Chaos in Topological Modules

Abstract

Chaotic and pathological phenomena in topological modules are studied in this manuscript. In particular, constructions of noncontinuous linear functionals are provided for a wide variety of topological modules. In addition, constructions of balanced and absorbing sets which are not neighborhoods of zero are also given in an extensive class of topological modules. Finally, we construct a linearly open set with empty interior in a large amount of topological modules. All these constructions are related to each other. Prior to developing all these results, we provide an axiomatization of the topological concept of limit by introducing the limit operators in a similar context as hull operators or closure operators are defined.

1. Introduction

The study of balanced and absorbing sets received an important boost with the development of the Theory of Topological Vector Spaces [1,2,3,4,5]. Important notions like barrelledness [6,7,8,9], which are crucial to understand the geometry of real or complex topological vector spaces, are directly linked to balancedness and absorbance. The existence of nonbarrelled spaces, that is, topological vector spaces containing a barrel (closed, convex, balanced, and absorbing set) that is not a zero-neighborhood, triggered the study of balanced and absorbing sets which are not zero-neighborhoods. A motivating result for this study was provided in [10] (Theorem 2.1), where it is shown that, if a finite-dimensional Hausdorff real or complex topological vector space has a convex subset that is a generator system and contains zero, then it must have nonempty interior. In [11] (Theorem 3.2), an absorbing and balanced set of void topological interior is constructed in each separable real or complex normed space with dimensions strictly greater than 1. By relying on this result, in [12] (Theorem 1.1), an absorbing and balanced set of void topological interior is constructed in any Hausdorff locally convex real or complex topological vector space with dimensions strictly greater than 1. Transporting these results to the scope of topological modules is not an easy task. A first attempt was given in [13] (Corollary 8). The main obstacle in extending [11] (Theorem 3.2) and [12] (Theorem 1.1) to the ambience of topological modules is the fact that balancedness and absorbance cannot be defined in topological modules without previously endowing the underlying topological ring with something similar to the unit interval $[-1,1]$ or to the unit circle $\overline{\mathbb{D}}$. This is why “unit neighborhoods of zero” came into play when they were conceived and introduced for the first time in [14]. That notion allows for extending balancedness and absorbance to topological modules even if the underlying topological ring is not seminormed or absolutely semivalued. After the notion of unit neighborhood of zero, the one of “unit segment” [15] (Definition 3.2) was originated and posed to provide a natural definition of convexity in topological modules [15] (Definition 3.4), different from the one provided in [16] for $C^{*}$-algebras.

The tendency described above simply corresponds to the natural generalization of Operator Theory to the more general setting provided by the topological modules over topological rings [17,18,19,20,21,22,23,24]. The main contributions of this manuscript can be summarized as follows:

• Axiomatization of limit operators and properties of limits of prefilters in topological modules (Theorems 2, 3 and 13);
• Existence (by explicit construction) of noncontinuous linear functionals for a wide variety of topological modules (Theorem 9);
• Existence (by explicit construction) of B-balanced and B-absorbing sets which are not zero-neighborhoods in many topological modules (Theorem 10);
• Existence (by explicit construction) of a linearly open set with empty interior in a large amount of topological modules (Theorem 11);
• Existence (by explicit construction) of additively symmetric and absorbing sets which are not zero-neighborhoods in many topological modules (Theorem 12).

The above constructions are related to each other, and their motivation strongly relies on the discrete topology. Indeed, if a ring and a module over that ring are both endowed with the discrete topology, then all linear functionals are continuous, all balanced and absorbing sets are zero-neighborhoods, and all linearly open sets are topologically open. This fact highlights the importance of our constructions.

2. Methodology

Throughout this manuscript and unless otherwise stated, all monoid actions considered will be left, all rings will be associative, unitary, and nonzero, and all modules over rings will be unital and nonzero. All ring morphisms, and in particular ring extensions, will be assumed unital. We refer the reader to [25,26,27,28,29,30] for a wide perspective on topological rings, modules, and algebras.

We will strongly rely on the concept of internal point [1] (TVS II.26 p.66), which finds its origins probably in [31] for real or complex topological vector spaces. Internal points were transported to the scope of topological modules in [32]. Let M be a module over a topological ring R. Let $A\subseteq M$. A point $a\in A$ is said to be an internal point of A provided for every $m\in M$ there exists a neighborhood V of 0 in R such that $a+Vm\subseteq A$. The set of internal points is usually denoted by $\mathrm{inter}\left(A\right)$.

Another concept on which we will strongly rely is the one of practical ring [33] (Definition 6). A topological ring R is said to be practical provided that $0\in \mathrm{cl}\left(\mathcal{U}\right(R\left)\right)$, that is, 0 belongs to the closure of the invertibles $\mathcal{U}\left(R\right)$ of R. Practical rings serve to extend the classical Operator Theory to the topological module setting. An extensive study on practical topological rings can be found in [33].

If X is a topological space and $U,B\subseteq X$, then U is called regular open provided that $U=\mathrm{int}\left(\mathrm{cl}\right(U\left)\right)$, and B is called regular closed provided that $B=\mathrm{cl}\left(\mathrm{int}\right(B\left)\right)$. If $U\subseteq X$ is open and $B\subseteq X$ is closed, then $\mathrm{cl}\left(U\right)$ and $\mathrm{int}\left(B\right)$ are regular closed and regular open, respectively. With the previous concepts in hand, we can recall the notion of unit neighborhood of zero [14] (Definition 2.2). A regular open neighborhood of zero in a topological ring is said to be an open unit neighborhood of zero if it is additively symmetric, multiplicatively idempotent, and its closure contains the unity. A closed unit neighborhood of zero is a regular closed neighborhood of zero whose interior is an open unit neighborhood of zero.

The above definitions are chosen properly and none of their conditions can be removed. For instance, if R is a discrete topological ring, then $U:=\{-1,1\}$ is a regular open, additively symmetric, and multiplicatively idempotent subset of R such that $1\in \mathrm{cl}\left(U\right)$. However, it is not a neighborhood of 0.

A closed unit neighborhood of zero B in a topological ring R is said to be total whenever, for every $u\in \mathcal{U}\left(R\right)$, either $u\in B$ or $u^{-1}\in B$. In addition, B is called left-feasible whenever $\{uB:u\in \mathcal{U}(R\left)\right\}$ is a basis of neighborhoods of 0 in R. Similarly, right-feasibility is defined. B is called feasible if it is left- and right-feasible. Whenever a topological ring R is endowed with a (total) (right-)(left-)feasible closed unit zero-neighborhood, we will say that R is c (totally) (right-)(left-)feasible. Observe that every right- or left-feasible topological ring is trivially practical.

A subset A of a topological module M over a topological ring R is said to be B-balanced, where $B\subseteq R$ is a closed unit neighborhood of zero, provided that $BA=A$, and A is said to be B-absorbing whenever, for every $m\in M$, there exists an invertible $u\in \mathcal{U}\left(R\right)$ satisfying that $Bm\subseteq uA$. This is how balancedness and absorbance are transported to the scope of topological modules, that is, by means of the unit zero-neighborhoods. Without relying on unit zero-neighborhoods, additive symmetricity is considered instead of B-balancedness. The way to define the notion of absorbing set regardless of unit zero-neighborhoods is the following: A subset A of a topological module M over a topological ring R is called absorbing provided that $0\in \mathrm{inter}\left(A\right)$. This definition is consistent with [24] (Proposition 75), where it is shown that every B-absorbing set is absorbing, and conversely, every absorbing set is B-absorbing whenever B is left-feasible.

We refer to [13,14,15,24,32,33,34,35] for a wider perspective on the above notions.

3. Results

Before stating and proving results on the chaotic behaviour of the geometry of topological modules in comparison to the geometry of real or complex topological vector spaces, an axiomatization of several notions related to the topological concept of limit is necessary.

3.1. Limit Operators

The notion of hull operator [36] is original from Linear Algebra and Operator Theory, although it can also be generalized to Category Theory. Let Z be a set. A function $H:\mathcal{P}\left(Z\right)\to \mathcal{P}\left(Z\right)$ is called a hull operator if it verifies the following three conditions for $A,B\in \mathcal{P}\left(Z\right)$:

• Extensivity: $A\subseteq H\left(A\right)$;
• Increasingness: $A\subseteq B\Rightarrow H\left(A\right)\subseteq H\left(B\right)$;
• Idempotency: $H\left(H\right(A\left)\right)=H\left(A\right)$.

