On [r, s]-Superporosity

Abstract

The main goal of the paper is to characterize families of $[r,s]$-(upper) superporous subsets of $\mathbb{R}$, which generalize well-known notions of superporosity and strong superporosity of subsets of $\mathbb{R}$. Definitions and properties of $[r,s]$-superporosity are symmetric to definitions and properties of superporosity and strong superporosity. The purpose in all cases is to define small subsets of the line using the notion of porosity. Superporous sets preserve positive porosity, and strongly superporous sets preserve strong porosity; i.e., if E is superporous (correspondingly, E is strongly superporous), then for every $x\in E$ and for every F such that porosity of F at x is greater than 0 (correspondingly, is equal to 1), porosity of $E\cup F$ at x is greater than 0 (correspondingly, is equal to 1). Taking arbitrary positive porosity, instead of 0 or 1, we obtain the symmetric definition as follows: $[r,s]$-superporosity for $0<r\le s<1$ transfers s-porosity to r-porosity; i.e., if E is $[r,s]$-superporous, then for every $x\in E$ and for every F such that porosity of F at x is not less than s, porosity of $E\cup F$ at x is not less than r. Even though the definition and properties of $[r,s]$-superporosity, superporosity and strong superporosity are symmetric and all of them consist of very small sets, the families of these sets are essentially different. In the paper, we focus on relationships between $[r,s]$-superporous sets for different indices $[r,s]$. Furthermore, we compare $[r,s]$-superporosity to superporosity and strong superporosity. We apply the notion of $[r,s]$-superporosity to find multipliers and adders of porouscontinuous functions.

1. Introduction

In [1,2] V. Kelar and L. Zajíček defined the notions of superporosity and strong superporosity, i.e., sets which preserve some aspects of porosity by taking a union of sets. The basic concept of porosity was introduced by E. P. Dolženko in [3] and is now being considered from very different points of view, see for example [4,5,6]. Properties of superporous and strongly superporous subsets of $\mathbb{R}$ were intensively studied [1,2,7,8]. In [1,2], using the notion of superporosity, topologies p, s, $p^{*}$ and $s^{*}$ are defined. Topologies p and s consist of complements of superporous and strongly superporous sets, respectively, and $p^{*}$ and $s^{*}$ consist of sets which are different from elements of p and s, respectively, by a meager set. In [2], L. Zajíček showed that $p^{*}$ consists of $\mathcal{I}$-dense sets.

Some types of generalized continuity connected with the notions of porosity, superporosity and strong superporosity were studied. In [9], J. Borsík and J. Holos defined families of porouscontinuous functions ${\mathcal{S}}_r$, ${\mathcal{P}}_r$ for $r\in [0,1)$ and ${\mathcal{M}}_r$, ${\mathcal{N}}_r$ for $r\in (0,1]$. Families of porouscontinuous functions are examples of so-called $\mathcal{A}$-continuous functions and path continuous functions with respect to the family $\mathcal{A}$. In [10], one can find many examples of generalized continuities and path continuities defined in this way. In the case of porouscontinuous functions, the family $\mathcal{A}$ is defined in terms of porosity. Using topologies p and s, maximal additive and maximal multiplicative classes for porouscontinuous functions are described [7].

In the paper, we introduce families $S(r,s)$ and $S^{*}(r,s)$ of $[r,s]$ and ${[r,s]}^{*}$-superporous sets. These concepts are symmetric to notions of superporosity and strong superporosity. As in the case of superporosity and strong superporosity, the $[r,s]$-superporous sets are very small and preserve some aspects of porosity under a union of sets. We also apply $[r,s]$-superporous sets to find multipliers and adders for porouscontinuous functions.

2. Preliminaries

As we mentioned before, in [1,2], V. Kelar and L. Zajíček defined the notions of superporosity and strong superporosity, i.e., sets which preserve some aspects of porosity by taking a union of sets. For a set $A\subset \mathbb{R}$ and an interval $I\subset \mathbb{R}$, let $\Lambda (A,I)$ denote the length of the largest open subinterval of I having an empty intersection with A. Then, according to [2,8,9], the right-handed (upper) porosity of the set A at $x\in \mathbb{R}$ is defined as

$$p^{+}(A,x)=\underset{h\to 0^{+}}{lim\; sup}{\displaystyle \frac{\Lambda (A,(x,x+h\left)\right)}{h}},$$

the left-handed (upper) porosity of the set A at x is defined as

$$p^{-}(A,x)=\underset{h\to 0^{+}}{lim\; sup}{\displaystyle \frac{\Lambda (A,(x-h,x\left)\right)}{h}},$$

and the (upper) porosity of A at x is defined as

$$p(A,x)=max\left\{p^{-}(A,x),p^{+}(A,x)\right\}.$$

We say that the set A is porous at $x\in \mathbb{R}$ if $p(A,x)>0$. The set A is called porous if A is porous at each point $x\in A$. We say that A is strongly porous at x if $p(A,x)=1$ and A is called strongly porous if A is strongly porous at each $x\in A$. Obviously, every strongly porous set is porous, and every porous set is nowhere dense. Moreover, none of the reverse inclusions is true.

In a similar way, taking the lower limit instead of upper limit, we can define lower porosity $\underline{p}(A,x)$.

We say that $E\subset \mathbb{R}$ is superporous at $x\in \mathbb{R}$ if porosity of $E\cup F$ is positive for every $F\subset \mathbb{R}$ with positive porosity at x. E is superporous if it is superporous at every point. Similarly, we say that $E\subset \mathbb{R}$ is strongly superporous at $x\in \mathbb{R}$ if porosity of $E\cup F$ is equal to 1 for every $F\subset \mathbb{R}$ with porosity 1 at x. E is strongly superporous if it is strongly superporous at every point.

