On the Structure of Topological Spaces

Abstract

The structure of topological spaces is analysed here through the lenses of fibrous preorders. Each topological space has an associated fibrous preorder and those fibrous preorders which return a topological space are called spatial. A special class of spatial fibrous preorders consisting of an interconnected family of preorders indexed by a unitary magma is called Cartesian and is studied here. Topological spaces that are obtained from those fibrous preorders, with a unitary magma I, are called I-Cartesian and are characterized. The characterization reveals a hidden structure on such spaces. Several other characterizations are obtained, and special attention is drawn to the case of a monoid equipped with a topology. A wide range of examples is provided, as well as general procedures to obtain topologies from other data types such as groups and their actions. Metric spaces and normed spaces are considered as well.

1. Introduction

The definition of a topological space as we know it today has a long history (see, e.g., [1]), and it is so rich and full of twists and turns that the simple task of tracking down its origins is transformed into an overwhelming undertaking (see, e.g., [2]). It arguably starts at the beginning of the twentieth century with seminal works by Dedekind [3], Lebesgue [4], Riesz [5], de la Vallé Poussin [6], and Frechét [7,8], whose primary interests were still focused on generalizing results from the previous century. The first few decades were characterized by several further improvements and alternative definitions, most notably by Hausdorff [9,10], Carathéodory [11], Kuratowski [12], Tietze [13], and Aleksandrov [14], while the vision of which definitions would be adopted as primitive and which concepts ought to be derived was nothing but a blur. It was only in the mid 1930s that the works by Aleksandrov and Hopf [15], Sierpinski [16], Kuratowski [17], and Lefschetz [18,19] started to become influential. The current established notion has finally settled with the wide dissemination of the classical works by Bourbaki [20] and Kelley [21]. From this modern point of view, a topological space is presented as a pair $(X,\tau )$, where X is a set and $\tau \subseteq \mathcal{P}\left(X\right)$ is a topology on X, i.e., a collection of so-called open sets that are nothing but subsets of X, closed under finite intersections and arbitrary unions, and moreover, at least the empty set and the set X itself must be open.

In spite of all the advantages of such an abstract definition, which has unquestionably led to great progress over the last more than hundred years not only in mathematics but also in physics and in other areas of knowledge, there are still a few difficulties in the treatment of topological spaces at this level of abstraction. The trouble is that the process of simplifying the definition of topological space to its bare bones has two undesirable effects: one is the appearance of redundant information, another one is that its true structure becomes hidden. Let us mention two concrete examples of this phenomenon.

If X is a finite set, then, as is well known [22], a topology on X is nothing but a preorder, which is simply a reflexive and transitive relation. Should this not be a simple observation that would follow from the definition of a topological space?

If X is a group and if the map $X\times X\to X$; $(x,y)\mapsto x{y}^{-1}$ is required to be continuous (with the product topology on its domain and the group operation used to form $x{y}^{-1}$), then we get a topological group [23] and the range of possible topologies is severely restricted. Surprisingly, a topological group is presented as a group equipped with an arbitrary topology (which is required to be compatible with the group operation), and it is clear that such topologies must be simpler than arbitrary ones. This is mainly because the simplification of the structure of those topologies is not apparent from its definition. Should there not be a way of identifying which key features of an arbitrary topology gives compatibility with a group operation?

Some attempts have been made to overcome these difficulties, most notably by Brown in the book Topology and Groupoids [24]. More recent works [25,26] account for great progress in the field, and yet there is no simple description for the category of topological groups. The aim of this work is to develop the tools for establishing a categorical equivalence between topological groups and a category whose objects are groups equipped with a subset (interpreted as a neighbourhood of identity) of which Proposition 4 and Theorem 7 are the first steps. The desired result would be comparable to the well-known equivalence between categorical groups and crossed modules [27]. A description of metric spaces in terms of a lax-left-associative Mal’tsev operation is obtained as a byproduct in Section 5, whereas in Section 6, a procedure that transforms a monoid B with an indexed family of subsets $\left(S_n\right)$ into a topological space $(B,\tau )$ in which each $S_n$ is an open neighbourhood around the origin is detailed. Concrete examples are provided in Section 7.

In [28], the notion of spatial fibrous preorder was used to structure the category of topological spaces so that it becomes apparent that every preorder gives rise to a topology, and moreover, for finite sets, there are no possibilities other than that. Here, we use some of those ideas in the context of a better understanding of the structure of metric spaces and topological groups. Of course, continuous morphisms must be taken into account, but we choose not to include their study here because the classical notion of continuity is seen as a property, whereas we plan to consider it as an extra structure. While doing so, we are able to consider different levels of continuity, such as uniform continuity and related concepts. However, at this moment, that study would lead us too far astray.

We choose to work with spatial fibrous preorders due to their intuitive interpretation as modified preorders and due to the categorical equivalence between spatial fibrous preorders and topological spaces [28]. Nevertheless, a reader who is not familiar with spatial fibrous preorders in the first place may fail to consider it an appealing structure. For that reason, we have decided to present here a simpler version that is nonetheless still sufficient for our purposes. In order to distinguish it from the general case, we will call it Cartesian spatial fibrous preorder (Section 2).

A fibrous preorder [28] is a sequence $R\to A\to B$ with $R\subseteq A\times B$ satisfying some conditions. If the set A is the Cartesian product of a set I and the set B, then we will speak of a Cartesian fibrous preorder. Not every fibrous preorder is realizable as a topological space; only the spatial ones are so [28]. In the same manner, we will restrict Cartesian fibrous preorders to spatial ones and simplify their structure by considering a unitary magma as its indexing set, I, rather than the slightly more general structure considered in [28]. Note that when $I=\left\{1\right\}$ is a singleton set, $A=I\times B$ can be identified with B and the relation $R\subseteq B\times B$ becomes precisely a preorder. A characterization of these topological spaces that are obtained from Cartesian spacial fibrous preorders is given in Section 3, whereas examples are provided in Section 4.

2. Cartesian Spatial Fibrous Preorders and Unitary Magmas

Let $(I,·,1)$ be a unitary magma that is a set I together with a distinguished element $1\in I$ and a binary operation

$$I\times I\to I;(i,j)\mapsto i·j,$$

such that $i·1=i=1·i$ for all $i\in I$. Sometimes, we write $i·j$ simply as $ij$. A unitary magma is the same as a monoid when the associativity condition $i\left(jk\right)=\left(ij\right)k$ holds true for all $i,j,k\in I$.

Definition 1.

