(\mathbb{R})$
defined for all matrices
$A=\left(a_{i,j}\right)_{\begin{subarray}{c}i=1...n\\\
j=1..n\end{subarray}}\in\mathcal{M}_{n}(\mathbb{R})$ by:

$\displaystyle\Phi_{p}(A)$
$\displaystyle=\big{(}\varphi_{p}(a_{i,j})\big{)}_{\begin{subarray}{c}i=1...n\\\
j=1...n\end{subarray}}=\left({a_{i,j}}^{2p+1}\right)_{\begin{subarray}{c}i=1...n\\\
j=1...n\end{subarray}}.$ (21)

Its reciprocal is the map
$\Phi_{p}^{-1}:\mathcal{M}_{n}(\mathbb{R})\longrightarrow\mathcal{M}_{n}(\mathbb{R})$
defined by:

$\displaystyle\Phi_{p}^{-1}(A)$
$\displaystyle=\left(\varphi_{p}^{-1}(a_{i,j})\right)_{\begin{subarray}{c}i=1...n\\\
j=1...n\end{subarray}}=\left({a_{i,j}}^{\frac{1}{2p+1}}\right)_{\begin{subarray}{c}i=1...n\\\
j=1...n\end{subarray}}.$ (22)

$\Phi_{p}$ is a natural extension of the map $\phi_{p}$ from $\mathbb{R}^{n}$
to $\mathcal{M}_{n}(\mathbb{R})$. $\Phi_{p}(A)$ is the $2p+1$ Hadamard power
of matrix $A$. In the following we introduce the matrix product:

$A\stackrel{{\scriptstyle
p}}{{.}}x=\phi_{p}^{-1}\big{(}\Phi_{p}(A).\phi_{p}(x)\big{)}.$ (23)

It is easy to see that the map $x\mapsto A\stackrel{{\scriptstyle
p}}{{\cdot}}x$ is $\varphi_{p}$-linear that is linear with respect to the
algebraic operation $\stackrel{{\scriptstyle p}}{{+}}$ and the usual scalar
multiplication. Conversely, if $g$ is a $\varphi_{p}$-linear map then it can
be represented by a matrix $A$ such that $g(x)=A\stackrel{{\scriptstyle
p}}{{\cdot}}x$ for all $x\in\mathbb{R}^{n}$.

For all $n\times n$ matrices $A$, let $|A|$ denotes its determinant. Let us
introduce the following definition of $\varphi_{p}$-determinant

$|A|_{p}=\varphi_{p}^{-1}|\Phi_{p}(A)|.$ (24)

Using the Leibnitz formula yields

$|A|_{p}=\Big{(}\sum_{\sigma\in
S_{n}}\operatorname*{\mathrm{sgn}}(\sigma)\prod_{i\in[n]}a_{i,\sigma(i)}^{2p+1}\Big{)}^{\frac{1}{2p+1}}.$
(25)

Along this line, one can define a determinant in limit as follows:

$\lim_{p\longrightarrow\infty}|A|_{p}:=|A|_{\infty}=\operatorname*{\text{\Huge${}_{\boxplus}$}}_{\sigma\in
S_{n}}\big{(}\operatorname*{\mathrm{sgn}}(\sigma)\prod_{i\in[n]}a_{i,\sigma(i)}\big{)}.$
(26)

This notion is compared in [10] to that of permanent originally defined in the
framework of Max-Plus algebras (se [6] and [5]).

###  1.3 An Idempotent and Non-associative Convex Structure

In [11], $\mathbb{B}$-convexity is introduced as a limit of linear
convexities. The $\Phi_{p}$-convex hull of a finite set
$A=\\{x^{(1)},...,x^{(m)}\\}\subset\mathbb{R}^{n}$ is defined by:

$Co^{\phi_{p}}(A)=\Big{\\{}\stackrel{{\scriptstyle\varphi_{p}}}{{\sum_{i\in[m]}}}t_{i}\stackrel{{\scriptstyle\varphi_{p}}}{{\cdot}}x^{(i)}:\stackrel{{\scriptstyle\varphi_{p}}}{{\sum_{i\in[m]}}}t_{i}=\varphi_{p}^{-1}(1),\varphi_{p}(t_{i})\geq
0,i\in[m]\Big{\\}}$ (27)

which can be rewritten:

$Co^{\phi_{p}}(A)=\Bigg{\\{}\phi_{p}^{-1}\Big{(}\sum_{i\in[m]}t_{i}^{2p+1}{\cdot}\phi_{p}(x^{(i)})\Big{)}:\big{(}\sum_{i\in[m]}t_{i}^{2p+1}\big{)}^{\frac{1}{2p+1}}=1,t_{i}\geq
0,i\in[m]\Bigg{\\}}.$ (28)

Moreover, for all $L\subset\mathbb{R}$, we simplify the notations denoting
$Co^{p}(L)=Co^{\phi_{p}}(L)$. Definition of $\mathbb{B}$-convex sets is based
on the definition of the Kuratowski-Painlevé limit of a sequence of
$\Phi_{p}$-convex sets. For a non-empty finite subset $A\subset\mathbb{R}^{n}$
we let
$Co^{\infty}(A)=\operatorname*{\mathrm{Ls}}_{p\longrightarrow\infty}Co^{p}(A)$.
A subset $C$ of $\mathbb{R}^{n}$ is $\mathbb{B}$-convex if for all finite
subset $A$ of $C$, $Co^{\infty}(A)\subset C$. When
$A\subset\mathbb{R}_{+}^{n}$, it was shown in [11] that:

$Co^{\infty}(A)=\Big{\\{}\bigvee_{i\in[m]}t_{i}x^{(i)}:t_{i}\in[0,1],\max_{i\in[m]}t_{i}=1\Big{\\}}=\operatorname*{\mathrm{Lim}}_{p\longrightarrow\infty}Co^{p}(A).$

Moreover it is also proved that the sequence is Haussdorf convergent.

If $L\subset\mathbb{R}_{+}^{n}$ is $\mathbb{B}$-convex, then
$L=\operatorname*{\mathrm{Lim}}_{p\longrightarrow\infty}Co^{p}(L)=\bigcap_{r=0}^{\infty}Co^{p}(L)$.
For all $S\subset\mathbb{R}_{+}^{n}$, we denote
$Co^{\infty}(S)=\operatorname*{\mathrm{Lim}}_{p\longrightarrow\infty}Co^{p}(S)$.
Properties of this operation are developed in [11]. In particular, for all
$S\subset\mathbb{R}_{+}^{n}$, $Co^{\infty}(S)$ is $\mathbb{B}$-convex.

An important limitation of this definition of $\mathbb{B}$-convexity is that
its algebraic form is restricted to the non-negative orthant
$\mathbb{R}_{+}^{n}$, over which the idempotent Max-Times semiring is
considered. To circumvent this difficulty an extended definition of
$\mathbb{B}$-convexity other $\mathbb{R}^{n}$ was suggested in [7]. A subset
of $\mathbb{R}^{n}$ $C$ is $\mathbb{B}^{\sharp}$-convex if for all $x,y\in C$
and all $t\in[0,1]$,

$x\boxplus ty\in C.$ (29)

It was shown in [7] that the sequence of subsets
$\\{Co^{(p)}(x,y)\\}_{p\in\mathbb{N}}$ of $\mathbb{R}^{n}$ has a Painlevé-
Kuratowski limit denoted $Co^{\infty}(x,y)$. These points can be expressed
trough an explicit algebraic formula in term of the idempotent and non-
associative algebraic structure defined on $(\mathbb{R}^{n},\boxplus,\cdot)$.

$x^{(1)}$$y^{(1)}$$x^{(2)}$$y^{(2)}$$x^{(5)}$$y^{(5)}$$y^{(6)}$$x^{(6)}$$x^{(4)}$$y^{(4)}$$x^{(3)}$$y^{(3)}$$0$$x_{2}$$x_{1}$Figure
4.1: Examples of sets $Co^{\infty}(x,y)$.

Moreover, it was established in [7] that this limit entirely characterizes a
$\mathbb{B}^{\sharp}$-convex set. A subset $C$ of $\mathbb{R}^{n}$ is
$\mathbb{B}^{\sharp}$-convex if and only if for all $x,y\in\mathbb{R}^{n}$
$Co^{\infty}(x,y)\subset C$. More recently, it has been established in [9]
that for all $x,y\in\mathbb{R}^{n}$, $Co^{\infty}(x,y)$ is
$\mathbb{B}^{\sharp}$-convex and thereby $Co^{\infty}(x,y)$ is the smallest
$\mathbb{B}^{\sharp}$-convex that contains $\\{x,y\\}$.

$x_{1}$$x_{2}$$0$Figure: Examples of $\mathbb{B}^{\sharp}$-convex sets

##  2 Idempotent Symmetric Ultra-Metric Spaces

The so-called notion of $\mathbb{B}$-space was defined in [13]. However, it is
limited to the non-negative $n$-dimensional orthant $\mathbb{R}_{+}^{n}$. In
the following, we will extend it to the whole Euclidean vector space. The new
definition will be useful in the following to analyze the algebraic structure
of a line defined on $(\mathbb{R}^{n},\boxplus,\cdot)$.

###  2.1 Idempotent Pseudo-Fields and Symmetric Spaces

A finite-dimensional $\mathbb{B}$-space is, by definition, a subset $X$ of
$\mathbb{R}_{+}^{n}$ such that $0\in X$, for all $t\geq 0$ and all $x\in X$,
$tx\in X$ and for all $x,y\in X$, $x\vee ty\in X$.

We first define a suitable notion of idempotent pseudo-field. By comparing
this definition with that of a field, note that we consider a weakened for of
associativity and idempotence is added.

###### Definition 2.1.1

A triple $(K,\boxplus_{K},\cdot_{K})$ is an idempotent pseudo-field if:

$(a)$ There exits a binary operation $\boxplus_{K}:K\times K\longrightarrow K$
that satisfies the following properties: $(i)$ For all $\lambda\in K$ we have
$\lambda\boxplus_{K}\lambda=\lambda$ (idempotence); $(ii)$ For all
$(\lambda,\mu)\in K\times K$ we have
$\lambda\boxplus_{K}\mu=\mu\boxplus_{K}\lambda$ (commutativity); $(iii)$ There
exists a neutral element $0_{K}\in K$ such that
$\lambda\boxplus_{K}0_{K}=0_{K}\boxplus_{K}\lambda=\lambda$ for all
$\lambda\in K$; $(iv)$ Every $\lambda\in K$ has a symmetrical element
$-\lambda$ such that
$(-\lambda)\boxplus_{K}\lambda=\lambda\boxplus_{K}(-\lambda)=0_{X}$
(symmetry). $(v)$ For any mutually non-symmetric triple of elements
$(\lambda,\mu,\eta)\in K^{3}$,
$(\lambda\boxplus_{K}\mu)\boxplus_{K}\eta=\lambda\boxplus_{K}(\mu\boxplus_{K}\eta)$
(weakened form of associativity).

$(b)$ There exists a scalar multiplication $\cdot_{K}:K\times K\longrightarrow
K$ that satisfies the following properties: $(i)$ For all $\lambda\in K$ and
all $(\lambda,\mu)\in K\times K$, we have
$\lambda\cdot_{K}\mu=\mu\cdot_{K}\lambda$ (commutativity); $(ii)$ For all
$\lambda,\mu,\eta\in K\times K\times K$ we have
$(\lambda\cdot_{K}\mu)\cdot_{K}\eta=\lambda\cdot_{K}(\mu\cdot_{K}\eta)$
(associativity); $(iii)$ There is an element $1_{K}\in K$ such that for all
$\lambda\in K$ we have $1_{K}\cdot_{K}\lambda=\lambda$; $(iv)$ For all
$\lambda\in K\backslash\\{0_{K}\\}$ there exists an element $\lambda^{-1}$
such that $\lambda\cdot_{K}\lambda^{-1}=1_{K}$.

$(c)$ The multiplication $\cdot_{K}$ is distributive over the operation
$\boxplus_{K}$: for all $\lambda\in K$ and all $(\mu,\nu)\in K\times K$, we
have

$\lambda\cdot_{K}(\mu\boxplus_{K}\nu)=(\lambda\cdot_{K}\mu)\boxplus_{K}(\lambda\cdot_{K}\nu).$

###### Example 2.1.2

From [7], the triple $(\mathbb{R},\boxplus,\cdot)$, where $\boxplus$ is the
binary operation defined in (4) is an idempotent pseudo-field.

In the following, when we use the $\boxplus$ symbol without further
specification on $\mathbb{R}^{n}$ this will mean that we are considering the
binary operation defined in equation (4) and its related $n$-ary extension.

###### Example 2.1.3

We consider the case of the non-associative symmetrisation of the Max-Plus
semi-module analysed in [10]. Let $\mathbb{M}=\mathbb{R}\cup\\{-\infty\\}$ and
let us denote $(\mathbb{M},\oplus,\otimes)$ the Maslov’s semi-module where we
replace the operations $\vee$ with $\oplus$ and $+$ with $\otimes$. More
precisely one can define on $\mathbb{M}$ the operations $\oplus$ and $\otimes$
respectively as $x\oplus y=\max\\{x,y\\}$ and $x\otimes y=x+y$, where
$-\infty$ is the neutral element of the operation $\oplus$. Suppose now that
$x\in\mathbb{R}_{-}$ and let us extend the logarithm function to the whole set
of real numbers. This we do by introducing the set

$\widetilde{\mathbb{M}}=\mathbb{M}\cup(\mathbb{R}+\mathrm{i}\pi)$ (30)

where $\mathrm{i}$ is the complex number such that $\mathrm{i}^{2}=-1$ and
$\mathbb{R}+\mathrm{i}\pi=\\{x+\mathrm{i}\pi:x\in\mathbb{R}\\}$. In the
following we extend the logarithmic function to $\widetilde{\mathbb{M}}$. Let
$\psi_{\ln}:{\mathbb{R}}\longrightarrow\widetilde{\mathbb{M}}$ be the map
defined by:

$\psi_{\ln}(x)=\left\\{\begin{matrix}\ln(x)&\text{ if }x>0\\\ -\infty&\text{
if }x=0\\\ \ln(-x)+\mathrm{i}\pi&\text{ if }x<0.\end{matrix}\right.$ (31)

The map $x\mapsto\psi_{\ln}(x)$ is an isomorphism from $\mathbb{\mathbb{R}}$
to $\widetilde{\mathbb{M}}$. Let
$\psi_{\exp}(x):\widetilde{\mathbb{M}}\longrightarrow\mathbb{\mathbb{R}}$ be
its inverse. Notice that $\psi_{\ln}(-1)=\mathrm{i}\pi$. By definition we have
$z\widetilde{\boxplus}u=\psi_{\ln}\big{(}\psi_{\exp}(z)\boxplus\psi_{\exp}(u)\big{)}$.
Moreover, we have
$z\widetilde{\otimes}u=\psi_{\ln}\big{(}\psi_{\exp}(z)\otimes\psi_{\exp}(u)\big{)}$.
It follows that $(\widetilde{\mathbb{M}},\tilde{\oplus},\tilde{\otimes})$ is
isomorphic to $(\mathbb{R},\boxplus,\cdot)$ and is also an idempotent pseudo-
field with the neutral elements $0_{\tilde{\mathbb{M}}}=-\infty$ and
$1_{\tilde{\mathbb{M}}}=0$.

In Chapter 3 of [19] it was shown that Max-Plus algebra can also be viewed as
a limit algebraic structure via the dequantization principle. For all
$z\in\widetilde{\mathbb{M}}^{n}$, let us denote:

$\widetilde{\operatorname*{\text{\Huge${}_{\boxplus}$}}_{i\in[n]}}z_{i}=\psi_{\ln}\big{(}\operatorname*{\text{\Huge${}_{\boxplus}$}}_{i\in[n]}\psi_{\exp}(z_{i})\big{)}.$
(32)

From [10], it follows that for all $z\in\widetilde{\mathbb{M}}^{n}$, we have

$\widetilde{\operatorname*{\text{\Huge${}_{\boxplus}$}}_{i\in[n]}}z_{i}=\lim_{p\longrightarrow\infty}\frac{1}{2p+1}\psi_{\ln}\Big{(}\sum_{i\in[n]}\psi_{\exp}\big{(}(2p+1)z_{i}\big{)}\Big{)}.$

In the following, we give another example arising in a Min -Times context (see
[1] for more details. )

###### Example 2.1.4

Let us denote $\mathbb{R}^{-1}=\mathbb{R}\backslash\\{0\\}\cup\\{\infty\\}$.
Let us consider the binary operation
$\boxplus^{-1}:\mathbb{\mathbb{R}}^{-1}\times\mathbb{R}^{-1}\longrightarrow\mathbb{R}^{-1}$
defined as:

$x\boxplus^{-1}\\!y=\left\\{\begin{matrix}x&\text{if}&|x|<|y|\\\
y&\text{if}&|x|>|y|\\\ +\infty&\text{if}&x+y=0\\\
x&\text{if}&x=y.\end{matrix}\right.$ (33)

This operation is idempotent, has $+\infty$ as neutral element and is
symmetrical. Moreover, The standard product of non-zero numbers in
$\mathbb{R}^{-1}$ satisfies the desired condition.

###### Remark 2.1.5

In [27], an idempotent algebraic structure was derived from a suitable
dequantization principle defined over the complex scalar field. Let us
consider the binary operation
$\boxplus_{\mathbb{C}}:\mathbb{C}\times\mathbb{C}\longrightarrow\mathbb{C}$
defined as:

$z\boxplus_{\mathbb{C}}w=\left\\{\begin{matrix}z&\text{if}&|z|>|w|\\\
w&\text{if}&|z|<|w|\\\ 0&\text{if}&z+w=0\\\
\rho\frac{z+w}{|z+w|}&\text{if}&|z|=|w|=\rho>0.\end{matrix}\right.$ (34)

It is easy to check that this operation is idempotent, has $0$ as neutral
element and is symmetrical. Moreover, the standard product of complex numbers
fulfils all the desired conditions. In addition note that this product is
distributive over the operation $\boxplus_{\mathbb{C}}$. In particular, if
$|z|=|w|=\rho>0$ then we have for all $u\in\mathbb{C}$:

$u\cdot(z\boxplus_{\mathbb{C}}w)=u\rho\frac{z+w}{|z+w|}=\rho\frac{uz+uw}{|z+w|}=\rho|u|\frac{uz+uw}{|uz+uw|}=(uz)\boxplus_{\mathbb{C}}(uw).$

However, it does not satisfy the weakened form of associativity of an
idempotent pseudo-field.