J. Borsík and J. Holos defined families of porouscontinuous functions ${\mathcal{S}}_r$, ${\mathcal{P}}_r$ for $r\in [0,1)$ and ${\mathcal{M}}_r$, ${\mathcal{N}}_r$ for $r\in (0,1]$. $f:\mathbb{R}\to \mathbb{R}$ is ${\mathcal{S}}_r$ (${\mathcal{N}}_r$)-continuous at $x\in \mathbb{R}$ if the complement of every pre-image $f^{-1}\left((f\left(x\right)-\epsilon ,f\left(x\right)+\epsilon )\right)$ has porosity greater than r (not less than r). Similarly, f is ${\mathcal{P}}_r$ (${\mathcal{M}}_r$)-continuous at $x\in \mathbb{R}$ if there exists $E\subset \mathbb{R}$ such that $x\in E$, $p(\mathbb{R}\setminus E,x)>r$ ($p(\mathbb{R}\setminus E,x)\ge r$) and f restricted to E is continuous at x. Using topologies p and s, maximal additive and maximal multiplicative classes for porouscontinuous functions are described [7]. Porouscontinuous functions are examples of $\mathcal{A}$-continuous functions and path continuous functions with respect to the family $\mathcal{A}$, $\mathcal{A}=\{A_x:x\in \mathbb{R}\}$, where $A_x$ is a family of "large" subsets of $\mathbb{R}$ containing x. Then, the continuity of the function $f:\mathbb{R}\to \mathbb{R}$ at $x\in \mathbb{R}$ means that pre-image $f^{-1}\left((f\left(x\right)-\epsilon ,f\left(x\right)+\epsilon )\right)$ contains an element of $A_x$ for every $\epsilon >0$. Meanwhile, path continuity at $x\in \mathbb{R}$ means that there is a set $E\in A_x$ such that f restricted to E is continuous in ordinary sense at x. Of course, path continuity with respect to $\mathcal{A}$ implies $\mathcal{A}$-continuity, and sometimes, they are equivalent. In the case of porouscontinuous functions, the family $\mathcal{A}$ is defined in terms of porosity.

In the paper, we study the properties of $S(r,s)$ and $S^{*}(r,s)$ of $[r,s]$ and ${[r,s]}^{*}$-superporous sets. These concepts are close and symmetric to notions of superporosity and strong superporosity; i.e., $[r,s]$-superporous sets are very small and preserve some values of porosity under taking a union of sets. Nevertheless, despite symmetric definitions and properties, there are huge differences between $[r,s]$-superporosity, superporosity and strong superporosity. The families of $[r,s]$ and ${[r,s]}^{*}$-superporous sets $S(r,s)$ and $S^{*}(r,s)$ for $0<r\le s<1$ form monotone families of subsets of the real line, i.e., $S(r_2,s)\subset S(r_1,s)$ and $S^{*}(r_2,s)\subset S^{*}(r_1,s)$ for every $0<r_1<r_2<1$ and $s\ge r_2$ and $S(r,s_1)\subset S(r,s_2)$ and $S^{*}(r,s_1)\subset S^{*}(r,s_2)$ for every $0<s_1<s_2<1$ and $r\le s_1$. Furthermore, if $s<1-{(1-r)}^2$, then $S(r,s)$ and $S^{*}(r,s)$ are trivial, i.e., they consist of all sets with empty derivatives. In the final part of the paper, we use $[r,s]$-superporous sets to find multipliers and adders for porouscontinuous functions.

In Section 3, we present basic properties of porosity of subsets of $\mathbb{R}$. Moreover, we proved a few technical lemmas used in the sequel.

The main part of the paper, Section 4, is devoted to definition and properties of families $S(r,s)$ and $S^{*}(r,s)$ of subset of $\mathbb{R}$ consisting of $[r,s]$-superporous and ${[r,s]}^{*}$-superporous sets, respectively. $E\subset \mathbb{R}$ is $[r,s]$-superporous at $x\in E$ if porosity of $E\cup F$ is not less than r for every $F\subset \mathbb{R}$ with porosity not less than s at x. Similarly, $E\subset \mathbb{R}$ is ${[r,s]}^{*}$-superporous at $x\in E$ if porosity of $E\cup F$ is greater than r for every $F\subset \mathbb{R}$ with porosity greater than s at x. Directly from the definition, r and s play a symmetric role in these definitions, i.e., $S(r_1,s)\subset S(r_2,s)$ and $S(r,s_2)\subset S(r,s_1)$ for every $r_1<r_2$ and $s_1<s_2$. It turned out that $S(r,s)$ and $S^{*}(r,s)$ have quite complicated structure, and we focus on investigating relationships between these families for different values of r and s. In Theorem 5, the structures of $S(r,s)$ and $S^{*}(r,s)$ are described. If $r>1-\sqrt{1-s}$, then $S(r,s)$ and $S^{*}(r,s)$ consist of discrete sets, and if $r\le 1-\sqrt{1-s}$, then $S(r,s)$ and $S^{*}(r,s)$ contain sets with nonempty derivative. For every $0<r<1$ and $s\ge 1-{(1-r)}^2$, we have $S^{*}(r,s)⫋S(r,s)$ (Theorem 7). Moreover, we show relationships between $[r,s]$-superporous sets for different r and s (Theorems 8 and 10). At the end of the section, we present necessary and sufficient conditions for belonging to $S(r,s)$ and $S^{*}(r,s)$ formulated in terms of lower porosity, and we show lack of dependencies between superporosity, strong superporosity, $[r,s]$-superporosity and ${[r,s]}^{*}$-superporosity.

