On Topologies Generated by Lower Porosity

Abstract

In the paper we study properties of a lower porosity of a set in a normed space $(X,\Vert \;\Vert )$. Two topologies ${\underline{p}}(X,\Vert \;\Vert )$ and ${\underline{s}}(X,\Vert \;\Vert )$ on X generated by the lower porosity are defined. Relationships between these topologies and, previously defined by V. Kelar and L. Zajíček, topologies $p(X,\Vert \;\Vert )$ and $s(X,\Vert \;\Vert )$ are studied. Applying topologies ${\underline{p}}(X,\Vert \;\Vert )$ and ${\underline{s}}(X,\Vert \;\Vert )$ we characterize maximal additive class of lower porouscontinuous functions. Some relevant properties of defined topologies are considered.

1 Preliminaries

Porosity of a set, defined in [4], is the notion of smallness more restrictive than nowhere density and meagerness. It can be defined in arbitrary metric space. The main idea is that we modify the ”ball” definition of nowhere density by the request that the sizes of holes should be estimated. Usually, the notion of the (upper) porosity of sets is used in many aspects, see for example [4,5,6, 10, 12, 13]. We deal with the lower porosity, which also be considered in some papers, [11, 12]. It is known that there are big differences between the lower and the upper porosities. In [12, 13] some properties of the lower porosity in metric spaces are presented, whereas in [11] some properties of the lower porosity on ${\mathbb {R}}^2$ and of lower porouscontinuous functions $f:{\mathbb {R}}^2\rightarrow {\mathbb {R}}$ are studied.

Let ${\mathbb {N}}$ and ${\mathbb {R}}$ denote the set of all positive integers and the set of all real numbers, respectively. For $f:X\rightarrow Y$ and $Z\subset X$, by $f_{\upharpoonright Z}$ we mean the restriction of f to Z. For the whole paper $(X,\Vert \;\Vert )$ denotes a normed space. The symbols ${{\,\mathrm{cl}\,}}_{\mathcal {T}} Z$, ${{\,\mathrm{int}\,}}_{\mathcal {T}} Z$ and ${{\,\mathrm{bd}\,}}_{\mathcal {T}} Z$ denote the closure, the interior and the boundary of $Z\subset X$ with respect to a topology $\mathcal {T}$ in X. The open ball in $(X,\Vert \;\Vert )$ with the center $x\in X$ and the radius $\varrho >0$ is denoted by $B(x,\varrho )$. Similarly, by $S(x,\varrho )$ and ${\overline{B}}(x,\varrho )$ we denote the sphere and the closed ball with the center x and the radius $\varrho$, respectively. By $0_X$ we denote the zero element of X.

We will also consider spaces $({\mathbb {R}}^n, \Vert \;\Vert _n)$, $({\mathbb {R}}^n, \Vert \;\Vert _{\max })$ for $n\ge 1$ and $(l_\infty , \Vert \;\Vert _{\sup })$, where $\Vert \;\Vert _n$ is the natural norm in ${\mathbb {R}}^n$, $\Vert \;\Vert _{\max }$ is a norm in ${\mathbb {R}}^n$ defined by $\Vert (x_1,\ldots ,x_n)\Vert _{\max }=\max \{|x_i|:i=1,\ldots , n\}$, $l_\infty$ is the space of all bounded real sequences and $\Vert \;\Vert _{\sup }$ is a norm in $l_\infty$ defined by $\Vert (x_1,\ldots ,x_n)\Vert _{\sup }=\sup \{|x_i|:i\ge 1\}$. Spaces $({\mathbb {R}}^n, \Vert \;\Vert _{\max })$ and $(l_\infty , \Vert \;\Vert _{\sup })$ play important role in our paper and we will need some their properties.

For each $n\ge 1$ and $\zeta \in \{-1,1\}^n$, $\zeta =(\zeta _1,\ldots ,\zeta _n)$, we define

$$\begin{aligned} H^\zeta =\left\{ x=(x_1,\ldots ,x_n)\in {\mathbb {R}}^n:\forall i\le n\; \left( x_i\le 0 \text { if } \; \zeta _i=-1 \; \text { and } \; x_i\ge 0 \text { if } \; \zeta _i=1 \right) \right\} . \end{aligned}$$

For each $\zeta \in \{-1,1\}^\omega$, $\zeta =(\zeta _1,\zeta _2,\ldots )$, we define

$$\begin{aligned} H^\zeta =\left\{ x=(x_1,x_2,\ldots )\in l_\infty :\forall i\ge 1\; \left( x_i\le 0 \text { if } \; \zeta _i=-1 \; \text { and } \; x_i\ge 0 \text { if } \; \zeta _i=1 \right) \right\} . \end{aligned}$$

Lemma 1.1

Let $(X,\Vert \;\Vert )$ be $({\mathbb {R}}^n,\Vert \;\Vert _{\max })$ or $(l_\infty ,\Vert \;\Vert _{\sup })$. For each $R>0$ and for each ball $B(x,\eta )\subset B(0_X,R)$ such that $\eta \in \left( \frac{R}{4},\frac{R}{2}\right)$ there exist $y\in X$ and $\zeta \in \{-1,1\}^n$ or $\zeta \in \{-1,1\}^\omega$ such that

$$\begin{aligned} B(y,\eta -\tfrac{R}{4})\subset B(x,\eta )\cap H^\zeta \cap B(0_X,\tfrac{R}{2}). \end{aligned}$$

Proof

(See Fig. 1) it is enough to show that for each interval $[\alpha ,\beta ]\subset [-R,R]$, where $\beta -\alpha =2\eta$, there exists $(\alpha _1,\beta _1)$ such that $\beta _1-\alpha _1=2\eta -\frac{R}{2}$ and $(\alpha _1,\beta _1)\subset \left( \frac{R}{2},0\right) \cup \left( 0,\frac{R}{2}\right)$.

Fig. 1

Construction in $({\mathbb {R}}^2,\Vert \;\Vert _{\max })$

If $0\in [\alpha ,\beta ]$ then $\beta \ge \eta$ or $\alpha \le -\eta$. Therefore as $(\alpha _1,\beta _1)$ we take $(0,\beta )$ or $(\alpha ,0)$, because $\eta \ge \eta +(\eta -\frac{R}{2})=2\eta -\frac{R}{2}$. If $0\not \in [\alpha ,\beta ]$ then $(R-2\eta ,\frac{R}{2})\subset (\alpha ,\beta )$ or $(-\frac{R}{2},2\eta -R)\subset (\alpha ,\beta )$. Then $\frac{R}{2}-(R-2\eta )=2\eta -\frac{R}{2}$, which completes the proof. $\square$

Let $(X,\Vert \;\Vert )$ be a normed space. By $\mathcal {T}_{\Vert \;\Vert }$ we denote a topology in X generated by $\Vert \;\Vert$. Sometimes we consider another topology $\mathcal {T}$ in X. We say that $f:X\rightarrow {\mathbb {R}}$ is $\mathcal {T}$-continuous at some $x\in X$ if f is continuous as a function $f:(X,\mathcal {T})\rightarrow ({\mathbb {R}}, \mathcal {T}_{\Vert \;\Vert _1})$.

Now, we recall definitions of the (upper) porosity and the lower porosity in a normed space. These notions can be defined in an arbitrary metric space but we present them only for a normed space $(X,\Vert \;\Vert )$, because only such the case will be considered in the paper. Let $U\subset X$, $x\in X$ and $R>0$. Then, according to [4, 12], by $\gamma (x,R,U)$ we denote the supremum of the set of all $\varrho >0$ for which there exists $y\in X$ such that $B(y,\varrho )\subset B(x,R)\setminus U$. The number $p(U,x)=2\limsup _{R\rightarrow 0^+}\frac{\gamma (x,R,U)}{R}$ is called the (upper) porosity of U at x. Obviously, $p(U,x)= p({{\,\mathrm{cl}\,}}_{\mathcal {T}_{\Vert \;\Vert }} U,x)$ for $U\subset X$ and $x\in X$.

Similarly, the number ${\underline{p}}(U,x)=2\liminf _{R\rightarrow 0^+}\frac{\gamma (x,R,U)}{R}$ is called the lower porosity of U at x. Clearly, ${\underline{p}}(U,x)=\underline{p}({{\,\mathrm{cl}\,}}_{\mathcal {T}_{\Vert \;\Vert }} U,x)$ and ${\underline{p}}(U,x)\le p(U,x)$ for $U\subset X$ and $x\in X$. Moreover, for $U\subset V\subset X$ we have ${\underline{p}}(V,x)\le {\underline{p}}(U,x)$, ${\underline{p}}(U,x)\in [0,2]$ and ${\underline{p}}(U,x)\in [0,1]$ if $x\in {{\,\mathrm{cl}\,}}_{\mathcal {T}_{\Vert \;\Vert }} U$. We say that $U\subset X$ is (upper) porous (lower porous) at $x\in X$ if $p(U,x)>0$ ($\underline{p}(U,x)>0$). Similarly, we say that $A\subset X$ is (upper) strongly porous (lower strongly porous) at $x\in X$ if $p(U,x)=1$ ($\underline{p}(U,x)=1$).

Theorem 1.2

Let $(X,\Vert \;\Vert )$ be a normed space, $A\subset X$, $x\in X$ and ${\underline{p}}(A,x)>0$. Then there exists a sequence of closed balls $\left( {\overline{B}}(x_n,\varrho _n)\right) _{n\ge 1}$, not necessary pairwise disjoint, such that $\lim _{n\rightarrow \infty }x_n=x$, $\varrho _n\le \frac{1}{n}$ for $n\ge 1$, $\bigcup _{n=1}^\infty {\overline{B}}(x_n,\varrho _n)\cap A=\emptyset$ and

$$\begin{aligned} {\underline{p}}(A,x)={\underline{p}}\left( X\setminus \bigcup _{n=1}^\infty {\overline{B}}(x_n,\varrho _n),x\right) =\liminf _{n\rightarrow \infty } 2n\varrho _n. \end{aligned}$$

Proof

For every $n\ge 1$ put $\gamma _n=\sup \left\{ \varrho :\exists _{y\in X}\; \left( {\overline{B}}(y,\varrho )\subset \overline{ B}(x,\frac{1}{n})\setminus A\right) \right\}$ and choose a closed ball ${\overline{B}}(x_n,\varrho _n)\subset \overline{B}(x,\frac{1}{n})\setminus A$ such that $\varrho _n>\gamma _n\left( 1-\frac{1}{n^2}\right)$. Denote $B=X\setminus \bigcup _{n=1}^\infty {\overline{B}}(x_n,\varrho _n)$. Since $A\subset B$, we get ${\underline{p}}(B,x)\le {\underline{p}}(A,x)$. Fix $n>1$ and choose any $R\in \left( \frac{1}{n+1},\frac{1}{n}\right]$. Then

$$\begin{aligned} \frac{2\gamma (x,R,A)}{R}\le \frac{2\gamma _n}{\frac{1}{n+1}}<\frac{2\varrho _n}{\left( 1-\frac{1}{n^2}\right) \frac{1}{n+1}}= \frac{2\varrho _nn^2}{n-1}. \end{aligned}$$

On the other hand,

$$\begin{aligned} \gamma (x,R,B)\ge \gamma (x, \tfrac{1}{n+1},B)\ge \varrho _n-\tfrac{1}{2} (\tfrac{1}{n}-\tfrac{1}{n+1})=\varrho _n-\frac{1}{2n(n+1)} \end{aligned}$$

and

$$\begin{aligned} \frac{2\gamma (x,R,B)}{R}\ge \frac{2\varrho _n-\frac{1}{n(n+1)}}{\frac{1}{n}}=2n\varrho _n-\frac{1}{n+1}. \end{aligned}$$

We have showed that for each $n>1$ and for each $R\in \left( \frac{1}{n+1},\frac{1}{n}\right]$ the following inequalities

$$\begin{aligned} \frac{2\gamma (x,R,A)}{R}<2\varrho _nn\frac{n}{n-1} \quad \text { and } \quad \frac{2\gamma (x,R,B)}{R}\ge 2n\varrho _n-\frac{1}{n+1} \end{aligned}$$

are true. Hence

$$\begin{aligned} {\underline{p}}(A,x)\le \liminf _{n\rightarrow \infty }2n\varrho _nn\frac{n}{n-1}= \liminf _{n\rightarrow \infty }2n\varrho _n\cdot \lim _{n\rightarrow \infty }\frac{n}{n-1}=\liminf _{n\rightarrow \infty }2n\varrho _n \end{aligned}$$

and

$$\begin{aligned} {\underline{p}}(B,x)=\liminf _{R\rightarrow 0}\frac{2\gamma (x,R,B)}{R}\ge \liminf _{n\rightarrow \infty }\left( 2n\varrho _n-\frac{1}{n+1}\right) =\liminf _{n\rightarrow \infty }2n\varrho _n. \end{aligned}$$

Finally, ${\underline{p}}(A,x)=\underline{p}(B,x)=\liminf _{n\rightarrow \infty }2n\varrho _n.$ $\square$

In [13] and [7] L. Zajíček and V. Kelar introduce two topologies using the notion of (upper) porosity and (upper) strong porosity.

Definition 1.3

[13]. Let $A\subset X$ and $x\in X$. We say that A is (upper) superporous at x if $A\cup B$ is (upper) porous at x whenever B is (upper) porous at x. A set A is said to be p-open (porosity open) if $X\setminus A$ is (upper) superporous at any point of A.

Definition 1.4

[7]. Let $A\subset X$ and $x\in X$. We say that A is (upper) strongly superporous at x if $A\cup B$ is (upper) porous at x whenever B is (upper) strongly porous at x. A set A is said to be s-open (strongly porosity open) if $X\setminus A$ is (upper) strongly superporous at any point of A.

The system of all p-open sets in $(X,\Vert \;\Vert )$ forms a topology $p(X,\Vert \;\Vert )$, which will also be called the p-topology or the porosity topology, [13]. The system of all s-open sets forms a topology $s(X,\Vert \;\Vert )$, which will be called s-topology or the strong porosity topology, [7]. Obviously $p(X,\Vert \;\Vert )$ and $s(X,\Vert \;\Vert )$ are finer than the initial topology. On a non-trivial normed space neither $s(X,\Vert \;\Vert )$ is finer than $p(X,\Vert \;\Vert )$ nor $p(X,\Vert \;\Vert )$ is finer than $s(X,\Vert \;\Vert )$, [7]. The both topologies are completely regular, [7].