In parallel with the definition of idempotent spaces provided in [20], we
consider the following definition of an idempotent symmetric space.

###### Definition 2.1.6

A triple $(X,\boxplus_{X},\cdot_{K})$ is an idempotent symmetric space defined
over a idempotent pseudo-field $K$ if:

$(a)$ There exits a binary operation $\boxplus_{X}:X\times X\longrightarrow X$
that satisfies the following properties: $(i)$ For all $x\in X$ we have
$x\boxplus_{X}x=x$; $(ii)$ For all $(x,y)\in X\times X$ we have
$x\boxplus_{X}y=y\boxplus_{X}x$; $(iii)$ There exists a neutral element
$0_{X}\in X$ such that $x\boxplus_{X}0_{X}=0_{X}\boxplus_{X}x$ for all $x\in
X$; $(iv)$ Every $x\in X$ has a symmetrical element $-x$ such that
$(-x)\boxplus_{X}x=x\boxplus_{X}(-x)=0_{X}$.

$(b)$ There exists a scalar multiplication $\cdot_{K}:\mathbb{R}\times
X\longrightarrow X$ that satisfies the following properties: $(i)$ For all
$\lambda\in\mathbb{R}$ and all $(x,y)\in X\times X$, we have
$\lambda\cdot_{K}(x\boxplus_{X}y)=(\lambda\cdot_{K}x)\boxplus_{X}(\lambda\cdot_{K}y)$;
$(ii)$ For all $\lambda,\mu\in\mathbb{R}\times\mathbb{R}$ we have
$(\lambda\boxplus_{K}\mu)\cdot_{K}x=(\lambda\cdot_{K}x)\boxplus_{X}(\mu\cdot_{K}x)$,
$(iii)$ For all $\lambda,\mu\in\mathbb{R}\times\mathbb{R}$ we have
$(\lambda\cdot_{K}\mu)\cdot_{K}x=\lambda\cdot_{K}(\mu\cdot_{K}x)$; $(iv)$ For
all $x\in X$ we have $1\cdot_{K}x=x$ and $-x=(-1)\cdot_{K}x$.

A subset $Y$ of $X$ is an idempotent symmetric subspace if for all $x,y\in Y$
and all $t\in\mathbb{R}$, $x\boxplus_{X}ty\in Y$. Notice that, in this
definition, associativity has been replaced with idempotence. Clearly,
$(\mathbb{R},\boxplus,\cdot)$ is an idempotent symmetric space.

###### Lemma 2.1.7

Suppose that $(X,\boxplus_{X},\cdot_{K})$ is an idempotent symmetric space
defined over an idempotent pseudo-field $K$. Then, for all $x\in X$, one has:

$(a)$ $\lambda\cdot_{K}0_{X}=0_{X}$

$(b)$ $0_{K}\cdot_{K}x=0_{X}$

$(c)$ $\big{(}(-y)\boxplus_{X}(-x)\big{)}\boxplus_{X}(x\boxplus_{X}y)=0_{X}$.

Proof: $(a)$ By definition, for all $x\in X$ and all $\lambda\in\mathbb{R}$,
$\lambda\cdot_{K}0_{X}=\lambda\cdot_{K}\big{(}x\boxplus_{X}(-x)\big{)}$ for
all $x\in X$. Using axioms $(b_{1})$, $(b_{2})$ and $(b_{4})$ yields
$\lambda\cdot_{K}0_{X}=(\lambda\cdot_{K}x)\boxplus_{X}(\lambda\cdot_{K}(-x))=(\lambda\cdot_{K}x)\boxplus_{X}\big{(}\lambda\cdot_{K}(-1)\cdot_{K}x)\big{)}=(\lambda\cdot_{K}x)\boxplus_{X}\big{(}(-1)\cdot_{K}\lambda\cdot_{K}x)\big{)}=(\lambda\cdot_{K}x)\boxplus_{X}\big{(}-(\lambda\cdot_{K}x)\big{)}=0_{X}$.
$(b)$ We have for all $x\in X$,
$0_{K}\cdot_{K}x=\big{(}1_{K}\boxplus(-1_{K})\big{)}x=(1_{K}\cdot_{K}x)\boxplus_{X}\big{(}(-1_{K})\cdot_{K}x\big{)}=x\boxplus_{X}(-x)=0_{X}$.
$(c)$ For all $x,y\in X$,
$\big{(}(-y)\boxplus_{X}(-x)\big{)}\boxplus_{X}(x\boxplus_{X}y)=\big{(}(-1_{K})\cdot_{K}y\boxplus_{X}(-1_{K})\cdot
x\big{)}\boxplus_{X}(x\boxplus_{X}y)=\big{(}(-1_{K})\cdot(y\boxplus_{X}x)\big{)}\boxplus_{X}(x\boxplus_{X}y)=-(x\boxplus_{X}y)\boxplus_{X}(x\boxplus_{X}y)=0_{X}$.$\Box$

###### Proposition 2.1.8

Let $X$ be an idempotent symmetric space. The intersection of two idempotent
symmetric subspaces of $X$ is an idempotent symmetric subspace.

###### Example 2.1.9

$(\mathbb{R}^{n},\boxplus,\cdot)$ is an idempotent symmetric space.
$Y=\\{(x_{1},x_{2},x_{3},x_{4})\in\mathbb{R}^{4}:x_{1}=x_{2},x_{3}=x_{4}\\}$
is an idempotent symmetric subspace of $\mathbb{R}^{4}$.

###### Example 2.1.10

$(\widetilde{\mathbb{M}}^{n},\tilde{\boxplus},\tilde{\otimes})$ is an
idempotent symmetric space.

###### Example 2.1.11

$(\mathbb{R}^{-n},\boxplus^{-1},\cdot)$ is an idempotent symmetric space, with
$\mathbb{R}^{-n}=\big{(}\mathbb{R}^{-1}\big{)}^{n}$.

###### Example 2.1.12

Suppose that $I$ is a subset of $\mathbb{R}$ and let $\lambda$ be the Lebesgue
measure. Let $L_{p}(\lambda)$ be the set of the real valued maps
$f:I\longrightarrow\mathbb{R}$ such that

$\int_{I}|f|^{p}d\lambda<+\infty$

for $p\in]1,+\infty[$. Let us consider the binary operation defined over
$L_{p}(\lambda)$ by:

$f\boxplus g:x\mapsto f(x)\boxplus g(x)=\left\\{\begin{matrix}f(x)\ &\hbox{ if
}&|f(x)|&>&|f(y)|\\\ \frac{1}{2}\big{(}f(x)+f(y)\big{)}&\hbox{ if
}&|f(x)|&=&|f(y)|\\\ f(y)&\hbox{ if }&|f(x)|&<&|f(y)|.\end{matrix}\right.$
(35)

Clearly, one has $|f(x)\boxplus g(x)|\leq|f(x)|+|g(x)|$. It follows that

$\int_{I}|f\boxplus
g|^{p}d\lambda<\int_{I}|f|^{p}d\lambda+\int_{I}|g|^{p}d\lambda<\infty.$

Let $\cdot$ be the standard product of a real valued function by a real
scalar, it follows that $(L_{p}(\lambda),\boxplus,\cdot)$ is an idempotent
symmetric space.

###### Remark 2.1.13

Notice that that Examples 2.1.10 and 2.1.11 are constructed from an
homeomorphic transformation of $(\mathbb{R}^{n},\boxplus,\cdot)$. It is
therefore easy to define two convex structures on
$(\widetilde{\mathbb{M}}^{n},\tilde{\boxplus},\tilde{\otimes})$ and
$(\mathbb{R}^{-n},\boxplus^{-1},\cdot)$ respectively. Namely a subset $M$ of
$\widetilde{\mathbb{M}}$ is $\mathbb{M}$-convex if for all $x,y\in M$ and all
$t\in\mathbb{M}$, $x\widetilde{\boxplus}ty\in C$. Paralleling [1], a subset
$C$ of $\mathbb{R}^{-n}$ is inverse-$\mathbb{B}$-convex if for all $x,y\in C$
and all $t\in[1,+\infty]$, $x\boxplus^{-1}ty\in C.$

The next result sepfically concerns the algebraic surture considered in [7].
However it has some immediate implications for the idempotent symmetrical
spaces defined in examples 2.1.10 and 2.1.11.

###### Proposition 2.1.14

For all subsets $Y$ of $\mathbb{R}^{n}$, the following claims are equivalent:

$(a)$ $Y$ is an idempotent symmetric subspace of
$(\mathbb{R}^{n},\boxplus,\cdot)$.

$(b)$ For all $(x_{1},...,x_{m})\in Y^{m}$ and all $t\in\mathbb{R}^{m}$, we
have
$\operatorname*{\text{\Huge${}_{\boxplus}$}}\limits_{i\in[m]}t_{i}x_{i}\in Y.$

Proof: Let us prove that $(a)$ implies $(b)$. If $Y$ is an idempotent
symmetric subspace, this property is true for $m=2$. Suppose it is true at
rank $m$ and let us prove that it is true at rank $m+1$. In other words,
assume that for all $(x_{1},...,x_{m})\in Y^{m}$ we have:
$\operatorname*{\text{\Huge${}_{\boxplus}$}}_{i\in[m]}t_{i}x_{i}\in Y$ for all
$t\in\mathbb{R}^{m}$, we need to prove that if $(x_{1},...,x_{m},x_{m+1})\in
Y^{m+1}$ then for all $t\in\mathbb{R}^{m+1}$ we have
$\operatorname*{\text{\Huge${}_{\boxplus}$}}_{i\in[m+1]}t_{i}x_{i}\in Y$. To
establish this property, we use Equation (10) which implies that if
$(x_{1},...,x_{m},x_{m+1})\in Y^{m+1}$ then, for all $t\in\mathbb{R}^{m+1}$,
we have

$\operatorname*{\text{\Huge${}_{\boxplus}$}}_{i\in[m+1]}t_{i}x_{i}=\operatorname*{\text{\Huge${}_{\boxplus}$}}_{i\in[m+1]}\Big{[}t_{i}x_{i}\boxplus\Big{(}\operatorname*{\text{\Huge${}_{\boxplus}$}}_{j\in[m+1]\backslash\\{i\\}}t_{j}x_{j}\Big{)}\Big{]}.\quad(\star)$

For all $i$, the vectors
$z_{i}=t_{i}x_{i}\boxplus\Big{(}\operatorname*{\text{\Huge${}_{\boxplus}$}}_{j\in[m+1]\backslash\\{i\\}}t_{j}x_{j}\Big{)}$
are copositive. Thus the associativity holds true for the $m$-tuple
$(z_{1},...,z_{m})$. However since we assume that the property holds true at
rank $m$, $z_{i}\in Y$ for all $i$. Thus we deduce that
$\operatorname*{\text{\Huge${}_{\boxplus}$}}_{i\in[m+1]}t_{i}x_{i}\in Y$.
Therefore, the first part of the proof is proven. Since $Y$ is a subset of
$\mathbb{R}^{n}$ that is an idempotent symmetric space the converse inclusion
is immediate. $\Box$

The following property is an immediate consequence.

###### Proposition 2.1.15

Let $Y$ be an idempotent symmetric subspace of
$(\mathbb{R}^{n},\boxplus,\cdot)$. Then $Y$ is $\mathbb{B}^{\sharp}$-convex.

###  2.2  Generalized Norms and Ultra-Metric in Limit.

Let $\|\cdot\|$ be the Euclidean norm defined for all $x\in\mathbb{R}^{n}$ as
$\|x\|=\big{(}\sum_{i\in[n]}{x_{i}}^{2}\big{)}^{\frac{1}{2}}$. Let $d$ be the
Euclidean distance defined for all $x,y\in\mathbb{R}^{n}$ as $d(x,y)=\|x-y\|$.
For all positive natural numbers $p$, we introduce the map
$\|\cdot\|_{\varphi,p}:\mathbb{R}^{n}\longrightarrow\mathbb{R}_{+}$ defined as

$\|x\|_{\varphi,p}=\varphi_{p}^{-1}\big{(}\|\phi_{p}(x)\|\big{)}.$ (36)

The following properties are immediate: $\|x\|_{\varphi,p}=0\iff x=0$;
$\|\alpha x\|_{\varphi,p}=|\alpha|\|x\|_{\varphi,p}$;
$\|x\stackrel{{\scriptstyle
p}}{{+}}y\|_{\varphi,p}\leq\|x\|_{\varphi,p}\stackrel{{\scriptstyle
p}}{{+}}\|y\|_{\varphi,p}$. We deduce the following statements:

###### Proposition 2.2.1

For all $x,y\in\mathbb{R}^{n}$ and all $\alpha\in\mathbb{R}$, we have the the
following properties:
$(a)$ $\|x\|_{\infty}=\lim_{p\longrightarrow\infty}\|x\|_{\varphi,p}$
$(b)$ $\|x\boxplus y\|_{\infty}\leq\max\\{\|x\|_{\infty},\|y\|_{\infty}\\}$.

Proof: $(a)$ For all $x\in\mathbb{R}^{n}$, we have

$\|x\|_{\varphi,p}=\Big{(}\sum_{i\in[n]}|x_{i}|^{2(2p+1)}\Big{)}^{\frac{1}{2(2p+1)}}.$

Taking the limit yields $(a)$. $(b)$ For all natural numbers $p$, we have:

$\|x\stackrel{{\scriptstyle
p}}{{+}}y\|_{\varphi,p}\leq\|x\|_{\varphi,p}\stackrel{{\scriptstyle
p}}{{+}}\|y\|_{\varphi,p}.\quad(A)$

Suppose that $\\{z^{(p)}\\}_{p\in\mathbb{N}}$ is a sequence of
$\mathbb{R}_{+}^{n}$ which converges to some $z\in\mathbb{R}_{+}^{n}$. It
follows that we have
$\lim_{p\longrightarrow\infty}\|z^{(p)}\|_{\varphi,p}=\|z\|_{\infty}$. For all
$x\in\mathbb{R}^{n}$, let us denote $|x|=(|x_{1}|,...,|x_{n}|)$. By
definition, we have

$\|x\stackrel{{\scriptstyle
p}}{{+}}y\|_{\varphi,p}=\||x\stackrel{{\scriptstyle p}}{{+}}y|\|_{\varphi,p}.$

Moreover, $\lim_{p\longrightarrow\infty}|x\stackrel{{\scriptstyle
p}}{{+}}y|=|x\boxplus y|$. It follows that:

$\lim_{p\longrightarrow\infty}\|x\stackrel{{\scriptstyle
p}}{{+}}y\|_{\varphi,p}=\||x\boxplus y|\|_{\infty}=\|x\boxplus y\|_{\infty}.$

Since $\|x\|_{\varphi,p}\geq 0$ and $\|y\|_{\varphi,p}\geq 0$ taking the limit
on both sides in $(A)$ yields the result. $\Box$

In the following, for all $\alpha,\beta\in\mathbb{R}$, we adopt the notation
$\alpha\boxminus\beta=\alpha\boxplus(-\beta)$. Let us consider the map
$d_{\boxplus}:\mathbb{R}^{n}\times\mathbb{R}^{n}\rightarrow\mathbb{R}_{+}$
defined as:

$d_{\boxplus}(x,y)=\max_{i\in[n]}|x_{i}\boxminus y_{i}|$ (37)

In addition, let us consider the map
$d_{\varphi,p}:\mathbb{R}^{n}\times\mathbb{R}^{n}\longrightarrow\mathbb{R}_{+}$
defined by:

$d_{\varphi,p}(x,y)=\|x\stackrel{{\scriptstyle p}}{{-}}y\|_{\varphi,p}.$ (38)

This function is a distance relative to the algebraic structure obtained from
the binary operation $\stackrel{{\scriptstyle p}}{{+}}$. It defines a metric
on any vector space $\varphi_{p}$.

###### Lemma 2.2.2

For all $x,y\in\mathbb{R}^{n}$, we have

$\lim_{p\longrightarrow\infty}d_{\varphi,p}(x,y)=d_{\boxplus}(x,y).$

Proof: For all $x\in\mathbb{R}^{n}$, we have

$\|x\stackrel{{\scriptstyle
p}}{{-}}y\|_{\varphi,p}=\Big{(}\sum_{i\in[n]}\Big{(}\Big{|}\big{(}{x_{i}}^{2p+1}-{y_{i}}^{2p+1}\big{)}^{\frac{1}{2p+1}}\Big{|}^{2}\Big{)}^{2p+1}\Big{)}^{\frac{1}{2(2p+1)}}.$

For all $i\in[n]$ and all natural numbers $p$, set
$z_{i}^{(p)}=\big{|}\big{(}{x_{i}}^{2p+1}-{y_{i}}^{2p+1}\big{)}^{\frac{1}{2p+1}}\big{|}^{2}$.
For all $i$, we have:
$\lim_{p\longrightarrow\infty}z_{i}^{(p)}=|x_{i}\boxminus y_{i}|^{2}$. Since
for all $p$, $z_{i}^{(p)}\geq 0$, it follows that

$\lim_{p\longrightarrow\infty}\|x\stackrel{{\scriptstyle
p}}{{-}}y\|_{\varphi,p}=\Big{(}\max_{i\in[n]}|x_{i}\boxminus
y_{i}|^{2}\Big{)}^{\frac{1}{2}}=d_{\boxplus}(x,y).\quad\Box$

###### Proposition 2.2.3

The map
$d_{\boxplus}:\mathbb{R}^{n}\times\mathbb{R}^{n}\longrightarrow\mathbb{R}_{+}$
defines an ultra-metric on $\mathbb{R}^{n}$. For all $x,y,z\in\mathbb{R}^{n}$:
$(1)$ $d_{\boxplus}(x,y)=d_{\boxplus}(y,x)$;
$(2)$ $d_{\boxplus}(x,y)=0$ if and only if $x=y$;
$(3)$ $d_{\boxplus}(x,y)\leq\max\\{d_{\boxplus}(x,z),d_{\boxplus}(z,y)\\}.$

Proof: $(1)$ is immediate. $(2)$ follows from the fact that for all $i$ we
have $x_{i}\boxminus y_{i}=0\iff x_{i}=-y_{i}$. $(3)$ For all
$x,y,z\in\mathbb{R}^{n}$ and for all natural numbers $p$, we have

$d_{\varphi,p}(x,y)\leq d_{\varphi,p}(x,z)\stackrel{{\scriptstyle
p}}{{+}}d_{\varphi,p}(z,y).$

Since for all $x,y\in\mathbb{R}^{n}$ we have $d_{\varphi,p}(x,y)\geq 0$ and
$\lim_{p\longrightarrow\infty}d_{\varphi,p}(x,y)=d_{\boxplus}(x,y)$, we deduce
the result. $\Box$

Note that any ultrametric defines a metric since the inequality
$d(x,y)\leq\max\\{d(x,z),d(y,z)\\}$ implies the triangular inequality
$d(x,y)\leq d(x,z)+d(y,z)$. In particular for all $x,y,z\in E$, at least one
of the three equalities $d(x,y)=d(y,z)$ or $d(x,z)=d(y,z)$ or $d(x,y)=d(z,x)$
holds. From the above definition, one can conclude several typical properties
of ultrametrics.