In Section 5, we present definitions and basic properties of porouscontinuous functions, multipliers and adders. Given two families of real function X and Y, we can consider all functions $g:\mathbb{R}\to \mathbb{R}$ such that the product $f·g$ (the sum $f+g$) belongs to X whenever f belongs to Y. Such a function g is said to be a multiplier (an adder) of the set X over the set Y. Multipliers are a generalization of the maximal multiplicative classes, and adders are a generalization of the notion of maximal additive classes of a given family of functions. Theorems 13, 14 and 15 show equalities ${\mathcal{S}}_{r,s}^{*}=\mathcal{A}({\mathcal{S}}_r/{\mathcal{S}}_s)$, ${\mathcal{S}}_{r,s}=\mathcal{A}({\mathcal{M}}_r/{\mathcal{M}}_s)$ and inclusion ${\mathcal{S}}_{r,s}^p\subset \mathcal{A}({\mathcal{P}}_r/{\mathcal{P}}_s)\subset {\mathcal{S}}_{r,s}^{*}$ for every $0<r\le s<1$, where ${\mathcal{S}}_{r,s}$ and ${\mathcal{S}}_{r,s}^{*}$ are families of $\mathcal{A}$-continuous, where $\mathcal{A}=S(r,s)$ or $\mathcal{A}=S^{*}(r,s)$, respectively, and ${\mathcal{S}}_{r,s}^p$ is a family of path continuous functions with respect to ${\mathcal{S}}_{r,s}^{*}$. To find multipliers of porouscontinuous functions, we need to introduce the concept of $[r,s]\left(A\right)$-continuity for $A\subset \mathbb{R}$. Theorems 16, 17 and Corollary 3 describe multipliers for porouscontinuous functions.

3. Porosity

Let $\mathbb{N}$ and $\mathbb{R}$ denote the set of all positive integers and the set of all real numbers, respectively. By $cl\left(A\right)$ and $A^d$, we denote a closure and a set of accumulation points of a set $A\subset \mathbb{R}$. By $f_{\upharpoonright A}$, we denote the restriction of f to $A\subset \mathbb{R}$.

Now, we show a few useful technical lemmas connected with porosity.

Theorem 1.

Let $x\in \mathbb{R}$, $x<\dots <b_{n+1}<a_n<b_n<\dots <b_1$, ${lim}_{n\to \infty }{a}_n=x$, $E=\left\{x\right\}\cup {\bigcup }_{n=1}^{\infty }\left\{a_n\right\}$ and $F=\left\{x\right\}\cup {\bigcup }_{n=1}^{\infty }[a_n,b_n]$. Then, $p^{+}(E,x)={lim\; sup}_{n\to \infty }{\displaystyle \frac{a_n-a_{n+1}}{a_n-x}}$ and $p^{+}(F,x)={lim\; sup}_{n\to \infty }{\displaystyle \frac{a_n-b_{n+1}}{a_n-x}}$.

Proof. 

Choose $n\ge 1$ and $t\in [a_{n+1},a_n]$. If $\Lambda (E,(x,t))=t-a_{n+1}$ then ${\displaystyle \frac{\Lambda (E,(x,t\left)\right)}{t-x}}\le {\displaystyle \frac{a_n-a_{n+1}}{a_n-x}}$ and if $\Lambda (E,(x,t))=\Lambda (E,(x,a_{n+1}))$ then ${\displaystyle \frac{\Lambda (E,(x,t\left)\right)}{t-x}}\le {\displaystyle \frac{a_k-a_{k+1}}{a_k-x}}$ for some $k>n$. Similarly, if $\Lambda (F,(x,t))=t-b_{n+1}$, then ${\displaystyle \frac{\Lambda (F,(x,t\left)\right)}{t-x}}\le {\displaystyle \frac{a_n-b_{n+1}}{a_n-x}}$, and if $\Lambda (F,(x,t))=\Lambda (F,(x,a_{n+1}))$, then ${\displaystyle \frac{\Lambda (F,(x,t\left)\right)}{t-x}}\le {\displaystyle \frac{a_k-b_{k+1}}{a_k-x}}$ for some $k>n$. Therefore,

$$p^{+}(E,x)=\underset{t\to x^{+}}{lim\; sup}{\displaystyle \frac{\Lambda (E,(x,t\left)\right))}{t-x}}=\underset{n\to \infty }{lim\; sup}{\displaystyle \frac{a_n-a_{n+1}}{a_n-x}}$$

and

$$p^{+}(F,x)=\underset{t\to x^{+}}{lim\; sup}{\displaystyle \frac{\Lambda (F,(x,t\left)\right)}{t-x}}=\underset{n\to \infty }{lim\; sup}{\displaystyle \frac{a_n-b_{n+1}}{a_n-x}}.$$

□

Definition 1.

We say that an open interval $(a,b)$ is maximal in relation to $E\subset \mathbb{R}$ if $(a,b)\cap E=\varnothing$ and $(a-\epsilon ,b)\cap E\ne \varnothing \ne (a,b+\epsilon )\cap E$ for every $\epsilon >0$.

Theorem 2.

For every $E\subset \mathbb{R}$ and $x\in E$ with $p^{+}(E,x)>0$, there exists a sequence ${\left((a_n,b_n)\right)}_{n\ge 1}$ of open intervals maximal in relation to E such that $x<\dots <b_{n+1}<a_n<\dots <b_1$, ${lim}_{n\to \infty }{a}_n=x$, $p^{+}(E,x)={lim\; sup}_{n\to \infty }{\displaystyle \frac{b_n-a_n}{b_n-x}}$ and $p^{+}(E\cup {\bigcup }_{n=1}^{\infty }(a_n,b_n),x)=0$.

Proof. 

Let $\mathbb{A}$ be a family of all open intervals $(a,b)\subset (x,x+1)$ maximal in relation to E such that $x<a$ and ${\displaystyle \frac{b-a}{b-x}}\ge {(a-x)}^2$. Obviously, $\mathbb{A}$ consists of pairwise disjoint open intervals and $(x,y)$ intersects only a finite number of elements of $\mathbb{A}$ for every $y\in (x,x+1)$. Therefore, $\mathbb{A}=\{(a_n,b_n):n\ge 1\}$, where $x<\dots <b_{n+1}<a_n<b_n<\dots <b_1$, $\Lambda (E,(x,b_n))=b_k-a_k$ for some $k\ge n$ and ${\displaystyle \frac{\Lambda (E,(x,t\left)\right)}{t-x}}\le {\displaystyle \frac{\Lambda (E,(x,b_n))}{b_n-x}}$ for every $t\in [b_n,a_{n-1}]$.