The aim of our paper is to describe the properties of topologies ${\underline{s}}(X,\Vert \;\Vert )$ and ${\underline{p}}(X,\Vert \;\Vert )$ which are generated by the lower porosity in a similar way as $s(X,\Vert \;\Vert )$ and $p(X,\Vert \;\Vert )$ were generated by the standard (upper) porosity. Section 2 describes relationships between topologies $s(X,\Vert \;\Vert )$, $p(X,\Vert \;\Vert )$, ${\underline{s}}(X,\Vert \;\Vert )$, ${\underline{p}}(X,\Vert \;\Vert )$ and $\mathcal {T}_{\Vert \;\Vert }$, which are quite interesting. First of all, we show that $\mathcal {T}_{\Vert \;\Vert }\subset {\underline{s}}(X,\Vert \;\Vert )$. We give examples of spaces in which this inclusion is proper and examples of spaces in which we have equality. The more, we prove that there are two equivalent norms in ${\mathbb {R}}^n$ such that in the first we have a proper inclusion and in the second we have equality. Namely, $\mathcal {T}_{\Vert \;\Vert _n}={\underline{s}}({\mathbb {R}}^n,\Vert \;\Vert _n)$ and $\mathcal {T}_{\Vert \;\Vert _{\max }}\subsetneqq {\underline{s}}({\mathbb {R}}^n,\Vert \;\Vert _{\max })$. In particular, ${\underline{s}}({\mathbb {R}}^n,\Vert \;\Vert _n)\ne {\underline{s}}({\mathbb {R}}^n,\Vert \;\Vert _{\max })$ although $\mathcal {T}_{\Vert \;\Vert _{\max }}=\mathcal {T}_{\Vert \;\Vert _n}$. Then we show that the inclusion $p(X,\Vert \;\Vert )\subset {\underline{p}}(X,\Vert \;\Vert )$ holds in every normed space. Next, we define two geometrical conditions (A) and (B) such that the condition (B) implies the condition (A) and every considered by the authors normed space satisfies the condition (A). We prove that the inclusion ${\underline{s}}(X,\Vert \;\Vert )\subset p(X,\Vert \;\Vert )$ holds under the condition (A) and ${\underline{s}}(X,\Vert \;\Vert )\subset s(X,\Vert \;\Vert )$ holds under the condition (B). It turns out that the condition (B) is not necessary. There is no other general relationships between considered topologies. Some other examples and properties are presented. Crucial keys in presented examples play spaces $({\mathbb {R}}^n, \Vert \;\Vert _{\max })$ and $(l_{\infty }, \Vert \;\Vert _{\sup })$.

The last section presents some applications of topologies ${\underline{s}}(X,\Vert \;\Vert )$ and ${\underline{p}}(X,\Vert \;\Vert )$. Namely, we define lower porouscontinuous functions, following ideas of J. Borsík and J. Holos from [1], and we describe maximal additive classes for some types of lower porouscontinuity in terms of topologies ${\underline{s}}(X,\Vert \;\Vert )$ and ${\underline{p}}(X,\Vert \;\Vert )$.

At the end of the section we prove a useful technical lemma.

Lemma 1.5

Let $\beta >0$ and $\delta \in \left( 0,\frac{\beta }{4}\right)$. Then for every $\alpha >0$ we have

$$\begin{aligned} \sup \left\{ \frac{b-a}{b}:[a,b]\subset [\delta ,\beta ]\setminus (\alpha , 2\alpha )\right\} \ge 1- \sqrt{\tfrac{2\delta }{\beta }}. \end{aligned}$$

Proof

Let us consider three cases. First, consider the case where $2\alpha \ge \beta$. Then $[\delta , \frac{\beta }{2}]\cap (\alpha ,2\alpha )=\emptyset$ and $\frac{\frac{\beta }{2}-\delta }{\frac{\beta }{2}}=1-\frac{2\delta }{\beta }$. Similarly, if $\alpha \le \delta$ then $[2\alpha , \beta ]\cap (\alpha ,2\alpha )=\emptyset$ and $\frac{\beta -2\alpha }{\beta }\ge 1-\frac{2\delta }{\beta }$. Finally, in the case where $\alpha \in (\delta , \frac{\beta }{2})$ we have $([\delta ,\alpha ]\cup [2\alpha ,\beta ])\cap (\alpha ,2\alpha )=\emptyset$. Consider two functions $\varphi , \psi :(\delta , \frac{\beta }{2})\rightarrow {\mathbb {R}}$ defined by $\varphi (\alpha )=\frac{\alpha -\delta }{\alpha }=1-\frac{\delta }{\alpha }$ and $\psi (\alpha )=\frac{\beta -2\alpha }{\beta }=1-\frac{2\alpha }{\beta }$. Then $\varphi$ is increasing, $\psi$ is decreasing and $\varphi (\alpha )=\psi (\alpha )$ if $\frac{\delta }{\alpha }=\frac{2\alpha }{\beta }$, i.e. $\alpha =\sqrt{\frac{\delta \beta }{2}}$. Moreover, $\varphi \left( \sqrt{\frac{\delta \beta }{2}}\right) =\psi \left( \sqrt{\frac{\delta \beta }{2}}\right) =1-\frac{\delta }{\sqrt{\frac{\delta \beta }{2}}}=1-\sqrt{\frac{2\delta }{\beta }}$. Since $1-\sqrt{\frac{2\delta }{\beta }}\le 1-\frac{2\delta }{\beta }$, the proof is completed. $\square$

2 Relationships between topologies generated by porosities

Analogously, as in the case of the standard (upper) porosity we can define lower superporosity, lower strong superporosity and topologies ${\underline{p}}(X,\Vert \;\Vert )$ and ${\underline{s}}(X,\Vert \;\Vert )$.

Definition 2.1

Let $A\subset X$ and $x\in X$. We say that A is lower superporous at x if $A\cup B$ is lower porous at x whenever B is lower porous at x. A set A is said to be ${\underline{p}}$-lower open (lower porosity open) if $X\setminus A$ is lower superporous at any point of A.

Definition 2.2

Let $A\subset X$ and $x\in X$. We say that A is lower strongly superporous at x if $A\cup B$ is lower porous at x whenever B is lower strongly porous at x. A set A is said to be ${\underline{s}}$-lower open (lower strongly porosity open) if $X\setminus A$ is lower strongly superporous at any point of A.

A simple check shows that the system of all ${\underline{p}}$-lower open sets in $(X,\Vert \;\Vert )$ forms a topology ${\underline{p}}(X,\Vert \;\Vert )$, which will also be called the ${\underline{p}}$-topology or the lower porosity topology. The system of all ${\underline{s}}$-lower open sets forms a topology ${\underline{s}}(X,\Vert \;\Vert )$, which will be called ${\underline{s}}$-topology or the lower strong porosity topology.

Remark 2.3

Let $(X,\Vert \;\Vert )$ be a normed space. For every $E\subset [0,\infty )$ let $A_E=\{x\in X:\Vert x\Vert \in E\}$. Then $p(A_E,0_X)=p(E\cup -E,0)$ and ${\underline{p}}(A_E,0_X)={\underline{p}}(E\cup -E,0)$.

Example 2.4

Let $(X,\Vert \;\Vert )$ be a normed space and $E=[0,\infty )\setminus \bigcup _{n=1}^\infty \left\{ \frac{1}{2^n}\right\}$. We claim that $A_{E}\in ({\underline{p}}{(X,\Vert \;\Vert )}\cap p{(X,\Vert \;\Vert )})\setminus (s(X,\Vert \;\Vert )\cup {\underline{s}}(X,\Vert \;\Vert )\cup \mathcal {T}_{\Vert \;\Vert })$. Clearly, $p(X\setminus A_E, 0_X)=\frac{1}{2}$ and ${\underline{p}}(X\setminus A_E, 0_X)<\frac{1}{2}$. The more, $X\setminus A_E$ is neither strongly superporous nor lower strongly superporous at $0_X$. Hence, $A_{E}\notin s(X,\Vert \;\Vert )\cup {\underline{s}}(X,\Vert \;\Vert )$. Obviously, $A_E\notin \mathcal {T}_{\Vert \;\Vert }$.

On the other hand, it is clear that $X\setminus A_E$ is superporous and lower superporous at every $x\in A_E$, $x\ne 0_X$. Moreover, for every $B\subset X$ and $B(x,\eta )\subset X\setminus B$ we can find $y\in B(x,\eta )$ such that $B(y,\min \{\frac{\eta }{2},\frac{1}{4}\Vert y\Vert \})\subset (X\setminus B)\cap A_E$. Therefore, $X\setminus A_E$ is superporous and lower superporous at $0_X$. Finally, $A_{E}\in {\underline{p}}(X,\Vert \;\Vert )\cap p(X,\Vert \;\Vert )$.

Example 2.5

Let $(X,\Vert \;\Vert )$ be a normed space and $E=[0,\infty )\setminus \bigcup _{n=2}^\infty [\frac{1}{n!},\frac{2}{n!}]$. We claim that $A_{E}\in (s(X,\Vert \;\Vert )\cap {\underline{p}}(X,\Vert \;\Vert ))\setminus (p(X,\Vert \;\Vert )\cup {\underline{s}}(X,\Vert \;\Vert )\cup \mathcal {T}_{\Vert \;\Vert })$. Clearly,

$$\begin{aligned} {\underline{p}}(X\setminus A_E, 0_X)\le \lim _{n\rightarrow \infty } \frac{2\gamma (0_X,\frac{2}{n!}, X\setminus A_E)}{\frac{2}{n!}}\le \frac{1}{2}<1. \end{aligned}$$

Hence, $X\setminus A_E$ is not lower strongly superporous at $0_X$ and $A_E\notin {\underline{s}}(X,\Vert \;\Vert )$. Let $B=A_{E}$. Then $p(B,0_X)=\lim _{n\rightarrow \infty }\frac{2\gamma (0_X,\frac{2}{n!},B)}{\frac{2}{n!}}=\frac{1}{2}$ and $p(B\cup (X\setminus A_E),0_X)=p(X,0_X)=0$. Thus $X\setminus A_E$ is not superporous at $0_X$ and $A_E\notin p(X,\Vert \;\Vert )$. Obviously, $A_E\notin \mathcal {T}_{\Vert \;\Vert }$.

It is clear that $X\setminus A_E$ is lower superporous and strongly superporous at every $x\in A_E$, $x\ne 0_X$. Take any $B\subset X$ satisfying ${\underline{p}}(B,0_X)=2c>0$. Choose $n_0,k_0\ge 2$ such that $\frac{4}{k_0}<c$, $k_0<n_0$ and $\frac{2\gamma (0_X,R,B)}{R}>c$ for every $R<\frac{1}{(n_0-1)!}$. Take any $n>n_0$ and $R\in [\frac{1}{n!},\frac{1}{(n-1)!}]$. If $R\in [\frac{1}{n!},\frac{k_0}{n!}]$ then

$$\begin{aligned} \frac{2\gamma (0_X,R,B\cup (X\setminus A_E))}{R}\ge \frac{2\gamma (0_X,\frac{1}{n!},B)- \frac{2}{(n+1)!}}{k_0\frac{1}{n!}}>\frac{c}{k_0}-\frac{2}{k_0(n+1)} \end{aligned}$$

and if $R\in [\frac{k_0}{n!},\frac{1}{(n-1)!}]$ then

$$\begin{aligned} \frac{2\gamma (0_X,R,B\cup (X\setminus A_E))}{R}\ge \frac{2\gamma (0_X,R,B)-\frac{2}{n!}}{R}>c-\frac{2}{k_0}>\frac{c}{2}. \end{aligned}$$

Therefore, ${\underline{p}}(B\cup (X\setminus A_E),0_X)\ge \frac{c}{k_0}$, $X\setminus A_E$ is lower superporous at $0_X$ and $A_E\in {\underline{p}}(X,\Vert \;\Vert )$.

Take any $B\subset X$ satisfying $p(B,0_X)=1$ and $R\in (0,1)$. Let $R\in \left[ \frac{1}{(n+1)!},\frac{1}{n!}\right)$. Choose $B(x_R,\eta _R)\subset X\setminus (B\cup \{0_X\})$ such that $\Vert x_R\Vert +\eta _R<R$, $\eta _R >\gamma (0_X,R, B)-\frac{R}{n}$ and $\Vert x_R\Vert -\eta _R>\frac{2}{(n+2)!}$. Since $p(B,0_X)=1$, we have $\limsup _{R\rightarrow 0^+}\frac{2\eta _R}{R}=1$, $\limsup _{R\rightarrow 0^+}\frac{2\eta _R}{\Vert x_R\Vert +\eta _R}=1$ and $\liminf _{R\rightarrow 0^+}\frac{\Vert x_R\Vert -\eta _R}{\Vert x_R\Vert +\eta _R}=0$. Applying Lemma 1.5 with $\beta =\Vert x_R\Vert +\eta _R$ and $\delta =\Vert x_R\Vert -\eta _R$ we obtain $\frac{2\gamma (0_X, \Vert x_R\Vert +\eta _R,B\cup (X\setminus A_E))}{\Vert x_R\Vert +\eta _R}\ge 1-\sqrt{\frac{2(\Vert x_R\Vert -\eta _R)}{\Vert x_R\Vert +\eta _R}}$. Therefore

$$\begin{aligned} p(B\cup (X\setminus A_E),0_X)=\limsup _{R\rightarrow 0^+}\frac{2\eta _R}{\Vert x_R\Vert +\eta _R}\ge 1-\liminf _{R\rightarrow 0^+}\sqrt{\frac{2(\Vert x_R\Vert -\eta _R)}{\Vert x_R\Vert +\eta _R}}=1. \end{aligned}$$

Hence $X\setminus A_E$ is strongly superporous at $0_X$ and $A_E\in s(X,\Vert \;\Vert )$.

Theorem 2.6

Let $\Vert \;\Vert _{\max }$ be the maximum norm in ${\mathbb {R}}^n$ and $\mathcal {T}_{N}$ be the natural topology in $\mathbb {R}^n$. Then ${\underline{s}}(\mathbb {R}^n,\Vert \;\Vert _{\max })\supsetneqq \mathcal {T}_N$.