A Cartesian spatial fibrous preorder, indexed by the unitary magma $(I,·,1)$, is a system $(X,{\left({\le }^i\right)}_{i\in I},{\left({\partial }^i\right)}_{i\in I})$ where X is a set and for every $i\in I$, ${\le }^i$ is a binary relation on X, whereas ${\partial }^i$ is a partial map $X\times X\to I$, which is defined for all pairs $(x,y)$ such that $x{\le }^{i}y$. Moreover, the following conditions must hold for all $x,y,z\in X$ and $i,j\in I$:

(C1) 

$x{\le }^{i}x$;

(C2) 

if $x{\le }^{i}y$, ${\partial }^i(x,y)=j$ and $y{\le }^{j}z$, then $x{\le }^{i}z$;

(C3) 

if $x{\le }^{ij}y$ then $x{\le }^{i}y$ and $x{\le }^{j}y$.

A metric space is the best example to illustrate the structure while providing useful intuition further on.

Example 1.

Let $I=\mathbb{N}$ be the unitary magma of natural numbers with $1\in \mathbb{N}$ as the neutral element and the usual multiplication as a binary operation. Let X be any metric space with metric $d:X\times X\to [0,+\infty ]$. Under these assumptions, we put $x{\le }^{n}y$ if and only if $d(x,y)<\frac{1}{n}$ and choose ${\partial }^n(x,y)=k$ to be such that $\frac{1}{k}\le \frac{1}{n}-d(x,y)$, with $x,y\in X$ and $n,k\in \mathbb{N}$. It is not difficult to see that conditions (C1)–(C3) are satisfied.

As for metric spaces, Cartesian spatial fibrous preorders give rise to topological spaces.

Proposition 1.

Every Cartesian spatial fibrous preorder

$$(X,{\left({\le }^i\right)}_{i\in I},{\left({\partial }^i\right)}_{i\in I})$$

gives rise to a topological space $(X,\tau )$ with τ defined as

$$\mathcal{O}\in \tau \iff \forall x\in \mathcal{O},\exists i\in I,N(i,x)\subseteq \mathcal{O}$$

(1)

where $N(i,x)=\{y\in X\mid x{\le }^{i}y\}$.

Proof. 

This is a special case of the equivalence betwen spatial fibrous preorders and topological spaces [28]. With a Cartesian spatial fibrous preorder $(X,{\left({\le }^i\right)}_{i\in I},{\left({\partial }^i\right)}_{i\in I})$, we define a spatial fibrous preorder $(R,A,B,\partial ,p,s,m)$ (see [28]) as:

$$\begin{array}{c}\hfill R\subseteq I\times X\times X,A=I\times X,B=X\end{array}$$

(2)

with R, $\partial :R\to A$, $p:A\to B$, $s:B\to A$ and $m:A{\times }_{B}A\to A$, defined as follows:

$$\begin{array}{c}\hfill R=\{(i,x,y)\mid x{\le }^{i}y\},\\ \hfill \partial (i,x,y)=({\partial }^i(x,y),y),\\ \hfill p(i,x)=x,\\ \hfill s\left(x\right)=(1,x),\\ \hfill m(i,j,x)=(i·j,x).\end{array}$$

A direct proof is easily obtained by showing that $N(i,x)$ is a system of open neighbourhoods instead. □

It is clear that when a Cartesian spatial fibrous preorder is obtained from a metric space, then its induced topology is the same as the usual topology generated by the metric.

3. A Characterization of $\mathit{I}$-Cartesian Spaces

The following result characterizes those topological spaces that are obtained from a Cartesian spatial fibrous preorder. Such spaces are called I-Cartesian when I is the indexing unitary magma. Most of the time, we will consider an arbitrary unitary magma, I; however, it is useful from time to time to recall that our motivating example is the unitary magma of natural numbers. For that reason, we will use the letters $i,j,k$ to represent elements in the set I as well as the letters $n,m,k$.

Theorem 2.

Let $(I,·,1)$ be a unitary magma and $(X,\tau )$ a topological space. The following conditions are equivalent:

(a) 

The space X is I-Cartesian.

(b) 

The topology τ is determined by a Cartesian spatial fibrous preorder $(X,{\left({\le }^i\right)}_{i\in I},{\left({\partial }^i\right)}_{i\in I})$ as

$$\mathcal{O}\in \tau \iff \forall x\in \mathcal{O},\exists i\in I,N(i,x)\subseteq \mathcal{O}$$

(3)

with $N(i,x)=\{y\in X\mid x{\le }^{i}y\}$.

(c) 

There exists a map $N:I\times X\to \mathcal{P}\left(X\right)$ such that

(i) 

$x\in N(n,x)$, for all $n\in I$, $x\in X$

(ii) 

$N(n,x)\subseteq \{y\in X\mid \exists k\in I,N(k,x)\subseteq N(n,x\left)\right\}$,

(iii) 

$N(n·m,x)\subseteq N(n,x)\cap N(m,x)$, $n,m\in I$, $x\in X$

for which τ is determined as

$$\mathcal{O}\in \tau \iff \forall x\in \mathcal{O},\exists n\in I,N(n,x)\subseteq \mathcal{O}.$$

(d) 

There exists a ternary relation $R\subseteq I\times X\times X$ together with a map $p:R\to I$ such that

(i) 

$(n,x,x)\in R$, for all $n\in I$, $x\in X$

(ii) 

if $(n,x,y)\in R$ and $\left(p\right(n,x,y),y,z)\in R$ then $(n,x,z)\in R$

(iii) 

if $(n·m,x,y)\in R$ then $(n,x,y)\in R$ and $(m,x,y)\in R$

for which τ is determined as

$$\mathcal{O}\in \tau \iff \forall x\in \mathcal{O},\exists n\in I,N_R(n,x)\subseteq \mathcal{O}$$

with $N_R(n,x)=\{y\in X\mid (n,x,y)\in R\}$.

(e) 

There are maps $\eta :I\times X\to \tau$ and $\gamma :\left\{\right(U,x)\mid x\in U\in \tau \}\to I$ such that:

(i) 

$x\in \eta (n,x)$ for every $n\in I$ and $x\in X$,

(ii) 

$\eta \left(\gamma \right(U,x),x)\subseteq U$, for all $x\in U\in \tau$,

(iii) 

$\eta (n·m,x)\subseteq \eta (n,x)\cap \eta (m,x)$, for all $n,m\in I$, and $x\in X$.

Proof. 

Conditions $\left(a\right)$ and $\left(b\right)$ are equivalent by definition.

To prove that $\left(b\right)$ implies $\left(c\right)$, we start with a Cartesian spatial fibrous preorder and define $N(n,x)=\{y\in X\mid x{\le }^{n}y\}$. Conditions $\left(c\right)\left(i\right)$ and $\left(c\right)\left(iii\right)$ follow, respectively, from axioms $\left(C1\right)$ and $\left(C3\right)$. To prove $\left(c\right)\left(ii\right)$, we start with $y\in N(n,x)$, that is, $x{\le }^{n}y$, and observe that there exists $k={\partial }^n(x,y)\in I$ such that $N(k,y)\subseteq N(n,x)$. Indeed, if $z\in N(k,y)$, that is, $y{\le }^{k}z$, then by $\left(C2\right)$ we have $x{\le }^{n}z$, which is the same as saying $z\in N(n,x)$. This shows that the map $N:I\times X\to \mathcal{P}\left(X\right)$ satisfies $\left(c\right)\left(ii\right)$. It remains to show that $\tau$ is determined by it. This follows from Proposition 1 and the assumption that $\tau$ is determined as in Equation (3).