Fix $n\ge 1$. Take any $t\in [a_{n+1},b_{n+1}]$. If $t-a_{n+1}=\Lambda (E,(x,t))$, then ${\displaystyle \frac{\Lambda (E,(x,t\left)\right)}{t-x}}\le {\displaystyle \frac{\Lambda (E,(x,b_{n+1}))}{b_{n+1}-x}}={\displaystyle \frac{b_{n+1}-a_{n+1}}{b_{n+1}-x}}$, and if $t-a_{n+1}<\Lambda (E,(x,t))$, then there exists $k\ge n$ such that ${\displaystyle \frac{\Lambda (E,(x,t\left)\right)}{t-x}}={\displaystyle \frac{b_k-a_k}{t-x}}\le {\displaystyle \frac{b_k-a_k}{b_k-x}}$. If $t\in [b_{n+1},a_n]$, then ${\displaystyle \frac{\Lambda (E,(x,t\left)\right)}{t-x}}\le {\displaystyle \frac{\Lambda (E,(x,b_{n+1}))}{b_{n+1}-x}}\le {\displaystyle \frac{b_k-a_k}{b_k-x}}$ for some $k\ge n$. Hence, for every $t\in [a_{n+1},a_n]$, we can find $k\ge n$ such that ${\displaystyle \frac{\Lambda (E,(x,t\left)\right)}{t-x}}\le {\displaystyle \frac{b_k-a_k}{b_k-x}}$. Therefore, $p^{+}(E,x)={lim\; sup}_{t\to x^{+}}{\displaystyle \frac{\Lambda (E,(x,t\left)\right)}{t-x}}={lim\; sup}_{n\to \infty }{\displaystyle \frac{b_n-a_n}{b_n-x}}$.

Finally, if $(a,b)$ is any open interval disjoint from $E\cup {\bigcup }_{n=1}^{\infty }(a_n,b_n)$, then ${\displaystyle \frac{b-a}{b-x}}<a-x$. Hence, $p^{+}(E\cup {\bigcup }_{n=1}^{\infty }(a_n,b_n),x)=0$. □

Lemma 1.

Let $x<a<b<c$, $\alpha ={\displaystyle \frac{b-a}{b-x}}$, $\beta ={\displaystyle \frac{c-b}{c-x}}$ and $\gamma ={\displaystyle \frac{c-a}{c-x}}$. Then, $1-\gamma =(1-\alpha )(1-\beta )$, i.e., $\gamma =1-(1-\alpha )(1-\beta )=\alpha +\beta -\alpha \beta$.

Proof. 

$$(1-\alpha )(1-\beta )={\displaystyle \frac{a-x}{b-x}}·{\displaystyle \frac{b-x}{c-x}}={\displaystyle \frac{a-x}{c-x}}=1-{\displaystyle \frac{c-a}{c-x}}=1-\gamma ,$$

see Figure 1. □

Figure 1. $1-\gamma =(1-\alpha )(1-\beta )$.

Corollary 1.

(1) 

Let $x<a<b<c$, $\alpha ={\displaystyle \frac{b-a}{b-x}}={\displaystyle \frac{c-b}{c-x}}$ and $\gamma ={\displaystyle \frac{c-a}{c-x}}$. Then, $1-\gamma ={(1-\alpha )}^2$, i.e., $\gamma =1-{(1-\alpha )}^2=2\alpha -{\alpha }^2$.

(2) 

Let $x<a<b<c<d$, $\alpha ={\displaystyle \frac{b-a}{b-x}}$, $\beta ={\displaystyle \frac{c-b}{c-x}}$, $\gamma ={\displaystyle \frac{d-c}{d-x}}$ and $\delta ={\displaystyle \frac{d-a}{d-x}}$. Then, $1-\delta =(1-\alpha )(1-\beta )(1-\gamma )$.

(3) 

Let $x<a<b<c<d$, $\alpha ={\displaystyle \frac{b-a}{b-x}}={\displaystyle \frac{d-c}{d-x}}$, $\beta ={\displaystyle \frac{c-b}{c-x}}$ and $\gamma ={\displaystyle \frac{d-a}{d-x}}$. Then, $1-\gamma ={(1-\alpha )}^2(1-\beta )$.

Lemma 2.

For every $x<t$ and $r\in (0,1)$, there exists an open interval $(a,b)$ such that $x<a<t<b$ and ${\displaystyle \frac{t-a}{t-x}}={\displaystyle \frac{b-t}{b-x}}=r$. Moreover, $\sqrt{(a-x)(b-x)}=t-x$ and ${\displaystyle \frac{b-a}{b-x}}=1-{(1-r)}^2$.

Proof. 

Fix $x<t$ and $r\in (0,1)$. Put $a=t-r(t-x)$ and $b=t+{\displaystyle \frac{r}{1-r}}(t-x)$. Then, $x<a<t<b$, ${\displaystyle \frac{t-a}{t-x}}={\displaystyle \frac{r(t-x)}{t-x}}=r$ and ${\displaystyle \frac{b-t}{b-x}}={\displaystyle \frac{\frac{r}{1-r}(t-x)}{t-x+\frac{r}{1-r}(t-x)}}={\displaystyle \frac{\frac{r}{1-r}}{1+\frac{r}{1-r}}}=r$. □

Lemma 3.

Let $t\in (0,\infty )$. For every $s\in (0,1)$ and every open interval $(a,b)$ such that $0<a<t<b$ and ${\displaystyle \frac{b-a}{b}}=s$, the following inequality $max\{{\displaystyle \frac{t-a}{t}},{\displaystyle \frac{b-t}{b}}\}\ge 1-\sqrt{1-s}$ holds. Moreover, if $a=t\sqrt{1-s}$ and $b={\displaystyle \frac{t}{\sqrt{1-s}}}$, then ${\displaystyle \frac{b-a}{b}}=s$ and ${\displaystyle \frac{t-a}{t}}={\displaystyle \frac{b-t}{b}}=1-\sqrt{1-s}$.