Proof

The inclusion ${\underline{s}}(\mathbb {R}^n,\Vert \;\Vert _{\max })\supset \mathcal {T}_{\Vert \;\Vert _{\max }}=\mathcal {T}_N$ is obvious. Let us take $U=\left( {\mathbb {R}}^n\setminus ({\mathbb {R}}\times \{0_{n-1}\}\right) \cup \{0_n\}$, where $0_n=(0,\ldots , 0)\in {\mathbb {R}}^n$. Certainly, $U\notin \mathcal {T}_{N}$. We claim that $U\in {\underline{s}}(\mathbb {R}^n,\Vert \;\Vert _{\max })$. It is easy to see that ${\mathbb {R}}^n\setminus U$ is lower strongly superporous at every ${x}\in U\setminus \{0_n\}$. Take any $V\subset {\mathbb {R}}^n$ such that V is lower strongly porous at $0_n$. For every $R>0$ choose $B({x}_R,\eta _R)\subset B(0_n, 2R)\setminus V$ such that $\eta _R>\gamma (0_n,2R,V)-(2R)^2$. By Lemma 1.1, we can find ${y}_R\in {\mathbb {R}}^n$ and $\xi \in \{-1,1\}^n$ such that

$$\begin{aligned} B({y}_R,\eta _R-\tfrac{R}{2})\subset B({x}_R,\eta _R)\cap H^{\xi }\cap B(0_n,R). \end{aligned}$$

Therefore $\gamma (0_n, R, V\cup ({\mathbb {R}}^n\setminus U))\ge \eta _R-\frac{R}{2}$. Hence

$$\begin{aligned} {\underline{p}}(V\cup ({\mathbb {R}}^n\setminus U),0_n)&=\lim _{R\rightarrow 0^+}\frac{2\gamma (0_n, R, V\cup ({\mathbb {R}}^n\setminus U))}{R} \\&\ge \lim _{R\rightarrow 0^+}\frac{4\gamma (0_n,2R,V)-4(2R)^2-2R}{2R}=1-\lim _{R\rightarrow 0^+}8R=1. \end{aligned}$$

Thus ${\mathbb {R}}^n\setminus U$ is lower strongly superporous at $0_n$ and $U\in {\underline{s}}(\mathbb {R}^n,\Vert \;\Vert _{\max })$, which completed the proof. $\square$

Remark 2.7

Repeating arguments from the proof of Theorem 1.2 one can prove that $\mathcal {T}_{\Vert \;\Vert _{\sup }}$ is a proper subset of ${\underline{s}}(l_\infty ,\Vert \;\Vert _{\sup })$.

We will show the equality ${\underline{s}}(\mathbb {R}^n,\mathcal {T}_{N})= \mathcal {T}_N$, but first we need two technical lemmas.

Lemma 2.8

Let $U=[0,\infty )\times {\mathbb {R}}^{n-1}$ be a subset of $({\mathbb {R}}^n,\Vert \;\Vert _n)$. Then $\gamma (0_n,\frac{3}{2}t,({\mathbb {R}}^n\setminus U)\cup \{(t,0,\ldots ,0)\})\le \frac{2}{3}t$ for every $t\in (0,\infty )$.

Proof

Choose any $t\in (0,\infty )$ and let $y_t=(t,0,0,\ldots ,0)\in {\mathbb {R}}^n$. Suppose to the contrary that there exists $B(x,\eta )\subset U\setminus \{y_t\}$ such that $\eta >\frac{2}{3}t$ (see Fig. 2). Then

• (1) $x_1>\frac{2}{3}t$,

• (2) $\sqrt{\sum _{i=1}^nx_i^2}<\frac{3}{2}t-\frac{2}{3}t=\frac{5}{6}t$,

• (3) $\Vert x-y_t\Vert _n>\frac{2}{3}t$,

where $x=(x_1,x_2,\ldots ,x_n)$. Then $x_1^2-(x_1-t)^2>\frac{4}{9}t^2-\frac{1}{9}t^2=\frac{1}{3}t^2$, by (1).

Fig. 2

$B(x,\eta )$ in $({\mathbb {R}}^2,\Vert \;\Vert _{2})$

Therefore, by (1) and (3) we obtain

$$\begin{aligned} \sum _{i=1}^nx_i^2=x_1^2-(x_1-t)^2+(x_1-t)^2+\sum _{i=2}^nx_i^2>\tfrac{1}{3}t^2+\tfrac{4}{9}t^2=\tfrac{28}{36}t^2> \left( \tfrac{5}{6}t\right) ^2, \end{aligned}$$

which contradicts (2). $\square$

In the sequel we will need the notions of cone and halfspace in ${\mathbb {R}}^n$. Let $a,b\in {\mathbb {R}}^n$, $a\ne b$ and $\varphi \in (0,\frac{\pi }{2})$. The cone $c(a,b,\varphi )$ with vertex a, angle $\varphi$ and axis ab is defined as $c(a,b,\varphi )=\{x\in {\mathbb {R}}^n:|\sphericalangle (ab,ax)|<\varphi \}$. Moreover, by h(a, b) we denote the halfspace $h(a,b)=\{x\in {\mathbb {R}}^n:|\sphericalangle (ab,ax)|>\frac{\pi }{2}\}$.

Lemma 2.9

Let $\varphi \in (0,\frac{\pi }{2})$ satisfy $\tan \varphi <\frac{1}{24}$. If $({a}_k)_{k\in {\mathbb {N}}}\subset c({0}_n,(1,0,\ldots ,0),\varphi )$ is a sequence in $({\mathbb {R}}^n,\Vert \;\Vert _n)$ converging to ${0}_n$ then

$$\begin{aligned} {\underline{p}}\left( (-\infty ,0)\times {\mathbb {R}}^{n-1}\cup \textstyle { \bigcup _{k=1}^\infty \{{a}_k\}},0_n\right) <1. \end{aligned}$$

Proof

Let ${a}_k^\prime =(a_1^k,0,\ldots , 0)$, where ${a}_k=(a_1^k,a_2^k,\ldots , a_n^k)$ for $k\ge 1$. Then $\frac{\Vert {a}_m^\prime -{a}_m\Vert _n}{\Vert {a}_m^\prime \Vert _n}=\tan \sphericalangle (0_n{a}_m,0_n{a}_m^\prime )<\tan \varphi <\frac{1}{24}$ and

$$\begin{aligned} \gamma (0_n,\textstyle {\frac{3}{2}}\Vert {a}_m^\prime \Vert _n,&(-\infty ,0)\times {\mathbb {R}}^{n-1}\cup \textstyle { \bigcup _{k=1}^\infty \{{a}_k\}})\\&\le \gamma (0_n,\textstyle {\frac{3}{2}}\Vert {a}_m^\prime \Vert _n,(-\infty ,0)\times {\mathbb {R}}^{n-1}\cup \{{a}_m\}) \\&\le \gamma (0_n,\textstyle {\frac{3}{2}}\Vert {a}_m^\prime \Vert _n,(-\infty ,0)\times {\mathbb {R}}^{n-1}\cup \{{a}_m^\prime \})+\Vert {a}_m^\prime -{a}_m\Vert _n \end{aligned}$$

for every $m\ge 1$. By Lemma 2.8,

$$\begin{aligned} \frac{\gamma (0_n,\textstyle {\frac{3}{2}}\Vert {a}_m^\prime \Vert _n,(-\infty ,0)\times {\mathbb {R}}^{n-1}\cup \textstyle { \bigcup _{k=1}^\infty \{{a}_k\}})}{\textstyle {\frac{3}{2}}\Vert {a}_m^\prime \Vert _n}\le \frac{\frac{2}{3}\Vert {a}_m^\prime \Vert _n+\frac{1}{24}\Vert {a}_m^\prime \Vert _n}{\frac{3}{2}\Vert {a}_m^\prime \Vert _n}= \frac{\frac{17}{24}}{\frac{3}{2}}= \tfrac{17}{36} \end{aligned}$$

for every $m\ge 1$. Hence ${\underline{p}}\left( (-\infty ,0)\times {\mathbb {R}}^{n-1}\cup \textstyle { \bigcup _{k=1}^\infty \{{a}_k\}},0_n\right) \le \frac{2\cdot 17}{36}<1,$ which completed the proof. $\square$

Since porosity does not change under any isometry, we obtain the following corollary.

Corollary 2.10

Let $\varphi \in (0,\frac{\pi }{2})$ satisfy $\tan \varphi <\frac{1}{24}$ and ${a},{b}\in {\mathbb {R}}^n$, ${a}\ne {b}$. If $({a}_k)_{k\in {\mathbb {N}}}\subset c({a},{b},\varphi )$ is a sequence converging to a then

$$\begin{aligned} {\underline{p}}\left( h({a},{b})\cup \textstyle { \bigcup _{k=1}^\infty \{{a}_k\}},0_n\right) <1. \end{aligned}$$

Theorem 2.11

Let $\Vert \;\Vert _{n}$ and $\mathcal {T}_{N}$ be the natural norm and the natural topology in $\mathbb {R}^n$, respectively. Then ${\underline{s}}(\mathbb {R}^n,\Vert \;\Vert _{n})= \mathcal {T}_N$.

Proof

Obviously, $\mathcal {T}_N\subset {\underline{s}}(\mathbb {R}^n,\Vert \;\Vert _{n})$. Let us take any $U\notin \mathcal {T}_N$. There exist ${a}_0\in U$ and a sequence $({a}_k)_{k\ge 1}\subset {\mathbb {R}}^n\setminus U$ converging to ${a}_0$. Then we can find ${b}\in {\mathbb {R}}^n$ and $\varphi \in (0,\frac{\pi }{2})$ such that ${a}\ne {b}$, $\tan \varphi <\frac{1}{24}$ and $c({a},{b},\varphi )$ contains infinitely many elements of $({a}_k)_{k\ge 1}$. Let $V=h({a},{b})$. Obviously ${\underline{p}}(V,{a})=1$. But

$$\begin{aligned} {\underline{p}}\Big (V\cup ({\mathbb {R}}^n\setminus U),{a}\Big )\le {\underline{p}}\left( h({a},{b})\cup \bigcup _{k=1}^\infty \{{a}_k\},{a}\right) < 1 \end{aligned}$$

by Corollary 2.10. Hence $U\notin {\underline{s}}(\mathbb {R}^n,\mathcal {T}_{N})$, which completed the proof. $\square$

Corollary 2.12

For every $n\ge 2$ there exist equivalent norms $\Vert \;\Vert ^1,\Vert \;\Vert ^2$ in $\mathbb {R}^n$ such that ${\underline{s}}(\mathbb {R}^n,\Vert \;\Vert ^1)=\mathcal {T}_N$ and ${\underline{s}}(\mathbb {R}^n,\Vert \;\Vert ^2)\supsetneqq \mathcal {T}_N$. In particular, ${\underline{s}}(\mathbb {R}^n,\Vert \;\Vert ^1)\subsetneqq {\underline{s}}(\mathbb {R}^n,\Vert \;\Vert ^2)$.

Question 2.13

For which norm $\Vert \;\Vert$ in $\mathbb {R}^n$ the equality ${\underline{s}}(\mathbb {R}^n,\Vert \;\Vert )=\mathcal {T}_N$ holds?

Question 2.14

For which normed space $(X,\Vert \;\Vert )$ the equality ${\underline{s}}(X,\Vert \;\Vert )=\mathcal {T}_{\Vert \;\Vert }$ holds?

Lemma 2.15

Let $(X,\Vert \;\Vert )$ be a normed space. If $U\in p(X,\Vert \;\Vert )\cup s(X,\Vert \;\Vert )$ then ${\underline{p}}(X\setminus U,x)>0$ for every $x\in U$.

Proof

Take any $U\subset X$ such that ${\underline{p}}(X\setminus U,x)=0$ for some $x\in U$. There exists a sequence $(R_n)_{n\ge 1}$ such that $\lim _{n\rightarrow \infty }\frac{2\gamma (x,R_n,X\setminus U)}{R_n}=0$. In particular, $R_n>4\gamma (x,R_n,X\setminus U)$ for almost all n. Without loss of generality we may assume that this is true for all $n\ge 1$. Take any $y\in X$ such that $\Vert x-y\Vert =1$. Let $x_n=x+\frac{R_n+4\gamma (x,R_n,X\setminus U)}{2}(y-x)$ and $\eta _n=\frac{R_n-4\gamma (x,R_n,X\setminus U)}{2}$ for $n\ge 1$. Define $A=X\setminus \bigcup _{n=1}^\infty B\left( x+\frac{3R_n}{4}(y-x),\frac{R_n}{4}\right)$ and $B=X\setminus \bigcup _{n=1}^\infty B\left( x_n,\eta _n\right)$. Taking a subsequence if necessary, we may assume that $X\setminus A$ and $X\setminus B$ consist of pairwise disjoint balls. Then $p(A,x)=\frac{1}{2}$ and $p(B,x)=1$. We claim that $p(A\cup (X\setminus U),x)=0$ and $p(B\cup (X\setminus U),x)<1$.

Fix $n\ge 1$ and $R\in [R_{n+1},R_n]$. If $R\in [\frac{R_n}{2}, R_n]$ then

$$\begin{aligned} \frac{2\gamma (x,R,A\cup (X\setminus U))}{R}\le \frac{2\gamma (x,R_n,A\cup (X\setminus U))}{\frac{R_n}{2}}\le \frac{4\gamma (x,R_n,X\setminus U)}{R_n} \end{aligned}$$

and if $R\in [ R_{n+1},\frac{R_n}{2}]$ then

$$\begin{aligned} \frac{2\gamma (x,R,A\cup (X\setminus U))}{R}\le \frac{2\gamma (x,R_{n+1},A\cup (X\setminus U))}{R_{n+1}} \le \frac{2\gamma (x,R_{n+1},X\setminus U)}{R_{n+1}}. \end{aligned}$$

Since $\lim _{n\rightarrow \infty } \frac{2\gamma (x,R_n,X\setminus U)}{R_n}=0$, we obtain $p(A\cup (X\setminus U),x)=0$.