Notice that hull operators do not necessarily map ⌀ to ⌀. Indeed, the linear span in a vector space is a hull operator mapping the empty set to the null subspace.

Remark 1.

Let Z be a set and let $H:\mathcal{P}\left(Z\right)\to \mathcal{P}\left(Z\right)$ be a hull operator. Notice that $⌀\subseteq A$ for every $A\in \mathcal{P}\left(Z\right)$; therefore, $H(⌀)\subseteq H\left(A\right)$ for every $A\in \mathcal{P}\left(Z\right)$, meaning that $H(⌀)={\bigcap }_{A\in \mathcal{P}\left(Z\right)}H\left(A\right)$. In case $H(⌀)\ne ⌀$, it is easy to see that $H(⌀)={\bigcap }_{A\in \mathcal{P}\left(Z\right)\setminus \{⌀\}}H\left(A\right)$ since $H(⌀)\in \mathcal{P}\left(Z\right)\setminus \{⌀\}$ and $H\left(H\right(⌀\left)\right)=H(⌀)$.

The following proposition allows for constructing hull operators in an easy manner. This proposition is a generalization of [24] (Proposition 3).

Proposition 1.

Let Z be a set. Let $\mathcal{C}\subseteq \mathcal{P}\left(Z\right)$ be closed under nonempty intersections and such that $Z\in \mathcal{C}$. The map

$$\begin{array}{cccc}\hfill H_{\mathcal{C}}:& \hfill \mathcal{P}\left(Z\right)& \to & \mathcal{P}\left(Z\right)\hfill \\ & \hfill A& \mapsto & H_{\mathcal{C}}\left(A\right):=\bigcap \{C\in \mathcal{C}:A\subseteq C\}\hfill \end{array}$$

(1)

satisfies the following:

• $H_{\mathcal{C}}$ is a hull operator.
• If $A\subseteq \bigcap \mathcal{C}$, then $H_{\mathcal{C}}\left(A\right)=\bigcap \mathcal{C}$. In particular, $H_{\mathcal{C}}(⌀)=\bigcap \mathcal{C}$.
• $\mathcal{C}\setminus \{⌀\}=\left\{A\in \mathcal{P}\left(Z\right)\setminus \{⌀\}:A=H_{\mathcal{C}}\left(A\right)\right\}$.
• $\bigcap \mathcal{C}=⌀$ if and only if $H_{\mathcal{C}}(⌀)=⌀$.

Proof. 

We will prove each item:

• We will check all three properties that characterize hull operators are verified:
  • Extensivity: $A\subseteq H_{\mathcal{C}}\left(A\right)$. This holds by definition.
  • Increasingness: $A\subseteq B\Rightarrow H_{\mathcal{C}}\left(A\right)\subseteq H_{\mathcal{C}}\left(B\right)$. This also holds by definition.
  • Idempotency: $H_{\mathcal{C}}\left(H_{\mathcal{C}}\left(A\right)\right)=H_{\mathcal{C}}\left(A\right)$. In the first place, note that, if $A\in \mathcal{C}$, then $H_{\mathcal{C}}\left(A\right)=A$. Now, since $\mathcal{C}$ is closed under nonempty intersections, if $A\in \mathcal{P}\left(Z\right)\setminus \{⌀\}$, then $H_{\mathcal{C}}\left(A\right)\in \mathcal{C}$ and thus $H_{\mathcal{C}}\left(H_{\mathcal{C}}\left(A\right)\right)=H_{\mathcal{C}}\left(A\right)$. Finally, $H_{\mathcal{C}}(⌀)=\bigcap \mathcal{C}$. Since every $C\in \mathcal{C}$ satisfies that $\bigcap \mathcal{C}\subseteq C$, we conclude that

    $$H_{\mathcal{C}}\left(H_{\mathcal{C}}(⌀)\right)=H_{\mathcal{C}}\left(\bigcap \mathcal{C}\right)=\bigcap \mathcal{C}=H_{\mathcal{C}}(⌀).$$
• If $A\subseteq \bigcap \mathcal{C}$, then, by definition, we know that $H_{\mathcal{C}}\left(A\right)=\bigcap \mathcal{C}$. In particular, $⌀\subseteq \bigcap \mathcal{C}$, so $H_{\mathcal{C}}(⌀)=\bigcap \mathcal{C}$ as we already knew from the previous item.
• If $A\in \mathcal{C}\setminus \{⌀\}$, then $A=\bigcap \{C\in \mathcal{C}:A\subseteq C\}=H_{\mathcal{C}}\left(A\right)$ by definition of $H_{\mathcal{C}}$. Conversely, if $A\in \mathcal{P}\left(Z\right)\setminus \{⌀\}$ is such that $A=H_{\mathcal{C}}\left(A\right)$, then $A\in \mathcal{C}$ because $\mathcal{C}$ is closed under nonempty intersections.
• It is a direct consequence of Proposition 1(2).

□

The next theorem serves as converse for the previous proposition since it shows that every hull operator is of the form described in Proposition 1.

Theorem 1.

Let Z be a set. Let $H:\mathcal{P}\left(Z\right)\to \mathcal{P}\left(Z\right)$ be a hull operator. Let ${\mathcal{C}}_H:=\{A\in \mathcal{P}\left(Z\right):A=H\left(A\right)\}$. The following conditions are satisfied:

• $Z\in {\mathcal{C}}_H$;
• ${\mathcal{C}}_H$ is closed under arbitrary intersections;
• $H=H_{{\mathcal{C}}_H}$, where $H_{{\mathcal{C}}_H}$ is the hull operator induced by ${\mathcal{C}}_H$ given in (1).

Proof. 

We will prove each item:

• Note that H is extensive, so $Z\subseteq H\left(Z\right)$. Hence, $Z=H\left(Z\right)$ and $Z\in {\mathcal{C}}_H$.
• Let ${\left(A_i\right)}_{i\in I}\subseteq {\mathcal{C}}_H$ be an arbitrary family and let $A:={\bigcap }_{i\in I}{A}_i$. We have to prove that $A=H\left(A\right)$. Since H is extensive, it suffices to show that $H\left(A\right)\subseteq A$. For every $i\in I$, $A\subseteq A_i$, so $H\left(A\right)\subseteq H\left(A_i\right)=A_i$, meaning that $H\left(A\right)\subseteq {\bigcap }_{i\in I}{A}_i=A$.
• Fix an arbitrary $A\in \mathcal{P}\left(Z\right)$. Since H is idempotent, $H\left(H\right(A\left)\right)=H\left(A\right)$, so $H\left(A\right)\in {\mathcal{C}}_H$. The extensivity of H assures that $A\subseteq H\left(A\right)$; therefore, $H_{{\mathcal{C}}_H}\left(A\right)\subseteq H\left(A\right)$. Finally, $A\subseteq H_{{\mathcal{C}}_H}\left(A\right)$ because $H_{{\mathcal{C}}_H}$ is also a hull operator in view of Proposition 1(1), which implies that $H\left(A\right)\subseteq H\left(H_{{\mathcal{C}}_H}\left(A\right)\right)$. However, $H_{{\mathcal{C}}_H}\left(A\right)\in {\mathcal{C}}_H$ because ${\mathcal{C}}_H$ is closed under arbitrary intersections, meaning that $H_{{\mathcal{C}}_H}\left(A\right)=H\left(H_{{\mathcal{C}}_H}\left(A\right)\right)$. This shows that $H\left(A\right)\subseteq H_{{\mathcal{C}}_H}\left(A\right)$. As a consequence, $H\left(A\right)=H_{{\mathcal{C}}_H}\left(A\right)$, and the arbitrariness of $A\in \mathcal{P}\left(Z\right)$ allows for concluding the proof.

□

Closure operators [37,38] are well known in Topology since they serve to characterize topologies. We will use them here to construct topologies by means of limit functions. Let X be a set. A closure operator is a map $C:\mathcal{P}\left(X\right)\to \mathcal{P}\left(X\right)$ satisfying the following conditions for all $A,B\in \mathcal{P}\left(X\right)$:

• Nullity: $C(⌀)=⌀$;
• Extensionality: $A\subseteq C\left(A\right)$;
• Additivity: $C(A\cup B)=C\left(A\right)\cup C\left(B\right)$;
• Idempotency: $C\left(A\right)=C\left(C\right(A\left)\right)$.