###### Proposition 2.2.4

Let $Y$ be an idempotent symmetric subspace of $\mathbb{R}^{n}$.
$(Y,d_{\boxplus})$ is an ultra-metric space.

Proof: Since $Y$ is an idempotent symmetric space the metric $d_{\boxplus}$ is
well defined on $Y$. From Proposition 2.2.3, the result is immediate. $\Box$

In the following we say that a sequence $\\{x^{(k)}\\}_{k\in\mathbb{N}}$ is
$\digamma$-convergent to $x\in\mathbb{R}^{n}$ if
$\lim_{k\longrightarrow\infty}d_{\boxplus}(x^{(k)},x)=0$.

###### Lemma 2.2.5

Suppose that $\\{x^{(k)}\\}_{k\in\mathbb{N}}$ is $\digamma$-convergent to
$x\in\mathbb{R}^{n}$. Let $I(x)=\\{i\in[n]:x_{i}\not=0\\}$. Then there exists
some natural number $k_{0}$ such that for all $k\geq k_{0}$ and for all $i\in
I(x)$ we have $x_{i}^{(k)}=x_{i}$. Moreover, for all $i\notin I(x)$,
$\lim_{k\longrightarrow\infty}x_{i}^{(k)}=0$.

Proof: Suppose that
$\lim_{k\longrightarrow\infty}d_{\boxplus}(x_{i}^{(k)},x)=0$. Assume that
there exits an increasing subsequence $\\{k_{q}\\}_{q\in\mathbb{N}}$ and some
$i_{0}\in I(x)$ such that $x_{i}^{(k_{q})}\not=x_{i}$ for all $q$ and let us
show a contradiction. In such a case for all $q\in\mathbb{N}$ we have
$|x_{i_{0}}^{(k_{q})}\boxminus
x_{i_{0}}|=\max\\{|x_{i_{0}}^{(k_{q})}|,|x_{i_{0}}|\\}\not=0$. This implies
that

$d_{\boxplus}(x^{(k_{q})},x)=\max_{i\in I(x)}\\{|x_{i}^{(k_{q})}\boxminus
x_{i}|\\}\geq|x_{i_{0}}^{(k_{q})}\boxminus
x_{i_{0}}|=\max\\{|x_{i_{0}}^{(k_{q})}|,|x_{i_{0}}|\\}>0$

that is a contradiction. Hence, there is some $k_{0}$ such that for all $k\geq
k_{0}$ and for all $i\in I(x)$ we have $x_{i}^{(k)}=x_{i}$. If $i\notin I(x)$
then $|x_{i}^{(k)}\boxminus x_{i}|=|x_{i}^{(k)}\boxminus 0|=|x_{i}^{(k)}|$ and
$\lim_{k\longrightarrow\infty}d_{\boxplus}(x^{(k)},x)=0$. Therefore
$\lim_{k\longrightarrow\infty}x_{i}^{(k)}=0$. $\Box$

It is now interesting to see that the idempotent symmetrized space
$(\widetilde{\mathbb{M}}^{n},\tilde{\boxplus},\tilde{\otimes})$ can also be
endowed with a suitable form of ultrametric. Let
$d_{\widetilde{\boxplus}}:\widetilde{\mathbb{M}}^{n}\times\widetilde{\mathbb{M}}^{n}\longrightarrow\mathbb{R}\cup\\{-\infty\\}$
defined as:

$d_{\widetilde{\boxplus}}(z,w)=\widetilde{\operatorname*{\text{\Huge${}_{\boxplus}$}}_{i\in[n]}}\big{|}z_{i}\widetilde{\boxminus}w_{i}\big{|}_{\widetilde{\mathbb{M}}},$
(39)

where for all $\gamma\in\widetilde{\mathbb{M}}$,
$|\gamma\big{|}_{\widetilde{\mathbb{M}}}=\psi_{\ln}\big{(}|\psi_{\exp}(\gamma)|\big{)}$
and for all $i$
$z_{i}\widetilde{\boxminus}w_{i}=\psi_{\ln}\Big{(}\psi_{\exp}(z_{i})\boxminus\psi_{\exp}(w_{i})\Big{)}$.
It is therefore easy to check that: $(i)$ For all
$(z,w)\in\widetilde{\mathbb{M}}^{n}\times\widetilde{\mathbb{M}}^{n}$,
$d_{\widetilde{\boxplus}}(z,w)=-\infty\iff z=w$; $(ii)$ For all
$z,t,w\in\widetilde{\mathbb{M}}^{n}$,
$d_{\widetilde{\boxplus}}(z,w)\leq\max\\{d_{\widetilde{\boxplus}}(z,t),d_{\widetilde{\boxplus}}(t,w)\\}$.
Therefore, paralleling Proposition 2.2.4,
$(\widetilde{\mathbb{M}},\tilde{\boxplus},\tilde{\otimes})$ can also be
endowed with a topological structure. In particular, we can construct a
standard ultrametric by defining for all $z,w\in\widetilde{\mathbb{M}}^{n}$
the distance
$\bar{d}_{\widetilde{\boxplus}}(z,w)=\psi_{\exp}\Big{(}d_{\widetilde{\boxplus}}(z,w)\Big{)}$.

###### Proposition 2.2.6

Let $Z$ be an idempotent symmetric subspace of $\widetilde{\mathbb{M}}^{n}$.
Then $(Z,\bar{d}_{\widetilde{\boxplus}})$ is an ultra-metric space.

###  2.3 Geometry of the Ultrametric Ball

In the following, let
$B_{\boxplus}(x,r)=\\{z\in\mathbb{R}^{n}:d_{\boxplus}(x,z)\leq r\\}$ and
$B_{\boxplus}(x,r]=\\{z\in\mathbb{R}^{n}:d_{\boxplus}(x,z)\leq r\\}$
respectively define the open and closed ball centered at $x$ of radius $r$.

For all $x,y\in\mathbb{R}^{n}$ and all subsets $I$ of $[n]$, let us consider
the map $\mathcal{A}_{x}:I\longrightarrow\mathbb{R}_{+}$ defined for all $i\in
I$ by $\mathcal{A}_{x}(i)=|x_{i}|.$ This map associates to each index $i$ the
absolute value of the $i$-th coordinate of $x$. For all
$\alpha\in\mathbb{R}_{+}$, $\mathcal{A}_{x}^{-1}(\alpha)=\\{i\in
I:|x_{i}|=\alpha\\}$.

For all $x,y\in\mathbb{R}^{n}$ let $\mathcal{L}(x,y)$ be a subset of $I$
defined by

$\mathcal{L}(x,y)=\Big{\\{}i\in[n]:x_{i}\not=y_{i}\Big{\\}}.$ (40)

$\mathcal{L}(x,y)$ is obtained by dropping from $[n]$ all the $i$’s such that
$x_{i}=y_{i}$. It follows that

$d_{\boxplus}(x,y)=\left\\{\begin{matrix}\max\limits_{i\in\mathcal{L}(x,y)}|x_{i}|&\text{
if
}&\max\limits_{i\in\mathcal{L}(x,y)}|x_{i}|\geq\max\limits_{i\in\mathcal{L}(x,y)}|y_{i}|\\\
\max\limits_{i\in\mathcal{L}(x,y)}|y_{i}|&\text{ if
}&\max\limits_{i\in\mathcal{L}(x,y)}|x_{i}|\leq\max\limits_{i\in\mathcal{L}(x,y)}|y_{i}|\\\
0&\text{ if }&x=y.\end{matrix}\right.$ (41)

In the following, we say that $x$ is subordinate to $y$ for the absolute value
if
$\max\limits_{i\in\mathcal{L}(x,y)}|x_{i}|<\max\limits_{i\in\mathcal{L}(x,y)}|y_{i}|$.
If $x$ and $y$ are mutually not subordinate, then there are
$j,k\in[n]\backslash\mathcal{L}(x,y)$ such that $|x_{j}|=|y_{k}|$. The number
of elements of the range of $[n]$ that is
$\mathcal{A}_{x}\big{(}\,[n]\,\big{)}$, is smaller than $n$ since many
components of $x$ may have the same value. In the next statement we describe
the geometry of any ball centered at $x\in\mathbb{R}^{n}$.

###### Proposition 2.3.1

Let $x\in\mathbb{R}^{n}$. Let
$\mathcal{A}_{x}(\,[n]\,)=\\{\alpha_{j}\\}_{j\in[m]}$ with
$\alpha_{j}<\alpha_{j+1}$ for all $j\in[m-1]$ and $\alpha_{j}\geq 0$ for all
$j$.

$(i)$ If $x=0$ and $\mathcal{A}_{x}(\,[n]\,)=\\{0\\}$, then for all
$\alpha\geq 0$, $B_{\boxplus}(0,\alpha]=B_{\infty}(0,\alpha]$.

$(ii)$ If either $\alpha=0$ or $\alpha\in[0,\alpha_{1}[$ then
$B_{\boxplus}(x,\alpha]=\\{x\\}$.

$(iii)$ If $\alpha\in[\alpha_{j},\alpha_{j+1}[$ then $z\in
B_{\boxplus}(x,\alpha]$ if and only if:

$|z_{k}|\leq\alpha,\text{ for all
}k\in\mathcal{A}_{x}^{-1}\big{(}[0,\alpha_{j}]\big{)}\text{ and
}z_{k}=x_{k}\text{ for all
}k\in\mathcal{A}_{x}^{-1}\big{(}[\alpha_{j+1},\alpha_{m}]\big{)}.$

$(iv)$ If $\alpha\geq\alpha_{m}=\|x\|_{\infty}$ then
$B_{\boxplus}(x,\alpha]=B_{\infty}(0,\alpha].$

Proof:  $(i)$ is immediate. $(ii)$ If $\alpha=0$, then
$B_{\boxplus}(x,0\,]=\\{x\\}$. If $\alpha_{1}=0$ the case $\alpha<\alpha_{1}$
is excluded. Suppose that $\alpha_{1}>0.$ For all $z\not=x$,
$\max_{i\in[n]}|x_{i}\boxplus(-z_{i})|\geq\alpha_{1}$. Therefore
$\alpha<\alpha_{1}$ implies that $B_{\boxplus}(x,\alpha\,]=\\{x\\}$. $(iii)$
Suppose that $\alpha\in[\alpha_{1},\alpha_{2}]$. If there is some
$i\in\mathcal{A}_{x}^{-1}\big{(}[\alpha_{2},\alpha_{m}]\big{)}$ such that
$z_{i}\not=x_{i}$, then $d_{\boxplus}(x,z)\geq\alpha_{2}>\alpha_{1}$.
Therefore we should have $z_{i}=x_{i}$ for all
$i\in\mathcal{A}_{x}^{-1}\big{(}[\alpha_{2},\alpha_{m}]\big{)}$. Moreover, for
all $i\in\mathcal{A}_{x}^{-1}\big{(}\\{\alpha_{1}\\}\big{)}$ we should have
$|z_{i}|\leq\alpha.$ Let us extend this result for an arbitrary $j\leq m-1$.
Suppose that $\alpha\in[\alpha_{j},\alpha_{j+1}]$. If there is some
$i\in\mathcal{A}_{x}^{-1}\big{(}[\alpha_{j+1},\alpha_{m}]\big{)}$ such that
$z_{i}\not=x_{i}$, then $d_{\boxplus}(x,z)\geq\alpha_{j+1}>\alpha_{j}$.
Therefore we should have $z_{i}=x_{i}$ for all
$i\in\mathcal{A}_{x}^{-1}\big{(}[\alpha_{j+1},\alpha_{m}]\big{)}$. Moreover,
for all $i\in\mathcal{A}_{x}^{-1}\big{(}[\alpha_{1},\alpha_{j}]\big{)}$ we
should have $|z_{i}|\leq\alpha.$ $(iv)$ Suppose now that
$\alpha\geq\|x\|_{\infty}$. $z\in B_{\boxplus}(x,\alpha]$ if and only if
$|z_{i}|\leq\|x\|_{\infty}$, which ends the proof. $\Box$

###### Corollary 2.3.2

For all $\alpha\geq 0$, and all $x\in\mathbb{R}^{n}$, $B_{\boxplus}(x,\alpha]$
is closed with respect to the norm topology.

Proof: Let $\mathcal{A}_{x}(\,[n]\,)=\\{\alpha_{j}\\}_{j\in[m]}$ with
$\alpha_{j}<\alpha_{j+1}$ for all $j\in[m-1]$ and $\alpha_{j}\geq 0$ for all
$j$. The result is immediate if either $\alpha=0$ or $\alpha\in[0,\alpha_{1}[$
since $B_{\boxplus}(x,\alpha]=\\{x\\}$. From Proposition 2.3.1, if
$\alpha\in[\alpha_{j},\alpha_{j+1}[$, then

$B_{\boxplus}(x,\alpha]=\Big{\\{}z\in\mathbb{R}^{n}\\!\\!\\!:|z_{k}|\leq\alpha,k\in\mathcal{A}_{x}^{-1}\big{(}[0,\alpha_{j}]\big{)}\Big{\\}}\cap\Big{\\{}z\in\mathbb{R}^{n}\\!\\!\\!:z_{k}=x_{k},k\in\mathcal{A}_{x}^{-1}\big{(}[\alpha_{j+1},\alpha_{m}]\big{)}\Big{\\}}.$

Since the intersection of two closed sets is a closed set, we deduce that
$B_{\boxplus}(x,\alpha]$ is closed. If $\alpha\geq\|x\|_{\infty}$, the result
is immediate. $\Box$

$0$$0$$0$$x_{1}$$x_{1}$$x_{1}$$x_{1}$$x_{2}$$x_{2}$$x_{2}$$x_{2}$$0\leq\alpha<2$$\alpha=3$$2\leq\alpha<3$$\alpha>3$$3$$3$$3$$3$$2$$2$$2$$2$$B_{\boxplus}\big{(}(3,2),\alpha\big{]}$$B_{\boxplus}\big{(}(3,2),\alpha\big{]}$$B_{\boxplus}\big{(}(3,2),\alpha\big{]}$$B_{\boxplus}\big{(}(3,2),\alpha\big{]}$Figure
2.1:Ultra-metric Ball.

###### Corollary 2.3.3

For all $\alpha\geq 0$, and all $x\in\mathbb{R}^{n}$, $B_{\boxplus}(x,\alpha]$
is $\mathbb{B}^{\sharp}$-convex.

Proof: From Proposition 2.3.1, $B_{\boxplus}(x,\alpha]$ is the cartesian
product between a box and a singleton. Since any box is
$\mathbb{B}^{\sharp}$-convex, the result immediately follows. $\Box$

In [7] the following formulation of a $\mathbb{B}$-polytope was established in
the case of two points.222We use the convention that for all
$(\alpha,\beta,\gamma,\delta)\in\mathbb{R}^{4}$,
$\alpha\boxplus\beta\boxplus\gamma\boxplus\delta=\digamma_{[4]}(\alpha,\beta,\gamma,\delta)$.
For all $x,y\in\mathbb{R}^{n}$,

$Co^{\infty}(x,y)=\Big{\\{}tx\boxplus rx\boxplus sy\boxplus
wy:\max\\{t,r,s,w\\}=1,t,r,s,w\geq 0\Big{\\}}.$ (42)

In classical convexity, taking into account the same point of a finite set
twice does not affect its convex hull. Here this is not the case due to the
particular properties of the non-associative operation we consider. This
formulation is useful to prove the next decomposition result over
$(\mathbb{R}^{n},\boxplus,\cdot)$.

###### Proposition 2.3.4

For all $x,y\in\mathbb{R}^{n}$, and all $z\in Co^{\infty}(x,y)$ we have:

$d_{\boxplus}(x,y)=\max\\{d_{\boxplus}(x,z),d_{\boxplus}(z,y)\\}.$

Proof: If $x=y$, the result is immediate. Let us assume that $x\not=y$.
Suppose that $z\in Co^{\infty}(x,y)$, in such a case there are some
$r,s,t,w\in[0,1]$ with $\max\\{r,s,t,w\\}=1$ such that

$z=rx\boxplus sx\boxplus ty\boxplus wy.$

Suppose that $\alpha=d_{\boxplus}(x,y)$. Let
$\mathcal{L}(x,y)=\\{i\in[n]:x_{i}\not=y_{i}\\}$. By construction

$d_{\boxplus}(x,y)=\max\\{|x_{i}|,|y_{i}|:i\in\mathcal{L}(x,y)\\}.$

For all $i\notin\mathcal{L}(x,y)$ if $|x_{i}|>d_{\boxplus}(x,y)$ then
$x_{i}=y_{i}$, thus $z_{i}=rx_{i}\boxplus sx_{i}\boxplus ty_{i}\boxplus
wy_{i}=x_{i}=y_{i}$. It follows that $x_{i}\boxminus z_{i}=y_{i}\boxminus
z_{i}=0$. Hence, we have:

$d_{\boxplus}(x,z)=\max\\{|x_{i}\boxminus z_{i}|:i\in\mathcal{L}(x,y)\\}$

and

$d_{\boxplus}(y,z)=\max\\{|y_{i}\boxminus z_{i}|:i\in\mathcal{L}(x,y)\\}.$

Now, notice that for all $i$

$|z_{i}|=|rx_{i}\boxplus sx_{i}\boxplus ty_{i}\boxplus
wy_{i}|\leq\max\\{|x_{i}|,|y_{i}|\\}$

Since, for all $i$, $|x_{i}\boxminus z_{i}|\leq\max\\{|x_{i}|,|y_{i}|\\}$ and
$|y_{i}\boxminus z_{i}|\leq\max\\{|x_{i}|,|y_{i}|\\}$, we deduce that

$d_{\boxplus}(x,z)\leq\max\\{|x_{i}|,|y_{i}|:i\in\mathcal{L}(x,y)\\}=d_{\boxplus}(x,y)$

and

$d_{\boxplus}(y,z)\leq\max\\{|x_{i}|,|y_{i}|:i\in\mathcal{L}(x,y)\\}=d_{\boxplus}(x,y).$

Thus

$\max\\{d_{\boxplus}(x,z),d_{\boxplus}(z,y)\\}\leq d_{\boxplus}(x,y).$

However, from the triangular inequality, we have

$d_{\boxplus}(x,y)\leq\max\\{d_{\boxplus}(x,z),d_{\boxplus}(z,y)\\}.$

Since the converse inequality holds, this completes the proof. $\Box$

The next result is an immediate consequence.