Again, fix $n\ge 1$ and $R\in [R_{n+1},R_n]$. If $R\in [4\gamma (x,R_n,X\setminus U), R_n]$ then

$$\begin{aligned} \frac{2\gamma (x,R,B\cup (X\setminus U))}{R}\le \frac{2\gamma (x,R_n,B\cup (X\setminus U))}{4\gamma (x,R_n,X\setminus U)}\le \frac{2\gamma (x,R_n,X\setminus U)}{4\gamma (x,R_n,X\setminus U)}=\frac{1}{2} \end{aligned}$$

and if $R\in [ R_{n+1},4\gamma (x,R_n,X\setminus U)]$ then

$$\begin{aligned} \frac{2\gamma (x,R,B\cup (X\setminus U))}{R}\le \frac{2\gamma (x,R_{n+1},B\cup (X\setminus U))}{R_{n+1}}\le \frac{2\gamma (x,R_{n+1},X\setminus U)}{R_{n+1}}. \end{aligned}$$

Since $\lim _{n\rightarrow \infty } \frac{2\gamma (x,R_n,X\setminus U)}{R_n}=0$, we obtain $p(B\cup (X\setminus U),x)\le \frac{1}{2}<1$. $\square$

Theorem 2.16

Let $(X,\Vert \;\Vert )$ be a normed space. Then $p(X,\Vert \;\Vert )\subsetneqq {\underline{p}}(X,\Vert \;\Vert )$.

Proof

Let us take any $U\notin {\underline{p}}(X,\Vert \;\Vert )$. Then $X\setminus U$ is not lower superporous at some $x_0\in U$. Hence there is $V\subset X$ satisfying ${\underline{p}}(V,x_0)=2c>0$ and ${\underline{p}}(V\cup (X\setminus U),x_0)=0$. If ${\underline{p}}(X\setminus U,x_0)=0$ then, by Lemma 2.15, $U\notin p(X,\Vert \;\Vert )$ at once. Therefore we may assume ${\underline{p}}(X\setminus U,x_0)>0$. Hence $x_0 \in {{\,\mathrm{cl}\,}}_{\mathcal {T}_{\Vert \;\Vert }} (V)$. Moreover, there exist a decreasing sequence of reals $(R_n)_{n\ge 1}$ tending to 0 such that

$$\begin{aligned} {\underline{p}}(V\cup (X\setminus U),x_0)=\lim _{n\rightarrow \infty }\frac{2\gamma (x_0,R_n,V\cup (X\setminus U))}{R_n}=0. \end{aligned}$$

Since ${\underline{p}}(V,x_0)=c$, for every $n\ge 1$ we can find an open ball $B(x_n,\eta _n)$ for which $B(x_n,\eta _n)\subset B(x_0,R_n)\setminus V$ and $\eta _n>\gamma (x_0,R_n,V)-\frac{R_n}{n}$. Since $x_0 \in {{\,\mathrm{cl}\,}}_{\mathcal {T}_{\Vert \;\Vert }} (V)$, we obtain $\eta _n\le \Vert x_0-x_n\Vert$.

Define $A=X\setminus \bigcup _{n=1}^\infty B(x_n,\frac{\eta _n}{2})$. Then

$$\begin{aligned} p(A,x_0)\ge \limsup _{n\rightarrow \infty }\frac{\eta _n}{R_n}\ge \limsup _{n\rightarrow \infty } \frac{\gamma (x_0,R_n,V)-\frac{R_n}{n}}{R_n}\ge \frac{{\underline{p}}(V,x_0)}{2}=c. \end{aligned}$$

Without loss of generality we may assume that $\frac{\eta _n}{R_n}>\frac{2c}{3}$ for every n. Thus $R_n<\frac{3\eta _n}{2c}$ and $\frac{\eta _n}{2}+\Vert x_n-x_0\Vert \ge \frac{3}{2}\eta _n>cR_n$. On the other hand, $V\subset A$ and

$$\begin{aligned} \lim _{n\rightarrow \infty }\frac{2\gamma (x_0,R_n,A\cup (X\setminus U))}{R_n}\le \lim _{n\rightarrow \infty }\frac{2\gamma (x_0,R_n,V\cup (X\setminus U))}{R_n}=0. \end{aligned}$$

Therefore

$$\begin{aligned}\textstyle { \limsup \limits _{n\rightarrow \infty }\frac{2\gamma (x_0,\frac{\eta _n}{2}+\Vert x_n-x_0\Vert ,A\cup (X\setminus U))}{\frac{\eta _n}{2}+\Vert x_n-x_0\Vert }}&\textstyle {\le \limsup \limits _{n\rightarrow \infty }\frac{2\gamma (x_0,R_n,A\cup (X\setminus U))}{\frac{\eta _n}{2}+\Vert x_n-x_0\Vert }\le } \\&\textstyle {\le \limsup \limits _{n\rightarrow \infty }\frac{2\gamma (x_0,R_n,A\cup (X\setminus U))}{cR_n}=0.} \end{aligned}$$

Fix $n\ge 1$ and take any $R\in [R_{n+1},R_n)$. Then

$$\begin{aligned} \textstyle { \frac{2\gamma (x_0,R,A\cup (X\setminus U))}{R}=\frac{2\gamma (x_0,\frac{\eta _n}{2}+\Vert x_n-x_0\Vert ,A\cup (X\setminus U))}{R}\le \frac{2\gamma (x_0,\frac{\eta _n}{2}+\Vert x_n-x_0\Vert ,A\cup (X\setminus U))}{\frac{\eta _n}{2}+\Vert x_n-x_0\Vert }} \end{aligned}$$

for $R\in [\frac{\eta _n}{2}+\Vert x_n-x_0\Vert , R_n]$,

$$\begin{aligned} \textstyle {\frac{2\gamma (x_0,R,A\cup (X\setminus U))}{R}\le \frac{2\gamma (x_0,\frac{\eta _n}{2}+\Vert x_n-x_0\Vert ,A\cup (X\setminus U))}{\Vert x_n-x_0\Vert -\frac{\eta _n}{2}}\le \frac{2\gamma (x_0,\frac{\eta _n}{2} +\Vert x_n-x_0\Vert ,A\cup (X\setminus U))}{\frac{1}{4}(\eta _n+\Vert x_n-x_0\Vert )}} \end{aligned}$$

for $R\in [\Vert x_n-x_0\Vert -\frac{\eta _n}{2},\Vert x_n-x_0\Vert +\frac{\eta _n}{2}]$ and

$$\begin{aligned} \textstyle {\frac{2\gamma (x_0,R,A\cup (X\setminus U))}{R}\le \frac{2\gamma (x_0,\frac{\eta _{n+1}}{2}+\Vert x_{n+1}-x_0\Vert ,A\cup (X\setminus U))}{\Vert x_{n+1}-x_0\Vert +\frac{\eta _{n+1}}{2}}} \end{aligned}$$

for $R\in [R_{n+1},\Vert x_n-x_0\Vert -\frac{\eta _n}{2}]$. Hence, $p(A\cup (X\setminus U),x_0)=0$ and $X\setminus U$ is not superporous at $x_0$. Therefore $U\notin p(X,\Vert \;\Vert )$. Thus $p(X,\Vert \;\Vert )\subset {\underline{p}}(X,\Vert \;\Vert )$. By Example 2.5, $p(X,\Vert \;\Vert )\ne {\underline{p}}(X,\Vert \;\Vert )$. $\square$

Lemma 2.17

Let $(X,\Vert \;\Vert )$ be a normed space and $x_0\in X$ and let $(B(x_n,\eta _n))_{n\ge 1}\subset X\setminus \{x_0\}$ be a sequence of balls such that $\lim _{n\rightarrow \infty } x_n=x_0$, $\eta _n>\eta _{n+1}$ and $\Vert x_n-x_0\Vert >\Vert x_{n+1}-x_0\Vert$ for every $n\ge 1$. Then

• (1) if $\lim _{n\rightarrow \infty }\frac{\Vert x_n-x_0\Vert -\eta _n}{\Vert x_{n+1}-x_0\Vert +\eta _{n+1}}=0$ then ${\underline{p}}\left( X\setminus \bigcup _{n=1}^\infty B(x_n,\eta _n),x_0\right) =1$;

• (2) if $\limsup _{n\rightarrow \infty }\frac{\Vert x_n-x_0\Vert -\eta _n}{\Vert x_{n+1}-x_0\Vert +\eta _{n+1}}<1$ then ${\underline{p}}\left( X\setminus \bigcup _{n=1}^\infty B(x_n,\eta _n),x_0\right) >0$.

Proof

Let $A=X\setminus \bigcup _{n=1}^\infty B(x_n,\eta _n)$, $\alpha _n=\Vert x_{n}-x_0\Vert -\eta _{n}$ and $\beta _n=\Vert x_{n}-x_0\Vert +\eta _{n}$ for $n\ge 1$. In both cases $\limsup _{n\rightarrow \infty }\frac{\alpha _n}{\beta _{n+1}}<1$, by assumption. Therefore $\beta _{n+1}>\alpha _n$ for almost every n and we may assume that this is true for every n. Fix $n\ge 1$ and take any $R\in [\beta _{n+1},\beta _{n}]$. Then $\gamma (x_0,\beta _{n+1} ,A)\ge \frac{\beta _{n+1}-\alpha _{n}}{2}$ and $\gamma (x_0,R, A)\ge \frac{\beta _{n+1}-\alpha _{n}}{2}+\frac{R-\beta _{n+1}}{2}$. Therefore,

$$\begin{aligned} \frac{2\gamma (x_0,R, A)}{R}\ge \frac{\beta _{n+1}-\alpha _{n}+R-\beta _{n+1}}{\beta _{n+1}+(R-\beta _{n+1})}\ge \frac{\beta _{n+1}-\alpha _{n}}{\beta _{n+1}}. \end{aligned}$$

(Observe that inequality $\frac{a+c}{b+c}\ge \frac{a}{b}$ holds for every $0<a\le b$ and $c>0$.) Hence

$$\begin{aligned} {\underline{p}}(A,x_0)=\liminf _{R\rightarrow 0^+}\frac{2\gamma (x_0,R,A)}{R}\ge \liminf _{n \rightarrow \infty }\frac{\beta _{n+1}-\alpha _n}{\beta _{n+1}}=1-\limsup _{n \rightarrow \infty }\frac{\alpha _n}{\beta _{n+1}}, \end{aligned}$$

which completed the proof. $\square$

Lemma 2.18

Let $(X,\Vert \;\Vert )$ be a normed space, $x_0\in X$, $\beta \in (0,1)$ and let $(B(x_n,\eta _n))_{n\ge 1}\subset X\setminus \{x_0\}$ be a sequence of balls such that $\lim _{n\rightarrow \infty }\frac{\eta _n}{\Vert x_n-x_0\Vert }=1$ and $\beta \Vert x_{n}-x_0\Vert >\Vert x_{n+1}-x_0\Vert -\eta _{n+1}$ for every $n\ge 1$. Then we can find a sequence $(B(y_n,\delta _n))_{n\ge 1}\subset X\setminus \{x_0\}$ of balls satisfying the following three conditions:

• (1) ${\underline{p}}\left( X\setminus \bigcup _{n=1}^\infty B(y_n,\delta _n),x_0\right) =1$,

• (2) $\Vert y_{n+1}-x_0\Vert +\delta _{n+1}=\beta \Vert x_n-x_0\Vert$ for every $n> 1$,

• (3) for every $n\ge 1$ points $x_0, x_n$ and $y_n$ are collinear and $\Vert y_n-x_0\Vert -\delta _n=\Vert x_n-x_0\Vert -\eta _n$.

Proof

Fix $\beta >0$. Let $\delta _1=\eta _1$ and $\delta _n=\frac{\beta \Vert x_{n-1}-x_0\Vert -\Vert x_n-x_0\Vert +\eta _n}{2}$ for $n>1$. By assumptions, $\delta _n>0$ for every n. Similarly, let $y_1=x_1$ and $y_n=x_n+(\delta _n-\eta _n)\frac{x_n-x_0}{\Vert x_n-x_0\Vert }$ for $n>1$, (see Fig. 3).

Fig. 3

Construction of $B(y_n,\delta _n)$ in $({\mathbb {R}}^2,\Vert \;\Vert _2)$

Clearly, $x_0, x_n$ and $y_n$ are collinear, $\Vert y_n-x_0\Vert -\delta _n=\Vert x_n-x_0\Vert -\eta _n$ and

$$\begin{aligned} \Vert y_{n+1}-x_0\Vert +\delta _{n+1}=\Vert x_{n+1}-x_0\Vert +2\delta _{n+1}-\eta _n =\beta \Vert x_{n}-x_0\Vert \end{aligned}$$

for every $n>1$. Therefore

$$\begin{aligned} \lim _{n\rightarrow \infty }\frac{\Vert y_n-x_0\Vert -\delta _n}{\Vert y_{n+1}-x_0\Vert +\delta _{n+1}}=\lim _{n\rightarrow \infty } \frac{\Vert x_n-x_0\Vert -\eta _n}{\beta \Vert y_n-x_0\Vert }=1. \end{aligned}$$

Hence ${\underline{p}}\left( X\setminus \bigcup _{n=1}^\infty B(y_n,\delta _n),x_0\right) =1$, by Lemma 2.17, which completed the proof. $\square$

Definition 2.19

Let $\alpha \in (0,\frac{1}{4})$ and x, y, z be collinear points in a normed space $(X,\Vert \;\Vert )$ such that y is between x and z. By $\Phi (x,y,z,\alpha )$ (see Fig. 4) we denote a set

$$\begin{aligned} \Phi&(x,y,z,\alpha )=\\&= \left\{ t\in X:\Vert t-x\Vert< \Vert x-y\Vert +\alpha \Vert x-y\Vert \text { and } \Vert t-z\Vert <\Vert z-y\Vert +\alpha \Vert x-y\Vert \right\} . \end{aligned}$$

Fig. 4

$\Phi (x,y,z,\alpha )$ in $({\mathbb {R}}^2,\Vert \;\Vert _2)$

Definition 2.20

We say that a normed space $(X,\Vert \;\Vert )$ satisfies condition (A) if for every $\varepsilon >1$ there exists $\alpha \in (0,1)$ such that for every collinear $x,y,z\in X$, where y is between x and z, we have

$$\begin{aligned} \Phi (x,y,z,\alpha )\subset B\left( y,\varepsilon \Vert x-y\Vert \right) . \end{aligned}$$

Remark 2.21

All considered by the authors normed spaces satisfy condition (A). Is there a normed space that does not satisfy condition (A)?