In order to prove that $\left(c\right)$ implies $\left(d\right)$, we define

$$R=\left\{\right(n,x,y)\in I\times X\times X\mid y\in N(n,x\left)\right\}$$

and put $p(n,x,y)=k$ for some $k\in I$ such that $N(k,y)\subseteq N(n,x)$, which exists by assumption on condition $\left(c\right)\left(ii\right)$. Once again, conditions $\left(d\right)\left(i\right)$ and $\left(d\right)\left(iii\right)$ are direct consequences of $\left(c\right)\left(i\right)$ and $\left(c\right)\left(iii\right)$, respectively. To see that $\left(d\right)\left(ii\right)$ is satisfied, we observe that if $y\in N(n,x)$ and $z\in N\left(p\right(n,x,y),y)$, then, by definition of $p(n,x,y)$, we have $N(k,y)\subseteq N(n,x)$. It follows that $z\in N(n,x)$ and hence $\left(d\right)\left(ii\right)$ is satisfied. Having R, we define $N_R=\{y\mid (n,x,y)\in R\}=\{y\in N(n,x)\}=N(n,x)$, and so the topology $\tau$ is obtained by $N_R=N$.

To prove that $\left(d\right)$ implies $\left(e\right)$, define $\eta (i,x)=\{y\in X\mid (n,x,y)\in R\}=N_R(n,x)$ and put

$$\gamma \left(\right(U,x\left)\right)=k$$

for some $k\in I$ such that $N_R(k,x)\subseteq U$, which exists by the assumption that $\tau$ is generated by $N_R$. The map $\eta :I\times X\to \tau$ is well defined because each $\eta (n,x)\in \tau$. Indeed, if $y\in \eta (n,x)$, that is $(n,x,y)\in R$, then there exists $k=p(n,x,y)\in I$ for which $N_R(k,y)\subseteq N_R(n,x)$. This is a consequence of $\left(d\right)\left(ii\right)$. If $z\in N_R(k,y)\iff (k,y,z)\in R$, then, given that $(n,x,y)\in R$ and $k=p(n,x,y)$, we have $(n,x,y)\in R$ or, in other words, $z\in N_R(n,x)$. This shows that each $\eta (n,x)$ is open in $\tau$. Conditions $\left(e\right)\left(i\right)$ and $\left(e\right)\left(iii\right)$ are a direct consequence of $\left(d\right)\left(i\right)$ and $\left(d\right)\left(iii\right)$, respectively. To show $\left(e\right)\left(ii\right)$, we observe that if $y\in \eta \left(\gamma \right(U,x),x)$, then $\left(\gamma \right(U,x),x,y)\in R$, but $\gamma (U,x)=k$, for some $k\in I$ such that $N_R(k,x)\subseteq U$. Since $y\in N_R(k,x)\iff (k,x,y)\in R$, we conclude that $y\in U$. This shows that $\eta \left(\gamma \right(U,x),x)\subseteq U$.

Finally, we prove that $\left(e\right)$ implies $\left(b\right)$. Having $\eta$ and $\gamma$, it is not difficult to see that a Cartesian spatial fibrous preorder is obtained if we let

$$\begin{array}{c}\hfill x{\le }^{i}y\iff y\in \eta (i,x)\\ \hfill {\partial }^i(x,y)=\gamma (\eta (i,x),y)\end{array}$$

with $i\in I$ and $x,y\in X$. Indeed, axioms $\left(C1\right)$ and $\left(C3\right)$ follow, respectively, from $\left(e\right)\left(i\right)$ and $\left(e\right)\left(iii\right)$. For $\left(C2\right)$, let us suppose $x{\le }^{i}y$, that is, $y\in \eta (i,x)$, and let us suppose $y{\le }^{k}z$ with $k={\partial }^i(x,y)=\gamma \left(\eta (i,x,y)\right)$. This means that $z\in \eta (k,y)$ and by condition $\left(e\right)\left(ii\right)$, we have $\eta \left(\gamma \right(\eta (i,x),y),y)\subseteq \eta (n,x)$; thus we have $z\in \eta (i,x)$, or $x{\le }^{i}z$ as desired.

It remains to show that the topology $\tau$ is recovered as prescribed in (3). On the one hand, if $\mathcal{O}\in \tau$ and if $x\in \mathcal{O}$, then there is $k=\gamma (\mathcal{O},x)$ with $N(k,x)\subseteq \mathcal{O}$. This means that every open set in $\tau$ is generated as in condition (3). To see the converse, let us consider any subset $\mathcal{O}\subseteq X$ and suppose it has the property that for all $x\in \mathcal{O}$ there is some $k=k\left(x\right)\in I$ with $N(k,x)=\eta (k,x)\subseteq \mathcal{O}$. We have to show that $\mathcal{O}\in \tau$. This follows because

$$\mathcal{O}=\bigcup _{x\in \mathcal{O}}N(k\left(x\right),x)$$

and every $N(k,x)=\eta (k,x)\in \tau$. □

We immediately observe some interesting special cases, namely when $I=\left\{1\right\}$ is the trivial unitary magma, or when $I=(\mathbb{N},·,1)$ is the unitary magma (monoid) of natural numbers with the usual multiplication. Furthermore, as we will see in Section 6, the map $p:R\to I$ of condition $\left(d\right)$ in the previous result can sometimes be decomposed as $p(n,x,y)=\beta (x,y)·n$ in which $\beta (x,y)=\gamma \left(\eta \right(1,x),y)$ with $\eta$ and $\gamma$ as in condition $\left(e\right)$ above.

In a sequel to this work, our attention will be turned to morphisms between fibrous preorders and on how they can be defined internally to any category with finite limits.

For the moment, let us briefly mention that a morphism between Cartesian spatial fibrous preorders, say from

$$(X,{\left({\le }^i\right)}_{i\in I},{\left({\partial }^i\right)}_{i\in I})$$

to

$$(Y,{\left({\le }^i\right)}_{i\in I},{\left({\partial }^i\right)}_{i\in I}),$$

consists of a map $f:X\to Y$ and a family of maps ${(g_j:X\to I)}_{j\in I}$ such that

$$x{\le }^{g_j\left(x\right)}y\Rightarrow f\left(x\right){\le }^{j}f\left(y\right)$$

(4)

for all $x,y\in X$ and $j\in I$. Note that when each $g_j$ can be chosen independently of x, the family ${\left(g_j\right)}_{j\in I}$ may be seen as a single map $g:I\to I$, and this is comparable to the notion of uniform continuity (see the example of metric spaces presented above). It is also clear that morphisms between Cartesian spatial fibrous preorders may be considered for different indexing unitary magmas. This and other considerations will be explored in a sequel to this work.