###### Proposition 2.3.5

Let $A=\\{x^{(1)},...,x^{(m)}\\}$ be a finite subset of $\mathbb{R}^{n}$.
Suppose moreover that $(i)$
$Co^{\infty}\big{(}x^{(1)},x^{(m)}\big{)}=\bigcup_{i\in[m-1]}Co^{\infty}\big{(}x^{(i)},x^{(i+1)}\big{)}$;
$(ii)$ for all $i$ $Co^{\infty}\big{(}x^{(i-1)},x^{(i)}\big{)}\cap
Co^{\infty}\big{(}x^{(i)},x^{(i+1)}\big{)}=\\{x^{(i)}\\}$. Then we have:

$d_{\boxplus}\big{(}x^{(1)},x^{(m)}\big{)}=\max_{i\in[m-1]}d_{\boxplus}\big{(}x^{(i)},x^{(i+1)}\big{)}.$

Proof: Since $x^{(2)}\in Co^{\infty}\big{(}x^{(1)},x^{(m)}\big{)}$, we have:

$d_{\boxplus}\big{(}x^{(1)},x^{(m)}\big{)}=\max\big{\\{}d_{\boxplus}\big{(}x^{(1)},x^{(2)}\big{)},d_{\boxplus}\big{(}x^{(2)},x^{(m)})\big{)}\big{\\}}.$

From the decomposition property established in [9] that
$Co^{\infty}\big{(}x^{(k)},x^{(m)})=Co^{\infty}\big{(}x^{(k)},x^{(k+1)})\cup
Co^{\infty}\big{(}x^{(k+1)},x^{(m)})$. It follows that for any $k\in[m-1]$,
$x^{(k+1)}\in Co^{\infty}\big{(}x^{(k)},x^{(m)}\big{)}$. Hence:

$d_{\boxplus}\big{(}x^{(k)},x^{(m)}\big{)}=\max\Big{\\{}d_{\boxplus}\big{(}x^{(k)},x^{(k+1)}\big{)},d_{\boxplus}\big{(}x^{(k+1)},x^{(m)}\big{)}\Big{\\}}.$

By recurrence the proof is then immediate. $\Box$

###### Proposition 2.3.6

Let $x,y\in\mathbb{R}^{n}$. If $x=-y$, then for all $z\in
Co^{\infty}(x,y)\backslash\\{x,y\\}$,
$d_{\boxplus}(x,y)=d_{\boxplus}(x,z)=d_{\boxplus}(z,y).$

Proof: Suppose that $z\in Co^{\infty}\big{(}x,y)$, in such a case there are
some $r,s,t,w\in[0,1]$ with $\max\\{r,s,t,w\\}=1$ such that $z=rx\boxplus
sx\boxplus ty\boxplus wy.$ Since $z\notin\\{x,y\\}$ we have
$\max\\{r,s\\}=\max\\{t,w\\}=1$, and $\min\\{r,s\\}=\max\\{t,w\\}=1$,
otherwise either $z=x$ or $z=y$. If $r=s=t=w=1$, then $z=0$ and the result is
immediate. Consequently $\max\\{r,s\\}=\max\\{t,w\\}=1$, $\min\\{r,s\\}<1$ and
$\min\\{t,w\\}<1$. Dropping the symmetrical terms, it follows that

$z=rx\boxplus sx\boxplus ty\boxplus wy=\min\\{r,s\\}x\boxplus\min\\{t,w\\}y.$

Since $\min\\{r,s\\}<1$ and $\min\\{t,w\\}<1$, this implies that
$d_{\boxplus}(x,y)=d_{\boxplus}(x,z)=d_{\boxplus}(z,y).$$\Box$

##  3  Some Trigonometric Properties in Limit

In this section, we show that certain properties associated with the notion of
orthogonality can be derived within the framework of the idempotent and
symmetrical algebraic structure that we consider.

###  3.1 Orthogonality and Pythagorean Properties in Limit

In the following, for all $p\in\mathbb{N}$, we say that $x,y\in\mathbb{R}^{n}$
are $\varphi_{p}$-orthogonal if:

$\langle x,y\rangle_{p}=0.$ (43)

In addition, we say that $x,y\in\mathbb{R}^{n}$ are $\digamma$-orthogonal if:

$\langle x,y\rangle_{\infty}=0.$ (44)

In the sequel, the earlier results are extended to the general case of a
triple $[x,y,z]$ of $\mathbb{R}^{n}$. Notice that we have for all natural
numbers $p$

$\big{\langle}x\stackrel{{\scriptstyle p}}{{-}}z,y\stackrel{{\scriptstyle
p}}{{-}}z\big{\rangle}_{p}=\stackrel{{\scriptstyle\varphi_{p}}}{{\sum_{i\in[n]}}}x_{i}y_{i}\stackrel{{\scriptstyle
p}}{{+}}\stackrel{{\scriptstyle\varphi_{p}}}{{\sum_{i\in[n]}}}(-x_{i}z_{i})\stackrel{{\scriptstyle
p}}{{+}}\stackrel{{\scriptstyle\varphi_{p}}}{{\sum_{i\in[n]}}}(-y_{i}z_{i})\stackrel{{\scriptstyle
p}}{{+}}\stackrel{{\scriptstyle\varphi_{p}}}{{\sum_{i\in[n]}}}{z_{i}}^{2}.$
(45)

Therefore:

$\displaystyle\lim$
${}_{p\longrightarrow\infty}\big{\langle}x\stackrel{{\scriptstyle
p}}{{-}}z,y\stackrel{{\scriptstyle
p}}{{-}}z\big{\rangle}_{p}=\digamma_{[4n]}\big{(}x_{1}y_{1},...,x_{n}y_{n},-x_{1}z_{1},...,-x_{n}z_{n},-y_{1}z_{1},...,-y_{n}z_{n},{z_{1}}^{2},...,{z_{n}}^{2}\big{)}.$
(46)

In the following, we introduce the operation
$\langle\\!\langle\cdot,\cdot,\cdot\rangle\\!\rangle_{\boxplus}:\mathbb{R}^{n}\times\mathbb{R}^{n}\times\mathbb{R}^{n}\longrightarrow\mathbb{R}$
defined as:

$\langle\\!\langle
x,y,z\rangle\\!\rangle_{\infty}=\lim_{p\longrightarrow\infty}\big{\langle}x\stackrel{{\scriptstyle
p}}{{-}}z,y\stackrel{{\scriptstyle p}}{{-}}z\big{\rangle}_{p}.$ (47)

For all $x,y,z\in\mathbb{R}^{n}$, $[x,y,z]$ is $\digamma$-right-angled in $z$
if

$\langle\\!\langle x,y,z\rangle\\!\rangle_{\infty}=0.$ (48)

In general, a right-angled triangle is not $\digamma$-right-angled triple.
This is shown in the next example.

###### Example 3.1.1

Let $x=(1,-2,-1)$,$y=(2,3,-2)$ and $z=(3,2,-3)$. We have $x-z=(-2,-4,2)$ and
$y-z=(-1,1,1)$. We have $\langle x-z,y-z\rangle=0$. Therefore $[x,y,z]$ is a
right-angle triangle.However $\digamma(2,-6,2,3,2,-3,-6,-6,-6,9,4,9)=9\not=0$.
Hence $[x,y,z]$ is not a $\digamma$-right-angle triple.

The next result was established in [9]. It plays an important role in the
remainder. It means that if the limit sum is zero then the
$\phi_{p}$-generalized sum is zero for any $p$.

###### Lemma 3.1.2

Suppose that there is some $x=(x_{1},...,x_{n})\in\mathbb{R}^{n}$ such that
$\operatorname*{\text{\Huge${}_{\boxplus}$}}_{i\in[n]}x_{i}=0$. Then for all
$p\in\mathbb{N}$,
$\stackrel{{\scriptstyle\varphi_{p}}}{{\sum_{i\in[n]}}}x_{i}=0$.

###### Proposition 3.1.3

Let $x,y,z\in\mathbb{R}^{n}$, and let $[x,y,z]$ be a triple. The triple
$[x,y,z]$ is $\digamma$-right-angled in $z$ all if and only if for all
$p\in\mathbb{N}$, $[x,y,z]$ is $\varphi_{p}$-right-angled in $z$, i.e.
$\big{\langle}x\stackrel{{\scriptstyle p}}{{-}}z,y\stackrel{{\scriptstyle
p}}{{-}}z\big{\rangle}_{p}=0$.

Proof: If $[x,y,z]$ is a $\digamma$-right-angled triple in $z$ then:

$\langle\\!\langle x,y,z\rangle\\!\rangle_{\infty}=0.$

It follows that

$\big{\langle}x\stackrel{{\scriptstyle p}}{{-}}z,y\stackrel{{\scriptstyle
p}}{{-}}z\big{\rangle}_{p}=0$

for all $p\in\mathbb{N}$ and the first implication follows.

Conversely if for all $p$, $[x,y,z]$ is a $\varphi_{p}$-right-angled triple in
$z$, then for all $p$

$\big{\langle}x\stackrel{{\scriptstyle p}}{{-}}z,y\stackrel{{\scriptstyle
p}}{{-}}z\big{\rangle}_{p}=0.$

Hence

$\lim_{p\longrightarrow\infty}\big{\langle}x\stackrel{{\scriptstyle
p}}{{-}}z,y\stackrel{{\scriptstyle
p}}{{-}}z\big{\rangle}_{p}=\langle\\!\langle
x,y,z\rangle\\!\rangle_{\infty}=0,$

which proves the reciprocal. $\Box$

$x_{1}$$x_{2}$$0$$a$$b$$c$$e$$d$$f$$h$$g$Figure 3.2.1.1.a:$\digamma$-right-
angled triple with convex lines.
$x_{1}$$x_{2}$$0$$a$$b$$c$$e$$d$$f$$h$$g$Figure 3.2.1.1.b:$\digamma$-right-
angled triple with $\mathbb{B}^{\sharp}$-convex lines.

###### Remark 3.1.4

A $\digamma$-right-angled triple is a right-angled triangle. However, the
converse is not true. In Figure 3.2.1.1.a, $[a,b,c]$ is right-angled but it is
not $\digamma$-right-angled. $[d,f,e]$ and $[g,0,h]$ are $\digamma$-right-
angled. In Figure 3.2.1.1.b they are represented with
$\mathbb{B}^{\sharp}$-convex lines. Clearly, the right-angle is only preserved
for the triples $[d,f,e]$ and $[g,0,h]$.

In the following, for all subsets $I$ of $[n]$ and all $x\in\mathbb{R}^{n}$,
let us denote $x_{[I]}=\sum_{i\in I}x_{i}e_{i}$, where $\\{e_{1},...,e_{n}\\}$
is the canonic basis of $\mathbb{R}^{n}$.

###### Proposition 3.1.5

Let $x,y,z\in\mathbb{R}^{n}$, and let $[x,y,0]$ is a $\digamma$-right-angled
triple in $0$. Then there exists a partition $\\{I_{1},...,I_{m},J\\}$ of
$[n]$ with $m=E[\frac{n}{2}]$,
$\operatorname*{\mathrm{Card}}\,(J)\in\\{0,1\\}$ and such that for all $k$:

$\Big{[}x_{[I_{k}]},y_{[I_{k}]},0_{[I_{k}]}\Big{]}$

is a right-angled triangle in $0_{[I_{k}]}$ of $\mathbb{R}^{I_{k}}$ whith
$\operatorname*{\mathrm{Card}}\,(I_{k})=2$. Moreover, if $J=\\{j\\}$ is
nonempty then $x_{j}y_{j}=0$.

Proof: If $[x,y,z]$ is a $\digamma$-right-angled triple in $0$ then:

$\operatorname*{\text{\Huge${}_{\boxplus}$}}_{i\in[n]}x_{i}y_{i}=0.$

$\operatorname*{\text{\Huge${}_{\boxplus}$}}_{i\in[n]}x_{i}y_{i}=0$ implies
that there exists a partition $\\{I_{1},...,I_{m},J\\}$ of $[n]$ with
$m=E[\frac{n}{2}]$, $\operatorname*{\mathrm{Card}}\,(I_{k})=2$ for all $k$ and
$\operatorname*{\mathrm{Card}}\,(J)\in\\{0,1\\}$ such that:

$\sum_{i\in[I_{k}]}w_{i}=\operatorname*{\text{\Huge${}_{\boxplus}$}}_{i\in[I_{k}]}x_{i}y_{i}=0,$

and $x_{j}y_{j}=0$ if $J=\\{j\\}$ is a nonempty set. Consequently

$\sum_{i\in[I_{k}]}x_{i}y_{i}=0.$

Suppose that $I_{k}=\\{i^{\prime},i^{\prime\prime}\\}$. Then

$x_{i^{\prime}}y_{i^{\prime}}\boxminus
x_{i^{\prime\prime}}y_{i^{\prime\prime}}=0\iff
x_{i^{\prime}}y_{i^{\prime}}-x_{i^{\prime\prime}}y_{i^{\prime\prime}}=0.$

It follows that $\langle x_{[I_{k}]},y_{[I_{k}]}\rangle_{p}=0$ for all $p.$
$\Box$

###### Corollary 3.1.6

Suppose that $n=2$. If $[x,y,0]$ is a right-angled triangle in $z$ of
$\mathbb{R}^{2}$, then $[x,y,0]$ is $\digamma$-right-angled in $0$ and it is
$\varphi_{p}$-right-angled in $0$ for all natural numbers $p$.

The following result is an idempotent and non-associative analogue of the
Pythagorean theorem.

###### Proposition 3.1.7

Let $x,y,z\in\mathbb{R}^{n}$, and let $[x,y,z]$ is a $\digamma$-right-angled
triple in $z$. Then for all $p\in\mathbb{N}$:

$d^{2}_{\varphi,p}(x,y)=d^{2}_{\varphi,p}(x,z)\stackrel{{\scriptstyle
p}}{{+}}d^{2}_{\varphi,p}(y,z),$

and

$d_{\boxplus}(x,y)=\max\\{d_{\boxplus}(x,z),d_{\boxplus}(y,z)\\}.$

Proof: If $[x,y,z]$ is a $\digamma$-right-angled triple in $z$ then

$\langle\\!\langle x,y,z\rangle\\!\rangle_{\infty}=0.$

Thus, for all $p$, $\langle x\stackrel{{\scriptstyle
p}}{{-}}z,y\stackrel{{\scriptstyle p}}{{-}}z\rangle_{p}=0$. It follows that

$\langle\phi_{p}(x)-\phi_{p}(z),\phi_{p}(y)-\phi_{p}(z)\rangle=0.$

Hence

$\|\phi_{p}(x)-\phi_{p}(y)\|^{2}=\|\phi_{p}(x)-\phi_{p}(z)\|^{2}+\|\phi_{p}(y)-\phi_{p}(z)\|^{2}.$

We deduce that

$\displaystyle\Big{\|}\phi_{p}\Big{(}\phi_{p}^{-1}\big{(}\phi_{p}(x)-\phi_{p}(y)\big{)}\Big{)}\Big{\|}^{2}=\varphi_{p}\Big{(}\varphi_{p}^{-1}(\|\phi_{p}\Big{(}\phi_{p}^{-1}\big{(}\phi_{p}(x)-\phi_{p}(z)\big{)}\Big{)}\|^{2})\Big{)}+\varphi_{p}\Big{(}\varphi_{p}^{-1}(\|\phi_{p}\Big{(}\phi_{p}^{-1}\big{(}\phi_{p}(y)-\phi_{p}(z)\big{)}\Big{)}\|^{2}\big{)}\Big{)}.$

Thus:

$d^{2}_{\varphi,p}(x,y)=\|x\stackrel{{\scriptstyle
p}}{{-}}y\|_{\varphi,p}^{2}=\|x\stackrel{{\scriptstyle
p}}{{-}}z\|_{\varphi,p}\stackrel{{\scriptstyle
p}}{{+}}\|y\stackrel{{\scriptstyle
p}}{{-}}z\|_{\varphi,p}=d^{2}_{\varphi,p}(x,z)\stackrel{{\scriptstyle
p}}{{+}}d^{2}_{\varphi,p}(y,z).$

Taking the limit, yields:

$d^{2}_{\boxplus}(x,y)=\lim_{p\longrightarrow\infty}d^{2}_{\varphi,p}(x,y)=\lim_{p\longrightarrow\infty}d^{2}_{\varphi,p}(x,z)\stackrel{{\scriptstyle
p}}{{+}}d^{2}_{\varphi,p}(y,z)=\max\\{d^{2}_{\boxplus}(x,z),d^{2}_{\boxplus}(y,z)\\}.$

The final statement immediately follows. $\Box$

###  3.2 Cosine, Sine in Limit and some Parametrization of the Unit Square

Paralleling the usual definition of a cosine, let us define the
$\varphi_{p}$-pseudo-cosine of the angle between two vectors as:

$\cos_{p}(x,y)=\frac{\langle
x,y\rangle_{p}}{\|x\|_{\varphi,p}\|y\|_{\varphi,p}}.$ (49)

Taking the limit we define the pseudo $\digamma$-cosine as:

$\cos_{\infty}(x,y)=\frac{\langle
x,y\rangle_{\infty}}{\|x\|_{\infty}\|y\|_{\infty}}.$ (50)

It is now natural to define a pseudo sine function. In [10] a $\varphi_{p}$
exterior product is defined for all $v_{1},v_{2},...,v_{n}\in\mathbb{R}^{n}$
as:

$\big{(}v_{1}\stackrel{{\scriptstyle p}}{{\wedge}}v_{2}\stackrel{{\scriptstyle
p}}{{\wedge}}\cdots\stackrel{{\scriptstyle
p}}{{\wedge}}v_{n}\big{)}=|v_{1},v_{2},...,v_{n}|_{p}\big{(}e_{1}{\wedge}e_{2}{\wedge}\cdots{\wedge}e_{n}\big{)}.$
(51)

where:

$\displaystyle\big{(}v_{1}\stackrel{{\scriptstyle
p}}{{\wedge}}v_{2}\stackrel{{\scriptstyle
p}}{{\wedge}}\cdots\stackrel{{\scriptstyle p}}{{\wedge}}v_{n}\big{)}$
$\displaystyle=\varphi_{p}^{-1}\big{(}\phi_{p}(v_{1}){\wedge}\phi_{p}(v_{2}){\wedge}\cdots{\wedge}\phi_{p}(v_{n})\
\big{)}$ (52)
$\displaystyle=\varphi_{p}^{-1}\Big{(}|\phi_{p}(v_{1}),\phi_{p}(v_{2}),...,\phi_{p}(v_{n})|\Big{)}\big{(}e_{1}{\wedge}e_{2}{\wedge}\cdots{\wedge}e_{n}\big{)}.$
(53)

Taking the limit

$\big{(}v_{1}\stackrel{{\scriptstyle\infty}}{{\wedge}}v_{2}\stackrel{{\scriptstyle\infty}}{{\wedge}}\cdots\stackrel{{\scriptstyle\infty}}{{\wedge}}v_{n}\big{)}=|v_{1},v_{2},...,v_{n}|_{\infty}\big{(}e_{1}\stackrel{{\scriptstyle\infty}}{{\wedge}}e_{2}\stackrel{{\scriptstyle\infty}}{{\wedge}}\cdots\stackrel{{\scriptstyle\infty}}{{\wedge}}e_{n}\big{)}.$
(54)

In the case of two vectors the exterior product can be identified to the
exterior product and the oriented area of the cross product can be interpreted
as the positive area of the parallelogram having $x$ and $y$ as sides:

${x\wedge y=\left|x,y\right|=\left\|x\right\|\left\|y\right\|\sin(x,y).}$ (55)

The $\varphi_{p}$-sine function for two vectors $x,y\in\mathbb{R}^{2}$ is
defined as

$\sin_{p}(x,y)=\frac{|x,y|_{p}}{{\|x\|_{\varphi,p}\|y\|_{\varphi,p}}}=\frac{\varphi_{p}^{-1}\big{(}\varphi_{p}(x_{1})\varphi_{p}(y_{2})-\varphi_{p}(x_{2})\varphi_{p}(y_{1})\big{)}}{{\|x\|_{\varphi,p}\|y\|_{\varphi,p}}}.$
(56)

Along this line, let us define

$\sin_{\infty}(x,y)=\frac{|x,y|_{\infty}}{\|x\|_{\infty}\|y\|_{\infty}}=\lim_{p\longrightarrow\infty}\sin_{p}(x,y)=\frac{x_{1}y_{2}\boxminus
x_{2}y_{1}}{\max\\{|x_{1},x_{2}|\\}\max\\{|y_{1},y_{2}|\\}}.$ (57)

###### Lemma 3.2.1

For all $x,y\in\mathbb{R}^{n}$, we have the following properties.
$(a)$
$\cos_{p}(x,y)=\varphi_{p}^{-1}\big{(}\cos\big{(}\phi_{p}(x),\phi_{p}(y))\big{)}$
and
$\sin_{p}(x,y)=\varphi_{p}^{-1}\big{(}\sin(\phi_{p}(x),\phi_{p}(y))\big{)}.$
$(b)$ ${\cos_{p}(x,y)}^{2}\stackrel{{\scriptstyle
p}}{{+}}{\sin_{p}(x,y)}^{2}=1$;
$(c)$ $\max\\{|\cos_{\infty}(x,y)|,|\sin_{\infty}(x,y)|\\}=1$.

Proof: $(a)$ The first statement is immediate. To prove the second part note
that:

$\sin_{p}(x,y)=\frac{|x,y|_{p}}{{\|x\|_{\varphi,p}\|y\|_{\varphi,p}}}=\frac{\varphi_{p}^{-1}\big{(}|\phi_{p}(x),\phi_{p}(y)|)}{{\varphi_{p}^{-1}(\|\phi_{p}(x)\|)\varphi_{p}^{-1}(\|\phi_{p}(y)\|)}}.$

Therefore

$\sin_{p}(x,y)=\varphi_{p}^{-1}\Big{(}\frac{|\phi_{p}(x),\phi_{p}(y)|}{{\|\phi_{p}(x)\|\|\phi_{p}(y)\|}}\Big{)}=\varphi_{p}^{-1}\big{(}\sin(\phi_{p}(x),\phi_{p}(y))\big{)}.$

$(b)$ We have

$\displaystyle{\cos_{p}(x,y)}^{2}\stackrel{{\scriptstyle
p}}{{+}}{\sin_{p}(x,y)}^{2}$
$\displaystyle=\varphi_{p}\Big{(}\big{[}\varphi_{p}^{-1}\big{(}\cos\big{(}\phi_{p}(x),\phi_{p}(y))\big{)}\big{]}^{2}\Big{)}+\varphi_{p}\Big{(}\big{[}\varphi_{p}^{-1}\big{(}\sin\big{(}\phi_{p}(x),\phi_{p}(y))\big{)}\big{]}^{2}\Big{)}$
$\displaystyle=\varphi_{p}\Big{(}\varphi_{p}^{-1}\big{[}\big{(}\cos\big{(}\phi_{p}(x),\phi_{p}(y))\big{]}^{2}\big{)}\Big{)}+\varphi_{p}\Big{(}\varphi_{p}^{-1}\big{(}\big{[}\sin\big{(}\phi_{p}(x),\phi_{p}(y))\big{]}^{2}\big{)}\Big{)}$
$\displaystyle=\big{(}\cos\big{(}\phi_{p}(x),\phi_{p}(y)))\big{)}^{2}+\big{(}\sin\big{(}\phi_{p}(x),\phi_{p}(y))\big{)}^{2}=1.$

$(c)$ From $(a)$ and $(b)$, we have

$\max\\{|\cos_{\infty}(x,y)|^{2},|\sin_{\infty}(x,y)|^{2}\\}=\lim_{p\longrightarrow\infty}{\cos_{p}(x,y)}^{2}\stackrel{{\scriptstyle
p}}{{+}}{\sin_{p}(x,y)}^{2}=1.$

Therefore,

$\max\\{|\cos_{\infty}(x,y)|,|\sin_{\infty}(x,y)|\\}=1.\quad\Box$

Let
$C_{\infty}(0,1]=\big{\\{}(x_{1},x_{2})\in\mathbb{R}^{2}:\max\\{|x_{1}|,|x_{2}|\\}=1\big{\\}}$
be the circle defined with respect to the $\|\,\|_{\infty}$ metric in
$\mathbb{R}^{2}$. Let us consider the points $a=(1,0)$, $b=(1,1)$, $c=(0,1)$,
$d=(-1,1)$, $e=(-1,0)$, $f=(-1,-1)$, $g=(0,-1)$, $h=(1,-1)$. By construction:

$C_{\infty}(0,1]=C_{\boxplus}(0,1]=[a,b]\cup[b,d]\cup[d,f]\cup[f,h]\cup[h,a].$
(58)

$0$$x_{1}$$x_{2}$$\mathrm{pcos}(\theta)$Figure 3.2.1.2 Pseudo
Cosine.$a$$b$$c$$d$$e$$f$$g$$h$$C_{\infty}(0,1)$

In the following we will identify the angles of the square with the length of
the segments of its circumference. Let us consider the map
$\alpha:C_{\infty}(0,1]\longrightarrow[0,8]$ defined by:

$\alpha(x_{1},x_{2})=\left\\{\begin{matrix}x_{2}&\text{ if }&x\in[a,b]\\\
2-x_{1}&\text{ if }&x\in[b,d]\\\ 4-x_{2}&\text{ if }&x\in[d,f]\\\
6+x_{1}&\text{ if }&x\in[f,h]\\\ 8+x_{2}&\text{ if
}&x\in[h,a]\end{matrix}\right..$ (59)

Clearly, we have $\alpha(a)=0$, $\alpha(b)=1$, $\alpha(c)=2$, $\alpha(d)=3$,
$\alpha(e)=4$, $\alpha(f)=5$, $\alpha(g)=6$, $\alpha(h)=7.$

It is easy to prove that this map is a bijection from $C_{\infty}(0,1]$ to
$[0,8]$. Notice that $8$ is the perimeter of $C_{\infty}(0,1]$. For all
$\theta\in[0,8]$ we define the pseudo-cosine
$\mathrm{pcos}:\mathbb{R}\longrightarrow[-1,1]$ as a periodic function of
period 8:

$\mathrm{pcos}(\theta)=\cos_{\infty}\big{(}\alpha^{-1}(\theta),e_{1}\big{)}\quad\mod
8.$ (60)

It follows that for all integer $k\in\mathbb{Z}$

$\mathrm{pcos}(\theta+8k)=\left\\{\begin{matrix}1&\text{ if
}&\theta\in[0,1]\\\ 2-\theta&\text{ if }&\theta\in[1,3]\\\ -1&\text{ if
}&\theta\in[3,5]\\\ -6+\theta&\text{ if }&\theta\in[5,7]\\\ 1&\text{ if
}&\theta\in[7,8]\end{matrix}\right..$ (61)

$-1$$+1$$0$$1$$2$$3$$4$$-1$$-2$$-3$$-4$Figure 3.2.1.3: Pseudo Cosine Function

Paralleling the earlier definition, for all $\theta\in[0,8]$ we define the
pseudo-sine $\mathrm{psin}:\mathbb{R}\longrightarrow[-1,1]$ as a periodic
function of period 8:

$\mathrm{psin}(\theta)=\sin_{\infty}\big{(}\alpha^{-1}(\theta),e_{1}\big{)}\quad\mod
8.$ (62)

It follows that for all integer $z\in\mathbb{Z}$

$\mathrm{psin}(\theta+8k)=\left\\{\begin{matrix}\theta&\text{ if
}&\theta\in[0,1]\\\ 1&\text{ if }&\theta\in[1,3]\\\ 4-\theta&\text{ if
}&\theta\in[3,5]\\\ -1&\text{ if }&\theta\in[5,7]\\\ 8-\theta&\text{ if
}&\theta\in[7,8]\end{matrix}\right..$ (63)

For all $\theta\in\mathbb{R}$, we have from Lemma 3.2.1

$\max\\{\big{|}\mathrm{pcos}(\theta)\big{|},\big{|}\mathrm{psin}(\theta)\big{|}\\}=\max\Big{\\{}\big{|}\cos_{\infty}\big{(}\alpha^{-1}(\theta),e_{1}\big{)}\big{|},\big{|}\sin_{\infty}\big{(}\alpha^{-1}(\theta),e_{1}\big{)}\big{|}\Big{\\}}=1.$
(64)

###  3.3 Some Formalism on the Complex Scalar Field

Let us denote
$C_{\varphi,p}(0,1]=\\{x\in\mathbb{R}^{n}:\|x\|_{\varphi,p}=1\\}$. We have the
relations $\phi_{p}\Big{(}C_{\varphi,p}(0,1]\Big{)}=C(0,1]$. Now, let us
consider the subsets of $\mathbb{C}$ defined as
$C^{\natural}_{\varphi,p}(0,1]=\\{z=a+\mathrm{i}b\in\mathbb{C}:\|(a,b)\|_{\varphi,p}=1\\}$
and
$C^{\natural}_{\infty}(0,1]=\\{z=a+\mathrm{i}b\in\mathbb{C}:\|(a,b)\|_{\infty}=1\\}$.
Let us consider the bijective map
$\varphi^{\natural}:\mathbb{C}\longrightarrow\mathbb{C}$ defined as:

$\varphi_{p}^{\natural}(a+\mathrm{i}b)=\varphi_{p}(a)+\mathrm{i}\varphi_{p}(b).$
(65)

Its reciprocal is the map
${\varphi^{\natural}}^{-1}:\mathbb{C}\longrightarrow\mathbb{C}$ defined as
${\varphi_{p}^{\natural}}^{-1}(a+\mathrm{i}b)=\varphi_{p}^{-1}(a)+\mathrm{i}\varphi_{p}^{-1}(b).$
The map $\varphi_{p}^{\natural}$ is homogenous on the real scalar field, i.e.
for all $\rho\in\mathbb{R}$,
$\varphi_{p}^{\natural}\big{(}\rho(a+\mathrm{i}b)\big{)}=\rho\big{(}\varphi_{p}(a)+\mathrm{i}\varphi_{p}(b)\big{)}=\rho\varphi_{p}\big{(}a+\mathrm{i}b\big{)}.$
Therefore for all $z=\rho e^{\mathrm{i}\theta}\in\mathbb{C}$, with $\rho=|z|$,
we have

${\varphi_{p}^{\natural}}^{-1}(z)=\rho\big{(}\varphi_{p}^{-1}(\cos\theta)+\mathrm{i}\varphi_{p}^{-1}(\sin\theta)\big{)}.$
(66)

For all $z,w\in\mathbb{C}^{2}$, let us define:

$z\stackrel{{\scriptstyle
p}}{{\cdot}}w={\varphi_{p}^{\natural}}^{-1}\Big{(}\varphi_{p}^{\natural}(z)\varphi_{p}^{\natural}(w)\Big{)}$
(67)

It follows that if $z=a+\mathrm{i}\,b$ and $w=c+\mathrm{i}d$, then:

$z\stackrel{{\scriptstyle
p}}{{\cdot}}w=\varphi_{p}^{-1}\Big{(}\varphi_{p}(a)\varphi_{p}(b)-\varphi_{p}(b)\varphi_{p}(d)\Big{)}+\mathrm{i}\,\varphi_{p}^{-1}\Big{(}\varphi_{p}(a)\varphi_{p}(c)+\varphi_{p}(b)\varphi_{p}(c)\Big{)}.$
(68)

Equivalently

$z\stackrel{{\scriptstyle p}}{{\cdot}}w=(ab\stackrel{{\scriptstyle
p}}{{-}}bd)+\mathrm{i}\,(ad\stackrel{{\scriptstyle p}}{{+}}bc).$ (69)

Along this line, taking the limit with respect to the product topology of
$\mathbb{C}^{2}$ we define the product:

$z\boxtimes w=\lim_{p\longrightarrow\infty}z\stackrel{{\scriptstyle
p}}{{\cdot}}w=(ab\boxminus bd)+\mathrm{i}\,(ad\boxplus bc).$ (70)

Note that this product is equivalent to the matrix product defined in [10] for
matrices representing complex numbers. We have for all complex numbers of the
form $z=a+\mathrm{i}\,b$ and $w=c+\mathrm{i}\,d$:

$\begin{pmatrix}a&-b\\\ b&a\end{pmatrix}\boxtimes\begin{pmatrix}c&-d\\\
d&c\end{pmatrix}=\begin{pmatrix}ac\boxminus bd&-(ad\boxplus bc)\\\ ad\boxplus
bc&ac\boxminus bd\end{pmatrix}.$ (71)

Moreover, for all $p\in\mathbb{N}$, and all $z=a+\mathrm{i}b\in\mathbb{C}$ we
define the $\varphi_{p}$-module of $z$ as $|z|_{p}=\|(a,b)\|_{\varphi,p}$.
Similarly, $|z|_{\infty}=\|(a,b)\|_{\infty}=\max\\{|a|,|b|\\}$. In the
following, we also extend the operation $\boxplus$ to the complex numbers. For
all $z=a+\mathrm{i}b$ and $w=c+\mathrm{i}d$, we have :

$z\boxplus w=(a\boxplus c)+\mathrm{i}(b\boxplus d).$ (72)

We can the establish the following relations.

###### Lemma 3.3.1

We have the following properties:
$(a)$ For all $z\in C$, there exists some $\theta\in[0,8]$ such that

$z=\|z\|_{\infty}(\mathrm{pcos}(\theta)+\mathrm{i}\,\mathrm{psin}(\theta))$

$(b)$ If $z=a+\mathrm{i}\,b$ then
$\sqrt{z\boxtimes\bar{z}}=|z|_{\infty}=\max\\{|a|,|b|\\}$.
$(c)$ If $z\not=0$, then $z\cdot\frac{1}{|z|_{\infty}^{2}}\bar{z}=1.$
$(d)$ Suppose that $z,w\in C^{\natural}_{\infty}(0,1]$, then $z\boxtimes w\in
C^{\natural}_{\infty}(0,1]$.
$(e)$ Suppose that $z,w\in\mathbb{C}$, then $|z\boxtimes
w|_{\infty}=|z|_{\infty}|w|_{\infty}$.

Proof: $(a)$ Suppose that $z=a+\mathrm{i}\,b$. If $z=0$, the result is
trivial. If $z\not=0$ then

$z=\max\\{|a|,|b|\\}\frac{a+\mathrm{i}\,b}{\max\\{|a|,|b|\\}}=|z|_{\infty}\Big{(}\frac{a}{\max\\{|a|,|b|\\}}+\mathrm{i}\frac{b}{\max\\{|a|,|b|\\}}\Big{)}.$

However
$\Big{(}\frac{a}{\max\\{|a|,|b|\\}}+\mathrm{i}\frac{b}{\max\\{|a|,|b|\\}}\Big{)}\in
C_{\infty}^{\natural}(0,1].$ Hence there is some $\theta\in[0,8]$ such that

$\Big{(}\frac{a}{\max\\{|a|,|b|\\}}+\mathrm{i}\frac{\,b}{\max\\{|a|,|b|\\}}\Big{)}=\mathrm{pcos}\theta+\mathrm{i}\,\mathrm{psin}\theta.$

Hence

$z=\|z\|_{\infty}\big{(}\mathrm{pcos}(\theta)+\mathrm{i}\,\mathrm{psin}(\theta)\big{)}.$

$(b)$ We have
$z\boxtimes\bar{z}=(a+\mathrm{i}\,b)\boxtimes(a-\mathrm{i}\,b)=(a^{2}\boxplus
b^{2})+\mathrm{i}\,(ab\boxminus ab)=a^{2}\boxplus b^{2}=\max\\{|a|,|b|\\}^{2}$
which yields the result. $(c)$ is immediate from $(b)$.