Theorem 2.22

If a normed space $(X,\Vert \;\Vert )$ satisfies condition (A) then ${\underline{s}}(X,{\Vert \;\Vert })\subset p(X,\Vert \;\Vert )$.

Proof

Let us take any $U\notin p(X,\Vert \;\Vert )$. Then $X\setminus U$ is not superporous at some $x_0\in U$. Hence there is $V\subset X$ satisfying $p(V,x_0)=c>0$ and $p(V\cup (X\setminus U),x_0)=0$. Since $p(V,x_0)=c$, there exist a sequence of pairwise disjoint open balls $(B(x_n,\eta _n))_{n\ge 1}$ such that $\lim _{n\rightarrow \infty }x_n=x_0$, $B(x_n,\eta _n)\cap V=\emptyset$ and $\lim _{n\rightarrow \infty }\frac{2\eta _n}{\Vert x_0-x_n\Vert +\eta _n}=p(V,x_0)=c$ and since $p(V\cup (X\setminus U),x_0)=0$, $\lim _{n\rightarrow \infty }\frac{2\gamma (x_0, \Vert x_0-x_n\Vert +\eta _n,V\cup (X\setminus U))}{\Vert x_0-x_n\Vert +\eta _n}=0$. Without loss of generality we may assume that $\eta _n>\frac{3}{8}c\Vert x_n-x_0\Vert$ and $\frac{2\gamma (x_0, \Vert x_0-x_n\Vert +\eta _n,V\cup (X\setminus U))}{\Vert x_0-x_n\Vert +\eta _n}<\frac{c}{8}$ for every $n\ge 1$.

Since $(X,\Vert \;\Vert )$ satisfies condition (A), we can find $\alpha _0\in (0,\frac{c}{8})$ such that for every collinear $x,y,z\in X$, where y lies between x and z, we have $\Phi (x,y,z,\alpha )\subset B\left( y,(1+\frac{c}{8})\Vert x-y\Vert \right)$ for every $\alpha <\alpha _0$.

Let $\eta _n^\prime =\Vert x_n-x_0\Vert (1-\frac{1}{n+1})$. If need be taking a subsequence, we may assume that

$$\begin{aligned} \Vert x_{n+1}-x_0\Vert +\eta _{n+1}^\prime <\tfrac{c}{4}\Vert x_{n}-x_0\Vert \qquad \text {for every } n. \end{aligned}$$

(2.1)

Obviously, $p(X\setminus \bigcup _{n=1}^\infty B(x_n,\eta _n^\prime ),x_0)=1$.

By Lemma 2.18, there exists a sequence $(B(y_n,\delta _n))_{n\ge 1}\subset X\setminus \{x_0\}$ of balls such that ${\underline{p}}\left( X\setminus \bigcup _{n=1}^\infty B(y_n,\delta _n),x_0\right) =1$, $x_0,x_n, y_n$ are collinear, $x_n$ is between $x_0$ and $y_n$, $\Vert y_{n}-x_0\Vert -\delta _{n}= \Vert x_{n}-x_0\Vert -\eta _{n}^\prime$ and

$$\begin{aligned} \Vert y_{n+1}-x_0\Vert +\delta _{n+1}=\tfrac{c}{4}\Vert x_n-x_0\Vert \quad \text { for every } n\ge 1. \end{aligned}$$

(2.2)

Let $z_n=x_0-(y_{n-1}-x_0)\frac{\Vert y_n-x_0\Vert }{\Vert y_{n-1}-x_0\Vert }$ for $n>1$. Then $x_0, z_n,y_{n-1}$ are collinear, $x_0$ lies between $z_n$ and $y_{n-1}$ and $\Vert x_0-z_n\Vert =\Vert x_0-y_n\Vert$. Let $A=X\setminus \bigcup _{n=1}^\infty (B(y_{2n-1},\delta _{2n-1})\cup B(z_{2n},\delta _{2n}))$. Obviously, ${\underline{p}}(A,x_0)=1$. Since $x_0, z_{2n},y_{2n-1}$ are collinear and $x_0$ is between $y_{2n-1}$ and $z_{2n}$, $B(y_{2n-1},\delta _{2n-1})\cap B(z_{2n},\delta _{2n})=\emptyset$. Moreover, by (2.1) and (2.2) if any ball is contained in $(X\setminus A)\cap B(x_0,\Vert x_{2n-1}-x_0\Vert +\eta ^\prime _{2n-1})$ then it is contained either in $B(y_{2n-1},\delta _{2n-1})$ or in $B(x_0,\frac{c}{4}\Vert x_{2n-1}-x_0\Vert )$.

We claim that ${\underline{p}}(A\cup (X\setminus U), x_0)<1$. Let $R_n=\Vert x_{n}-x_0\Vert +\eta _{n}^\prime$ for $n\ge 1$. Obviously, $\lim _{n\rightarrow \infty }\frac{R_n}{2\Vert x_{n}-x_0\Vert }=1$. Fix $n\ge 1$. Assume that there exists a ball $B(z,\varrho )$ contained in $B(x_0,R_{2n-1})$, disjoint from $A\cup (X\setminus U)$ and such that $\varrho > \Vert x_{2n-1}-x_0\Vert -\frac{\alpha _0}{2}\Vert x_{2n-1}-x_0\Vert$. Then $B(z, \varrho )\not \subset B(x_0,\frac{c}{4}\Vert x_{2n-1}-x_0\Vert )$ and therefore, $B(z, \varrho )\subset B(y_{2n-1},\delta _{2n-1})\subset B(y_{2n-1},\Vert y_{2n-1}-x_0\Vert )$. Since $\varrho > (1-\frac{\alpha _0}{2})\Vert x_{2n-1}-x_0\Vert$, $\Vert z-y_{2n-1}\Vert <\Vert y_{2n-1}-x_{2n-1}\Vert +\frac{\alpha _0}{2}\Vert x_{2n-1}-x_0\Vert$.

On the other hand, since $B(z,\varrho )\subset B(x_0,R_{2n-1})$ and $\varrho > (1-\frac{\alpha _0}{2})\Vert x_{2n-1}-x_0\Vert$, we obtain $\Vert z-x_{0}\Vert <(1+\frac{\alpha _0}{2})\Vert x_{2n-1}-x_0\Vert$. Therefore, $z\in \Phi (x_0,x_{2n-1},y_{2n-1},\alpha _0)\subset B(x_{2n-1},(1+\frac{c}{8})\Vert x_{2n-1}-x_0\Vert )$. It follows that $B(z,\varrho )\cap B(x_{2n-1},\eta _{2n-1})$ contains a ball disjoint from $A\cup (X\setminus U)$ with radius at least $\frac{c}{8}\Vert x_{2n-1}-x_0\Vert$, a contradictions. Therefore any ball contained in $B(x_0,R_{2n-1})$ and disjoint from $A\cup (X\setminus U)$ has a radius less than $(1- \frac{\alpha _0}{2})\Vert x_{2n-1}-x_0\Vert$. Thus

$$\begin{aligned} {\underline{p}}(A\cup (X\setminus U), x_0)\le \liminf _{n\rightarrow \infty }\frac{2(1-\frac{\alpha _0}{2})\Vert x_{2n-1}-x_0\Vert }{R_n}=1-\tfrac{\alpha _0}{2}<1. \end{aligned}$$

Hence, $X\setminus U$ is not lower strongly superporous at $x_0$ and $U\notin {\underline{s}}(X,\Vert \;\Vert )$. The proof is completed. $\square$

Definition 2.23

We say that a normed space $(X,\Vert \;\Vert )$ satisfies condition (B) if for every $\varepsilon >0$ there exists $\alpha \in (0,1)$ such that for every collinear $x,y,z\in X$, where y lies between x and z, we have

$$\begin{aligned} \Phi (x,y,z,\alpha )\subset B\left( y,\varepsilon \Vert x-y\Vert \right) . \end{aligned}$$

Remark 2.24

Obviously, if a normed space $(X,\Vert \;\Vert )$ satisfies condition (B) then $(X,\Vert \;\Vert )$ satisfies condition (A).

Theorem 2.25

If a normed space $(X,\Vert \;\Vert )$ satisfies condition (B) then ${\underline{s}}(X,\Vert \;\Vert )\subset s(X,\Vert \;\Vert )$.

Proof

Let us take any $U\notin s(X,\Vert \;\Vert )$. Then $X\setminus U$ is not strongly superporous at some $x_0\in U$. Hence there is $V\subset X$ satisfying $p(V,x_0)=1$ and $p(V\cup (X\setminus U),x_0)=1-c<1$. Since $p(V,x_0)=1$, there exists a sequence of pairwise disjoint open balls $(B(x_n,\eta _n))_{n\ge 1}$ such that $\lim _{n\rightarrow \infty }x_n=x_0$, $B(x_n,\eta _n)\cap V=\emptyset$ for $n\ge 1$ and $\lim _{n\rightarrow \infty }\frac{2\eta _n}{\Vert x_0-x_n\Vert +\eta _n}=p(V,x_0)=1$. Moreover, since $p(V\cup (X\setminus U),x_0)=1-c<1$, $\lim _{n\rightarrow \infty }\frac{2\gamma (x_0, \Vert x_0-x_n\Vert +\eta _n,V\cup (X\setminus U))}{\Vert x_0-x_n\Vert +\eta _n}\le 1-c<1$. Without loss of generality we may assume that $\eta _n>(1-\frac{c}{8})\Vert x_n-x_0\Vert$ and $\frac{2\gamma (x_0, \Vert x_0-x_n\Vert +\eta _n,V\cup (X\setminus U))}{\Vert x_0-x_n\Vert +\eta _n}<1-\frac{7}{8}c$ for every $n\ge 1$.

Since $(X,\Vert \;\Vert )$ satisfies condition (B), we can find $\alpha _0\in (0,\frac{c}{8})$ such that for every collinear $x,y,z\in X$, where y lies between x and z, we have $\Phi (x,y,z,\alpha )\subset B\left( y,\frac{c}{8}\Vert x-y\Vert \right)$ for every $\alpha <\alpha _0$.

If need be taking a subsequence, we may assume that

$$\begin{aligned} \Vert x_{n+1}-x_0\Vert +\eta _{n+1}<\tfrac{c}{4}\Vert x_{n}-x_0\Vert \qquad \text {for every } n. \end{aligned}$$

(2.3)

Obviously, $p(X\setminus \bigcup _{n=1}^\infty B(x_n,\eta _n),x_0)=1$.

By Lemma 2.18, there exists a sequence $(B(y_n,\delta _n))_{n\ge 1}\subset X\setminus \{x_0\}$ of balls such that ${\underline{p}}\left( X\setminus \bigcup _{n=1}^\infty B(y_n,\delta _n),x_0\right) =1$, $x_0,x_n, y_n$ are collinear, $x_n$ lies between $x_0$ and $y_n$, $\Vert y_{n}-x_0\Vert -\delta _{n}= \Vert x_{n}-x_0\Vert -\eta _{n}$ and

$$\begin{aligned} \Vert y_{n+1}-x_0\Vert +\delta _{n+1}=\tfrac{c}{4}\Vert x_n-x_0\Vert \quad \text { for every } n\ge 1. \end{aligned}$$

(2.4)

Let $z_n=x_0-(y_{n-1}-x_0)\frac{\Vert y_n-x_0\Vert }{\Vert y_{n-1}-x_0\Vert }$ for $n>1$. Then $x_0, z_n,y_{n-1}$ are collinear, $x_0$ lies between $z_n$ and $y_{n-1}$ and $\Vert x_0-z_n\Vert =\Vert x_0-y_n\Vert$. Let $A=X\setminus \bigcup _{n=1}^\infty (B(y_{2n-1},\delta _{2n-1})\cup B(z_{2n},\delta _{2n}))$. Obviously, ${\underline{p}}(A,x_0)=1$. Since $x_0, z_{2n},y_{2n-1}$ are collinear, $B(y_{2n-1},\delta _{2n-1})\cap B(z_{2n},\delta _{2n})=\emptyset$. Moreover, by (2.3) and (2.4) if any ball is contained in $(X\setminus A)\cap B(x_0,\Vert x_{2n-1}-x_0\Vert +\eta _{2n-1})$ then it is contained either in $B(y_{2n-1},\delta _{2n-1})$ or in $B(x_0,\frac{c}{4}\Vert x_{2n-1}-x_0\Vert )$.

We claim that ${\underline{p}}(A\cup (X\setminus U), x_0)<1$. Let $R_n=\Vert x_{n}-x_0\Vert +\eta _{n}$ for $n\ge 1$. Obviously, $\lim _{n\rightarrow \infty }\frac{R_n}{2\Vert x_{n}-x_0\Vert }=1$. Fix $n\ge 1$. Assume that there exists a ball $B(z,\varrho )$ contained in $B(x_0,R_{2n-1})$, disjoint from $A\cup (X\setminus U)$ and such that $\varrho > \Vert x_{2n-1}-x_0\Vert -\frac{\alpha _0}{2}\Vert x_{2n-1}-x_0\Vert$. Then $B(z, \varrho )\not \subset B(x_0,\frac{c}{4}\Vert x_{2n-1}-x_0\Vert )$ and therefore, $B(z, \varrho )\subset B(y_{2n-1},\delta _{2n-1})\subset B(y_{2n-1},\Vert y_{2n-1}-x_0\Vert )$. Since $\varrho > (1-\frac{\alpha _0}{2})\Vert x_{2n-1}-x_0\Vert$, we obtain $\Vert z-y_{2n-1}\Vert <\Vert y_{2n-1}-x_{2n-1}\Vert +\frac{\alpha _0}{2}\Vert x_{2n-1}-x_0\Vert$.