To date, we have isolated one basic ingredient in the structure of those topological spaces that are obtained as Cartesian spatial fibrous preorders: a unitary magma. Note that every ascending chain on a set I makes it a unitary magma with the smallest element as neutral and as a binary operation the procedure of choosing the greatest between any two.

We now ask: Besides a unitary magma, is it possible that the structure of a system $(X,{\left({\le }^i\right)}_{i\in I},{\left({\partial }^i\right)}_{i\in I})$, inducing a topology on X can further be decomposed into simpler structures that are hence easier to analyse?

The answer to this question, as we will see, is yes, provided only that we restrict its generality a bit further. Nevertheless, we are still able to capture the majority of examples which we are interested in, namely preorders, metric spaces, and certain special classes of groups and monoids with a topology (see Section 5 and Section 6). First, let us analyse the concrete examples.

4. Some Examples of Cartesian Spatial Fibrous Preorders

Let us start with two simple cases of interest as unitary magmas. The trivial (singleton) unitary magma, and the unitary magma of natural numbers with usual multiplication as its binary operation.

It is clear that a Cartesian spatial fibrous preorder, indexed by a singleton unitary magma $I=\left\{1\right\}$, is nothing but a preorder. Indeed, in this case, we have ${\partial }^i(x,y)=i$ with $i=1$ and hence condition (C2) becomes transitive. Condition (C1) asserts reflexivity, while condition (C3) is trivial.

Proposition 3.

The category of preorders is equivalent to the category of 1-Cartesian topological spaces, which are the same as Aleksandrov or discrete spaces.

Proof. 

Preorders are precisely Cartesian spatial fibrous preorders indexed by a singleton unitary magma. Moreover, if $I=\left\{1\right\}$, then condition (4) asserts the monotonicity of the map f. The result follows from Theorem 2. It is well-known that Aleksandrov spaces, or discrete spaces, are the same as preorders [22] (see also [28]). □

When $I=(\mathbb{N},·,1)$ is the unitary magma of natural numbers with usual multiplication, we have several interesting cases.

A natural space [28] is essentially a first-countable topological space. It consists of a pair $(X,N)$ with X a set and $N:\mathbb{N}\times X\to \mathcal{P}\left(X\right)$, a map into the power-set of X satisfying the following three conditions for all $x\in X$, and $i,j\in \mathbb{N}$:

$$\begin{array}{}\mathrm{(5)}& x\in N(i,x)\mathrm{(6)}& N(i,x)\subseteq \{y\in X\mid \exists j\in \mathbb{N},N(j,y)\subseteq N(i,x\left)\right\}\mathrm{(7)}& N(ij,x)\subseteq N(i,x)\cap N(j,x).\end{array}$$

Natural spaces are precisely $\mathbb{N}$-Cartesian spaces (see item $\left(c\right)$ of Theorem 2).

In a similar manner, every metric space $(X,d)$ gives rise to a Cartesian spatial fibrous preorder indexed by the natural numbers as already presented. Put $x{\le }^{n}y$ whenever $d(x,y)<\frac{1}{n}$ and define ${\partial }^n(x,y)=k$ such that $\frac{1}{k}\le \frac{1}{n}-d(x,y)$.

Similarly, every normed vector space gives rise to a Cartesian spatial fibrous preorder indexed by the natural numbers. We simply put $x{\le }^{n}y$ whenever $\parallel y-x\parallel <\frac{1}{n}$ and define ${\partial }^n(x,y)=k\in \mathbb{N}$ as any natural number such that $\frac{1}{k}\le \frac{1}{n}-\parallel y-x\parallel$. However, if reformulated as:

$$\begin{array}{c}\hfill x{\le }^{n}y\iff n\parallel y-x\parallel <1\end{array}$$

(8)

and ${\partial }^n(x,y)=rn$ with $r\in \mathbb{N}$, any natural number such that for all $u\in X$

$$\begin{array}{c}\hfill r\parallel u\parallel <1\Rightarrow \parallel u+n(y-x)\parallel <1,\end{array}$$

(9)

then we observe that the needed topological information is reduced to:

• The open ball, $B_1$, of radius 1 and centred at the origin;
• A choice of a natural number $r=r\left(z\right)$, for every $z\in B_1$, such that for all $u\in X$

  $$\begin{array}{c}\hfill r\parallel u\parallel <1\Rightarrow \parallel u+z\parallel <1.\end{array}$$

  (10)

This is sufficient motivation to consider those topological groups $(X,0,+,\tau )$ for which there exists an open neighbourhood of the origin, say, $B\subseteq X$, satisfying the following condition:

$$\begin{array}{c}\hfill \forall x\in B,\exists n\in \mathbb{N},B_n+x\subseteq B,\end{array}$$

(11)

which, in other words, if defining $B_n=\{u\in X\mid nu\in B\}$, can be stated as

$$\begin{array}{c}\hfill \forall x\in B,\exists n\in \mathbb{N},\forall u\in X,nu\in B\Rightarrow u+x\in B.\end{array}$$

(12)

We observe, furthermore, that there is no need to start with a topological group. Any group with a subset B containing the origin and satisfying the previous condition, immediately satisfies conditions (C1) and (C2) in the definition of a Cartesian spatial fibrous preorder as soon as we put $x{\le }^{i}y$ whenever $n(y-x)\in B$ and define ${\partial }^n(x,y)=rn$ with $r\in \mathbb{N}$ any natural number such that

$$\begin{array}{c}\hfill \forall u\in X,ru\in B\Rightarrow u+n(y-x)\in B.\end{array}$$

(13)

Indeed, for every $n\in \mathbb{N}$, we get $x{\le }^{n}x$ if and only if $n(x-x)\in B$ or equivalently $0\in B$. Now, if $x{\le }^{n}y$ and $y{\le }^{rn}z$ with $r\in \mathbb{N}$ such that (13) holds, then as a consequence we have $x{\le }^{n}z$. To see this, let us take any $n(y-x)\in B$ and $rn(z-y)\in B$; then, by taking $u=n(z-y)\in X$, we observe $ru\in B$ and hence $u+n(y-x)\in B$ or equivalently $n(z-y)+n(y-x)\in B$, which is the same as $n(z-x)\in B$, precisely stating that $x{\le }^{n}z$.

In order to prove condition (C3), we need an extra assumption on the subset B, namely that $B_n\subseteq B$ for every $n\in \mathbb{N}$. This is clearly the case for normed vector spaces, and the previous considerations can be summarized as follows.

Proposition 4.