$(d)$ Suppose that $z=a+\mathrm{i}\,b$ and $w=c+\mathrm{i}\,d$. We first
assume that $z,w\in C^{\natural}_{\infty}(0,1].$ This implies that
$\max\\{|a|,|b|\\}=\max\\{|c|,|d|\\}=1$. By definition we have:

$z\boxtimes w=\big{(}ac\boxminus bd\big{)}+\mathrm{i}\,\big{(}ad\boxplus
bc\big{)}.$

We need to prove that $\max\big{\\{}|ac\boxminus bd|,|ad\boxplus
bc|\big{\\}}=1.$ We consider the following cases:
$(i)$ If $|a|<1$ and $|c|<1$ then $|bd|=1$ and consequently $|ac\boxminus
bd|=1$; $(ii)$ If $|a|<1$ and $|d|<1$ then $|bc|=1$, therefore $|ad\boxplus
bc|=1$. $(iii)$ If $|c|<1$ and $|b|<1$ then $|ad|=1$ and we deduce that
$|ad\boxplus bc|=1$. $(iv)$ If $|b|<1$ and $|d|<1$ then $|ac|=1$, hence we
have $|ac\boxminus bd|=1$.

Suppose now that $|a|=|b|=|c|=|d|=1$. All we need to show is that we cannot
have $ac\boxminus bd=ad\boxplus bc=0$. In such a case we have $ac=bd$ and
$ad=-bc$. However this implies that ${d^{2}}=-c^{2}$ that is a contradiction
since $|c|=|d|=1.$ Consequently we deduce that that if $z,w\in
C^{\natural}_{\infty}(0,1]$ then $z\boxtimes w\in C^{\natural}_{\infty}(0,1]$.

$(e)$ Suppose that
$z=|z|_{\infty}\big{(}\mathrm{pcos}(\theta)+\mathrm{i}\,\mathrm{psin}(\theta)\big{)}$
and
$w=|w|_{\infty}\big{(}\mathrm{pcos}(\alpha)+\mathrm{i}\,\mathrm{psin}(\alpha)\big{)}$.
We have:

$z\boxtimes
w=|z|_{\infty}|w|_{\infty}\big{(}\mathrm{pcos}(\theta)\mathrm{pcos}(\alpha)\boxminus\mathrm{psin}(\theta)\mathrm{psin}(\alpha)+\mathrm{i}\,(\mathrm{pcos}(\theta)\mathrm{psin}(\alpha)\boxplus\mathrm{psin}(\theta)\mathrm{pcos}(\alpha))\big{)}.$

Hence, since

$\left|\mathrm{pcos}(\theta)\mathrm{pcos}(\alpha)\boxminus\mathrm{psin}(\theta)\mathrm{psin}(\alpha)+\mathrm{i}\,(\mathrm{pcos}(\theta)\mathrm{psin}(\alpha)\boxplus\mathrm{psin}(\theta)\mathrm{pcos}(\alpha))\right|_{\infty}=1$

we deduce the result. $\Box$

##  4 Algebraic Formulation of the Kuratowski-Peano Limit of a Line

###  4.1 Algebraic Formulation of a Line

In the standard convex case it is easy to give an algebraic formulation of a
line passing trough two different points. In such a case one can define the
line passing trough $x$ and $y$ ($x\not=y$) as:

$\mathcal{D}(x,y)=\\{tx+sy:s+t=1,s,t\in\mathbb{R}\\}.$ (73)

Note that there is no restriction imposing the non-negativity of $s$ and $t$
which limits a line to a line-segment. Paralleling this definition one can
define a $\varphi_{p}$-line as:

$\mathcal{D}^{(p)}(x,y)=\big{\\{}tx\stackrel{{\scriptstyle
p}}{{+}}sy:s\stackrel{{\scriptstyle p}}{{+}}t=1,s,t\in\mathbb{R}\big{\\}}.$
(74)

It was pointed in [7] that the subset $\\{tx\boxplus
sy:\max\\{s,t\\}=1,s,t\geq 0\\}$ is not path-connected and does not describe
over $\mathbb{R}^{n}$ the $\mathbb{B}$-convex hull of $\\{x,y\\}$. To that end
it was shown in [7] that $Co^{\infty}(x,y)=\Big{\\{}tx\boxplus rx\boxplus
sy\boxplus wy:\max\\{t,r,s,w\\}=1,t,r,s,w\geq 0\Big{\\}}$. In what follows, we
take a parallel approach to give the algebraic description of a line. Consider
the following subset:

$\mathcal{D}^{\infty}(x,y):=\Big{\\{}tx\boxplus rx\boxplus sy\boxplus
wy:t\boxplus r\boxplus s\boxplus w=1,t,r,s,w\in\mathbb{R}\Big{\\}}.$ (75)

###### Lemma 4.1.1

For all $x,y\in\mathbb{R}^{n}$ with $x\not=y$,

$\mathcal{D}^{\infty}(x,y)\subset\operatorname*{\mathrm{Li}}_{p\longrightarrow\infty}\mathcal{D}^{(p)}(x,y).$

Proof: Suppose that $z\in\Big{\\{}tx\boxplus rx\boxplus sy\boxplus
wy:t\boxplus r\boxplus s\boxplus w=1,t,r,s,w\in\mathbb{R}\Big{\\}}$. By
definition there exists
$\alpha_{1},\alpha_{2},\beta_{1},\beta_{2}\in\mathbb{R}$ with
$\alpha_{1}\boxplus\alpha_{2}\boxplus\beta_{1}\boxplus\beta_{2}=1$ and such
that

$z=\alpha_{1}x\boxplus\alpha_{2}x\boxplus\beta_{1}y\boxplus\beta_{2}y.$

Since $\alpha_{1}\boxplus\alpha_{2}\boxplus\beta_{1}\boxplus\beta_{2}=1$ there
exists some $p_{0}\in\mathbb{N}$ such that, for all $p\geq p_{0}$,
$\alpha_{1}\stackrel{{\scriptstyle p}}{{+}}\alpha_{2}\stackrel{{\scriptstyle
p}}{{+}}\beta_{1}\stackrel{{\scriptstyle p}}{{+}}\beta_{2}\not=0$. For all
$p\geq p_{0}$, define

$z^{(p)}=\frac{1}{\alpha_{1}\stackrel{{\scriptstyle
p}}{{+}}\alpha_{2}\stackrel{{\scriptstyle
p}}{{+}}\beta_{1}\stackrel{{\scriptstyle
p}}{{+}}\beta_{2}}\big{(}\alpha_{1}x\stackrel{{\scriptstyle
p}}{{+}}\alpha_{2}x\stackrel{{\scriptstyle
p}}{{+}}\beta_{1}y\stackrel{{\scriptstyle p}}{{+}}\beta_{2}y\big{)}.$

By construction, for all $p\geq p_{0}$, $z^{(p)}\in\mathcal{D}^{(p)}(x,y)$.
Taking the limit on both sides yields:

$\displaystyle\lim_{p\longrightarrow\infty}z^{(p)}$
$\displaystyle=\frac{1}{\alpha_{1}\boxplus\alpha_{2}\boxplus\beta_{1}\boxplus\beta_{2}}\big{(}\alpha_{1}x\boxplus\alpha_{2}x\boxplus\beta_{1}y\boxplus\beta_{2}y\big{)}$
$\displaystyle=\alpha_{1}x\boxplus\alpha_{2}x\boxplus\beta_{1}y\boxplus\beta_{2}y=z.$

Consequently,
$z\in\operatorname*{\mathrm{Li}}_{p\longrightarrow+\infty}\mathcal{D}^{(p)}(x,y)$.
$\Box$

###### Lemma 4.1.2

Let $\big{\\{}(s^{(p)},t^{(p)})\big{\\}}_{p\in\mathbb{N}}$ be sequence of
$\mathbb{R}^{2}$. Suppose there is an increasing subsequence
$\\{p_{q}\\}_{q\in\mathbb{N}}$ of natural numbers such that
$s^{(p_{q})}\stackrel{{\scriptstyle p_{q}}}{{+}}t^{(p_{q})}=1$ and
$\lim\limits_{q\longrightarrow\infty}(s^{(p_{q})},t^{(p_{q})})=(s,t)$. We have
the following properties:
$(a)$ If $\max\\{\,|s|,|t|\,\\}>1$ then $s=-t$.
$(b)$ If $s=-t$ then $|s|=|t|\geq 1$.
$(c)$ If $\min\\{\,|s|,|t|\,\\}\leq 1$ then $\max\\{s,t\\}=1$.
$(d)$ If $s\not=-t$ then $\max\\{s,t\\}=1$.

Proof: $(a)$ In such a case if $|s|>1$ then
$t=\lim\limits_{q\longrightarrow\infty}{t_{p_{q}}}=\lim\limits_{q\longrightarrow\infty}1\stackrel{{\scriptstyle
p_{q}}}{{-}}{s_{p_{q}}}=-s$. The proof is symmetrical if $|t|>1$. $(b)$ Let us
prove that in such a case we have $|s|\geq 1$ and $|t|\geq 1$. If $|s|<1$ then
$t=\lim\limits_{q\longrightarrow\infty}{t_{p_{q}}}=\lim\limits_{q\longrightarrow\infty}1\stackrel{{\scriptstyle
p_{q}}}{{-}}{s_{p_{q}}}=1$, which contradicts $s=-t$. Similarly, if $|t|<1$ we
have a contradiction. Therefore $|s|\geq 1$ and $|t|\geq 1$. $(c)$ If $s=1$,
then the conclusion is obvious. If $|s|<1$, then
$t=\lim\limits_{q\longrightarrow\infty}{t_{p_{q}}}=\lim\limits_{q\longrightarrow\infty}1\stackrel{{\scriptstyle
p_{q}}}{{-}}{s_{p_{q}}}=1$ and $\max\\{s,t\\}=1$. To conclude if $s=-1$, then
$t=\lim\limits_{q\longrightarrow\infty}{t_{p_{q}}}=\lim\limits_{q\longrightarrow\infty}1\stackrel{{\scriptstyle
p_{q}}}{{+}}{1}=1$, which ends the proof of $(c)$. $(d)$ From $(a)$,
$s\not=-t$ implies that $\max\\{\,|s|,|t|\,\\}\leq 1$. Hence
$\min\\{\,|s|,|t|\,\\}\leq 1$ and the result is then immediate from $(c)$.
$\Box$

The next lemma is a preliminary result to show the upper limit of the sequence
$\\{\mathcal{D}\\}_{p\in\mathbb{N}}$ is identical to $\mathcal{D}^{\infty}$.

###### Lemma 4.1.3

Let $x,y\in\mathbb{R}^{n}$ with $x\not=y$ and suppose that
$z\in\operatorname*{\mathrm{Ls}}_{p\longrightarrow\infty}\mathcal{D}_{p}(x,y)$.
Then there exists a sequence
$\big{\\{}(s^{(p)},t^{(p)})\big{\\}}_{p\in\mathbb{N}}\subset\mathbb{R}^{2}$
and an increasing subsequence $\\{p_{k}\\}_{k\in\mathbb{N}}$ of natural
numbers such that $s^{(p_{k})}\stackrel{{\scriptstyle
p_{k}}}{{+}}t^{(p_{k})}=1$,
$\lim\limits_{q\longrightarrow\infty}(s^{(p_{k})},t^{(p_{k})})=(s,t)$ and

$z_{i}=sx_{i}\boxplus ty_{i},$

for all $i\in[n]$ such that $sx_{i}\not=-ty_{i}$.

Proof: Suppose that
$z\in\operatorname*{\mathrm{Ls}}_{p\longrightarrow\infty}\mathcal{D}^{(p)}(x,y)$.
In such a case there exists an increasing sequence of natural numbers
$\\{p_{k}\\}_{k\in\mathbb{N}}$ and sequence
$\\{z^{(p_{k})}\\}_{k\in\mathbb{N}}$ with
$z^{(p_{k})}\in\mathcal{D}^{p_{k}}(x,y)$ for all $k$ and
$\lim_{q\longrightarrow\infty}z^{(p_{k})}=z$. This implies that there exists a
sequence
$\\{s^{(p_{k})},t^{(p_{k})}\\}_{k\in\mathbb{N}}\subset\mathbb{R}\times\mathbb{R}$
such that

$z^{(p_{k})}=t^{(p_{k})}x\stackrel{{\scriptstyle p_{k}}}{{+}}s^{(p_{k})}y,$
(76)

with $s^{(p_{k})}\stackrel{{\scriptstyle p_{k}}}{{+}}t^{(p_{k})}=1$ for all
$k$. Let us prove that
$\big{\\{}(s^{(p_{k})},t^{(p_{k})})\big{\\}}_{q\in\mathbb{N}}$ is a bounded
sequence. Suppose that this is not the case and let us show a contradiction.
Suppose for example that $\\{s^{(p_{k})}\\}_{k\in\mathbb{N}}$ is not bounded.
An elementary calculus shows that since $t^{(p_{k})}=1\stackrel{{\scriptstyle
p_{k}}}{{-}}s^{(p_{k})}$. Thus:

$z^{(p_{k})}=x\stackrel{{\scriptstyle
p_{k}}}{{+}}s^{(p_{k})}(y\stackrel{{\scriptstyle p_{k}}}{{-}}x).$ (77)

However $\lim_{k\longrightarrow\infty}y\stackrel{{\scriptstyle
p_{k}}}{{-}}x=y\boxminus x$. Since $x\not=y$, $y\boxminus x\not=0$ and this
implies that $\lim_{k\longrightarrow\infty}\|z^{(p_{q})}\|=\infty$ that is a
contradiction. Consequently $\\{s^{(p_{k})}\\}_{k\in\mathbb{N}}$ is bounded.
Since $s^{(p_{k})}=1\stackrel{{\scriptstyle p_{k}}}{{-}}t^{(p_{k})}$ this also
implies that $\\{s^{(p_{k})}\\}_{k\in\mathbb{N}}$ is bounded. Therefore, we
deduce that the sequence $\\{(s^{(p_{k})},t^{(p_{k})})\\}_{k\in\mathbb{N}}$ is
bounded. Consequently one can extract an increasing subsequence
$\\{p_{k_{m}}\\}_{m\in\infty}$ such that

$\lim_{m\longrightarrow\infty}(s^{(p_{k_{m}})},t^{(p_{k_{m}})})=(s,p).$

Let us denote $I=\\{i\in[n]:sx_{i}\boxplus ty_{i}=0\\}$. For all $i\notin I$,
since $sx_{i}\not=-ty_{i}$ we have

$\lim_{m\longrightarrow\infty}s^{(p_{k_{m}})}x_{i}\stackrel{{\scriptstyle
p_{k_{m}}}}{{+}}t^{(p_{k_{m}})}y_{i}=sx_{i}\boxplus ty_{i}.$

Setting $p_{q}=p_{k_{m}}$ for alll $m$, yields the result. $\Box$

###### Proposition 4.1.4

For all $x,y\in\mathbb{R}^{n}$ with $x\not=y$,

$\operatorname*{\mathrm{Lim}}_{p\longrightarrow\infty}\mathcal{D}^{(p)}(x,y)=\mathcal{D}^{\infty}(x,y).$

Proof: To prove this result we establish that
$\operatorname*{\mathrm{Ls}}_{p\longrightarrow\infty}\mathcal{D}^{(p)}(x,y)\subset\Big{\\{}tx\boxplus
rx\boxplus sy\boxplus wy:t\boxplus r\boxplus s\boxplus
w=1,t,r,s,w\in\mathbb{R}\Big{\\}}$. Suppose that
$z\in\operatorname*{\mathrm{Ls}}_{p\longrightarrow\infty}\mathcal{D}^{(p)}(x,y)$.
We have shown that there exists an increasing sequence of natural numbers
$\\{p_{q}\\}_{q\in\mathbb{N}}$ and sequence
$\\{z^{(p_{q})}\\}_{q\in\mathbb{N}}$ with
$z^{(p_{q})}=t^{(p_{q})}x\stackrel{{\scriptstyle
p_{q}}}{{+}}s^{(p_{q})}y\in\mathcal{D}^{p_{q}}(x,y)$ for all $q$ and
$\lim_{q\longrightarrow\infty}z^{(p_{q})}=z$. Moreover,
$\lim_{p\longrightarrow\infty}(s^{(p_{q})},t^{(p_{q})})=(s,p)$ and

$\lim_{p\longrightarrow\infty}s^{(p_{q})}x_{i}\stackrel{{\scriptstyle
p_{q}}}{{+}}t^{(p_{q})}y_{i}=sx_{i}\boxplus ty_{i}.\quad(1)$

where $I=\\{i\in[n]:sx_{i}\boxplus ty_{i}=0\\}$. We consider three cases.

$(a)$ $I=\emptyset$. If $\min\\{|s|,|t|\\}>1$ then from Lemma 4.1.2.(a),
$s=-t$. Let us consider the vector

$sx\boxplus x\boxplus(1-\epsilon)y\boxplus ty,$

where $\epsilon>0$ is sufficiently small. Clearly, we have $s\boxplus
1\boxplus(1-\epsilon)\boxplus t=1\boxplus(1-\epsilon)=1$. Moreover since
$|s|=|t|>1$,for all $i\in[n]$, $sx_{i}\boxplus ty_{i}=sx_{i}\boxplus
x_{i}\boxplus(1-\epsilon)y_{i}\boxplus ty_{i}$. From equation $(1)$

$z=sx\boxplus ty=sx\boxplus x\boxplus(1-\epsilon)y\boxplus ty.$

If $\min\\{|s|,|t|\\}\leq 1$ then from Lemma 4.1.2, $\max\\{s,t\\}=1$. Suppose
that $t=1$. Then $sx\boxplus ty=sx\boxplus 0x\boxplus y\boxplus y$ and
$s\boxplus 0\boxplus 1\boxplus 1=1$. One can proceeds symmetrically if $s=1$.