On the other hand, since $B(z,\varrho )\subset B(x_0,R_{2n-1})$ and $\varrho > (1-\frac{\alpha _0}{2})\Vert x_{2n-1}-x_0\Vert$, we obtain $\Vert z-x_{0}\Vert <(1+\frac{\alpha _0}{2})\Vert x_{2n-1}-x_0\Vert$. Therefore, $z\in \Phi (x_0,x_{2n-1},y_{2n-1},\alpha _0)\subset B(x_{2n-1},\frac{c}{8}\Vert x_{2n-1}-x_0\Vert )$. It follows that $B(z,\varrho )\cap B(x_{2n-1},\eta _{2n-1})$ contains a ball disjoint from $A\cup (X\setminus U)$ with radius at least $(1-\frac{c}{4})\Vert x_{2n-1}-x_0\Vert$, a contradictions. Therefore any ball contained in $B(x_0,R_{2n-1})$ and disjoint from $A\cup (X\setminus U)$ has a radius less than $(1- \frac{\alpha _0}{2})\Vert x_{2n-1}-x_0\Vert$. Thus

$$\begin{aligned} {\underline{p}}(A\cup (X\setminus U), x_0)\le \liminf _{n\rightarrow \infty }\frac{2(1-\frac{\alpha _0}{2})\Vert x_{2n-1}-x_0\Vert }{R_n}=1-\tfrac{\alpha _0}{2}<1. \end{aligned}$$

Hence, $X\setminus U$ is not lower strongly superporous at $x_0$ and $U\notin {\underline{s}}(X,\Vert \;\Vert )$. The proof is completed. $\square$

Example 2.26

Let $\mathcal {T}_{\max }$ be a topology in $\mathbb {R}^n$ generated by the norm $\Vert \;\Vert _{\max }$. Let $x=0_n$, $y=(1,0,\ldots , 0)$ and $z=(10,0,\ldots ,0)$. Then

$$\begin{aligned} \Phi (x,y,z,\alpha )&= [1-\alpha ,1+\alpha ]\times [-1-\alpha ,1+\alpha ]^{n-1}\\&\supset [1-\alpha ,1+\alpha ]\times [-1,1]^{n-1} \end{aligned}$$

and $B(y,\frac{1}{2}\Vert x-y\Vert _{\max })=B((1,0,\ldots ,0), \frac{1}{2})=[\frac{1}{2},\frac{3}{2}]\times [-\frac{1}{2},\frac{1}{2}]^{n-1}\not \supset \Phi (x,y,z,\alpha )$ for any $\alpha >0$, (see Fig. 5).

Fig. 5

$\Phi ((0,0),(1,0),(10,0),\alpha )$ in $({\mathbb {R}}^2,\Vert \;\Vert _{\max })$

Therefore $({\mathbb {R}}^n, \Vert \;\Vert _{\max })$ does not satisfies condition (B) (and satisfies condition (A)).

Remark 2.27

Repeating arguments from the proof of Example 2.26 one can prove that $(l_\infty ,\Vert \;\Vert _{\sup })$ does not satisfies condition (B) and satisfies condition (A).

Theorem 2.28

Let $(X,\Vert \;\Vert )$ be either $({\mathbb {R}}^n,\Vert \;\Vert _{\max })$ or $(l_\infty ,\Vert \;\Vert _{\sup })$. Then ${\underline{s}}(X,\Vert \;\Vert )\subset s(X,\Vert \;\Vert )$, although neither $({\mathbb {R}}^n,\Vert \;\Vert _{\max })$ nor $(l_\infty ,\Vert \;\Vert _{\sup })$ satisfies condition (B).

Proof

Take any $U\notin s(X,\Vert \;\Vert )$. Then $X\setminus U$ is not strongly superporous at some $x_0\in U$. We may assume $x_0=0_X$. Hence we can find $A\subset X$ such that $p(A,0_X)=1$ and $p(A\cup (X\setminus U),0_X)<1$. Since $p(A,0_X)=1$, there exists a sequence of pairwise disjoint balls $(B(x_k,\eta _k))_{k\ge 1}$ such that $(\eta _k)_{k\ge 1}$ and $(\Vert x_k-x_0\Vert )_{k\ge 1}$ are decreasing and tend to 0, $A\cap \bigcup _{k=1}^\infty B(x_k,\eta _k)=\emptyset$ and $\lim _{k\rightarrow \infty }\frac{2\eta _k}{\Vert x_k-x_0\Vert +\eta _k}=1$. Let $R_k=\Vert x_k-x_0\Vert +\eta _k$ for $k\ge 1$. By Lemma 1.1, there are sequences $(y_k)_{k\ge 1}\subset X$ and $(\zeta _k)_{k\ge 1}\subset \{-1,1\}^k$ or $(\zeta _k)_{k\ge 1}\subset \{-1,1\}^{\omega }$ such that

$$\begin{aligned} B(y_k,\eta _k-\tfrac{R_k}{4})\subset B(x_k,\eta _k)\cap B(0_X,\tfrac{R_k}{2})\cap H^{\zeta _k} \end{aligned}$$

for every k. Let $\delta _k=\eta _k-\frac{R_k}{4}$. Then

$$\begin{aligned} \lim _{k\rightarrow \infty }\frac{2\delta _k}{\delta _k+\Vert y_k\Vert }\ge \lim _{k\rightarrow \infty }\frac{2\eta _k-\frac{R_k}{2}}{\frac{R_k}{2}}=\lim _{k\rightarrow \infty }(\tfrac{4\eta _k}{R_k}-1)=1 \end{aligned}$$

(2.5)

and $p(X\setminus \bigcup _{k=1}^\infty B(y_k,\delta _k),0_X)=1$. Since

$$\begin{aligned} p\Big ((X\setminus U)\cup \Big (X\setminus \bigcup _{k=1}^\infty B(y_k,\delta _k)\Big ),0_X\Big )\le p((X\setminus U)\cup A,0_X)<1, \end{aligned}$$

(2.6)

$\limsup _{k\rightarrow \infty } \frac{\gamma (0_X,\Vert y_k\Vert +\delta _k, (X\setminus U)\cup (X\setminus \bigcup _{k=1}^\infty B(y_k,\delta _k))}{\Vert y_k\Vert +\delta _k}<1$.

For any ball $B(x,\nu )$ let $\varphi (B(x,\nu ))=(x_1-\nu {{\,\mathrm{sgn}\,}}(x_1), x_2-\nu {{\,\mathrm{sgn}\,}}(x_2), \ldots )$. Observe that if $B(x,\nu )\subset H^{\zeta }$, then $\varphi (B(x,\nu ))\in H^{\zeta }$ too and a point $\varphi (B(x,\nu ))$ minimizes the distance between $0_X$ and ${\overline{B}}(x,\nu )$. Moreover, for any $B(x,\nu )\subset H^{\zeta }$ and $\varrho >0$ there exists a unique $y\in X$ such that $\varphi (B(x,\nu ))=\varphi (B(y,\varrho ))$ and $B(y,\varrho )\subset H^{\zeta }$ too (see Fig. 6).

Fig. 6

$\varphi (B(x,\nu ))$ and $B(y,\varrho )$ in $({\mathbb {R}}^2,\Vert \;\Vert _{\max })$

For $k>1$ let $\sigma _k=\frac{1}{2}\left( \Vert y_{k-1}\Vert -(\Vert y_k\Vert -\delta _k)\right)$ and $z_k$ be such that $\varphi (B(y_k,\delta _k))=\varphi (B(z_k,\sigma _k))$. Then $B(z_k,\sigma _k)\subset H^{\zeta _k}$ and $\Vert z_k\Vert +\sigma _k=\frac{1}{2}\Vert y_{k-1}\Vert$. Let $B=X\setminus \bigcup _{k=1}^\infty B(z_k,\sigma _k)$ and $r_k=\Vert y_k\Vert +\delta _k$ for $k\ge 1$. By Lemma 2.17, ${\underline{p}}(B,0_X)=1$.

On the other hand,

$$\begin{aligned} \gamma (0_X,r_k,(X\setminus U)\cup B )\le \min \big \{\tfrac{1}{4}\Vert y_k\Vert , \gamma (0_X,r_k,H^{\zeta _n}\cap ((X\setminus U)\cup B) )\big \} \end{aligned}$$

and

$$\begin{aligned} \gamma \big (0_X,r_k, H^{\zeta _k}\cap&((X\setminus U)\cup B)\big ) \\ \le&\gamma (0_X,r_k,(X\setminus U)\cup A) +{{\,\mathrm{diam}\,}}((H^{\zeta _k}\cap B(0_X,r_k))\setminus B(y_k,\delta _k)). \end{aligned}$$

By (2.5) and (2.6), we obtain

$$\begin{aligned} \limsup _{k\rightarrow \infty }\frac{\gamma (0_X,r_k,H^{\zeta _n}\cap ((X\setminus U)\cup B))}{r_k}<1. \end{aligned}$$

Therefore ${\underline{p}}((X\setminus U)\cup B,0_X)<1$, $X\setminus U$ is not lower strongly superporous at $0_X$ and $U\notin {\underline{s}}(X,\Vert \;\Vert )$. The proof is completed. $\square$

Theorem 2.29

$s(l_\infty ,{\Vert \;\Vert _{\sup }})\not \subset {\underline{p}}(l_\infty ,{\Vert \;\Vert _{\sup }})$.

Proof

Define $b_n=\left( \frac{3}{4}\right) ^{n-1}$, $a_n=\frac{b_n}{2}=\frac{1}{2}\left( \frac{3}{4}\right) ^{n-1}$, $c_n=\frac{a_n+b_n}{2}=\left( \frac{3}{4}\right) ^{n}$ for $n\ge 1$. Let ${0}_\infty \in l_\infty$, ${0}_\infty =(0,0,\ldots )$ and ${x}_n\in l_\infty$, ${x}_n=(-\frac{a_n}{2},-\frac{a_n}{2},\ldots ,-\frac{a_n}{2}, \underbrace{c_n}_{n-th term},-\frac{a_n}{2},\ldots )$ for $n\ge 1$. Finally, let $U=l_\infty \setminus \bigcup _{n=1}^\infty {\overline{B}}({x}_n,\frac{a_n}{2})$. Observe that ${\overline{B}}({x}_n,\tfrac{a_n}{2})\subset H^{\zeta _n}$, where $\zeta _n= (-1,-1,\ldots ,-1, \underbrace{1}_{n-th term},-1,\ldots )$.

Since $\Vert {x}_n\Vert _{\sup }-\frac{a_n}{2}=c_n-\frac{a_n}{2}=\frac{1}{2}\left( \frac{3}{4}\right) ^{n-1}$ and $\Vert {x}_{n+1}\Vert _{\sup }+\frac{a_{n+1}}{2}=c_{n+1}+\frac{a_{n+1}}{2}=b_{n+1}=\left( \frac{3}{4}\right) ^{n}$, we obtain

$$\begin{aligned} \limsup _{n\rightarrow \infty }\frac{\Vert {x}_n\Vert _{\sup }-\frac{a_n}{2}}{\Vert {x}_{n+1}\Vert _{\sup } +\frac{a_{n+1}}{2}}=\limsup _{n\rightarrow \infty }\frac{\frac{1}{2}\left( \frac{3}{4}\right) ^{n-1}}{\left( \frac{3}{4}\right) ^{n}}=\frac{2}{3}<1. \end{aligned}$$

Therefore, by Lemma 2.17, ${\underline{p}}(U,{0}_\infty )>0$. Moreover, ${\underline{p}}(U\cup (l_\infty \setminus U), {0}_\infty )={\underline{p}}(l_\infty ,{0}_\infty )=0$ and U is not lower superporous at ${0}_\infty \in E$. Therefore, $U\notin {\underline{p}}(l_\infty , {\Vert \;\Vert _{\sup }})$.

We claim that $U\in s(l_\infty ,{\Vert \;\Vert _{\sup }})$. Clearly, $l_\infty \setminus U$ is strongly superporous at every ${x}\in U\setminus \{{0}_\infty \}$. Take any $A\subset l_\infty$ such that $p(A,{0}_\infty )=1$. There is a sequence of pairwise disjoint balls $(B({y}_k,\eta _k))_{n\ge 1}\subset l_\infty \setminus (A\cup \{{0}_\infty \})$ such that $\lim _{n\rightarrow \infty }{y}_k={0}_\infty$ and $\lim _{n\rightarrow \infty }\frac{2\eta _k}{\Vert {y}_k\Vert _{\sup }+\eta _k}=1$. By Lemma 1.1, there exist sequences $(B(z_k,\mu _k))_{k\ge 1}\subset l_\infty$ and $(\varsigma _k)_{k\ge 1}\subset \{-1,1\}^{\omega }$ such that $B({z}_k,\mu _k)\subset B({y}_k,\eta _k)\cap H^{\varsigma _k}\cap B({0}_\infty ,\frac{1}{2}(\Vert y_k\Vert _{\sup }+\eta _k))$ and $\mu _k>\eta _k-\frac{1}{4}(\Vert y_k\Vert _{\sup }+\eta _k)$. Then

$$\begin{aligned} \lim _{k\rightarrow \infty }\frac{2\mu _k }{\Vert {z}_k\Vert _{\sup }+\mu _k}\ge \lim _{k\rightarrow \infty } \frac{2\eta _k-\frac{1}{2}(\Vert {y}_k\Vert _{\sup }+\eta _k)}{\frac{1}{2}(\Vert {y}_k\Vert _{\sup }+\eta _k)}=1 \end{aligned}$$

and $A\cap \bigcup _{k=1}^\infty B({z}_k,\mu _k)=\emptyset$.