The classical example of a closure operator is the closure on a topological space. Notice that a closure operator is necessarily increasing; thus, it is a hull operator. An example of a hull operator which is not a closure operator is the linear span in a vector space. By means of closure operators, topologies on a set can be characterized in the following sense: If $C:\mathcal{P}\left(X\right)\to \mathcal{P}\left(X\right)$ is a closure operator on a given set X, then there exists a unique topology on X whose closed sets are precisely the image of C. This topology is described as

$$\{B\subseteq X:C(X\setminus B)=X\setminus B\}.$$

(2)

We will define now limit operators, which is a novel concept from this work. These ideas are inherited from the well-known concept of Banach limit [39,40,41,42,43]. However, we will first need the following remark.

Remark 2.

Let X be a set. For every $x\in X$, we will let $\mathbf{x}$ denote the constant sequence of general term equal to x, hence $\mathbf{x}\in X^{\mathbb{N}}$. On the other hand, we define the following preorder on $X^{\mathbb{N}}$: $a\le b$ if and only if a is a subsequence of b.

Keep in mind that Banach limits need an algebraic structure to be defined. Nevertheless, our concept of limit operator does not have such needs.

Definition 1

(Limit operator). Let X be a set. Let $L:X^{\mathbb{N}}\to \mathcal{P}\left(X\right)$. We say that L is a limit operator when the following conditions are satisfied:

• $x\in L\left(\mathbf{x}\right)$ for every $x\in X$;
• If $a,b\in X^{\mathbb{N}}$ and $a\le b$, then $L\left(b\right)\subseteq L\left(a\right)$;
• If $b\in X^{\mathbb{N}}$, $x\in L\left(b\right)$, and for every $n\in \mathbb{N}$ there exists $a_n\in X^{\mathbb{N}}$ with $b\left(n\right)\in L\left(a_n\right)$, then there are two subsequences ${\left(n_k\right)}_{k\in \mathbb{N}},{\left(m_k\right)}_{k\in \mathbb{N}}\subseteq \mathbb{N}$ such that $x\in L\left(c\right)$, where $c\left(k\right):=a_{n_k}\left(m_k\right)$ for all $k\in \mathbb{N}$.

The limit of a sequence in a first-countable topological space is clearly a limit operator.

Example 1.

Let X be a first-countable topological space. The following is a limit operator:

$$\begin{array}{cccc}\hfill lim:& \hfill X^{\mathbb{N}}& \to & \mathcal{P}\left(X\right)\hfill \\ \hfill & \hfill {\left(x_n\right)}_{n\in \mathbb{N}}& \mapsto & {\displaystyle \underset{n\to \infty }{lim}x\left(n\right).}\hfill \end{array}$$

(3)

Every limit operator induces a closure operator.

Theorem 2.

Let X be a set. Let $L:X^{\mathbb{N}}\to \mathcal{P}\left(X\right)$ be a limit operator. The following is a closure operator:

$$\begin{array}{cccc}\hfill C_L:& \hfill \mathcal{P}\left(X\right)& \to & \mathcal{P}\left(X\right)\hfill \\ & \hfill A& \mapsto & {\displaystyle C_L\left(A\right):=\bigcup _{a\in A^{\mathbb{N}}}L\left(a\right).}\hfill \end{array}$$

(4)

Proof. 

We will check all four properties that characterize hull operators are verified:

• Nullity: Simply notice that $C_L(⌀)$ is a union indexed by the empty set, thus $C_L(⌀)=⌀$.
• Extensionality: If $x\in A$, then $x\in L\left(\mathbf{x}\right)\subseteq C_L\left(A\right)$, thus $A\subseteq C_L\left(A\right)$.
• Additivity: Notice that $A^{\mathbb{N}}\cup B^{\mathbb{N}}\subseteq {(A\cup B)}^{\mathbb{N}}$, hence $C_L\left(A\right)\cup C_L\left(B\right)\subseteq C_L(A\cup B)$. Fix an arbitrary $c\in {(A\cup B)}^{\mathbb{N}}$. We can assume without any loss of generality that there exists a subsequence $a\le c$ such that $a\in A^{\mathbb{N}}$. Then, $L\left(c\right)\subseteq L\left(a\right)\subseteq C_L\left(A\right)$. This shows that $C_L(A\cup B)\subseteq C_L\left(A\right)\cup C_L\left(B\right)$.
• Idempotency: The extensionality assures that $A\subseteq C_L\left(A\right)$, so $C_L\left(A\right)\subseteq C_L\left(C_L\left(A\right)\right)$. Fix arbitrary elements $b\in C_L{\left(A\right)}^{\mathbb{N}}$ and $x\in L\left(b\right)$. For every $n\in \mathbb{N}$, $b\left(n\right)\in C_L\left(A\right)$, there exists $a_n\in X^{\mathbb{N}}$ with $b\left(n\right)\in L\left(a_n\right)$. Then, we can find two subsequences ${\left(n_k\right)}_{k\in \mathbb{N}},{\left(m_k\right)}_{k\in \mathbb{N}}\subseteq \mathbb{N}$ such that $x\in L\left(c\right)$ where $c\left(k\right):=a_{n_k}\left(m_k\right)$ for all $k\in \mathbb{N}$. Notice that $c\in A^{\mathbb{N}}$, hence $L\left(c\right)\subseteq C_L\left(A\right)$. As a consequence, $C_L\left(C_L\left(A\right)\right)\subseteq C_L\left(A\right)$.

□

The topology induced by a closure operator defined by means of a limit operator is somehow compatible with the limit operator.

Theorem 3.

Let X be a set. Let $L:X^{\mathbb{N}}\to \mathcal{P}\left(X\right)$ be a limit operator. Consider the closure operator $C_L$ induced by L and given by (4). Suppose that X is endowed with the unique topology induced by $C_L$ as in (2). If $x\in X^{\mathbb{N}}$, then ${\displaystyle L\left(x\right)\subseteq \underset{n\to \infty }{lim}x\left(n\right)}$. As a consequence, every closed subset of X is sequentially closed.

Proof. 

Fix an arbitrary $z\in L\left(x\right)$ and suppose, on the contrary, that ${\displaystyle z\notin \underset{n\to \infty }{lim}x\left(n\right)}$. There exists a neighborhood U of z and a subsequence $y\le x$ such that $y\left(n\right)\notin U$ for all $n\in \mathbb{N}$. By our hypotheses, $z\in L\left(x\right)\subseteq L\left(y\right)\subseteq C_L\left(\left\{y\right(n):n\in \mathbb{N}\}\right)=\mathrm{cl}\left(\left\{y\right(n):n\in \mathbb{N}\}\right)$. This contradicts the fact that $y\left(n\right)\notin U$ for all $n\in \mathbb{N}$. Finally, let us prove that every closed subset of X is sequentially closed. Indeed, let $A\subseteq X$ be closed. By definition of the induced topology (2) by the closure operator $C_L$, we have that $A=C_L\left(A\right)$. Fix any arbitrary $x\in A$. There exists $a\in A^{\mathbb{N}}$ satisfying that ${\displaystyle x\in L\left(a\right)\subseteq \underset{n\to \infty }{lim}a\left(n\right)}$. □

3.2. Pathological Phenomena in Topological Modules

There are topological properties that are preserved by addition or linearity. However, there are other ones which are not. One of these pathological properties with respect to addition is the separation property ${\mathrm{T}}_{\mathtt{2}}$, also known as the Hausdorff separation property. First, we need to recall that, if X is a topological space and $x\in X$, then ${\mathcal{N}}_x\left(X\right)$ stands for the filter of all neighborhoods of x. The notation $\bigcap {\mathcal{N}}_x\left(X\right)$ stands for the intersection of all neighborhoods of x; in other words, $\bigcap {\mathcal{N}}_x\left(X\right)={\bigcap }_{U\in {\mathcal{N}}_x\left(X\right)}U$. If M is a topological module, then $\bigcap {\mathcal{N}}_0\left(M\right)$ is a submodule of M [33] (Theorem 2). In addition, M is Hausdorff if and only if $\bigcap {\mathcal{N}}_0\left(M\right)=\left\{0\right\}$, and M has the trivial topology if and only if $\bigcap {\mathcal{N}}_0\left(M\right)=M$. Finally, notice that a submodule N of M is Hausdorff if and only if $N\cap \bigcap {\mathcal{N}}_0\left(M\right)=\left\{0\right\}$.

Theorem 4.

Let M be a topological module over a division topological ring. If M is not Hausdorff and is not endowed with the trivial topology, then there are two Hausdorff submodules $P,Q$ of M such that $P\cap Q=\left\{0\right\}$ and $P+Q$ is not Hausdorff.

Proof. 