$(b)$ $I\not=\emptyset$ and for all $i\in I$, $x_{i}=y_{i}$. In such a case
for all $i\in I$, $z_{i}^{(p_{q})}=x_{i}=z_{i}$. For all $i\in I$,
$sx_{i}=-ty_{i}=-tx_{i}$ and this implies that $s=-t$. From Lemma Lemma
4.1.2.(b) we have $|s|\geq 1$ and $|t|\geq 1$. Let us consider the vector

$sx\boxplus x\boxplus(1-\epsilon)y\boxplus ty,$

where $\epsilon>0$ is sufficiently small. Clearly, we have $s\boxplus
1\boxplus(1-\epsilon)\boxplus t=1\boxplus(1-\epsilon)=1$.

If $i\in I$, then since $s=-t$ and $x_{i}=y_{i}$ one has $sx_{i}\boxplus
x_{i}\boxplus(1-\epsilon)y_{i}\boxplus
ty_{i}=x_{i}\boxplus(1-\epsilon)y_{i}=x_{i}=y_{i}=z_{i}.$

If $i\notin I$ then, since $sx_{i}\not=-tx_{i}$, $|s|\geq 1$ and $|t|\geq 1$,
we have $sx_{i}\boxplus x_{i}\boxplus(1-\epsilon)y_{i}\boxplus
ty_{i}=sx_{i}\boxplus ty_{i}.$

$(c)$ $I\not=\emptyset$ and there is some $i_{0}\in I$, such that
$x_{i_{0}}\not=y_{i_{0}}$. We first prove that in such a case we have
$s\not=-t$. Suppose that this is not the case. We have $sx_{i_{0}}\boxplus
ty_{i_{0}}=sx_{i_{0}}\boxplus(-s)y_{i_{0}}=0$. However this implies
$x_{i_{0}}=y_{i_{0}}$ that is a contradiction. Moreover, from Lemma 4.1.2.(c)
we have $\max\\{|s|,|t|\\}=1$ and $\max\\{s,t\\}=1$. Assume for example that
$t=1$.

Suppose that $i\in I$. We have $ty_{i}=y_{i}=-sx_{i}$ and therefore

$s^{(p_{q})}x_{i}\stackrel{{\scriptstyle
p_{q}}}{{+}}t^{(p_{q})}y_{i}=s^{(p_{q})}x_{i}\stackrel{{\scriptstyle
p_{q}}}{{-}}{s}t^{(p_{q})}x_{i}=\Big{(}s^{(p_{q})}\stackrel{{\scriptstyle
p_{q}}}{{-}}({s})t^{(p_{q})}\Big{)}x_{i}.$

By hypothesis $\lim_{k\longrightarrow\infty}z_{i}^{(p_{q})}=z_{i}$. Hence the
sequence $\big{\\{}s^{(p_{q})}\stackrel{{\scriptstyle
p_{q}}}{{-}}st^{(p_{q})}\Big{\\}}_{q\in\mathbb{N}}$ has a limit. Let us denote

$s^{\prime}=\lim_{q\longrightarrow\infty}s^{(p_{q})}\stackrel{{\scriptstyle
p_{q}}}{{-}}{s}t^{(p_{q})}.$

By hypothesis we have $z_{i}=s^{\prime}x_{i}.$ Note that since

$|s^{(p_{q})}\stackrel{{\scriptstyle
p_{q}}}{{-}}{s}t^{(p_{q})}|\leq|s^{(p_{q})}|\stackrel{{\scriptstyle
p_{q}}}{{+}}|{s}t^{(p_{q})}|$

we have taking the limit $|s^{\prime}|\leq\max\\{|s|,t|s|\\}\leq|s|$. We
consider two cases.

$(i)$ Suppose that $s^{\prime}\not=-s$. Let us consider the vector

$sx\boxplus s^{\prime}x\boxplus 0y\boxplus y.$

If $i\in I$, then, since $sx_{i}=-y_{i}$ we have $w_{i}=sx_{i}\boxplus
s^{\prime}x_{i}\boxplus 0y_{i}\boxplus y_{i}=s^{\prime}x_{i}=z_{i}$.

If $i\notin I$, since $s\not=-s^{\prime}$, we have either $s=s^{\prime}$ or
$|s^{\prime}|<|s|$. Hence $sx_{i}\boxplus s^{\prime}x_{i}\boxplus
0y_{i}\boxplus y_{i}=sx_{i}\boxplus y_{i}=z_{i}$.

It follows that $z=sx\boxplus s^{\prime}x\boxplus 0y\boxplus y.$ Moreover
since $-s\not=t=1$ and $|s^{\prime}|\leq|s|$, we have $s\boxplus
s^{\prime}\boxplus 0\boxplus 1=1$.

$(ii)$ Suppose that $s^{\prime}=-s$. Now let us consider the vector

$sx\boxplus 0x\boxplus y\boxplus y.$

If $i\in I$, then, since $sx_{i}=-y_{i}$ we have $sx_{i}\boxplus
0x_{i}\boxplus y_{i}\boxplus y_{i}=y_{i}$. However since $y_{i}=-sx_{i}$, we
deduce that $y_{i}=s^{\prime}x_{i}=z_{i}$.

If $i\notin I$, in such a case $sx_{i}\not=-y_{i}$, thus we have
$sx_{i}\boxplus 0x_{i}\boxplus y_{i}\boxplus y_{i}=sx_{i}\boxplus
y_{i}=z_{i}$.

Hence, we conclude that $z=sx\boxplus 0x\boxplus y\boxplus y.$ Moreover
$1\boxplus 0\boxplus 1\boxplus 1=1$.

Consequently, we have shown that $z\in\Big{\\{}tx\boxplus rx\boxplus
sy\boxplus wy:t\boxplus r\boxplus s\boxplus
w=1,t,r,s,w\in\mathbb{R}\Big{\\}}$. If $s=1$ the proof is symmetrical. Hence

$\operatorname*{\mathrm{Ls}}_{p\longrightarrow\infty}\mathcal{D}^{(p)}(x,y)\subset\Big{\\{}tx\boxplus
rx\boxplus sy\boxplus wy:t\boxplus r\boxplus s\boxplus
w=1,t,r,s,w\in\mathbb{R}\Big{\\}}.\quad\Box$

$p\in)0,\infty[$$p=0$$x_{1}$$x_{2}$$0$Figure 4.1Limit of a sequence of
$\varphi_{p}$-lines.$p=\infty$

In the next example it is shown that the formalism described above allows to
define subsets of $\mathbb{R}^{n}$ which are unbounded.

###### Example 4.1.5

Suppose that $n=3$ and let us consider the points $x=(3,-2,1)$ and
$y=(1,-1,1)$. Suppose that $t=1$, $r=\delta=-w$ and $s=0$ where $\delta$ is a
real number. Clearly, $t\boxplus r\boxplus s\boxplus
w=1\boxplus\delta\boxplus(-\delta)\boxplus 0=1$. It follows that

$z_{\delta}:=tx\boxplus rx\boxplus sy\boxplus wy=t\begin{pmatrix}3\\\ -2\\\
1\end{pmatrix}\boxplus\delta\begin{pmatrix}3\\\ -2\\\
1\end{pmatrix}\boxplus(-\delta)\begin{pmatrix}1\\\ -1\\\
1\end{pmatrix}\boxplus 0\begin{pmatrix}1\\\ -1\\\
1\end{pmatrix}=\begin{pmatrix}3\delta\\\ -2\delta\\\
1\end{pmatrix}\in\mathcal{D}^{\infty}(x,y).$

It follows that

$\lim_{|\delta|\longrightarrow\infty}\|z_{\delta}\|=+\infty.$

In the following it is shown that $\mathcal{D}^{\infty}(x,y)$ is path
connected.

###### Proposition 4.1.6

Let $x,y\in\mathbb{R}^{n}$ and suppose that $x\not=y$. Then
$\mathcal{D}^{\infty}(x,y)$ is path-connected.

Proof: Suppose that $u,v\in\mathcal{D}^{\infty}(x,y)$ with $u\not=v$. Then
there exists a sequence $\\{(u^{(p)},v^{(p)})\\}_{p\in\mathbb{N}}$ such that
for all $p\in\mathbb{N}$
$(u^{(p)},v^{(p)})\in\mathcal{D}^{(p)}(x,y)\times\mathcal{D}^{(p)}(x,y)$ and
$\lim_{p\longrightarrow\infty}(u^{(p)},v^{(p)})=(u,v)$. Let
$\mathcal{I}(u,v)=\\{i\in[n]:u_{i}v_{i}<0\\}$. Since
$\lim_{p\longrightarrow\infty}(u^{(p)},v^{(p)})=(u,v)$, there is some
$p_{0}\in\mathbb{N}$ such that for all $p\geq p_{0}$,
$\mathcal{I}(u,v)=\mathcal{I}(u^{(p)},v^{(p)})$. For all
$i\in\mathcal{I}(u,v)$ and all $p\geq p_{0}$, there is a unique point
$\gamma_{i}^{(p)}\in Co^{p}(x,y)$ such that
$\\{\gamma_{i}^{(p)}\\}=H_{i}\cap\mathcal{D}^{(p)}(x,y)$, where for all
$i\in[n]$ $H_{i}=\\{x\in\mathbb{R}^{n}:x_{i}=0\\}$. For all
$i\in\mathcal{I}(u,v)$,

$\operatorname*{\mathrm{Ls}}_{p\longrightarrow\infty}(H_{i}\cap\mathcal{D}^{(p)}(x,y))\subset
H_{i}\cap\mathcal{D}^{\infty}(x,y)$

that is a nonempty subset of $\mathcal{D}^{\infty}(x,y)$. Let
$\gamma_{i}\in\operatorname*{\mathrm{Ls}}_{p\longrightarrow\infty}(H_{i}\cap\mathcal{D}^{(p)}(x,y))$.
This implies that there is an increasing sequence $\\{p_{q}\\}$ such that

$\lim_{q\longrightarrow\infty}\gamma_{i}^{(p_{q})}=\gamma_{i}\in
H_{i}\cap\mathcal{D}^{\infty}(x,y).$

Let us denote $n(u,v)=\mathrm{Card}\,\mathcal{I}(u,v)$. For all $p\geq p_{0}$
one can find a sequence $\\{i_{k}\\}_{k\in[n(u,v)-1]}$ such that

$\gamma_{i_{k}}^{(p_{q})}\cdot\gamma_{i_{k+1}}^{(p_{q})}\geq 0.$

It follows that

$\operatorname*{\mathrm{Lim}}_{q\longrightarrow\infty}Co^{p_{q}}\Big{(}\gamma_{i_{k}}^{(p_{q})},\gamma_{i_{k+1}}^{(p_{q})}\Big{)}=Co^{\infty}\Big{(}\gamma_{i_{k}},\gamma_{i_{k+1}}\Big{)}.$

However

$\displaystyle\operatorname*{\mathrm{Ls}}_{p\longrightarrow\infty}Co^{p}(u^{(p)},v^{(p)})$
$\displaystyle=\bigcup_{k\in[{n(u,v)-1}]}\operatorname*{\mathrm{Ls}}_{q\longrightarrow\infty}Co^{p_{q}}\Big{(}\gamma_{i_{k}}^{(p_{q})},\gamma_{i_{k+1}}^{(p_{q})}\Big{)}$
$\displaystyle=\bigcup_{k\in[{n(u,v)-1}]}\operatorname*{\mathrm{Lim}}_{q\longrightarrow\infty}Co^{p_{q}}\Big{(}\gamma_{i_{k}}^{(p_{q})},\gamma_{i_{k+1}}^{(p_{q})}\Big{)}$
$\displaystyle=\bigcup_{k\in[{n(u,v)-1}]}Co^{\infty}\Big{(}\gamma_{i_{k}},\gamma_{i_{k+1}}\Big{)}.$

However, since any set
$Co^{\infty}\Big{(}\gamma_{i_{k}},\gamma_{i_{k+1}}\Big{)}$ is path-connected,
it follows that there exists a continuous map
$\psi:[0,1]\longrightarrow\mathcal{D}^{\infty}(x,y)$ such that $\psi(0)=u$ and
$\psi(1)=v$. $\Box$

$x_{1}$$x_{2}$$0$Figure 4.2$\digamma$-line spanned from $x$ and
$y$.$x$$y$$\mathcal{D}^{\infty}(x,y)$

$x_{1}$$x_{2}$$x_{3}$$0$$\gamma_{3}$$\gamma_{2}$$\gamma_{1}$$x$$y$$\mathcal{D}^{\infty}(x,y)$Figure
4.33-dimensional $\digamma$-line.

Figure 4.3 depicts the case of a 3-dimensional $\digamma$-line passing trough
$x$ and $y$. For $i=1,2,3$, $\gamma_{i}$ is the intersection between the
$\digamma$-line and the 2-dimensions plane with equation $x_{i}=0.$

###### Corollary 4.1.7

For all $x,y\in\mathbb{R}^{n}$ with $x\not=y$,

$Co^{\infty}(x,y)\subset\mathcal{D}^{\infty}(x,y).$

Proof: From [7]

$Co^{\infty}(x,y):=\Big{\\{}tx\boxplus rx\boxplus sy\boxplus wy:t\boxplus
r\boxplus s\boxplus w=1,t,r,s,w\in\mathbb{R}_{+}\Big{\\}}.$

The result is then immediate, since $Co^{\infty}(x,y)$ is the set of all the
$\mathbb{B}$-convex combinations such that $t,r,s,w\geq 0.$ $\Box$

Corollary 4.1.7 does not mean that $\mathcal{D}^{\infty}$ is
$\mathbb{B}^{\sharp}$-convex. In general, this is not true. This can be seen
in Figure 4.3, where $x\boxminus y$ and $y\boxminus x$ belong to
$\mathcal{D}^{\infty}(x,y)$. However, $(x\boxminus y)\boxplus(y\boxminus
x)=0\notin\mathcal{D}^{\infty}.$

###### Proposition 4.1.8

Let $Y$ be an idempotent symmetric subspace of $\mathbb{R}^{n}$. Then, for all
$x,y\in\mathbb{R}^{n}$ with $x\not=y$ we have
$\mathcal{D}^{\infty}(x,y)\subset Y$.

Proof: We have shown that

$\mathcal{D}^{\infty}(x,y)=\Big{\\{}tx\boxplus rx\boxplus sy\boxplus
wy:t\boxplus r\boxplus s\boxplus w=1,t,r,s,w\in\mathbb{R}\Big{\\}}.$

From Proposition 2.1.14 the result is then immediate. $\Box$

Notice that since $(\mathbb{R}^{n},\boxplus,\cdot)$ and
$(\widetilde{\mathbb{M}}^{n},\widetilde{\boxplus},\otimes)$ are isomorphic one
can deduce the algebraic form of a line passing trough $z$ and $u$. The
multiplicative neutral element of $\cdot$ is replaced with
$0=1_{\widetilde{\mathbb{M}}}$ and we have

${\widetilde{\mathcal{D}}}^{\infty}(z,u)=\Big{\\{}tz\,\widetilde{\boxplus}\,rz\,\widetilde{\boxplus}\,su\,\widetilde{\boxplus}\,wu:t\,\widetilde{\boxplus}\,r\widetilde{\boxplus}\,s\,\widetilde{\boxplus}\,w=0,t,r,s,w\in{\mathbb{M}}\Big{\\}}.$
(78)

###  4.2 Half-Lines and Inclusion Properties

In the next statement it is shown that for all $x,y\in\mathbb{R}^{n}$ with
$x\not=y$, the line $\mathcal{D}^{\infty}(x,y)$ contains two half lines
spanned from two specific points. $\mathcal{D}^{\infty}(x,y)$ is called the
$\digamma$-line spanned by $x$ and $y$.

###### Proposition 4.2.1

Let $x,y\in\mathbb{R}^{n}$ and suppose that $x\not=y$. Let
$I=\\{i\in[n]:x_{i}\not=y_{i}\\}$ and $J=[n]\backslash I$. We have the
following inclusions:
$(a)$ If $J=\emptyset$ then

$\\{t(y\boxminus x):t\in[1,\infty[\,\\}\subset\mathcal{D}^{\infty}(x,y)$

and

$\\{t(x\boxminus y):t\in[1,\infty[\,\\}\subset\mathcal{D}^{\infty}(x,y).$

$(b)$ If $J\not=\emptyset$ then

$x_{[J]}+\Big{\\{}t(y\boxminus
x):t\in[1,\infty[\,\Big{\\}}\subset\mathcal{D}^{\infty}(x,y)$

and

$x_{[J]}+\Big{\\{}t(x\boxminus
y):t\in[1,\infty[\,\Big{\\}}\subset\mathcal{D}^{\infty}(x,y),$

where $x_{[J]}=\sum\limits_{k\in J}x_{k}e_{k}=\sum\limits_{k\in
J}y_{k}e_{k}=y_{[J]}.$

Proof: $(a)$ We first note that since $J=\emptyset$ and $I=[n]$

$x\boxminus y=x\boxplus x\boxplus 0y\boxplus(-y)\in\Big{\\{}tx\boxplus
rx\boxplus sy\boxplus wy:t\boxplus r\boxplus s\boxplus
w=1,t,r,s,w\in\mathbb{R}\Big{\\}}.$

For all $t\geq 1$, let us consider the vector

$z=tx\boxplus(-t)y.$

Since, we have

$z=tx\boxplus x\boxplus 0y\boxplus(-t)y.$

Moreover $t\boxplus 1\boxplus 0\boxplus(-t)=1$. Clearly
$z\in\Big{\\{}tx\boxplus rx\boxplus sy\boxplus wy:t\boxplus r\boxplus
s\boxplus w=1,t,r,s,w\in\mathbb{R}\Big{\\}}$, which proves the first part of
$(a)$. The proof of the second part is symmetrical swapping $x$ and $y$. $(b)$
Let us denote $\mathbb{R}_{[I]}=\\{\sum_{k\in
I}x_{k}e_{k}:x\in\mathbb{R}^{n}\\}$. By hypothesis $x,y\in
x_{[J]}+\mathbb{R}_{[I]}$. Hence $\mathcal{D}^{\infty}(x,y)\subset
x_{[J]}+\mathbb{R}_{[I]}.$ Let $n_{I}=\mathrm{Card}(I)$ and let
$\mathcal{D}_{[I]}^{\infty}(x,y)$ be the canonical decomposition of
$\mathcal{D}^{\infty}(x,y)$ over $\mathbb{R}_{[I]}$. By definition for all
$i\in I$ $x_{i}\not=y_{i}$. Since $\mathbb{R}_{[I]}$ is isomorphic to its
restriction projection onto $\mathbb{R}^{n_{I}}$ we deduce from $(a)$ that

$\big{\\{}t(x_{[I]}\boxminus
y_{[I]}):t\in[1,\infty[\,\big{\\}}\subset\mathcal{D}_{[I]}^{\infty}(x,y).$

However $\mathcal{D}^{\infty}(x,y)=x_{[J]}+\mathcal{D}_{[I]}^{\infty}(x,y)$.
Moreover $x_{[J]}\boxminus y_{[J]}=0.$ Hence

$\displaystyle x_{[J]}+\big{\\{}t(x\boxminus y):t\in[1,\infty[\,\big{\\}}$
$\displaystyle=x_{[J]}+\big{\\{}t(x_{[I]}\boxminus
y_{[I]}):t\in[1,\infty[\,\big{\\}}$ $\displaystyle\subset
x_{[J]}+\mathcal{D}_{[I]}^{\infty}(x,y)$
$\displaystyle\subset\mathcal{D}^{\infty}(x,y).$

The proof of the second part of the statement is symmetrical. $\Box$

Notice that Proposition 4.2.1.(b) can be deduced from Proposition 4.2.1.(a) by
considering the canonical isomorphism $\psi_{I}$ defined from the set
$\\{z\in\mathbb{R}^{n}:z_{j}=0,\;j\notin I\\}$ to $\mathbb{R}^{m}$ where
$m=\mathrm{Card}(I)$.