We construct a new sequence of open balls $(B({v}_k,\nu _k))_{k\ge 1}$ such that $B({v}_k,\nu _k)\subset B({z}_k,\mu _k)\cap U$ and $p(l_\infty \setminus \bigcup _{k=1}^\infty B({v}_k,\nu _k),{0}_\infty )=1$. Fix $k\ge 1$. If $\varsigma _k\ne \zeta _n$ for $n\ge 1$, i.e. $B({z}_k,\mu _k)\subset U$ then we take $B({v}_k,\nu _k)=B({z}_k,\mu _k)$. Let us consider the case, where $\varsigma _k= \zeta _{n_k}$ for some $n_k\ge 1$ (obviously, $\varsigma _k\ne \zeta _{n}$ for $n\ne n_k$). Let $B({z}_{k},\mu _k)=(d_1^k,d_1^k+2\mu _k)\times (d_{2}^k,d_{2}^k+2\mu _k)\times \cdots$, where $d_{n_k}^k\ge 0$ and $d_{n}^k+2\mu _k\le 0$ for $n\ne n_k$. Since

$$\begin{aligned} \lim _{k\rightarrow \infty }\frac{2\mu _k }{\Vert {z}_k\Vert _{\sup }+\mu _k}\ge \lim _{k\rightarrow \infty } \frac{2\eta _k-\frac{1}{2}(\Vert {y}_k\Vert _{\sup }+\eta _k)}{\frac{1}{2}(\Vert {y}_k\Vert _{\sup }+\eta _k)}=1, \end{aligned}$$

we have $\lim _{n\rightarrow \infty }\frac{\sup \{|d_n^k|:n\ge 1\} }{\mu _k}=0$. By Lemma 1.5, we can find $(d,e)\subset (d_{n_k}^k,d_{n_k}^k+2\mu _k)$ such that $(d,e)\cap [a_{n_k},b_{n_k}]=\emptyset$ and $\frac{e-d}{e}>1-\sqrt{\frac{3|d_{n_k}^k|}{|d_{n_k}^k|+2\mu _k}}$. Then

$$\begin{aligned} (d_1^k,d_1^k+2\mu _k)&\times \cdots \times (d_{n_k-1}^k,d_{n_k-1}^k+2\mu _k)\times (d,e) \\&\times (d_{n_k+1}^k,d_{n_k+1}^k+2\mu _k)\times \cdots \subset U \end{aligned}$$

and we can find $B({v}_k,\frac{e-d}{2})\subset U$, i.e., we have $\nu _k=\frac{e-d}{2}$. Since

$$\begin{aligned} \lim _{k\rightarrow \infty }\left( 1-\sqrt{\frac{\sup \{3|d_{n}^k|:n\ge 1\}}{\inf \{|d_{n}^k|+2\mu _k:n\ge 1\}}}\right) =1, \end{aligned}$$

we finally obtain $p(l_\infty \setminus \bigcup _{k=1}^\infty B({v}_k,\nu _k),0_\infty )=1$. Therefore $l_\infty \setminus U$ is strongly superporous at ${0}_\infty$. Hence, $U\in s(l_\infty ,\Vert \;\Vert _{\sup })$, which completed the proof. $\square$

Question 2.30

Does there exist a normed space $(X,\Vert \;\Vert )$ such that $s(X,{\Vert \;\Vert })\subset {\underline{p}}(X,{\Vert \;\Vert })$?

We may present relationships between considered topologies in the following diagram.

$$\begin{aligned} \begin{array}{ccccccc} \mathcal {T}_{\Vert \;\Vert } &{} \subseteqq &{} {\underline{s}}(X,\Vert \;\Vert ) &{} \overset{(A)}{\varsubsetneqq } &{} p(X,\Vert \;\Vert ) &{} \varsubsetneqq &{} {\underline{p}}(X,\Vert \;\Vert ) \\ &{} &{} \overset{(B)}{\varsubsetneqq } &{} &{} &{} &{} \\ &{} &{} s(X,\Vert \;\Vert ) &{} &{} &{} &{} \end{array} \end{aligned}$$

Inclusion $\mathcal {T}_{\Vert \;\Vert } \subset {\underline{s}}(X,\Vert \;\Vert )$ is just equality in some normed spaces, inclusion ${\underline{s}}(X,\Vert \;\Vert )\varsubsetneqq p(X,\Vert \;\Vert )$ holds under condition (A) and inclusion ${\underline{s}}(X,\Vert \;\Vert )\varsubsetneqq s(X,\Vert \;\Vert )$ holds under condition (B). No other inclusion, in general, holds, although we know only one example of a normed space in which $s(X,\Vert \;\Vert )\not \subset {\underline{p}}(X,\Vert \;\Vert )$.

3 Lower porouscontinuity

In [1] J. Borsík and J. Holos defined families of porouscontinuous functions $f:{\mathbb {R}}\rightarrow {\mathbb {R}}$. Some properties of porouscontinuity can be found in [1, 2, 10]. Applying their ideas and replacing standard porosity in ${\mathbb {R}}$ by the lower porosity in X we transfer this concept for real functions defined on $(X,\Vert \;\Vert )$.

Definition 3.1

Let $(X,\Vert \;\Vert )$ be a normed space, $f:X\rightarrow R$ and $x\in X$.

Let $r\in [0,1)$.The function f will be called:

• $\underline{{\mathcal {P}}_r}$-continuous at x if there exists a set $U\subset X$ such that $x\in U$, ${\underline{p}}(X\setminus U,x)>r$ and $f_{\upharpoonright U}$ is continuous at x;

• $\underline{{\mathcal {S}}_r}$-continuous at x if for each $\varepsilon >0$ there exists a set $U\subset X$ such that $x\in U$, ${\underline{p}}(X\setminus U,x)>r$ and $f(U)\subset (f(x)-\varepsilon ,f(x)+\varepsilon )$;

Let $r\in (0,1]$.The function f will be called:

• $\underline{{\mathcal {M}}_r}$-continuous at x if there exists a set $U\subset X$ such that $x\in U$, ${\underline{p}}(X\setminus U,x)\ge r$ and $f_{\upharpoonright U}$ is continuous at x;

• $\underline{{\mathcal {N}}_r}$-continuous at x if for each $\varepsilon >0$ there exists a set $U\subset X$ such that $x\in U$, ${\underline{p}}(X\setminus U,x)\ge r$ and $f(U)\subset (f(x)-\varepsilon ,f(x)+\varepsilon )$;

By $\underline{{\mathcal {P}}_r}(f)$, $\underline{{\mathcal {S}}_r}(f)$, $\underline{{\mathcal {M}}_r}(f)$ and $\underline{{\mathcal {N}}_r}(f)$ we denote the sets of points at which f is $\underline{\mathcal P_r}$-continuous, $\underline{{\mathcal {S}}_r}$-continuous, $\underline{{\mathcal {M}}_r}$-continuous and $\underline{\mathcal N_r}$-continuous, respectively.

Proposition 3.2

Let $f:X\rightarrow {\mathbb {R}}$ and $x\in X$. Then

• (1) $x\in \underline{\mathcal {S}_{r}}(f)$ if and only if ${\underline{p}}(X\setminus \{t:|f(t)-f(x)|<\varepsilon \},x)>r$ for every $\varepsilon >0$;

• (2) $x\in \underline{\mathcal {N}_{r}}(f)$ if and only if ${\underline{p}}(X\setminus \{t:|f(t)-f(x)|<\varepsilon \},x)\ge r$ for every $\varepsilon >0$.

for corresponding r.

Similarly as in [1], we can easily check that f is $\underline{{\mathcal {M}}_r}$-continuous at x if and only if it is $\underline{{\mathcal {N}}_r}$-continuous at x.

If f is $\underline{{\mathcal {P}}_r}$-continuous, $\underline{{\mathcal {S}}_r}$-continuous, $\underline{\mathcal M_r}$-continuous at every point of X for some corresponding r then we say that f is $\underline{{\mathcal {P}}_r}$-continuous, $\underline{{\mathcal {S}}_r}$-continuous, $\underline{\mathcal M_r}$-continuous, respectively. All of these functions are called lower porouscontinuous functions.

Obviously, if f is continuous at some x then f is lower porouscontinuous (in each sense) at x. We introduce for corresponding r the following notations:

• $\underline{{\mathcal {M}}_r}=\underline{{\mathcal {N}}_r}=\{f:\underline{{\mathcal {M}}_r}(f)=X\}$;

• $\underline{{\mathcal {P}}_r}=\{f:\underline{{\mathcal {P}}_r}(f)=X\}$;

• $\underline{{\mathcal {S}}_r}=\{f:\underline{{\mathcal {S}}_r}(f)=X\}$.

In the paper we focus on $\underline{{\mathcal {M}}_1}$ and $\underline{{\mathcal {S}}_0}$. In [11] some properties of lower porouscontinuous functions $\underline{{\mathcal {P}}_r}$, $\underline{{\mathcal {S}}_r}$, $\underline{{\mathcal {M}}_r}$ for $r\in (0,1)$ defined on ${\mathbb {R}}^2$ are presented.

Lemma 3.3

Let $(X,\Vert \;\Vert )$ be a normed space. Then

$$\begin{aligned} {{\,\mathrm{int}\,}}_{{\underline{p}}(X,\Vert \;\Vert )}E=\{x\in X:X\setminus E \text { is lower superporous at } x\} \end{aligned}$$

and

$$\begin{aligned} {{\,\mathrm{int}\,}}_{{\underline{s}}(X,\Vert \;\Vert )}E=\{x\in X:X\setminus E \text { is lower strongly superporous at } x\} \end{aligned}$$

for every $E\subset X$.

Proof

Denote $V={{\,\mathrm{int}\,}}_{{\underline{p}}(X,\Vert \;\Vert )}E$. Take $x_0\in V$. Then $V\in {\underline{p}}(X,\Vert \;\Vert )$ and $X\setminus V$ is lower superporous at $x_0$. Since $X\setminus E\subset X\setminus V$, the set $X\setminus E$ is lower superporous at $x_0$.

Now, let $x_0\in \{x\in X:X\setminus E \text { is lower superporous at } x\}$. Denote $V={{\,\mathrm{int}\,}}_{\mathcal {T}_{\Vert \;\Vert }}E\cup \{x_0\}$. Obviously, $X\setminus V$ is lower superporous at each point $x\in V\setminus \{x_0\}$. Moreover, for each set $A\subset X$ we obtain ${\underline{p}}((X\setminus V)\cup A, x_0)={\underline{p}}((X\setminus E)\cup A, x_0)$, because every open ball disjoint with $X\setminus E$ is contained in V. Since $X\setminus E$ is lower superporous at $x_0$, the set $X\setminus V$ is lower superporous at $x_0$, too. Therefore $V\in {\underline{p}}(X,\Vert \;\Vert )$. Finally $x_0\in {{\,\mathrm{int}\,}}_{{\underline{p}}(X,\Vert \;\Vert )}E$, because $V\subset E$.

The proof of the second statement is very similar and we omit it. $\square$

Lemma 3.4

Let A be a closed subset of a normed space $(X,\Vert \;\Vert )$ and $x_0\in A$. Then there exists $E\subset X\setminus A$ such that

• ${{\,\mathrm{cl}\,}}_{\mathcal {T}_{\Vert \;\Vert }} E\subset E\cup \{x_0\}$;

• E is discrete;

• for each $B\subset X$, if $E\subset B$ then ${\underline{p}}(B,x_0)={\underline{p}}(B\cup (X\setminus A),x_0)$.

Proof

Let $U_n={\overline{B}}(x_0,\frac{1}{n})\setminus B(x_0,\frac{1}{n+1})$ for $n\ge 1$. By the Zorn Lemma, for every n we can choose a discrete set $E_n\subset U_n\setminus A$ such that $U_n\setminus A\subset \bigcup _{x\in E_n} B\left( x,\frac{1}{(n+1)^2}\right)$ and $\Vert x_1-x_2\Vert \ge \frac{1}{(n+1)^2}$ for $x_1,x_2\in E_n$, $x_1\ne x_2$. Let

$$\begin{aligned} E=\bigcup _{n=1}^\infty E_n. \end{aligned}$$

Then E is discrete, $E\cap A=\emptyset$ and ${{\,\mathrm{cl}\,}}_{\mathcal {T}_{\Vert \;\Vert }} E\subset E\cup \{x_0\}$. Take any $B\subset X$ such that $E\subset B$. The inequality ${\underline{p}}(B,x_0)\ge {\underline{p}}(B\cup (X\setminus A),x_0)$ is obvious. If ${\underline{p}}(B,x_0)=0$ then certainly ${\underline{p}}(B\cup (X\setminus A),x_0)=0$. Let ${\underline{p}}(B,x_0)=\alpha >0$. Choose $\beta$, $\beta _1$ such that $0<\beta<\beta _1<\alpha$. We can find $n_0>1$ such that $\frac{1}{n_0}<\min \left\{ \frac{\beta _1-\beta }{4}, \frac{\varepsilon }{8}, \frac{\beta _1}{8}\right\}$. Since ${\underline{p}}(B,x_0)=\alpha >\beta _1$, we can find $R_0\in (0,\frac{1}{4n_0})$ such that $\frac{2\gamma (x_0,R,B)}{R}>\beta _1$ for $R\in (0,R_0)$. Choose any $R\in (0,R_0)$. There exists $B(y,\eta )$ such that $\frac{2\eta }{\eta +\Vert x_0-y\Vert }\ge \frac{2\eta }{R}>\beta _1$ and $B(y,\eta )\cap B=\emptyset$.