Let $(X,+,0)$ be a group, and suppose there are $B\subseteq X$ and $\alpha :B\to \mathbb{N}$, satisfying, for every $x\in B$, $u\in X$, and $n\in \mathbb{N}$, the following three conditions:

(i) 

$0\in B$

(ii) 

if $\alpha \left(x\right)u\in B$ then $u+x\in B$

(iii) 

if $nu\in B$ then $u\in B$

Then, the system $(X,{\le }^n,{\partial }^n)$, defined as $x{\le }^{n}y$ whenever $n(y-x)\in B$ and ${\partial }^n(x,y)=\alpha \left(n(y-x)\right)n$, is a Cartesian spatial fibrous preorder indexed by the natural numbers. Thus, giving rise to an $\mathbb{N}$-Cartesian space on the set X.

In the following section, we will analyse in more detail the example of a metric space. The example of a normed vector space, which has been generalized by the previous proposition, will be analysed further in Section 6.

5. The Structure of Metric Spaces

In this section, we will see how to decompose the structure of a metric space in terms of a Cartesian spatial fibrous preorder. First, instead of the non-negative real interval $[0,+\infty ]$, we can take any preorder $(E,\le )$. Secondly, instead of the formula $\frac{1}{n}-d(x,y)$, we will use maps $p(\alpha ,x,y)=\alpha -d(x,y)$ and $g\left(n\right)=\frac{1}{n}$ so that $\frac{1}{n}-d(x,y)$ is obtained by taking $p\left(g\right(n),x,y)$. These maps take their values on the preorder $(E,\le )$ and are required to satisfy some conditions. The definition of $x{\le }^{n}y$ is recovered as $g\left(k\right)\le p\left(g\right(n),x,y)$ for some $k\in I$, which is then used to define ${\partial }^n(x,y)=k$.

Let $(E,\le )$ be a preorder and B a set. We will say that a map $p:E\times B\times B\to E$ is a lax-left-associative Mal’tsev operation when the following three conditions are satisfied for all $a,b\in E$ and $x,y,z\in B$:

• $a\le p(a,x,x)$,
• $p\left(p\right(a,x,y),y,z)\le p(a,x,z)$,
• If $a\le b$ then $p(a,x,y)\le p(b,x,y)$.

Let $(I,·,1)$ be a unitary magma and $(E,\le )$ a preorder. We will say that a map $g:I\to E$ is a linking map if

$$g(n·(k·m\left)\right)\le g(n·m)$$

for all $n,m,k\in I$.

Proposition 5.

Let $(I,·,1)$ be a unitary magma and $(E,\le )$ a preorder. Every lax-left-associative Mal’tsev operation $p:E\times B\times B\to E$ together with a linking map $g:I\to E$ induces a topology τ on the set B determined by

$$\mathcal{O}\in \tau \iff \forall x\in \mathcal{O},\exists n\in I,N(n,x)\subseteq \mathcal{O}$$

with $N(n,x)=\{y\in B\mid \exists m\in I,g(m)\le p(g\left(n\right),x,y\left)\right\}$.

Proof. 

The proof makes use of Proposition 1 by showing that the system $(B,{\le }^n,{\partial }^n)$ is a Cartesian spatial fibrous preorder with

$$x{\le }^{n}y\iff \exists m\in I,g\left(m\right)\le p(g\left(n\right),x,y)$$

and ${\partial }^n(x,y)=m$.

Axiom (C1) holds because $g\left(n\right)\le p\left(g\right(n),x,x)$ for all $n\in I$ and $x\in B$.

In order to prove Axiom (C2), we consider $x{\le }^{n}y$ with ${\partial }^n(x,y)=m$ such that $g\left(m\right)\le p\left(g\right(n),x,y)$ and $y{\le }^{m}z$ for which there exists $m^{\prime }={\partial }^m(y,z)\in I$ such that

$$g\left(m^{\prime }\right)\le p(g\left(m\right),y,z).$$

Under these assumptions, keeping in mind that p is a lax-left-associative Mal’tsev operation, we observe

$$g\left(m^{\prime }\right)\le p(g\left(m\right),y,z)\le p(p(g\left(n\right),x,y),y,z)\le p(g\left(n\right),x,z)$$

which shows that $x{\le }^{n}z$.

Axiom (C3) is a consequence of g being a linking map from which, in particular, we obtain

$$g(k·m)\le g\left(m\right)$$

and

$$g(n·k)\le g\left(n\right).$$

In the former case, $n=1$, while in the latter, $m=1$. This allows us to conclude that if $x{\le }^{n·k}y$ then $x{\le }^{n}y$. Indeed, if there exists $m\in I$ such that $g\left(m\right)\le p\left(g\right(n·k),x,y)$ then

$$p\left(g\right(n·k),x,y)\le p\left(g\right(n),x,y)$$

and so $g\left(m\right)\le p\left(g\right(n),x,y)$, thus ensuring $x{\le }^{n}y$. Similarly, we prove that if $x{\le }^{k·m}y$, then $x{\le }^{m}y$. □

We remark that the reflexivity of the preorder $(E,\le )$ is never used, so the previous result still holds for a set E equipped with a transitive relation. So, for example, the above result holds true if we replace the preordered set $(E,\le )$ with a semigroup $(E,\circ )$ and define the transitive relation $x<y$ as $x\circ y=x$, as it is well known this relation fails to be reflexive when E is not an idempotent semigroup and fails to be anti-symmetric when E is not commutative. In addition, the requirement of the map g being a linking map could be replaced by the condition that $g(n·m)\le g\left(n\right)$ and $g(n·m)\le g\left(m\right)$.

The example of a metric space is obtained by letting E be the set of real numbers with the usual order and taking the maps $p(a,x,y)=a-d(x,y)$ and $g\left(n\right)=\frac{1}{n}$ with I the unitary magma of natural numbers with usual multiplication.

Another interesting example is obtained by taking E to be an ordered group $(E,+,0,\le )$ and $I=(I,·,1)$ a monoid. Suppose there exists a map $g:I\to E$ with $g\left(mkn\right)\le g\left(mn\right)$ for all $m,n,k\in I$. Let B be any set and consider a map

$$\delta :B\times B\to E$$

such that

$$\begin{array}{}\mathrm{(14)}& \delta (x,x)=0,\mathrm{for}\mathrm{all}x\in B\mathrm{(15)}& \delta (x,y)+\delta (y,z)\ge \delta (x,z),\mathrm{for}\mathrm{all}x,y,z\in B.\end{array}$$

This is a straightforward generalization of a metric space, and we have that $p(a,x,y)=a-\delta (x,y)$ is a lax-left-associative Mal’tsev operation.

Let us now suppose that B is a group. Then, with g and E as before, for every map $t:B\to E$ such that $t\left(0\right)=0$ and

$$\begin{array}{c}\hfill t\left(u\right)+t\left(v\right)\ge t(u+v),\mathrm{for}\mathrm{all}u,v\in B\end{array}$$

(16)

we get a lax-left-associative Mal’tsev operation with $p(a,x,y)=a-t(y-x)$. This is an immediate generalization for normed spaces. In particular, when t is a homomorphism, we get $p(a,x,y)=a+t\left(x\right)-t\left(y\right)$.