$x_{1}$$x_{2}$$0$Figure 4.3Inclusion of half-lines.$x=x\boxminus y$$y\boxminus
x$$y$$\mathcal{D}^{\infty}(x,y)$$\boxminus y$

Let $x,y\in\mathbb{R}^{n}$. The half lines
$\mathcal{D}_{+}(x,y)=\big{\\{}t(x\boxminus y):t\in[1,\infty[\,\big{\\}}$ and
$\mathcal{D}_{-}(x,y)=\big{\\{}t(y\boxminus x):t\in[1,\infty[\,\big{\\}}$ are
respectively called the upper and lower half-lines components of the
$\digamma$-line $\mathcal{D}^{\infty}(x,y)$.

###### Example 4.2.2

Let us consider the case proposed in 4.1.5 with $x=(3,-2,1)$ and $y=(1,-1,1)$.
In such a case $I=\\{1,2\\}$ and $J=\\{3\\}$. We have:

$x\boxminus y=\begin{pmatrix}3\\\ -2\\\
1\end{pmatrix}\boxminus\begin{pmatrix}1\\\ -1\\\
1\end{pmatrix}=\begin{pmatrix}3\\\ -2\\\ 0\end{pmatrix}$

and

$x_{\\{3\\}}=y_{\\{3\\}}=\begin{pmatrix}0\\\ 0\\\ 1\end{pmatrix}.$

It follows that

$D^{\infty}(x,y)\supset\begin{pmatrix}0\\\ 0\\\
1\end{pmatrix}+\left\\{t\begin{pmatrix}3\\\ -2\\\
0\end{pmatrix},t\in\mathbb{R}\right\\}=\left\\{\begin{pmatrix}3t\\\ -2t\\\
1\end{pmatrix}:t\in\mathbb{R}\right\\}=\mathcal{D}_{+}(x,y)$

and

$D^{\infty}(x,y)\supset\begin{pmatrix}0\\\ 0\\\
1\end{pmatrix}+\left\\{t\begin{pmatrix}-3\\\ 2\\\
0\end{pmatrix},t\in\mathbb{R}\right\\}=\left\\{\begin{pmatrix}-3t\\\ 2t\\\
1\end{pmatrix}:t\in\mathbb{R}\right\\}=\mathcal{D}_{-}(x,y).$

###  4.3 Parallel Lines

From Proposition 4.2.1 every $\digamma$-lines contain two half-lines
components. Let $\mathcal{D}^{\infty}(x,y)$ and $\mathcal{E}^{\infty}(u,v)$ be
two $\digamma$-lines respectively spanned from $x$ and $y$ and $u$ and $v$. We
say that $\mathcal{E}^{\infty}(u,v)$ and $\mathcal{D}^{\infty}(x,y)$ are
$\digamma$-parallel if $x\boxminus y\propto u\boxminus v$. Equivalently this
means that $x\boxminus y$ and $u\boxminus v$ are collinear, that is there is
some $\alpha\in\mathbb{R}\backslash\\{0\\}$ such that $(x\boxminus
y)=\alpha(u\boxminus v)$. From Proposition 4.2.1 there exist two half lines
components $\mathcal{E}_{-}(u,v)$ and $\mathcal{E}_{+}(u,v)$ of
$\mathcal{E}^{\infty}(u,v)$ and two half lines components
$\mathcal{D}_{-}(x,y)$ and $\mathcal{D}_{+}(x,y)$ of
$\mathcal{D}^{\infty}(x,y)$ such that
$\mathcal{E}_{+}(u,v)\supset(\subset)\;\mathcal{D}_{+}(x,y)$ and
$\mathcal{E}_{-}(u,v)\supset(\subset)\;\mathcal{D}_{-}(x,y)$.

$x_{1}$$x_{2}$$0$Figure 4.3Parallel
$\digamma$-lines.$\mathcal{D}^{\infty}$$\mathcal{E}_{\infty}$$\mathcal{E}_{-}$$\mathcal{D}_{+}$$\mathcal{D}_{-}$$\mathcal{E}_{+}$$u$$x$$v$$y$

Figure 4.3 depicts the case of two parallel lines. In Figure 4.4 we consider
the case of two secant lines.

$x_{1}$$x_{2}$$0$Figure 4.4Secant
$\digamma$-lines.$\mathcal{D}^{\infty}$$\mathcal{E}_{\infty}$$x$$y$$u$$v$

Traditional axioms of planar Euclidean geometry postulate that: $(a_{1})$ a
straight line may be drawn between any two points; $(a_{2})$ any terminated
straight line may be extended indefinitely; $(a_{3})$ a circle may be drawn
with any given point as center and any given radius, $(a_{4})$ all right
angles are equal; $(v)$ for any given point not on a given line, there is
exactly one line through the point that does not meet the given line.

It is well known that the fifth axiom does not follow from the first four.
Clearly the geometric implications of the algebraic structure proposed in this
paper do not conflict with the first 4 axioms. Not surprisingly, this is not
the case for the fifth. In particular:

$(i)$ Two distinct parallel lines may have an infinity of common points
(Figure 4.3);

$(ii)$ There may exist an infinity of $\digamma$-lines passing by two distinct
points $x$ and $y$ (Figure 4.3);

$(iii)$ Two secant lines may have an infinity of common points (Figure 4.4);

$(iv)$ A $\digamma$-line may not be spanned by two distinct points it contains
(Figure 4.4).

In the following a system of inequations characterizing a $\digamma$-line that
passes from two distinct points is provided. This we do by considering a
suitable notion of determinant introduced in [9]. The next intermediary result
establishes an algebraic description of the limit of a sequence of
$\varphi_{p}$-hyperplanes. The next result was established in [10].

###### Proposition 4.3.1

Let $V$ be the $n\times n$ matrix with $v_{i}$ as $i$-th column for each $i$.
Let $V_{(i)}$ be the matrix obtained from $V$ by replacing line $i$ with the
transpose of the unit vector $1\\!\\!1_{n}$. Suppose that
$|V|_{\infty}\not=0$. Then

$\operatorname*{\mathrm{Lim}}_{p\longrightarrow\infty}H_{p}(V)=\Big{\\{}x\in\mathbb{R}^{n}:\operatorname*{\text{\Large${\stackrel{{\scriptstyle-}}{{\smile}}}$}}_{i\in[n]}|V_{(i)}|_{\infty}x_{i}\leq|V|_{\infty}\leq\operatorname*{\text{\Large${\stackrel{{{}_{+}}}{{\smile}}}$}}_{i\in[n]}|V_{(i)}|_{\infty}x_{i}\Big{\\}}.$

Given two distinct points $x,y\in\mathbb{R}^{2}$, every points
$z\in\mathcal{D}(x,y)$ satisfy the relation:

$|z{-}x,z{-}y|=\left|\begin{matrix}z_{1}-x_{1}&z_{1}-y_{1}\\\
z_{2}-x_{2}&z_{2}-y_{2}\end{matrix}\right|=0.$ (79)

Equivalently, we have

$(x_{2}-y_{2})z_{1}+(y_{1}-x_{1})z_{2}=|x,y|.$ (80)

The following proposition establishes an analogous result in the case of the
$\digamma$-line $\mathcal{D}^{\infty}$. This algebraic characterization is an
immediate consequence of Proposition 4.3.1 (see [10]). Let
$x,y\in\mathbb{R}^{2}$ with $x\not=y$. Then:

$\mathcal{D}^{\infty}(x,y)=\Big{\\{}z\in\mathbb{R}^{2}:(x_{2}\boxminus
y_{2})z_{1}\stackrel{{\scriptstyle-}}{{\smile}}(y_{1}\boxminus
x_{1})z_{2}\leq|x,y|_{\infty}\leq(x_{2}\boxminus
y_{2})z_{1}\stackrel{{\scriptstyle+}}{{\smile}}(y_{1}\boxminus
x_{1})z_{2}\Big{\\}}.$ (81)

In the following, we show that the equations of parallel lines in the plane
obey algebraic rules analogous to those of the linear case.

###### Lemma 4.3.2

Let $\mathcal{D}^{\infty}$ and $\mathcal{E}^{\infty}$ be two parallel
$\digamma$-lines of $\mathbb{R}^{2}$. Then there exists $a\in\mathbb{R}^{2}$,
$c,d\in\mathbb{R}$ such that:

$\mathcal{D}^{\infty}=\\{z\in\mathbb{R}^{2}:\langle
a,z\rangle_{\infty}^{-}\leq c\leq\langle a,z\rangle_{\infty}^{+}\leq c\\}$

and

$\mathcal{E}^{\infty}=\\{z\in\mathbb{R}^{2}:\langle
a,z\rangle_{\infty}^{-}\leq d\leq\langle a,z\rangle_{\infty}^{+}\leq c\\}.$

Proof: Suppose now that $\mathcal{D}^{\infty}$ and $\mathcal{E}^{\infty}$ are
two $\digamma$-line respectively spanned from $x,y$ and $u,v$, where
$x,y,u,v\in\mathbb{R}^{2}$. By definition we have:

$\mathcal{E}^{\infty}=\Big{\\{}z\in\mathbb{R}^{2}:(u_{2}\boxminus
v_{2})z_{1}\stackrel{{\scriptstyle-}}{{\smile}}(v_{1}\boxminus
u_{1})z_{2}\leq|u,v|_{\infty}\leq(u_{2}\boxminus
v_{2})z_{1}\stackrel{{\scriptstyle+}}{{\smile}}(v_{1}\boxminus
u_{1})z_{2}\Big{\\}}.$

If $\mathcal{D}^{\infty}$ and $\mathcal{E}^{\infty}$ are parallel then
$(u\boxminus v)\propto(x\boxminus y)$. Therefore there is a real number
$\alpha\not=0$ such that $(u\boxminus v)=\alpha(x\boxminus y)$. Hence:

$\mathcal{E}^{\infty}=\Big{\\{}z\in\mathbb{R}^{2}:\alpha(x_{2}\boxminus
y_{2})z_{1}\stackrel{{\scriptstyle-}}{{\smile}}\alpha(y_{1}\boxminus
x_{1})z_{2}\leq|u,v|_{\infty}\leq\alpha(x_{2}\boxminus
y_{2})z_{1}\stackrel{{\scriptstyle+}}{{\smile}}\alpha(y_{1}\boxminus
x_{1})z_{2}\Big{\\}}.$

Suppose that $\alpha>0$, then:

$\mathcal{E}^{\infty}=\Big{\\{}z\in\mathbb{R}^{2}:(x_{2}\boxminus
y_{2})z_{1}\stackrel{{\scriptstyle-}}{{\smile}}(y_{1}\boxminus
x_{1})z_{2}\leq\alpha^{-1}|u,v|_{\infty}\leq(x_{2}\boxminus
y_{2})z_{1}\stackrel{{\scriptstyle+}}{{\smile}}(y_{1}\boxminus
x_{1})z_{2}\Big{\\}}.$

We deduce the result taking $a=x\boxminus y$, $c=|x,y|_{\infty}$, and
$d=\alpha^{-1}|u,v|_{\infty}$. If $\alpha<0$, then:

$\mathcal{E}^{\infty}=\Big{\\{}z\in\mathbb{R}^{2}:(x_{2}\boxminus
y_{2})z_{1}\stackrel{{\scriptstyle-}}{{\smile}}(y_{1}\boxminus
x_{1})z_{2}\geq\alpha^{-1}|u,v|_{\infty}\geq(x_{2}\boxminus
y_{2})z_{1}\stackrel{{\scriptstyle+}}{{\smile}}(y_{1}\boxminus
x_{1})z_{2}\Big{\\}}.$

Since $(x_{2}\boxminus
y_{2})z_{1}\stackrel{{\scriptstyle-}}{{\smile}}(y_{1}\boxminus
x_{1})=-\big{(}(y_{2}\boxminus
x_{2})z_{1}\stackrel{{\scriptstyle+}}{{\smile}}(x_{1}\boxminus y_{1})\big{)}$
and $(x_{2}\boxminus
y_{2})z_{1}\stackrel{{\scriptstyle+}}{{\smile}}(y_{1}\boxminus
x_{1})=-\big{(}(y_{2}\boxminus
x_{2})z_{1}\stackrel{{\scriptstyle-}}{{\smile}}(x_{1}\boxminus y_{1})\big{)}$
we deduce that:

$\mathcal{E}^{\infty}=\Big{\\{}z\in\mathbb{R}^{2}:(y_{2}\boxminus
x_{2})z_{1}\stackrel{{\scriptstyle-}}{{\smile}}(x_{1}\boxminus
y_{1})z_{2}\leq|\alpha|^{-1}|u,v|_{\infty}\leq(y_{2}\boxminus
x_{2})z_{1}\stackrel{{\scriptstyle+}}{{\smile}}(x_{1}\boxminus
y_{1})z_{2}\Big{\\}}.$

Now, note that since $\mathcal{D}^{\infty}(x,y)=\mathcal{D}^{\infty}(y,x)$ we
also have

$\mathcal{D}^{\infty}=\Big{\\{}z\in\mathbb{R}^{2}:(y_{2}\boxminus
x_{2})z_{1}\stackrel{{\scriptstyle-}}{{\smile}}(x_{1}\boxminus
y_{1})z_{2}\leq|y,x|_{\infty}\leq(y_{2}\boxminus
x_{2})z_{1}\stackrel{{\scriptstyle+}}{{\smile}}(x_{1}\boxminus
y_{1})z_{2}\Big{\\}}.$

Hence, taking $a=y\boxminus x$, $c=|y,x|_{\infty}$ and
$d=|\alpha|^{-1}|u,v|_{\infty}$ yields the result. $\Box$

###### Example 4.3.3

Suppose that $x=(3,1)$, $y=(1,-2)$, $u=(-2,4)$ and $v=(-6,1)$. We have:

$x\boxminus y=\begin{pmatrix}3\boxminus 1\\\
1\boxminus(-2)\end{pmatrix}=\begin{pmatrix}3\\\
2\end{pmatrix}\quad\text{and}\quad u\boxminus
v=\begin{pmatrix}-2\boxminus(-6)\\\ 4\boxminus
1\end{pmatrix}=\begin{pmatrix}6\\\ 4\end{pmatrix}.$

It follows that $u\boxminus v=2(x\boxminus y).$ Moreover

$|x,y|_{\infty}=3\cdot(-2)\boxminus 1\cdot
1=-6\quad\text{and}\quad|u,v|_{\infty}=(-2)\cdot 1\boxminus(-6)\cdot 4=24.$

Therefore:

$\mathcal{D}^{\infty}(x,y)=\Big{\\{}z\in\mathbb{R}^{2}:\big{(}1\boxminus(-2)\big{)}z_{1}\stackrel{{\scriptstyle-}}{{\smile}}(1\boxminus
3)z_{2}\leq-6\leq(1\boxminus\big{(}-2\big{)})z_{1}\stackrel{{\scriptstyle+}}{{\smile}}(1\boxminus
3)z_{2}\Big{\\}}.$

Hence:

$\mathcal{D}^{\infty}(x,y)=\Big{\\{}z\in\mathbb{R}^{2}:2z_{1}\stackrel{{\scriptstyle-}}{{\smile}}(-3)z_{2}\leq-6\leq
2z_{1}\stackrel{{\scriptstyle+}}{{\smile}}(-3)z_{2}\Big{\\}}.$

In addition we have:

$\mathcal{D}^{\infty}(u,v)=\Big{\\{}z\in\mathbb{R}^{2}:\big{(}4\boxminus(-1)\big{)}z_{1}\stackrel{{\scriptstyle-}}{{\smile}}\big{(}(-6)\boxminus(-2)\big{)}z_{2}\leq-6\leq\big{(}4\boxminus(-1)\big{)}z_{1}\stackrel{{\scriptstyle+}}{{\smile}}\big{(}(-6)\boxminus(-2)\big{)}z_{2}\Big{\\}}.$

Hence:

$\mathcal{D}^{\infty}(u,v)=\Big{\\{}z\in\mathbb{R}^{2}:(-4)z_{1}\stackrel{{\scriptstyle-}}{{\smile}}6z_{2}\leq
24\leq(-4)z_{1}\stackrel{{\scriptstyle+}}{{\smile}}6z_{2}\Big{\\}}.$

Equivalently, by changing the sign and simplifying by 2:

$\mathcal{D}^{\infty}(u,v)=\Big{\\{}z\in\mathbb{R}^{2}:2z_{1}\stackrel{{\scriptstyle-}}{{\smile}}(-3)z_{2}\leq-12\leq
2z_{1}\stackrel{{\scriptstyle+}}{{\smile}}(-3)z_{2}\Big{\\}}.$

We can notice that the coefficients multiplying $z_{1}$ and $z_{2}$ are the
same the difference relating only to the constant term.

## Conclusion

In this paper we have investigated some limit geometric properties arising
from the deformation of a vector space structure defined on a scalar field.
Among other things we have paid a special attention to the notions of lines
that can be viewed as special limits of algebraic varieties. It would be
useful in future works to alayse the connections between this notion and that
one thait is usualy derived from Max-Plus algebra traditionally used in the
formalism of tropical mathematics initiated in [25] and [23].

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