Since M is not Hausdorff, we can find $m_0\in \bigcap {\mathcal{N}}_0\left(M\right)\setminus \left\{0\right\}$. On the other hand, M is not endowed with the trivial topology, thus we can find $n_0\in M\setminus \bigcap {\mathcal{N}}_0\left(M\right)$. Take $l_0:=m_0-n_0$. Note that $l_0\notin \bigcap {\mathcal{N}}_0\left(M\right)$ since otherwise $n_0=m_0-l_0\in \bigcap {\mathcal{N}}_0\left(M\right)$. Consider the submodules of M given by $P:=R{l}_0$ and $Q:=R{n}_0$. Notice that $l_0+n_0=m_0\in \bigcap {\mathcal{N}}_0\left(M\right)\setminus \left\{0\right\}$, meaning that $P+Q=R{l}_0+R{n}_0$ is not Hausdorff since $(P+Q)\cap \bigcap {\mathcal{N}}_0\left(M\right)\ne \left\{0\right\}$. Let us prove that both $R{l}_0,R{n}_0$ are Hausdorff. Suppose, for instance, that there exists $r_0\in R$ with $r_0{l}_0\in \bigcap {\mathcal{N}}_0\left(M\right)\setminus \left\{0\right\}$. Then, $r_0\ne 0$, obtaining the contradiction that $l_0=r_0^{-1}\left(r_0{l}_0\right)\in \bigcap {\mathcal{N}}_0\left(M\right)$. As a consequence, $R{l}_0$ is Hausdorff, and, similarly, so is $R{n}_0$. Finally, let us prove that $P\cap Q=\left\{0\right\}$. Assume, on the contrary, that there exists $m\in \left(R{l}_0\cap R{n}_0\right)\setminus \left\{0\right\}$. Then, we can find $r_1,r_2\in R\setminus \left\{0\right\}$ satisfying that $m=r_1{l}_0=r_2{n}_0$. Therefore, $r_1(m_0-n_0)=r_2{n}_0$, in other words, $(r_2+r_1)n_0=r_1{m}_0$. If $r_2+r_1=0$, then $r_1{m}_0=0$, meaning the contradiction that $m_0=0$ because $r_1\ne 0$ so it is invertible. As a consequence, $r_1+r_2\ne 0$, hence $n_0={(r_1+r_2)}^{-1}{r}_1{m}_0\in \bigcap {\mathcal{N}}_0\left(M\right)$, which contradicts the choice of $n_0\in M\setminus \bigcap {\mathcal{N}}_0\left(M\right)$. □

The following result, which shows that proper submodules are free of internal points if the underlying ring is practical, improves [24] (Theorem 53).

Theorem 5.

Let R be a practical topological ring and M a topological R-module. If $N⊊M$ is a submodule of M, then N is free of internal points.

Proof. 

Assume, on the contrary, that $\mathrm{inter}\left(N\right)\ne ⌀$ and choose any $n\in \mathrm{inter}\left(N\right)$. Consider an arbitrary $m\in M.$ A neighborhood $W\subseteq R$ of 0 exists with $n+Wm\subseteq N.$ Since R is practical, an invertible $w\in \mathcal{U}\left(R\right)$ exists satisfying that $w\in W.$ Then, $n+wm\in n+Wm\subseteq N$, meaning that $m\in u^{-1}(N-n)=N.$ The arbitrariness of $m\in M$ implies that $M\subseteq N$, hence N is not proper. □

The next theorem is a generalization of [24] (Theorem 54) and shows that, if two linear operators coincide on a set which has 0 as an internal point, then the two operators are equal, provided that the underlying ring is practical.

Theorem 6.

Let R be a practical topological ring and $M,N$ topological R-modules. Let $A\subseteq M$ such that $0\in \mathrm{inter}\left(A\right)$. If $T,S:M\to N$ are linear maps such that ${T|}_A{=S|}_A$, then $T=S$.

Proof. 

Take any arbitrary $m\in M$. Since $0\in \mathrm{inter}\left(A\right)$, there can be found a 0-neighborhood $W\subseteq R$ verifying that $0+Wm\subseteq A$. The fact that R is practical allows for finding an invertible element $u\in \mathcal{U}\left(R\right)\cap W$. Then, $um\in Wm\subseteq A$. Finally,

$$T\left(m\right)=u^{-1}T\left(um\right)=u^{-1}S\left(um\right)=S\left(m\right).$$

The arbitrariness of $m\in M$ assures that $T=S$. □

Our next result is an improvement of [24] (Proposition 76).

Theorem 7.

Let M be a module over a topological ring R such that there exists $B\subseteq R$, a right-feasible closed unit neighborhood of 0. Let $A\subseteq M$ be absorbing, that is, satisfying that $0\in \mathrm{inter}\left(A\right)$. Then, every maximal B-balanced subset $D\subseteq A$ is also absorbing. In fact, there exists a maximal B-balanced subset $D\subseteq A$.

Proof. 

We will assume that A is not B-balanced, since otherwise there is nothing to prove. Let $\mathcal{L}:=\{C\subseteq A:C\mathrm{is}B-\mathrm{balanced}\}$. Clearly, $\mathcal{L}$ is nonempty because A contains 0 and $\left\{0\right\}$ is trivially B-balanced. Observe that $\mathcal{L}$ can be partially ordered by the inclusion. Every chain ${\left(C_i\right)}_{i\in I}$ of $\mathcal{L}$ satisfies that ${\bigcup }_{i\in I}{C}_i\in \mathcal{L}$. Then, Zorn’s Lemma assures the existence of a maximal element $D\in \mathcal{L}$. Let us finally prove that $0\in \mathrm{inter}\left(D\right)$. Fix an arbitrary $m\in M$. Since $0\in \mathrm{inter}\left(A\right)$, there exists a 0-neighborhood $V\subseteq R$ such that $0+Vm\subseteq A$. Since B is right-feasible, there exists $w\in \mathcal{U}\left(R\right)$ such that $Bw\subseteq V$. Notice that $Bwm$ is B-balanced and thus $D\cup Bwm$ is B-balanced as well. In addition, $Bwm\subseteq Vm\subseteq A$, so $D\subseteq D\cup Bwm\subseteq A$. By maximality, either $D\cup Bwm=A$ or $D\cup Bwm=D$. If $D\cup Bwm=A$, then A is B-balanced, obtaining a contradiction with the initial assumption that A is not B-balanced. Therefore, $D\cup Bwm=D$, meaning that $Bwm\subseteq D$. Since $Bw$ is a 0-neighborhood in R, we already conclude that $0\in \mathrm{inter}\left(D\right)$. □

As an immediate corollary of Theorem 7, we obtain [24] (Proposition 76).

Corollary 1.

Let M be a module over a topological ring R. Let $B\subseteq R$ be a feasible closed unit neighborhood of 0. If $A\subseteq M$ is B-absorbing, then there exists a maximal B-balanced and B-absorbing subset D of A.

Proof. 

Since A is B-absorbing, then $0\in \mathrm{inter}\left(A\right)$ by [24] (Proposition 75(4)). By applying Theorem 7, we conclude that every maximal B-balanced subset $D\subseteq A$ satisfies that $0\in \mathrm{inter}\left(D\right)$. If we apply now [24] (Proposition 75(4)), we deduce that D is B-absorbing. □

A slight modification in the proof of Theorem 7 allows for concluding the following theorem, the details of which we spare to the reader.

Theorem 8.

Let M be a module over a topological ring R. Let $A\subseteq M$ be absorbing. Then, every maximal additively symmetric subset $D\subseteq A$ is also absorbing. In fact, there exists a maximal additively symmetric subset $D\subseteq A$.

In [10] (Theorems 3.2 and 3.3), a construction of noncontinuous linear functionals on certain real or complex topological vector spaces is presented. We will show next that, on a considerably high amount of topological modules, a noncontinuous linear functional can be constructed. For this, we are in need of introducing a new notion that shares certain similarities with the concept of “large modules” introduced in [13] (Definition 7).

Definition 2.

Let R be a topological ring. Let M be a topological free R-module. We will say that M is ℓ-free if there exists a basis $\mathcal{B}$ of 0-neighborhoods in M and a basis B of M as a free R-module such that $\mathrm{card}\left(\mathcal{B}\right)\le \mathrm{card}\left(B\right)$.

The following proposition displays examples of ℓ-free modules. We refer to [13] (Example 9) together with [24] (Proposition 64) for the similar version of the next proposition for large modules.

Proposition 2.