A simple procedure to construct unitary magmas which in general are not associative is to start with an arbitrary non-empty set of indexes I, choose and element $1\in I$, and consider a family of endo-maps ${\mu }_n:I\to I$, indexed by the elements in I, such that ${\mu }_n\left(1\right)=1$ and ${\mu }_1\left(n\right)=n$, for all $n\in I$. A binary operation is thus obtained as $m·n={\mu }_m\left(n\right)$.

One final remark on notation. The notion of a linking map comes from the structure of a link [29]. The name lax-left-associative Mal’tsev operation is due to the fact that when $E=B$ and the preorder is the identity or discrete order, then a lax-left-associative Mal’tsev operation reduces to

$$\begin{array}{c}\hfill a=p(a,x,x)\\ \hfill p\left(p\right(a,x,y),y,z)=p(a,x,z).\end{array}$$

If adding the respective two similar identities on the right $p(x,x,a)=a$ and $p(z,x,a)=p(z,y,p(y,x,a\left)\right)$, which make sense because $E=B$, then we would obtain an associative Mal’tsev operation.

6. Monoids and Modules as Cartesian Spaces

An interesting special class of I-Cartesian spaces is obtained by imposing on the structuring maps $\gamma$ and $\eta$ of Theorem 2(e) the condition

$$\gamma \left(\eta \right(n,x),y)=\gamma \left(\eta \right(1,x),y)·n$$

for all $y\in \eta (n,x)$ and for all $n\in I$ and $x\in X$.

We will analyse this condition by considering a monoid structure on the set B. This will allow us to decompose $\eta (n,x)=x+\eta (n,0)$. We then consider an I-module structure on B thus providing a way to obtain $\eta (n,0)$ as an n-scaling of $\eta (1,0)$. Therefore, when B is an I-module, all the information is encompassed in $\eta (1,0)$ and $\gamma \left(\eta \right(1,0),z)$.

Proposition 6.

Let $(I,·,1)$ be a unitary magma. Every monoid $(B,+,0)$, equipped with a family of subsets $S_n\subseteq B$, together with maps

$${\alpha }_n:S_n\to I,n\in I$$

such that

(i) 

$0\in S_n$, for all $n\in I$

(ii) 

for each $n\in I$, if $a,a^{\prime }\in S_n$ and $a^{\prime }\in S_{{\alpha }_n\left(a\right)}$ then $a+a^{\prime }\in S_n$

(iii) 

$S_{n·m}\subseteq S_n\cap S_m$

induces a topology τ on B determined as

$$\mathcal{O}\in \tau \iff \forall x\in \mathcal{O},\exists n\in I,x+S_n\subseteq \mathcal{O}.$$

Proof. 

We make use of Proposition 1 by showing that the system $(B,{\left({\le }^n\right)}_{n\in I},{\left({\partial }^n\right)}_{n\in I})$ is a Cartesian spatial fibrous preorder with

$$x{\le }^{n}y\iff \exists a\in S_n,x+a=y$$

and ${\partial }^n(x,y)={\alpha }_n\left(a\right)·n$ while noting that $N(n,x)=\{y\in B\mid x{\le }^{n}y\}$ is the same as $x+S_n=\{x+a\mid a\in S_n\}$.

Axiom (C1) holds because $0\in S_n$ and hence $x{\le }^{n}x$ for all $x\in B$ and $n\in I$.

In proving axiom (C2), we observe that if $x{\le }^{n}y$, that is, $x+a=y$ for some $a\in S_n$, and if $y{\le }^{m}z$, with $m={\alpha }_n\left(a\right)·n$, which means $y+a^{\prime }=z$ for some $a^{\prime }\in S_m$, then by condition $\left(iii\right)$, we conclude that $a^{\prime }\in S_n$ and $a^{\prime }\in S_{{\alpha }_n\left(a\right)}$. Condition $\left(ii\right)$ now tells us that $a+a^{\prime }\in S_n$ and hence $x{\le }^{n}z$. Indeed, there exists $a^{″}=a+a^{\prime }\in S_n$ such that $x+a^{″}=z$.

Axiom (C3) is a straightforward consequence of condition $\left(iii\right)$. If $x{\le }^{n·m}y$, that is, $x+a=y$, for some $a\in S_{n·m}$, then $x{\le }^{n}y$ and $x{\le }^{m}y$, since, by $\left(iii\right)$, $a\in S_n$ and $a\in S_m$. □

In particular, if there exists an action of I on B, in the sense of a map $\xi :I\times B\to B$ such that for all $n,m\in I$ and $x,y\in B$,

• $\xi (1,x)=x$, $\xi (n,0)=0$,
• $\xi (n,x+y)=\xi (n,x)+\xi (n,y)$
• $\xi (n·m,x)=\xi (n,\xi (m,x\left)\right)=\xi (m,\xi (n,x\left)\right)$,

then we may wonder if each $S_n$ in the previous proposition is determined from $S_1$ as $S_n=\{\xi (n,a)\mid a\in S_1\}$.

A monoid $(B,+,0)$ equipped with an action $\xi$ of I on B in the sense above is said to be an I-module and represented as $(B,+,0,\xi )$.

Theorem 7.

Let $(I,·,1)$ be a unitary magma, $(B,+,0,\xi )$ an I-module, and $(B,\tau )$ a topological space. The following are equivalent:

(a) 

There exists $S\subseteq B$ and $\alpha :S\to I$ such that

(i) 

$0\in S$

(ii) 

$a+\xi (\alpha \left(a\right),a^{\prime })\in S$, for all $a,a^{\prime }\in S$

for which τ is determined as

$$\mathcal{O}\in \tau \iff \forall x\in \mathcal{O},\exists n\in I,x+S_n\subseteq \mathcal{O}$$

with $S_n=\{\xi (n,a)\in B\mid a\in S\}$.

(b) 

There exists $S\subseteq B$ such that

(i) 

$0\in S$

(ii) 

$S\subseteq \{y\in B\mid \exists n\in I,y+S_n\subseteq S\}$, with

$$S_n=\{\xi (n,a)\in B\mid a\in S\}$$

for which τ is determined as

$$\mathcal{O}\in \tau \iff \forall x\in \mathcal{O},\exists n\in I,x+S_n\subseteq \mathcal{O}.$$

(c) 

There exists $S\in \tau$ and $\gamma :\left\{\right(U,x)\mid x\in U\in \tau \}\to I$ such that

(i) 

$0\in S$

(ii) 

$x+S_n\in \tau$ for all $x\in B$ and $n\in I$ with

$$S_n=\{\xi (n,a)\in B\mid a\in S\}$$

(iii) 

if $x\in U\in \tau$ and $n=\gamma (U,x)$, then $x+S_n\subseteq U$.