Proof. 

Fix $s\in (0,1)$ and $t>0$. Put $c=t\sqrt{1-s}$ and $d={\displaystyle \frac{t}{\sqrt{1-s}}}$. Then, $c<t<d$, ${\displaystyle \frac{t-c}{t}}=1-\sqrt{1-s}$, ${\displaystyle \frac{d-t}{d}}=1-{\displaystyle \frac{t}{\frac{t}{\sqrt{1-s}}}}=1-\sqrt{1-s}$ and ${\displaystyle \frac{d-c}{d}}=1-(1-s)=s$. For every open interval $(a,b)$ such that $0<a<t<b$ and ${\displaystyle \frac{b-a}{b}}=s$, either $a\le c$ and ${\displaystyle \frac{t-a}{t}}\ge {\displaystyle \frac{t-c}{t}}=1-\sqrt{1-s}=s$ or $b\ge d$ and ${\displaystyle \frac{b-t}{b}}\ge {\displaystyle \frac{d-t}{d}}=1-\sqrt{1-s}=s$. □

Lemma 4.

Let $(c,d)\subset (0,\infty )$ be an arbitrary open interval. For every $s\in (0,1)$ and every open interval $(a,b)$ such that $0<a<c<d<b$, $s>{\displaystyle \frac{d-c}{d}}=\gamma$ and ${\displaystyle \frac{b-a}{b}}=s$, the following inequality $max\{{\displaystyle \frac{c-a}{c}},{\displaystyle \frac{b-d}{b}}\}\ge 1-\sqrt{{\displaystyle \frac{1-s}{1-\gamma }}}$ holds. Moreover, if $a=c\sqrt{{\displaystyle \frac{1-s}{1-\gamma }}}$ and $b={\displaystyle \frac{d}{\sqrt{\frac{1-s}{1-\gamma }}}}$, then ${\displaystyle \frac{b-a}{b}}=s$ and ${\displaystyle \frac{c-a}{c}}={\displaystyle \frac{b-d}{b}}=1-\sqrt{{\displaystyle \frac{1-s}{1-\gamma }}}$.

Proof. 

Fix $s,\gamma \in (0,1)$ and $0<c<d$ such that $\gamma <s$ and ${\displaystyle \frac{d-c}{d}}=\gamma$. Put $a^{\prime }=c\sqrt{{\displaystyle \frac{1-s}{1-\gamma }}}$ and $b^{\prime }={\displaystyle \frac{d}{\sqrt{\frac{1-s}{1-\gamma }}}}$. Then, $0<a^{\prime }<c<d<b^{\prime }$, ${\displaystyle \frac{c-a^{\prime }}{c}}=1-\sqrt{{\displaystyle \frac{1-s}{1-\gamma }}}$ and ${\displaystyle \frac{b^{\prime }-d}{b^{\prime }}}=1-\sqrt{{\displaystyle \frac{1-s}{1-\gamma }}}$. For every open interval $(a,b)$ such that $0<a<c<d<b$ and ${\displaystyle \frac{b-a}{b}}=s$ either $a\le a^{\prime }$ and ${\displaystyle \frac{c-a}{c}}\ge {\displaystyle \frac{c-a^{\prime }}{c}}=1-\sqrt{{\displaystyle \frac{1-s}{1-\gamma }}}$ or $b\ge b^{\prime }$ and ${\displaystyle \frac{b-d}{d}}\ge {\displaystyle \frac{b^{\prime }-d}{b^{\prime }}}=1-\sqrt{{\displaystyle \frac{1-s}{1-\gamma }}}$. □

4. $[r,s]$-Superporosity

We now introduce the main concept of our work, i.e., $[r,s]$-superporosity.

Definition 2.

Let $0<r\le s<1$. We say that a set $E\subset \mathbb{R}$ is $[r,s]$-superporous at $x\in E$ if for every $F\subset \mathbb{R}$ containing x, if $p(F,x)\ge s$, then $p(E\cup F,x)\ge r$. We say that E is $[r,s]$-superporous if it is $[r,s]$-superporous at each of its points. The family of all $[r,s]$-superporous subsets of $\mathbb{R}$ we denote by $S(r,s)$.

Definition 3.

Let $0<r\le s<1$. We say that a set $E\subset \mathbb{R}$ is ${[r,s]}^{*}$-superporous at $x\in E$ if for every $F\subset \mathbb{R}$ containing x, if $p(F,x)>s$, then $p(E\cup F,x)>r$. We say that E is ${[r,s]}^{*}$-superporous if it is ${[r,s]}^{*}$-superporous at each of its points. The family of all ${[r,s]}^{*}$-superporous subsets of $\mathbb{R}$ we denote by $S^{*}(r,s)$.

The following proposition follows directly from definitions of superporosity and shows the symmetric structure of families $S^{*}(r,s)$ and $S^{*}(r,s)$.

Proposition 1.

(1) 

$S(r_2,s)\subset S(r_1,s)$ and $S^{*}(r_2,s)\subset S^{*}(r_1,s)$ for every $0<r_1<r_2<1$ and $s\ge r_2$.

(2) 

$S(r,s_1)\subset S(r,s_2)$ and $S^{*}(r,s_1)\subset S^{*}(r,s_2)$ for every $0<s_1<s_2<1$ and $r\le s_1$.

Theorem 3.

If $E\subset \mathbb{R}$ and $x\in E$ is an accumulation point of $E\cap (x,\infty )$, then for every $s\in (0,1)$, there exists $F\subset \mathbb{R}$ such that $p(F,x)=s$ and $1-s={(1-p(E\cup F,x))}^2$, i.e., $p(E\cup F,x)=1-\sqrt{1-s}$.