Suppose that $B(y,\eta )\not \subset A$ and take any $z\in B(y,\eta )\setminus A$. There exists $n_1$ such that $\frac{1}{n_1+1}<\Vert z-x_0\Vert \le \frac{1}{n_1}$, i.e. $z\in U_n$. Since

$$\begin{aligned} \Vert z-x_0\Vert \le \Vert z-y\Vert +\Vert y-x_0\Vert<\eta +\Vert y-x_0\Vert<2 \Vert y-x_0\Vert <\frac{1}{2n_0}, \end{aligned}$$

we obtain $\frac{1}{n_1+1}<\frac{1}{2n_0}$ and $n_1>n_0$. By construction of E, there exists $v\in E_{n_1}$ such that $\Vert z-v\Vert \le \frac{1}{(n_1+1)^2}$. Observe that $v\in E\subset B$ and $B\cap B(y,\eta )=\emptyset$. Therefore $v\not \in B(y,\eta )$, i.e. $\Vert v-y\Vert \ge \eta$. Thus

$$\begin{aligned} \Vert z-y|\ge \Vert y-v\Vert -\Vert v-z\Vert \ge \eta -\frac{1}{(n_1+1)^2}. \end{aligned}$$

This means that $B\Big (y, \eta -\frac{1}{(n_1+1)^2}\Big )\cap (B\cup (X\setminus A))=\emptyset$. By inequality $\frac{2\eta }{\eta +\Vert x_0-y\Vert }>\beta _1$, we obtain $2\eta >\beta _1 \Vert x_0-y\Vert$. Hence

$$\begin{aligned} \eta>\frac{\beta _1 \Vert x_0-y\Vert }{2}>\frac{\frac{1}{2}\beta _1 \Vert x_0-z\Vert }{2}>\frac{\beta _1}{4(n_1+1)}>\frac{8}{n_0}\frac{1}{4(n_1+1)}>\frac{1}{n_1^2} \end{aligned}$$

and $\eta -\frac{1}{(n_1+1)^2}>0$. Moreover,

$$\begin{aligned} \begin{aligned}&\frac{2\left( \eta -\frac{1}{(n_1+1)^2}\right) }{\eta -\frac{1}{(n_1+1)^2}+\Vert x_0-y\Vert }> \frac{2\eta -\frac{2}{(n_1+1)^2}}{\eta +\Vert x_0-y\Vert }\\&\quad = \frac{2\eta }{\eta +\Vert x_0-y\Vert }-\frac{2}{(n_1+1)^2\left( \eta +\Vert x_0-y\Vert \right) }>\beta _1- \frac{2}{(n_1+1)^2\Vert x_0-z\Vert }\\&\quad> \beta _1- \frac{2}{(n_1+1)^2\frac{1}{n_1+1}}=\beta _1-\frac{2}{n_1+1}>\beta _1-\frac{4}{n_0}>\beta _1-(\beta _1-\beta )=\beta . \end{aligned} \end{aligned}$$

Since $\beta \in (0,\alpha )$ was chosen arbitrary, ${\underline{p}}(B\cup (X\setminus A),x_0)\ge \alpha$, which completed the proof. $\square$

Lemma 3.5

Let $(X,\Vert \;\Vert )$ be a normed space, $f:X\rightarrow {\mathbb {R}}$, $x_0\in X$ and $\varrho >0$. If f restricted to ${\overline{B}}(x_0,\varrho )$ is continuous then f is ${\underline{p}}(X,\Vert \;\Vert )$ and ${\underline{s}}(X,\Vert \;\Vert )$-continuous at every $x\in S(x_0,\varrho )$.

Proof

For every $x\in S(x_0,\varrho )$ and $\varepsilon >0$ there exists $R_0$ such that

$$\begin{aligned} B\left( x+\tfrac{R}{2}\tfrac{x_0-x}{\Vert x_0-x\Vert },\tfrac{R}{2}\right) \subset B(x_0,\varrho )\cap B(x,R)\cap \{t\in X:|f(x)-f(x_0)|<\varepsilon \} \end{aligned}$$

for every $R<R_0$, which completed the proof. $\square$

It is easily seen that result of addition and multiplication of functions from discussed classes of functions, in general, need not belong to these classes. Therefore we studied the following notion.

Definition 3.6

[3]. Let ${\mathcal {F}}$ be a family of real functions defined on $(X,\Vert \;\Vert )$. A set ${\mathfrak {M}}_a({\mathcal {F}})=\{g:X\rightarrow {\mathbb {R}}:\forall _{f\in {\mathcal {F}}} \left( f+g\in \mathcal F\right) \}$ is called the maximal additive class for $\mathcal F$.

Remark 3.7

Let $f:X\rightarrow {\mathbb {R}}$, $f(x)=0$ for $x\in X$ be a constant function. Clearly, if $f\in {\mathcal {F}}$ then ${\mathfrak {M}}_a(\mathcal {F}) \subset \mathcal {F}$.

Let ${\mathcal {C}}_{\mathcal {T}}$ denote the class of continuous functions $f:(X,\mathcal {T})\rightarrow {\mathbb {R}}$.

Theorem 3.8

${\mathfrak {M}}_a(\underline{{\mathcal {S}}_0})=\mathcal C_{{\underline{p}}(X,\Vert \;\Vert )}$. Moreover, ${\mathfrak {M}}_a(\underline{{\mathcal {P}}_0})\subset \mathcal C_{{\underline{p}}(X,\Vert \;\Vert )}$.

Proof

Let $f\in {\mathcal {C}}_{{\underline{p}}(X,\Vert \;\Vert )}$. Take $g\in \underline{{\mathcal {S}}_0}$, $x_0\in X$, $\varepsilon >0$.

Denote $E_\varepsilon =\left\{ x\in X:|g(x)-g(x_0)|<\frac{\varepsilon }{2}\right\}$. Then ${\underline{p}}(X\setminus E_\varepsilon , x_0)>0$. Since $f\in {\mathcal {C}}_{{\underline{p}}(X,\Vert \;\Vert )}$, there exists a set U such that $x_0\in U$, $U\in {\underline{p}}(X,\Vert \;\Vert )$ and $U\subset \left\{ x\in X:|f(x)-f(x_0)|<\frac{\varepsilon }{2}\right\}$. By the definition of topology ${\underline{p}}(X,\Vert \;\Vert )$ the set $(X\setminus U)\cup (X\setminus E_\varepsilon )$ is lower porous. Moreover, $(X\setminus U)\cup (X\setminus E_\varepsilon )=X\setminus (U\cap E_\varepsilon )$. Thus ${\underline{p}}(X\setminus (U\cap E_\varepsilon ),x_0)>0$ and $|f(x)+g(x)-f(x_0)-g(x_0)|<\varepsilon$ for each $x\in U\cap E_\varepsilon$. Therefore $f+g$ is $\underline{\mathcal S_0}$-continuous at $x_0$. Hence $f\in \mathfrak M_a(\underline{{\mathcal {S}}_0})$.

Suppose that $f\not \in {\mathcal {C}}_{{\underline{p}}(X,\Vert \;\Vert )}$. Then there exist $x_0\in X$ and $\varepsilon >0$ such that $x_0\not \in {{\,\mathrm{int}\,}}_{{\underline{p}}(X,\Vert \;\Vert )}E_\varepsilon$, where $E_\varepsilon =\{x\in X:|f(x)-f(x_0)|<\varepsilon \}$. The set $E_\varepsilon$ does not contain any ${\underline{p}}$-neighbourhood of point $x_0$. By Lemma 3.3, the set $X\setminus E_\varepsilon$ is not lower superporous at $x_0$. Therefore there exists a set $F\subset X$ such that ${\underline{p}}(F, x_0)>0$ and ${\underline{p}}((X\setminus E_\varepsilon )\cup F,x_0)=0$. There exists a sequence of closed balls $\left( {\overline{B}}(x_n,\delta _n)\right) _{n\ge 1}$ such that $\bigcup _{n=1}^\infty {\overline{B}}(x_n,\delta _n)\subset X\setminus F$, $x_0\not \in \bigcup _{n=1}^\infty {\overline{B}}(x_n,\delta _n)$, $\lim _{n\rightarrow \infty }x_n=x_0$ and ${\underline{p}}(F,x_0)={\underline{p}}\left( X\setminus \bigcup _{n=1}^\infty {\overline{B}}(x_n,\delta _n),x_0\right) >0$. Let $A=\{x_0\}\cup \bigcup _{n=1}^\infty {\overline{B}}(x_n,\delta _n)$. By Lemma 3.4, we can find $E\subset X\setminus A$ such that for every $B\subset X$ if $E\subset B$ then ${\underline{p}}(B,x_0)={\underline{p}}(B\cup (X\setminus A),x_0)$. Define ${\widetilde{g}}:(A\setminus \{x_0\})\cup E\rightarrow {\mathbb {R}}$ by ${\widetilde{g}}(x)=0$ for $x \in A\setminus \{x_0\}$ and ${\widetilde{g}}(x)=-f(x)+f(x_0)+\varepsilon$ for $x\in E$. Since $(A\setminus \{x_0\})\cup E$ is a closed subset of $X\setminus \{x_0\}$ and ${\widetilde{g}}$ is continuous, by the Tietze Theorem, there exists a continuous extension ${\widehat{g}}:X\setminus \{x_0\}\rightarrow {\mathbb {R}}$ of ${\widetilde{g}}$. Finally, let $g:X\rightarrow {\mathbb {R}}$ be defined by $g(x)={\widehat{g}}(x)$ for $x\ne x_0$ and $g(x_0)=0$.

Since g is continuous at every point except $x_0$, $g(x)=g(x_0)$ for $x\in A$ and ${\underline{p}}(X\setminus A,x_0)={\underline{p}}(F,x_0)>0$, we have $g\in \underline{\mathcal {P}_0}$. On the other hand, $E\subset \{x\in X:|(f+g)(x)-(f+g)(x_0)|\ge \varepsilon \}$ and

$$\begin{aligned}&{\underline{p}}(X\setminus \{x\in X:|(f+g)(x)-(f+g)(x_0)|<\varepsilon \},x_0)\\&\quad ={\underline{p}}((X\setminus A)\cup (X\setminus \{x\in X:|(f+g)(x)-(f+g)(x_0)|<\varepsilon \}),x_0)\\&\quad ={\underline{p}}(X\setminus \{x\in A:|(f+g)(x)-(f+g)(x_0)|<\varepsilon \},x_0)\\&\quad = {\underline{p}}(X\setminus (E_\varepsilon \cap A), x_0)={\underline{p}}((X\setminus E_\varepsilon )\cup (X{\setminus }A),x_0)\le {\underline{p}}((X{\setminus } E_\varepsilon )\cup F,x_0)=0. \end{aligned}$$

Therefore $X\setminus \{x\in X:|(f+g)(x)-(f+g)(x_0)|<\varepsilon \}$ is not lower superporous at $x_0$ and $f+g\notin \underline{\mathcal {S}_0}$. It implies ${\mathfrak {M}}_a(\underline{{\mathcal {S}}_0})\subset \mathcal C_{{\underline{p}}(X,\Vert \;\Vert )}$ and ${\mathfrak {M}}_a(\underline{{\mathcal {P}}_0})\subset \mathcal C_{{\underline{p}}(X,\Vert \;\Vert )}$. The proof is completed. $\square$

Theorem 3.9

${\mathfrak {M}}_a(\underline{{\mathcal {M}}_1})=\mathcal C_{{\underline{s}}(X,\Vert \;\Vert )}$.

Proof

The proof of inclusion ${\mathcal {C}}_{{\underline{s}}(X,\Vert \;\Vert )}\subset {\mathfrak {M}}_a(\underline{{\mathcal {M}}_1})$ is very similar to the proof of inclusion ${\mathcal {C}}_{{\underline{p}}(X,\Vert \;\Vert )}\subset {\mathfrak {M}}_a(\underline{{\mathcal {S}}_0})$ in the proof of Theorem 3.8 and we omit it.

Take any $f\not \in {\mathcal {C}}_{{\underline{s}}(X,\Vert \;\Vert )}$. Then there exist $x_0\in X$ and $\varepsilon >0$ such that $x_0\not \in {{\,\mathrm{int}\,}}_{{\underline{s}}(X,\Vert \;\Vert )}E_\varepsilon$, where $E_\varepsilon =\{x\in X:|f(x)-f(x_0)|<\varepsilon \}$. By Lemma 3.3, the set $X\setminus E_\varepsilon$ is not lower strongly superporous at $x_0$. Similarly as in the proof of Theorem 3.8, we can find $F\subset X$ and $A=\{x_0\}\cup \bigcup _{n=1}^\infty {\overline{B}}(x_n,\delta _n)$ such that ${\underline{p}}(F, x_0)=1$, ${\underline{p}}((X\setminus E_\varepsilon )\cup F,x_0)<1$, $A\subset X\setminus F$, $x_0\not \in \bigcup _{n=1}^\infty {\overline{B}}(x_n,\delta _n)$, $\lim _{n\rightarrow \infty }x_n=x_0$ and ${\underline{p}}(F,x_0)={\underline{p}}\left( X\setminus A,x_0\right) >0$. By Lemma 3.4, we can find $E\subset X\setminus A$ such that for every $B\subset X$ if $E\subset B$ then ${\underline{p}}(B,x_0)={\underline{p}}(B\cup (X\setminus A),x_0)$. Define ${\widetilde{g}}:(A\setminus \{x_0\})\cup E\rightarrow {\mathbb {R}}$ by ${\widetilde{g}}(x)=0$ for $x \in A\setminus \{x_0\}$ and ${\widetilde{g}}(x)=-f(x)+f(x_0)+\varepsilon$ for $x\in E$. Again, similarly as in the proof of Theorem 3.8, we can define $g:X\rightarrow {\mathbb {R}}$ such that g is continuous at every point except $x_0$, $g(x)=0$ for $x\in A$ and $g(x)=f(x_0)-f(x)+\varepsilon$ for $x\in E$. Since g is continuous at every point except $x_0$ and ${\underline{p}}(X\setminus A,x_0)={\underline{p}}(F,x_0)=1$, we have $g\in \underline{\mathcal {M}_1}$. On the other hand, $E\subset \{x\in X:|(f+g)(x)-(f+g)(x_0)|\ge \varepsilon \}$ and

$$\begin{aligned}&{\underline{p}}(X\setminus \{x\in X:|(f+g)(x)-(f+g)(x_0)|<\varepsilon \},x_0)\\&\quad ={\underline{p}}((X\setminus A)\cup (X\setminus \{x\in X:|(f+g)(x)-(f+g)(x_0)|<\varepsilon \}),x_0)\\&\quad = {\underline{p}}(X\setminus \{x\in A:|(f+g)(x)-(f+g)(x_0)|<\varepsilon \},x_0)\\&\quad = {\underline{p}}(X{\setminus }(E_\varepsilon \cap A), x_0)={\underline{p}}((X{\setminus } E_\varepsilon )\cup (X{\setminus } A),x_0)\le {\underline{p}}((X{\setminus } E_\varepsilon )\cup F,x_0)<1. \end{aligned}$$

Therefore, $X\setminus \{x\in X:|(f+g)(x)-(f+g)(x_0)|<\varepsilon \}$ is not lower strongly superporous at $x_0$ and $f+g\notin \underline{\mathcal {M}_1}$. It implies ${\mathfrak {M}}_a(\underline{{\mathcal {M}}_1})\subset \mathcal C_{{\underline{s}}(X,\Vert \;\Vert )}$. The proof is completed. $\square$

Remark 3.10

In a similar way, applying ${\underline{p}}(X,\Vert \;\Vert )$ and ${\underline{s}}(X,\Vert \;\Vert )$, we can describe maximal multiplicative classes for $\underline{\mathcal {S}_0}$ and $\underline{\mathcal {M}_1}$. But in this case we need a notion of topology extended by a set, see [8, 9].