Let R be a topological ring and M a topological free R-module. If M is first countable but not finitely generated, then M is ℓ-free.

Proof. 

Since M is first countable, there exists a basis $\mathcal{B}$ of 0-neighborhoods in M such that $\mathrm{card}\left(\mathcal{B}\right)\le {\aleph }_0$. Let $B\subseteq M$ be a basis of M as free R-module. Since $\mathrm{span}\left(B\right)=M$ and M is not finitely generated, we conclude that B is infinite. As a consequence, $\mathrm{card}\left(\mathcal{B}\right)\le {\aleph }_0\le \mathrm{card}\left(B\right)$. □

The following technical lemma is not explicitly used in the construction of a noncontinuous linear functional on ℓ-free modules; however, we keep it here because it displays the idea behind such construction.

Lemma 1.

Let R be a Hausdorff topological ring. If ${\left(u_i\right)}_{i\in I}\subseteq \mathcal{U}\left(R\right)$ is a net of invertibles converging to 0, then no subnet of ${\left(u_i^{-1}\right)}_{i\in I}$ is convergent.

Proof. 

Assume, on the contrary, that there exists a subnet ${\left(u_{i_j}^{-1}\right)}_{j\in J}$ which converges to some $r\in R$. Then, the net ${\left(u_{i_j}{u}_{i_j}^{-1}\right)}_{J\in J}$ converges to $0·r=0$. However, $u_{i_j}{u}_{i_j}^{-1}=1$ for every $j\in J$, reaching a contradiction with the facts that $1\ne 0$ and R is Hausdorff. □

The importance of the Hausdorff hypothesis in Lemma 1 is justified by the fact that, if a ring is endowed with the trivial topology, then all nets are converging to 0. Now, we are finally at the right position to construct noncontinuous linear functionals on ℓ-free modules.

Theorem 9.

Let R be a topological ring. Let M be a topological free R-module. Then:

• If R is Hausdorff and there exists a non-closed basis B of M as free R-module, then there exists a noncontinuous R-linear map $f:M\to R$.
• If R is practical and M is ℓ-free, then there exists a non-closed basis of M as a free R-module.

Proof. 

• Let $m\in \mathrm{cl}\left(B\right)\setminus B$. By considering the basis $B^{\prime }:=B-m$, we may assume without any loss of generality that $m=0$. We will begin by defining f on B as $f\left(B\right)=\left\{1\right\}$, that is, $f\left(b\right):=1$ for all $b\in B$. Since B is a basis of M as free R-module, we can extend f to the whole of M by R-linearity. Finally, f is not continuous because $1\ne 0$ and R is Hausdorff.
• Let $\mathcal{B}$ be a basis of 0-neighborhoods of M and let $B\subseteq M$ be a basis of M as free R-module such that $\mathrm{card}\left(\mathcal{B}\right)\le \mathrm{card}\left(B\right)$. Consider a one-to-one function

  $$\begin{array}{ccc}\hfill \mathcal{B}& \to & B\hfill \\ \hfill U& \mapsto & b_U.\hfill \end{array}$$

  (5)
  For every $U\in \mathcal{B}$, there exists a 0-neighborhood $V_U\subseteq R$ satisfying that $V_U{b}_U\subseteq U$. Since R is practical, we can find an invertible $v_U\in V_U$ for every $U\in \mathcal{B}$. Notice then that $v_U{b}_U\in V_U{b}_U\subseteq U$ for every $U\in \mathcal{B}$. As a consequence, the net ${\left(v_U{b}_U\right)}_{U\in \mathcal{B}}$ converges to 0. Finally, the new basis $B^{\prime }:=\left\{v_U{b}_U:U\in \mathcal{B}\right\}\cup B\setminus \left\{b_U:U\in \mathcal{B}\right\}$ satisfies that $0\in \mathrm{cl}\left(B^{\prime }\right)\setminus B^{\prime }$.

□

Corollary 2.

Let R be a Hausdorff practical topological ring. Let M be an ℓ-free topological R-module. There exists a non-closed basis of M as free R-module and a noncontinuous R-linear map $f:M\to R$.

Our next results in this manuscript are aimed at showing that there are absorbing and balanced sets which are not zero-neighborhoods. The existence of noncontinuous linear functionals helps in this matter.

Theorem 10.

Let R be a topological ring. Let $B\subseteq R$ be a closed unit 0-neighborhood. Let M be a topological R-module such that there exists a noncontinuous R-linear map $f:M\to R$. Then:

• $f^{-1}\left(B\right)$ is B-balanced;
• If B is left-feasible, then $f^{-1}\left(B\right)$ is B-absorbing;
• If B is left- or right-feasible, then $f^{-1}\left(B\right)$ is not a 0-neighborhood.

Proof. 

• Let us show first that $f^{-1}\left(B\right)$ is B-balanced. Indeed, for every $b\in B$ and every $m\in f^{-1}\left(B\right)$, $f\left(bm\right)=bf\left(m\right)\in BB=B$, meaning that $B{f}^{-1}\left(B\right)\subseteq f^{-1}\left(B\right)$, hence $f^{-1}\left(B\right)$ is B-balanced.
• Next, let us prove that $f^{-1}\left(B\right)$ is B-absorbing provided that B is left-feasible. Indeed, fix an arbitrary $m\in M$. Since B is a neighborhood of 0 in R, there exists a 0-neighborhood $V\subseteq R$ such that $Vf\left(m\right)\subseteq B$. Next, since B is left-feasible, we can find $u\in \mathcal{U}\left(R\right)$ such that $uB\subseteq V$. Finally, $uBm\subseteq f^{-1}\left(B\right)$ because $f\left(ubm\right)=ubf\left(m\right)\in uBf\left(m\right)\subseteq Vf\left(m\right)\subseteq B$ for all $b\in B$. This shows that $f^{-1}\left(B\right)$ is B-absorbing.
• We will be assuming that B is left-feasible because, if B is right-feasible, then a similar proof applies. Indeed, assume, on the contrary, that $f^{-1}\left(B\right)$ is a 0-neighborhood. We will show that f is continuous at 0, which automatically implies the contradiction that f is continuous. Fix any arbitrary 0-neighborhood $V\subseteq R$. Since B is left-feasible, there exists $u\in \mathcal{U}\left(R\right)$ such that $uB\subseteq V$. Then, $u{f}^{-1}\left(B\right)=f^{-1}\left(uB\right)\subseteq f^{-1}\left(V\right)$, meaning that $f^{-1}\left(V\right)$ is a 0-neighborhood in M because so is $u{f}^{-1}\left(B\right)$ since u is invertible.

□

The following corollary, which is a direct consequence of Theorems 9 and 10, improves [13] (Corollary 8).

Corollary 3.

Let R be a Hausdorff practical topological ring such that there exists a left-feasible closed unit 0-neighborhood $B\subseteq R$. If M is an ℓ-free topological R-module, then there exists a B-balanced and B-absorbing subset of M which is not a 0-neighborhood.

By bearing in mind Proposition 2, we obtain the following corollary as a direct consequence of Corollary 3.

Corollary 4.

Let R be a Hausdorff practical topological ring such that there exists a left-feasible closed unit 0-neighborhood $B\subseteq R$. If M is a first countable, but not finitely generated, topological free R-module, then there exists a B-balanced and B-absorbing subset of M which is not a 0-neighborhood.

We will illustrate an example of a topological module next satisfying the hypotheses of Corollary 4.

Example 2.

Let $R:=\mathbb{Q}\left[\pi \right]$ endowed with the inherited topology from $\mathbb{R}$. It is immediate that R is Hausdorff. In addition, R is a $\mathbb{Q}$-algebra, so it is clearly practical. Next, $B:=[-1,1]\cap R$ is a left-feasible closed unit 0-neighborhood of R since $[-1,1]$ is a left-feasible closed unit 0-neighborhood of $\mathbb{R}$ in view of [14] (Theorem 3.7). Finally, $M:={⨁}_{n\in \mathbb{N}}R$ is a free R-module, which is not finitely generated, but it is first countable when endowed with the product topology since R is first countable.

The notion of linearly open set [44] (Definition 2) has been around ever since in the literature of Topological Vector Spaces [1]. This notion is transported for the first time to the scope of topological modules here in this manuscript.

Definition 3.

Let M be a topological module over a topological ring R. A subset $A\subseteq M$ is called linearly open provided that $A=\mathrm{inter}\left(A\right)$.