Proof. 

Fix $s\in (0,1)$ and let $r=1-\sqrt{1-s}$. For every $t\in E\cap (x,\infty )$, define $a_t=t-r(t-x)$ and $b_t={\displaystyle \frac{t-rx}{1-r}}$, i.e., ${\displaystyle \frac{t-a_t}{t-x}}={\displaystyle \frac{b_t-t}{b_t-x}}=r$. Then, by Corollary 1 $\left(1\right)$, ${\displaystyle \frac{b_t-a_t}{b_t-x}}=s$. We can find the inductively decreasing sequence ${\left(t_n\right)}_{n\ge 1}$ such that $x<\dots <b_{t_{n+1}}<a_{t_n}<b_{t_n}<\dots <b_{t_1}$. Let $F=\mathbb{R}\setminus {\bigcup }_{n=1}^{\infty }(a_{t_n},b_{t_n})$. Since $F\cap (x,\infty )=[b_{t_1},\infty )\cup {\bigcup }_{n=1}^{\infty }\left[b_{t_{n+1}},a_{t_n}\right]$, by Theorem 1, $p(F,x)=p^{+}(F,x)={lim\; sup}_{n\to \infty }{\displaystyle \frac{b_{t_n}-a_{t_n}}{b_{t_n}-x}}=s$. On the other hand, $\mathbb{R}\setminus (E\cup F)\subset {\bigcup }_{n=1}^{\infty }\left((a_{t_n},t_n)\cup (t_n,b_{t_n})\right)\cup (b_{t_1},\infty )$. Again by Theorem 1, $p(E\cup F,x)=p^{+}(E\cup F,x)=r=1-\sqrt{1-s}$. □

Theorem 4.

For every $x\in \mathbb{R}$, there exists $\mathbf{A}\subset \mathbb{R}$ such that

• $x\in \mathbf{A}$ is an accumulation point of $\mathbf{A}\cap (x,\infty )$;
• $\mathbf{A}\in S(r,s)$ iff $s\ge 1-{(1-r)}^2$;
• $\mathbf{A}\in S^{*}(r,s)$ iff $s\ge 1-{(1-r)}^2$.

Proof. 

Fix $x\in \mathbb{R}$ and $r\in (0,1)$. Let $\mathbf{A}=\left\{x\right\}\cup {\bigcup }_{n=1}^{\infty }\left\{x_n\right\}$, where $x<\dots <x_{n+1}<x_n<\dots <x_1$ and ${\displaystyle \frac{x_n-x_{n+1}}{x_n-x}}={\displaystyle \frac{n-1}{n}}$, i.e., $x_{n+1}-x={\displaystyle \frac{x_n-x}{n}}$ for $n\ge 1$. Since ${\mathbf{A}}^d=\left\{x\right\}$, we only need to check $[r,s]$-superporosity at x. Choose any $s\ge r$. By Theorem 3, we can find $F\subset \mathbb{R}$ such that $p(F,x)=s$ and $p(\mathbf{A}\cup F,x)=1-\sqrt{1-s}$. Therefore, if $s<1-{(1-r)}^2$, then $\mathbf{A}\notin S(r,s)\cup S^{*}(r,s)$.

Assume $s\ge 1-{(1-r)}^2$ and let us take any $F\subset \mathbb{R}$ such that $p(F,x)=s$. If $p(F,x)=p^{-}(F,x)$, then, obviously, $p(\mathbf{A}\cup F,x)=s>r$. Let $p(F,x)=p^{+}(F,x)=s$. By Theorem 2, there exists a sequence ${\left((a_k,b_k)\right)}_{k\ge 1}$ of open intervals maximal in relation to F such that $x<\dots <b_{k+1}<a_k<\dots <b_1$, ${lim}_{k\to \infty }{a}_k=x$ and $s={lim\; sup}_{k\to \infty }{\displaystyle \frac{b_k-a_k}{b_k-x}}$. We may assume that $b_1<x_{n_0}$, where $n_0\in \mathbb{N}$ is such that ${\displaystyle \frac{n_0-1}{n_0}}>r$. Choose $k\ge 1$. If $\mathbf{A}\cap (a_k,b_k)=\varnothing$, then $(a_k,b_k)\cap (\mathbf{A}\cup F)=\varnothing$. If $\mathbf{A}\cap (a_k,b_k)$ contains at least two points, then $(x_{n+1},x_n)\cap (\mathbf{A}\cup F)=\varnothing$ for some $n\ge n_0$ and ${\displaystyle \frac{x_n-x_{n+1}}{x_n-x}}\ge {\displaystyle \frac{n_0-1}{n_0}}>r$. Finally, consider the case where $\mathbf{A}\cap (a_k,b_k)=\left\{x_n\right\}$ for some n. By Lemma 3, $(a_k,b_k)$ contains an interval $(c,d)$ disjoint from $\mathbf{A}\cup F$ and satisfying ${\displaystyle \frac{d-c}{d-x}}\ge 1-\sqrt{1-{\displaystyle \frac{b_k-a_k}{b_k-x}}}$. We have shown that every interval $(a_k,b_k)$ contains interval $(c,d)$ disjoint from $\mathbf{A}\cup F$ and such that ${\displaystyle \frac{d-c}{d-x}}\ge min\left\{{\displaystyle \frac{n_0-1}{n_0}},1-\sqrt{1-{\displaystyle \frac{b_k-a_k}{b_k-x}}}\right\}$. Therefore,

$$p(\mathbf{A}\cup F,x)\ge \underset{k\to \infty }{lim\; sup}min\left\{\frac{n_0-1}{n_0},1-\sqrt{1-\frac{b_k-a_k}{b_k-x}}\right\}\ge min\left\{\frac{n_0-1}{n_0},1-\sqrt{1-s}\right\}.$$

Hence, if $s\ge 1-{(1-r)}^2$, then $p(\mathbf{A}\cup F,x)\ge r$, and if $s>1-{(1-r)}^2$, then $p(\mathbf{A}\cup F,x)>r$. The proof is complete. □

Directly from Theorems 3 and 4, we obtain characterizations of $S(r,s)$ and $S^{*}(r,s)$.