Moreover, under these conditions, $(B,\tau )$ is an I-Cartesian space with $x{\le }^{n}y$ defined as $y=x+\xi (n,a)$ for some $a\in S$ and ${\partial }^n(x,y)=\alpha \left(a\right)·n$ if and only if $\xi (n,a)\in S$ for all $a\in S$ and $n\in I$.

Proof. 

$\left(a\right)\Rightarrow \left(b\right)$ We only need to prove that S is contained in

$$\{y\in B\mid \exists n\in I,y+S_n\subseteq S\}$$

with $S_n=\{\xi (n,a)\in B\mid a\in S\}$. This is the same as proving that for every $a\in S$ there exists $n\in I$ for which $a+S_n\subseteq S$. This follows from condition $\left(a\right)\left(ii\right)$ by choosing $n=\alpha \left(a\right)$. Indeed, for every $a\in S$, we have $a+S_{\alpha \left(a\right)}\subseteq S$ as a consequence of $S_{\alpha \left(a\right)}=\{\xi (\alpha \left(a\right),a^{\prime })\mid a^{\prime }\in S\}$ together with condition $\left(a\right)\left(ii\right)$.

$\left(b\right)\Rightarrow \left(c\right)$ Let $x\in B$ and $n\in I$. In order to show that $x+S_n\in \tau$, we consider $z=x+\xi (n,a)$ for some $a\in S$ and find $m\in I$ for which $z+S_m\subseteq x+S_n$. Using condition $\left(b\right)\left(ii\right)$ and given that $a\in S$, we obtain $k\in I$ such that $a+S_k\subseteq S$, or, in other words, $a+\xi (k,a^{\prime })\in S$ for all $a^{\prime }\in S$. We now take $m=k·n$ and show $z+S_m\subseteq x+S_n$.Indeed, for every $a^{\prime }\in S$, taking into account that $\xi$ is an action, we observe

$$\begin{array}{ccc}\hfill z+\xi (m,a^{\prime })& =& z+\xi (k·n,a^{\prime })=z+\xi (n,\xi (k,a^{\prime }))\hfill \\ & =& x+\xi (n,a)+\xi (n,\xi (k,a^{\prime }))\hfill \\ & =& x+\xi (n,a+\xi (k,a^{\prime }))\hfill \end{array}$$

showing that $z+\xi (m,a^{\prime })=x+\xi (n,a+\xi (k,a^{\prime }))$ is in $x+S_n$ because $a+\xi (k,a^{\prime })\in S$ for all $a^{\prime }\in S$. In particular, when $x=0$ and $n=1$, we obtain $S\in \tau$. Thus, so far we have shown the existence of S satisfying condition $\left(c\right)\left(ii\right)$. Condition $\left(c\right)\left(i\right)$ is clear, while Condition $\left(c\right)\left(iii\right)$ is obtained by observing that, if $x\in U\in \tau$, then, by the assumption on $\tau$, there exists $n\in I$ for which $x+S_n\subseteq U$, and hence we choose $\gamma (U,x)=n$ in order to satisfy condition $\left(c\right)\left(iii\right)$.

$\left(c\right)\Rightarrow \left(a\right)$ Given $S\in \tau$, the map $\alpha :S\to I$ is defined as $\alpha \left(a\right)=\gamma (S,a)$. From condition $\left(c\right)\left(iii\right)$, it follows that

$$a+S_{\alpha \left(a\right)}\subseteq S$$

which means that for every $a,a^{\prime }\in S$,

$$a+\xi (\alpha \left(a\right),a^{\prime })\in S.$$

This proves condition $\left(a\right)\left(ii\right)$. Condition $\left(a\right)\left(i\right)$ is clear. It remains to prove that $\tau$ is determined as prescribed. We observe, on the one hand, that if $\mathcal{O}\in \tau$, then for every $x\in \mathcal{O}$, there exists $m=\gamma (\mathcal{O},x)\in I$ such that $x+S_m\subseteq \mathcal{O}$ as asserted by condition $\left(c\right)\left(iii\right)$. On the other hand, if $\mathcal{O}\subseteq B$ is such that for all $x\in \mathcal{O}$ there exists $m\in I$ with $x+S_m\subseteq \mathcal{O}$, then, because $x+S_m\in \tau$ and

$$\mathcal{O}=\bigcup _{x\in \mathcal{O}}x+S_m$$

we conclude that $\mathcal{O}\in \tau$.

Moreover, under these conditions $(B,\tau )$ is I-Cartesian, where the map $\eta$ (see item $\left(e\right)$ in Theorem 2) is obtained as $\eta (n,x)=x+S_n$, if and only if $S_n\subset S$ for all $n\in I$. Indeed, if $S_n\subset S$ for all $n\in I$, we observe:

(i) 

$x\in \eta (n,x)$, since $0\in S$ and $\xi (n,0)=0$.

(ii) 

If $x\in U\in \tau$ and $m=\gamma (U,x)$ then $x+S_m\subseteq U$, which is the same as $\eta \left(\gamma \right(U,x),x)\subseteq U$.

(iii) 

$\eta (n·m,x)\subseteq \eta (n,x)\cap \eta (m,x)$ follows from $S_{n·m}\subseteq S_n\cap S_m$, which is a consequence of $\xi (n,a)\in S$ for all $n\in I$ and $a\in S$ and the fact that $\xi$, being an action, is such that $\xi (n·m,a)=\xi (n,\xi (m,a\left)\right)=\xi (m,\xi (n,a\left)\right)$.

Conversely, if $(B,\tau )$ is I-Cartesian with $\eta (n,x)=x+S_n$, then from $\eta (n·m,x)\subseteq \eta (n,x)\cap \eta (m,x)$ for all $m,n\in I$, we deduce $S_n\subseteq S$ when $m=1$. □

7. Examples

Some examples are presented so to illustrate the results of the previous section.

Theorem 7 can be specialized into the case when the unitary magma is the set of natural numbers with usual multiplication $I=(\mathbb{N},·,1)$, and $(B,+,0)$ is any monoid considered as an I-module with $\xi (n,x)=nx$. In this case, any subset $S\subseteq B$ together with a map $\alpha :S\to \mathbb{N}$ satisfying the three conditions:

(i) 

$0\in S$

(ii) 

$a+\alpha \left(a\right)a^{\prime }\in S$ for all $a,a^{\prime }\in S$

(iii) 

$na\in S$ for all $a\in S$ and $n\in \mathbb{N}$.

induces a topology $\tau$ on B generated by the system of open neighbourhoods $N(n,x)=\{x+na\mid a\in S\}$.

Note that condition $\left(iii\right)$ is necessary so that the system $(B,{\le }^n,{\partial }^n)$ with $x{\le }^{n}y$ whenever $y=x+na$ for some $a\in S$ and ${\partial }^n(x,y)=\alpha \left(a\right)n$ is a Cartesian spatial fibrous preorder. When B is a group, it offers a further comparison with Proposition 4, which will be deepened in a future work.