According to [24] (Theorem 68), $\mathrm{int}\left(A\right)\subseteq \mathrm{inter}\left(A\right)$ for every subset A of a topological module M over a topological ring R. As a direct consequence, every open subset of a topological module is automatically linearly open. To continue with the Results section of this manuscript, we present a simple way to construct linearly open sets with empty interiors in topological modules admitting a noncontinuous linear functional. Theorem 11 is a generalization of [44] (Theorem 9).

Theorem 11.

Let R be a Hausdorff topological ring. If M is a topological R-module such that there exists a noncontinuous R-linear map $f:M\to R$, then $M\setminus ker\left(f\right)$ is linearly open. Furthermore, if, in addition, R is a totally right-feasible topological division ring, then $M\setminus ker\left(f\right)$ has empty interior in M.

Proof. 

We will show first that $M\setminus ker\left(f\right)$ is linearly open. Indeed, take any $m\in M\setminus ker\left(f\right)$. Fix an arbitrary $n\in M$. Since R is Hausdorff and $f\left(m\right)\ne 0$, there exists an additively symmetric 0-neighborhood $U\subseteq R$ such that $f\left(m\right)\notin U$. There exists an additively symmetric 0-neighborhood $V\subseteq R$ such that $Vf\left(n\right)\subseteq U$. Let us show that $m+Vn\subseteq M\setminus ker\left(f\right)$. Suppose, on the contrary, that there exists $v\in V$ such that $f(m+vn)=0$. Then, $f\left(m\right)=-vf\left(n\right)\in Vf\left(n\right)\subseteq U$, which is a contradiction. As a consequence, $m+Vn\subseteq M\setminus ker\left(f\right)$, hence $m\in \mathrm{inter}(M\setminus ker(f\left)\right)$. Thus, $M\setminus ker\left(f\right)=\mathrm{inter}(M\setminus ker(f\left)\right)$, meaning that $M\setminus ker\left(f\right)$ is linearly open. Next, let us assume that R is a totally right-feasible topological division ring. According to [33] (Lemma 2), $ker\left(f\right)$ is a maximal submodule of M. Therefore, $\mathrm{cl}(ker(f\left)\right)$ is either $ker\left(f\right)$ or M. If $\mathrm{cl}(ker(f\left)\right)=ker\left(f\right)$, then $ker\left(f\right)$ is closed in M, hence f is continuous in view of [33] (Lemma 3), which is a contradiction. As a consequence, $M=\mathrm{cl}(ker(f\left)\right)$. This implies that $ker\left(f\right)$ is dense in M, meaning that $M\setminus ker\left(f\right)$ has an empty interior in M. □

To finalize the Results section of this manuscript, we will modify the construction given in [13] to obtain an additively symmetric absorbing set which is not a zero-neighborhood.

Lemma 2.

Let R be a topological ring such that there exists a 0-neighborhood $W\subseteq R$ such that $W\setminus \left\{0\right\}\subseteq \mathcal{U}\left(R\right)$. Then:

• If R is practical, then R is a division topological ring.
• If R is not discrete, then R is practical, hence a division topological ring.

Proof. 

• Fix an arbitrary $r\in R\setminus \left\{0\right\}$. There exists a 0-neighborhood $V\subseteq R$ such that $Vr\subseteq W$. Since R is practical, we can find $v\in \mathcal{U}\left(R\right)\cap V$. Let $w\in W$ such that $vr=w$. Notice that v is invertible, hence it cannot be a zero divisor, thus $w\ne 0$. Then, by hypothesis, $w\in \mathcal{U}\left(R\right)$. Finally, $r=v^{-1}w\in \mathcal{U}\left(R\right)$.
• Fix an arbitrary 0-neighborhood $U\subseteq R$. Notice that $U\cap W$ is a neighborhood of 0 in R. By hypothesis, R is not discrete, thus $U\cap W\ne \left\{0\right\}$. As a consequence, $(U\cap W)\setminus \left\{0\right\}\subseteq W\setminus \left\{0\right\}\subseteq \mathcal{U}\left(R\right)$. This shows that R is practical. By Lemma 2(1), R is a division topological ring.

□

The following example displays that in, Lemma 2(2), the hypothesis of not being discrete cannot be disregarded.

Example 3.

$\mathbb{Z}$ is a topological ring which is not a division ring but there exists a 0-neighborhood $W\subseteq R$ such that $W\setminus \left\{0\right\}\subseteq \mathcal{U}\left(R\right)$. Indeed, it suffices to take $W:=\{-1,0,1\}$.

The following technical lemma will be helpful for our purposes.

Lemma 3.

Let R be a topological ring such that $\mathcal{U}\left(R\right)$ is open. Let $W\subseteq R$ be a 0-neighborhood such that $1\notin \mathrm{int}\left(W\right)$. Let M be a topological R-module. For every subset $G\subseteq M$ and every $g\in \mathrm{inter}\left(G\right)$, there exists $s\in \mathcal{U}\left(R\right)\setminus W$ such that $sg\in G$.

Proof. 

There exists a 0-neighborhood $V_1\subseteq R$ verifying that $1+V_1\subseteq \mathcal{U}\left(R\right)$. There can be found a 0-neighborhood $V_2\subseteq R$ satisfying that $g+V_{2}g\subseteq G$. Consider $V:=V_1\cap V_2$. Then, $1+V\subseteq \mathcal{U}\left(R\right)$ and $g+Vg\subseteq G$. If $1+V\subseteq W$, then $1\in \mathrm{int}\left(W\right)$, reaching a contradiction with our hypothesis. As a consequence, $1+V⊈W$, meaning that there exists $v\in V$ with $1+v\notin W$. Finally, it only suffices to take $s:=1+v$ because $sg=g+vg\in g+Vg\subseteq G$. □

Recall that a maximal equivalence free [13] (Theorem 16) subset of a module M is a subset $P\subseteq M$ such that $0\notin P$ and, for every $m\in M\setminus \left\{0\right\}$, $P\cap \mathcal{U}\left(R\right)m$ is a singleton. We refer to [13] (Theorem 16) and to [24] (Theorem 73) for the existence of maximal equivalence free sets whose closure contains 0.

Theorem 12.

Let R be a topological ring and let M be a topological R-module. For every additively symmetric 0-neighborhood $W\subseteq R$ and every maximal equivalence free subset $P\subseteq M$, let $A(W,P):={\bigcup }_{p\in P}Wp$. The following holds:

• $A(W,P)$ is additively symmetric and absorbing.
• If $\mathcal{U}\left(R\right)$ is open, $0\in \mathrm{cl}\left(P\right)$, $1\notin \mathrm{int}\left(W\right)$, and $W\setminus \left\{0\right\}\subseteq \mathcal{U}\left(R\right)$, then $A(W,P)$ is not a neighborhood of 0.

Proof. 

• For every $w\in W$ and every $p\in P$, $-wp=(-w)p\in Wp\subseteq A(W,P)$, hence $A(W,P)$ is additively symmetric. Let us prove next that $A(W,P)$ is absorbing. Indeed, take any $m\in M\setminus \left\{0\right\}$. Since P is maximal equivalence free, there exists a unique $p\in P$ and an invertible $u\in \mathcal{U}\left(R\right)$ such that $p=um$. There exists a 0-neighborhood $V\subseteq R$ satisfying that $V{u}^{-1}\subseteq W$. Then, $0+Vm=0+\left(V{u}^{-1}\right)p\subseteq 0+Wp=Wp\subseteq A(W,P)$. This shows that $0\in \mathrm{inter}\left(A(W,P)\right)$, meaning that $A(W,P)$ is absorbing.
• Assume, on the contrary, that $A(W,P)$ is a neighborhood of 0. Since $0\in \mathrm{cl}\left(P\right)$, we can find $p_0\in P$ with $p_0\in \mathrm{int}\left(A(W,P)\right)$. By bearing in mind Lemma 3, we can find $s\in \mathcal{U}\left(R\right)\setminus W$ such that $s{p}_0\in A(W,P)$. There exist $w\in W$ and $p_1\in P$ such that $s{p}_0=w{p}_1$. Notice that $p_0,p_1\in P\cap \mathcal{U}\left(R\right)p_0$. By hypothesis, P is maximal equivalence free, meaning that $P\cap \mathcal{U}\left(R\right)p_0$ is a singleton. As a consequence, $p_0=p_1$. In other words, $s{p}_0=w{p}_0$, or equivalently, $(s-w)p_0=0$. At this stage, observe the fact that $0\notin P$, and $0\in \mathrm{cl}\left(P\right)$ implies that R is not discrete. Then, we call on Lemma 2 to conclude that R is a division topological ring. Since $s\notin W$, we have that $s-w\ne 0$, so we obtain the contradiction that $p_0={(s-w)}^{-1}(s-w)p_0=0$. As a consequence, $A(W,P)$ is not a neighborhood of 0.