Theorem 5.

Let $0<r\le s<1$.

(1) 

If $s<1-{(1-r)}^2$, i.e., $r>1-\sqrt{1-s}$, then $S(r,s)$ and $S^{*}(r,s)$ are trivial, i.e., they consist of sets with empty derivatives.

(2)

If $s\ge 1-{(1-r)}^2$, i.e., $r\le 1-\sqrt{1-s}$, then $S(r,s)$ and $S^{*}(r,s)$ are nontrivial, i.e., they contain sets with nonempty derivative.

Theorem 6.

For every $\alpha \in (0,1)$, there exists ${\mathbf{B}}_{\alpha }\subset \mathbb{R}$ such that

• $0\in {\mathbf{B}}_{\alpha }$ is an accumulation point of ${\mathbf{B}}_{\alpha }\cap (0,\infty )$;
• ${\mathbf{B}}_{\alpha }\in S(r,s)$ iff $r\le \alpha$ and $s\ge 1-{(1-r)}^2$;
• ${\mathbf{B}}_{\alpha }\in S^{*}(r,s)$ iff $r<\alpha$ and $s\ge 1-{(1-r)}^2$.

Proof. 

Fix $\alpha \in (0,1)$ and $r\in (0,1)$. Let ${\mathbf{B}}_{\alpha }=\left\{0\right\}\cup {\bigcup }_{n=1}^{\infty }\left\{x_n\right\}$, where $0<\dots <x_{n+1}<x_n<\dots <x_1$ and ${\displaystyle \frac{x_n-x_{n+1}}{x_n}}=\alpha$ for $n\ge 1$. Since ${\left({\mathbf{B}}_{\alpha }\right)}^d=\left\{0\right\}$, we only need to check $[r,s]$-superporosity at 0. By Theorem 1, $p^{+}({\mathbf{B}}_{\alpha },0)=\alpha$. Taking $F\subset \mathbb{R}$ with $p(F,0)=1$, we can conclude ${\mathbf{B}}_{\alpha }\notin S(r,s)$ for $r>\alpha$ and ${\mathbf{B}}_{\alpha }\notin S^{*}(r,s)$ for $r\ge \alpha$. Choose any $r\le \alpha$ and $s\in [r,1)$. By Theorem 3, we can find $F\subset \mathbb{R}$ such that $p(F,0)=s$ and $p({\mathbf{B}}_{\alpha }\cup F,0)=1-\sqrt{1-s}$. Therefore, if $s<1-{(1-r)}^2$, then ${\mathbf{B}}_{\alpha }\notin S(r,s)\cup S^{*}(r,s)$.

Assume $r\le \alpha$ and $s\ge 1-{(1-r)}^2$. Let us take any $F\subset \mathbb{R}$ such that $p(F,0)=s$. If $p(F,0)=p^{-}(F,0)$, then, obviously, $p({\mathbf{B}}_{\alpha }\cup F,0)=s>r$. Let $p(F,0)=p^{+}(F,0)=s$. By Theorem 2, there exists a sequence ${\left((a_k,b_k)\right)}_{k\ge 1}$ of open intervals maximal in relation to F such that $0<\dots <b_{k+1}<a_k<\dots <b_1$, ${lim}_{k\to \infty }{a}_k=x$ and $s={lim\; sup}_{k\to \infty }{\displaystyle \frac{b_k-a_k}{b_k}}$. Choose $k\ge 1$. If ${\mathbf{B}}_{\alpha }\cap (a_k,b_k)=\varnothing$, then $(a_k,b_k)\cap ({\mathbf{B}}_{\alpha }\cup F)=\varnothing$. If ${\mathbf{B}}_{\alpha }\cap (a_k,b_k)$ contains at least two points, then $(x_{n+1},x_n)\cap ({\mathbf{B}}_{\alpha }\cup F)=\varnothing$ for some n. Finally, consider the case where ${\mathbf{B}}_{\alpha }\cap (a_k,b_k)=\left\{x_n\right\}$ for some n. By Lemma 3, $(a_k,b_k)$ contains an interval $(c,d)$ disjoint from ${\mathbf{B}}_{\alpha }\cup F$ and satisfying ${\displaystyle \frac{d-c}{d}}\ge 1-\sqrt{1-{\displaystyle \frac{b_k-a_k}{b_k}}}$. We have shown that every interval $(a_k,b_k)$ contains interval $(c,d)$ disjoint from ${\mathbf{B}}_{\alpha }\cup F$ and such that ${\displaystyle \frac{d-c}{d}}\ge min\left\{\alpha ,1-\sqrt{1-{\displaystyle \frac{b_k-a_k}{b_k}}}\right\}$. Therefore,

$$p({\mathbf{B}}_{\alpha }\cup F,0)\ge \underset{k\to \infty }{lim\; sup}min\left\{\alpha ,1-\sqrt{1-\frac{b_k-a_k}{b_k}}\right\}\ge min\left\{\alpha ,1-\sqrt{1-s}\right\}.$$

Hence, if $s\ge 1-{(1-r)}^2$ and $r\le \alpha$, then $p({\mathbf{B}}_{\alpha }\cup F,0)\ge r$, and if $s>1-{(1-r)}^2$ and $r<\alpha$, then $p({\mathbf{B}}_{\alpha }\cup F,0)>r$. The proof is complete. □

Theorem 7.

$S^{*}(r,s)⫋S(r,s)$ for every $0<r<1$ and $s\ge 1-{(1-r)}^2$.

Proof. 