When the map $\alpha :S\to \mathbb{N}$ is constant with $\alpha \left(a\right)=1$, then S must be a submonoid.

Another example is obtained when $B=(B,+,·,0,1)$ is a semi-ring. In this case, we can take $I=({\mathbb{N}}_0,+,0)$ and choose any submonoid S for the additive structure, that is, $S\subseteq B$ with $0\in S$ and $a+a^{\prime }\in S$ for all $a,a^{\prime }\in S$. Now, every choice of an element $p\in S$ such that $p^{n}a\in S$, for all $a\in S$ and $n\in {\mathbb{N}}_0$, gives a topology on the set B generated by the system of open neighbourhoods

$$N(n,x)=\{x+p^{n}a\mid a\in S\}.$$

Indeed, we define an I-module with action $\xi (n,x)=p^{n}x$ by taking successive powers of p and considering $p^0=1$. In this case, the map $\alpha :S\to {\mathbb{N}}_0$ is the constant map $\alpha \left(a\right)=0$.

When S is not closed under addition, then we have to find for every $a\in S$ a non-negative integer $\alpha \left(a\right)\in {\mathbb{N}}_0$ such that $a+p^{\alpha \left(a\right)}{a}^{\prime }\in S$ for all $a^{\prime }\in S$. The simple example of the real numbers $(\mathbb{R},+,·,0,1)$ with usual addition and multiplication, when $S=[0,1]$ and $p=\frac{1}{2}$, illustrates this situation.

One more example with $I=({\mathbb{N}}_0,+,0)$ is constructed as follows.

Take any monoid $(X,+,0)$ together with $t:X\to X$, an endomorphism of X, and consider a subset $P\subseteq X$ with $0\in P$. Let B be the collection of all maps from X to X considered as a monoid with component-wise addition, that is, $B=(X^X,+,0_X)$. With $I=({\mathbb{N}}_0,+,0)$ we define an action on B as $\xi (n,f)=t^n\circ f$ for every non-negative integer n and every map $f:X\to X$. The action is defined for every selected endomorphism t of X.

In order to produce a topology on the set B, we need to find a subset $S\subseteq B$ together with a map $\alpha :S\to {\mathbb{N}}_0$ as in Theorem 7. We take

$$S=\{f\in B\mid \exists n\in {\mathbb{N}}_0,\forall x\in X,\forall y\in P,f\left(x\right)+t^n\left(y\right)\in P\}$$

and put $\alpha \left(f\right)=n$ where n is the smallest non-negative integer such that for all $x\in X$ and $y\in P$, $f\left(x\right)+t^n\left(y\right)\in P$.

Clearly, the constant null function $0_X:X\to X$ is in S and moreover $\alpha \left(0_X\right)=0$.

The reason why $f+t^{\alpha \left(f\right)}\circ g\in S$ for all $f,g\in S$ is mainly because t is an endomorphism and so $t^n\circ g+t^{n+m}=t^n\circ g+t^n\circ t^m=t^n\circ (g+t^m)$. Indeed, if $f,g\in S$ with $\alpha \left(f\right)=n$ and $\alpha \left(g\right)=m$, then there exists $k=n+m$ such that $(f+t^n\circ g)\left(x\right)+t^k\left(y\right)\in P$ for all $x\in X$ and $y\in P$.

The imposition that $t^n\circ f\in S$ for all $n\in {\mathbb{N}}_0$ and $f\in S$ is guaranteed as soon as the endomorphism $t:X\to X$ satisfies the condition $t\left(y\right)\in P$ for all $y\in P$. In that case, $\alpha (t^n\circ f)=n+\alpha \left(f\right)$.

A concrete simple example is the following. Take $X=\left(\right[0,1],·,1)$ to be the unit interval of real numbers with usual multiplication and not containing the element 0. Consider the endomorphism $t\left(x\right)=\sqrt{x}=x^{\frac{1}{2}}$ and let $P=[\frac{1}{2},1]$. It follows that $f\in S$ if and only if $f\left(x\right)\in P$ for all $x\in X$. For every $f\in S$, $\alpha \left(f\right)=n$ with n being the smallest non-negative integer for which

$$u·{\left(\frac{1}{2}\right)}^{\frac{1}{2^n}}>\frac{1}{2}$$

with $u=inf\left\{f\right(x)\mid x\in X\}$. Note that u is always greater than $\frac{1}{2}$, and

$$\underset{n\to +\infty }{lim}{\left(\frac{1}{2}\right)}^{\frac{1}{2^n}}=1.$$

As a consequence of Theorem 7, we obtain a topology on B generated from

$$N(n,u)=\{u+t^n\circ v\mid v\in S\}$$

as system of open neighbourhoods.

8. Conclusions

The notion of Cartesian spatial fibrous preorder has been introduced as a system $(X,{\left({\le }^i\right)}_{i\in I},{\left({\partial }^i\right)}_{i\in I})$ indexed over a unitary magma, I, and satisfying conditions (C1), (C2), and (C3). Every such structure gives rise to a topological space (Proposition 1) and the spaces that are thus obtained were called I-Cartesian. It has been proven (Theorem 2) that a space $(X,\tau )$ is I-Cartesian if and only if there exists a map $\eta :I\times X\to \tau$ such that $x\in \eta (i,x)$ and $\eta (ij,x)\subseteq \eta (i,x)\cap \eta (j,x)$ for all $i,j\in I$ and $x\in X$, together with a map

$$\gamma :\left\{\right(U,x)\mid x\in U\in \tau \}\to I$$

such that $\eta \left(\gamma \right(U,x),x)\subseteq U$. In a sequel to this work, we will concentrate our attention on the structure of morphisms between Cartesian spatial fibrous preorders, thus elevating the characterization of Theorem 2 to a categorical equivalence. The structure of metric spaces has been decomposed as an indexing unitary magma, a preorder, a lax-left-associative Mal’tsev operation, and a linking map (Proposition 5). The special case of a normed vector space has been treated in three different manners:

• In Proposition 4, as a group with a distinguished subset containing the origin and satisfying some conditions (intuitively a convex and bounded neighbourhood of the origin, namely the open ball of radius one) together with a map intuitively measuring how close to the boundary an element is.
• In Proposition 5, as a special case of a lax-left-associative Mal’tsev operation with $p(a,x,y)=a-\parallel y-x\parallel$.
• In Theorem 7, as an I-module with action $\xi (n,x)=\frac{1}{n}x$, $S=\{a\in B\mid \parallel a\parallel <1\}$ and $\alpha \left(a\right)$ such that $\frac{1}{\alpha \left(a\right)}\le 1-\parallel a\parallel$.

A list of examples has been presented as a way to illustrate possible applications. This is a first step for a systematic analysis of the structure of topological spaces.