□

We will next illustrate an example of a topological module satisfying the hypotheses of Theorem 12.

Example 4.

Let $R:=\mathbb{Q}\left[\sqrt{2}\right]$ endowed with the inherited topology from $\mathbb{R}$. Note that R is a division ring since $\sqrt{2}$ is algebraic over $\mathbb{Q}$. It is immediate that R is Hausdorff, meaning that $\mathcal{U}\left(R\right)=R\setminus \left\{0\right\}$ is open. In addition, R is a $\mathbb{Q}$-algebra, so it is clearly practical. Next, $W:=(-1,1)\cap R$ is an additively symmetric open 0-neighborhood of R satisfying that $1\notin W$ and $W\setminus \left\{0\right\}\subseteq \mathcal{U}\left(R\right)$. Finally, $M:={⨁}_{n\in \mathbb{N}}R$ is a free R-module, which is not finitely generated, but it is first countable when endowed with the product topology since R is first countable. According to [13] (Theorem 16), there exists a maximal equivalence free subset $P\subseteq M$ such that $0\in \mathrm{cl}\left(P\right)$.

We will conclude this manuscript by showing that the limit of prefilters (filter bases) in a topological module behaves in some sense like a limit operator satisfying interesting properties. First, observe that, if ${\mathcal{B}}_1,{\mathcal{B}}_2$ are filter bases of $\mathcal{P}\left(M\right)$, for M a module over a ring R, and $r\in R$, then ${\mathcal{B}}_1+{\mathcal{B}}_2:=\{A_1+A_2:A_1\in {\mathcal{B}}_1,A_2\in {\mathcal{B}}_2\}$ and $r{\mathcal{B}}_1:=\{r{A}_1:A_1\in {\mathcal{B}}_1\}$ are filter bases of $\mathcal{P}\left(M\right)$. On the other hand, if $M=R$, then ${\mathcal{B}}_1{\mathcal{B}}_2:=\{A_1{A}_2:A_1\in {\mathcal{B}}_1,A_2\in {\mathcal{B}}_2\}$ is also a filter base of $\mathcal{P}\left(R\right)$.

Remark 3.

Let M be a module over a ring R. Let $A,B,C$ be subsets of M. Then:

• If $A+B\subseteq C$ and $C-A\subseteq B$, then $A+B=C$.
• If $C-A\subseteq B$ and $C-B\subseteq A$, then $B=C-A$.

We will rely on Remark 3 to prove the linear properties of limits of prefilters in topological modules.

Theorem 13.

Let M be a topological module over a topological ring R. Let ${\mathcal{B}}_1,{\mathcal{B}}_2$ be filter bases of $\mathcal{P}\left(M\right)$ and consider $r\in R$. Then:

• $lim\left({\mathcal{B}}_1\right)+lim\left({\mathcal{B}}_2\right)=lim({\mathcal{B}}_1+{\mathcal{B}}_2)$.
• $rlim\left({\mathcal{B}}_1\right)\subseteq lim\left(r{\mathcal{B}}_1\right)$.
• If $r\in \mathcal{U}\left(R\right)$, then $rlim\left({\mathcal{B}}_1\right)=lim\left(r{\mathcal{B}}_1\right)$.
• If $M=R$, then $lim\left({\mathcal{B}}_1\right)lim\left({\mathcal{B}}_2\right)\subseteq lim\left({\mathcal{B}}_1{\mathcal{B}}_2\right)$.

Proof. 

• Take $m_i\in lim\left({\mathcal{B}}_i\right)$ for $i=1,2$. Fix an arbitrary 0-neighborhood $U\subseteq M$. Let $V\subseteq M$ another 0-neighborhood such that $V+V\subseteq U$. There are $A_i\in {\mathcal{B}}_i$, $i=1,2$, with $A_i\subseteq m_i+V$. Then, $A_1+A_2\in {\mathcal{B}}_1+{\mathcal{B}}_2$ and $A_1+A_2\subseteq (m_1+V)+(m_2+V)=(m_1+m_2)+(V+V)\subseteq (m_1+m_2)+U$. This shows that $m_1+m_2\in lim({\mathcal{B}}_1+{\mathcal{B}}_2)$. As a consequence, $lim\left({\mathcal{B}}_1\right)+lim\left({\mathcal{B}}_2\right)\subseteq lim({\mathcal{B}}_1+{\mathcal{B}}_2)$. Following a similar reasoning, it can be proved that $lim({\mathcal{B}}_1+{\mathcal{B}}_2)-lim\left({\mathcal{B}}_1\right)\subseteq lim\left({\mathcal{B}}_2\right)$. In view of Remark 3, we conclude that $lim\left({\mathcal{B}}_1\right)+lim\left({\mathcal{B}}_2\right)=lim({\mathcal{B}}_1+{\mathcal{B}}_2)$.
• Take $m_1\in lim\left({\mathcal{B}}_1\right)$. Fix an arbitrary 0-neighborhood $U\subseteq M$. Let $V\subseteq M$ another 0-neighborhood such that $rV\subseteq U$. There is $A_1\in {\mathcal{B}}_1$ with $A_1\subseteq m_1+V$. Then, $r{A}_1\in r{\mathcal{B}}_1$ and $r{A}_1\subseteq r(m_1+V)=r{m}_1+rV\subseteq r{m}_1+U$. This shows that $r{m}_1\in lim\left(r{\mathcal{B}}_1\right)$. As a consequence, $rlim\left({\mathcal{B}}_1\right)\subseteq lim\left(r{\mathcal{B}}_1\right)$.
• If $r\in \mathcal{U}\left(R\right)$, then $r^{-1}lim\left(r{\mathcal{B}}_1\right)\subseteq lim\left(r^{-1}\left(r{\mathcal{B}}_1\right)\right)=lim\left({\mathcal{B}}_1\right)$. Therefore, $lim\left(r{\mathcal{B}}_1\right)\subseteq rlim\left({\mathcal{B}}_1\right)$.
• Take $a_i\in lim\left({\mathcal{B}}_i\right)$ for $i=1,2$. Fix an arbitrary 0-neighborhood $U\subseteq R$. Let $W\subseteq R$ be another 0-neighborhood such that $W+W+W\subseteq U$. We can find 0-neighborhoods $V_1,V_2,V_3\subseteq R$ such that $V_1{a}_2\subseteq W$, $a_1{V}_2\subseteq W$, and $V_3{V}_3\subseteq W$. Take $V:=V_1\cap V_2\cap V_3$. There are $A_i\in {\mathcal{B}}_i$, $i=1,2$, with $A_i\subseteq a_i+V$. Then, $A_1{A}_2\in {\mathcal{B}}_1{\mathcal{B}}_2$ and $A_1{A}_2\subseteq (a_1+V)(a_2+V)\subseteq a_1{a}_2+(V{a}_2+a_{1}V+VV)\subseteq a_1{a}_2+(W+W+W)\subseteq a_1{a}_2+U$. This shows that $a_1{a}_2\in lim\left({\mathcal{B}}_1{\mathcal{B}}_2\right)$. As a consequence, $lim\left({\mathcal{B}}_1\right)lim\left({\mathcal{B}}_2\right)\subseteq lim\left({\mathcal{B}}_1{\mathcal{B}}_2\right)$.

□

By bearing in mind Theorem 13, we can obtain as immediate corollaries the linearity of limits of functions and nets on topological modules. First, a technical lemma is needed. Recall that, if $\mathcal{B}$ is a prefilter, then the filter generated by $\mathcal{B}$ is usually denoted by $\mathcal{J}\left(\mathcal{B}\right)$.

Lemma 4.

Let X be a nonempty set and $\mathcal{B}$ a filter base of $\mathcal{P}\left(X\right)$. Let M be a module over a ring R. Let $f_1,f_2:X\to M$ be functions. Then, $\mathcal{J}\left((f_1+f_2)\left(\mathcal{B}\right)\right)\supseteq \mathcal{J}\left(f_1\left(\mathcal{B}\right)+f_2\left(\mathcal{B}\right)\right)$. If $M=R$, then $\mathcal{J}\left(\left(f_1{f}_2\right)\left(\mathcal{B}\right)\right)\supseteq \mathcal{J}\left(f_1\left(\mathcal{B}\right)f_2\left(\mathcal{B}\right)\right)$.

Proof. 

Since $\mathcal{B}$ is a filter base, for every $A,C\in \mathcal{B}$, there exists $D\in \mathcal{B}$ with $D\subseteq A\cap C$, meaning that $(f_1+f_2)\left(D\right)\subseteq f_1\left(A\right)+f_2\left(C\right)$. Then, $f_1\left(A\right)+f_2\left(C\right)\in \mathcal{J}\left((f_1+f_2)\left(\mathcal{B}\right)\right)$. This shows that $f_1\left(\mathcal{B}\right)+f_2\left(\mathcal{B}\right)\subseteq \mathcal{J}\left((f_1+f_2)\left(\mathcal{B}\right)\right)$, hence $\mathcal{J}\left(f_1\left(B\right)+f_2\left(\mathcal{B}\right)\right)\subseteq \mathcal{J}\left((f_1+f_2)\left(\mathcal{B}\right)\right)$. In a similar way, it can be shown that, if $M=R$, then $\mathcal{J}\left(f_1\left(\mathcal{B}\right)f_2\left(\mathcal{B}\right)\right)\subseteq \mathcal{J}\left(\left(f_1{f}_2\right)\left(\mathcal{B}\right)\right)$. □

Corollary 5.

Let X be a nonempty set and $\mathcal{B}$ a filter base of $\mathcal{P}\left(X\right)$. Let M be a topological module over a topological ring R. Let $f_1,f_2:X\to M$ be functions and consider $r\in R$. Then:

• ${\displaystyle \underset{\mathcal{B}}{lim}\left(f_1\right)+\underset{\mathcal{B}}{lim}\left(f_2\right)=\underset{\mathcal{B}}{lim}(f_1+f_2)}$.
• ${\displaystyle r\underset{\mathcal{B}}{lim}\left(f_1\right)\subseteq \underset{\mathcal{B}}{lim}\left(r{f}_1\right)}$.
• If $r\in \mathcal{U}\left(R\right)$, then ${\displaystyle r\underset{\mathcal{B}}{lim}\left(f_1\right)=\underset{\mathcal{B}}{lim}\left(r{f}_1\right)}$.
• If $M=R$, then ${\displaystyle \underset{\mathcal{B}}{lim}\left(f_1\right)\underset{\mathcal{B}}{lim}\left(f_2\right)\subseteq \underset{\mathcal{B}}{lim}\left(f_1{f}_2\right)}$.

Proof. 

• According to Lemma 4, $\mathcal{J}\left((f_1+f_2)\left(\mathcal{B}\right)\right)\supseteq \mathcal{J}\left(f_1\left(\mathcal{B}\right)+f_2\left(\mathcal{B}\right)\right)$. As a consequence, by relying on Theorem 13(1),

  $$\begin{array}{ccc}\hfill \underset{\mathcal{B}}{lim}\left(f_1\right)+\underset{\mathcal{B}}{lim}\left(f_2\right)& =& lim\left(f_1\left(\mathcal{B}\right)\right)+lim\left(f_2\left(\mathcal{B}\right)\right)\hfill \\ & =& lim(f_1\left(\mathcal{B}\right)+f_2\left(\mathcal{B}\right))\hfill \\ & =& lim\left(\mathcal{J}\left(f_1\left(\mathcal{B}\right)+f_2\left(\mathcal{B}\right)\right)\right)\hfill \\ & \subseteq & lim\left(\mathcal{J}\left((f_1+f_2)\left(\mathcal{B}\right)\right)\right)\hfill \\ & =& lim\left((f_1+f_2)\left(\mathcal{B}\right)\right)\hfill \\ & =& \underset{\mathcal{B}}{lim}(f_1+f_2).\hfill \end{array}$$
  Following a similar reasoning, it can be proved that ${lim}_{\mathcal{B}}(f_1+f_2)-{lim}_{\mathcal{B}}\left(f_1\right)\subseteq {lim}_{\mathcal{B}}\left(f_2\right)$. In view of Remark 3, we conclude that ${\displaystyle \underset{\mathcal{B}}{lim}\left(f_1\right)+\underset{\mathcal{B}}{lim}\left(f_2\right)=\underset{\mathcal{B}}{lim}(f_1+f_2)}$.
• According to Theorem 13(2), ${\displaystyle r\underset{\mathcal{B}}{lim}\left(f_1\right)=rlim\left(f_1\left(\mathcal{B}\right)\right)\subseteq lim\left(r{f}_1\left(\mathcal{B}\right)\right)=\underset{\mathcal{B}}{lim}\left(r{f}_1\right)}$.
• If $r\in \mathcal{U}\left(R\right)$, then ${\displaystyle r^{-1}\underset{\mathcal{B}}{lim}\left(r{f}_1\right)\subseteq \underset{\mathcal{B}}{lim}\left(r^{-1}\left(r{f}_1\right)\right)=\underset{\mathcal{B}}{lim}\left(f_1\right)}$. Therefore, ${\displaystyle \underset{\mathcal{B}}{lim}\left(r{f}_1\right)\subseteq r\underset{\mathcal{B}}{lim}\left(f_1\right)}$.
• According to Lemma 4, $\mathcal{J}\left(\left(f_1{f}_2\right)\left(\mathcal{B}\right)\right)\supseteq \mathcal{J}\left(f_1\left(B\right)f_2\left(\mathcal{B}\right)\right)$. As a consequence, by relying on Theorem 13(4),

  $$\begin{array}{ccc}\hfill \underset{\mathcal{B}}{lim}\left(f_1\right)\underset{\mathcal{B}}{lim}\left(f_2\right)& =& lim\left(f_1\left(\mathcal{B}\right)\right)lim\left(f_2\left(\mathcal{B}\right)\right)\hfill \\ & \subseteq & lim\left(f_1\left(\mathcal{B}\right)f_2\left(\mathcal{B}\right)\right)\hfill \\ & =& lim\left(\mathcal{J}\left(f_1\left(\mathcal{B}\right)f_2\left(\mathcal{B}\right)\right)\right)\hfill \\ & \subseteq & lim\left(\mathcal{J}\left(\left(f_1{f}_2\right)\left(\mathcal{B}\right)\right)\right)\hfill \\ & =& lim\left(\left(f_1{f}_2\right)\left(\mathcal{B}\right)\right)\hfill \\ & =& \underset{\mathcal{B}}{lim}\left(f_1{f}_2\right).\hfill \end{array}$$

□

The following two corollaries are immediate consequences of Corollary 5 since the limit of functions at a limit point and the limit of nets are defined as limits of functions on a prefilter.

Corollary 6.

Let X be a topological space and $x_0$ a limit point of X. Let M be a topological module over a topological ring R. Let $f_1,f_2:X\setminus \left\{x_0\right\}\to M$ be functions and consider $r\in R$. Then:

• ${\displaystyle \underset{x\to x_0}{lim}\left(f_1\right)+\underset{x\to x_0}{lim}\left(f_2\right)=\underset{x\to x_0}{lim}(f_1+f_2)}$.
• ${\displaystyle r\underset{x\to x_0}{lim}\left(f_1\right)\subseteq \underset{x\to x_0}{lim}\left(r{f}_1\right)}$.
• If $r\in \mathcal{U}\left(R\right)$, then ${\displaystyle r\underset{x\to x_0}{lim}\left(f_1\right)=\underset{x\to x_0}{lim}\left(r{f}_1\right)}$.
• If $M=R$, then ${\displaystyle \underset{x\to x_0}{lim}\left(f_1\right)\underset{x\to x_0}{lim}\left(f_2\right)\subseteq \underset{x\to x_0}{lim}\left(f_1{f}_2\right)}$.

Corollary 7.

Let D be a directed set and M a topological module over a topological ring R. Let ${\gamma }_1,{\gamma }_2:D\to M$ be nets and consider $r\in R$. Then:

• ${\displaystyle lim\left({\gamma }_1\right)+lim\left({\gamma }_2\right)=lim({\gamma }_1+{\gamma }_2)}$.
• ${\displaystyle rlim\left({\gamma }_1\right)\subseteq lim\left(r{\gamma }_1\right)}$.
• If $r\in \mathcal{U}\left(R\right)$, then ${\displaystyle rlim\left({\gamma }_1\right)=lim\left(r{\gamma }_1\right)}$.
• If $M=R$, then ${\displaystyle lim\left({\gamma }_1\right)lim\left({\gamma }_2\right)\subseteq lim\left({\gamma }_1{\gamma }_2\right)}$.