Fix $r\in (0,1)$ and $s\in [1-{(1-r)}^2,1)$. Take any $E\in S^{*}(r,s)$ and $x\in E$. Assume that E is not $[r,s]$-superporous at x. Then, we can find $F\subset \mathbb{R}$ with $p(F,x)\ge s$ and $p(F\cup E,x)<r$. Without loss of generality, we may assume $p^{+}(F,x)\ge s$, and moreover, $p^{+}(F\cup E,x)<r$. By Theorem 2, there exists a sequence ${\left((a_n,b_n)\right)}_{n\ge 1}$ of open intervals maximal in relation to F such that $x<\dots <b_{n+1}<a_n<\dots <b_1$, ${lim}_{n\to \infty }{a}_n=x$ and $p^{+}(F,x)={lim\; sup}_{n\to \infty }{\displaystyle \frac{b_n-a_n}{b_n-x}}$. Inductively, we can find a sequence of open intervals ${\left((c_k,d_k)\right)}_{k\ge 1}$ such that $x<\dots <d_{k+1}<c_k<\dots <b_1$, ${lim}_{k\to \infty }{c}_k=x$, $p^{+}(F,x)={lim\; sup}_{k\to \infty }{\displaystyle \frac{d_k-c_k}{d_k-x}}$, ${\bigcup }_{k=1}^{\infty }(c_k,d_k)\subset {\bigcup }_{n=1}^{\infty }(a_n,b_n)$ and ${lim}_{k\to \infty }{\displaystyle \frac{c_k-d_{k+1}}{d_{k+1}-c_{k+1}}}=\infty$. In particular, all $(c_k,d_k)$ are disjoint from F. Define $F^{\prime }=\mathbb{R}\setminus {\bigcup }_{k=1}^{\infty }(c_k,d_k)$. Obviously, $p(F^{\prime },x)=p(F,x)\ge s$ and $\alpha =p(F^{\prime }\cup E,x)<r$. Let $\beta ={\displaystyle \frac{r-\alpha }{2}}>0$. Then, $1-(1-\alpha )(1-\beta )=\alpha +\beta -\alpha \beta <r$. For $k\ge 1$, let $e_k=c_k-\beta (c_k-x)$, i.e., ${\displaystyle \frac{c_k-e_k}{c_k-x}}=\beta$. Let $F_1=\mathbb{R}\setminus {\bigcup }_{k=1}^{\infty }(e_k,d_k)$. Then,

$$\begin{array}{cc}\hfill p(F_1,x)=& {\displaystyle \underset{k\to \infty }{lim\; sup}\frac{d_k-e_k}{d_k-x}=\underset{k\to \infty }{lim\; sup}\left(1-\left(1-\frac{d_k-c_k}{d_k-x}\right)(1-\beta )\right)=}\hfill \\ \hfill & 1-(1-p^{+}(F,x))(1-\beta )>s.\hfill \end{array}$$

Since $E\in S^{*}(r,s)$, $p(F_1\cup E,x)>r$. For every $k\ge 1$, let $(f_k,g_k)$ be a maximal open interval contained in $(e_k,d_k)$ and disjoint from E (if such an interval does not exist, we take $(f_k,g_k)=\varnothing$ and ${\displaystyle \frac{g_k-f_k}{g_k-x}}=0$). Then, ${lim\; sup}_{k\to \infty }{\displaystyle \frac{g_k-f_k}{g_k-x}}=p(F_1\cup E,x)>r$ and ${lim\; sup}_{k\to \infty }{\displaystyle \frac{g_k-c_k}{g_k-x}}=p(F^{\prime }\cup E,x)=\alpha$. On the other hand, $1-{\displaystyle \frac{g_k-f_k}{g_k-x}}=\left(1-{\displaystyle \frac{g_k-c_k}{g_k-x}}\right)\left(1-{\displaystyle \frac{c_k-f_k}{c_k-x}}\right).$ Therefore,

$${\displaystyle \frac{g_k-f_k}{g_k-x}}=1-\left(1-{\displaystyle \frac{g_k-c_k}{g_k-x}}\right)\left(1-{\displaystyle \frac{c_k-f_k}{c_k-x}}\right)\le 1-\left(1-{\displaystyle \frac{g_k-c_k}{g_k-x}}\right)\left(1-\beta \right)$$

and

$$p(F_1\cup E,x)=\underset{k\to \infty }{lim\; sup}\left(1-\left(1-{\displaystyle \frac{g_k-c_k}{g_k-x}}\right)\left(1-\beta \right)\right)=1-(1-\alpha )(1-\beta )<r,$$

contrary to assumption $p(E\cup F_1,x)>r$. Hence, inclusion $S^{*}(r,s)\subset S(r,s)$ is proved.

Let ${\mathbf{B}}_{\alpha }$ for $\alpha \in (0,1)$ be the set from Theorem 6. Then, ${\mathbf{B}}_r\in S(r,s)\setminus S^{*}(r,s)$ for all $s\ge 1-{(1-r)}^2$. The proof is completed. □

Theorem 8.

(1) 

If $r_1<r_2$ and $s_1\in (1-{(1-r_1)}^2,1)$, then $S(r_1,s_1)\not\subset S(r_2,s_2)$ and $S^{*}(r_1,s_1)\not\subset S^{*}(r_2,s_2)$ for every $s_2\in [r_2,1)$.

(2) 

If $r_1<r_2$ and $s\in (1-{(1-r_1)}^2,1)$, then $S(r_1,s)\not\subset S(r_2,s)$ and $S^{*}(r_1,s)\not\subset S^{*}(r_2,s)$.

Proof. 

Both assertions follow directly from Theorem 6. For any $\alpha \in (r_1,r_2)$, we have ${\mathbf{B}}_{\alpha }\in S^{*}(r_1,s_1)\setminus S(r_2,s_2)$ for every $s_1\in (1-{(1-r_1)}^2,1)$ and $s_2\in [r_2,1)$. Similarly, ${\mathbf{B}}_{\alpha }\in S^{*}(r_1,s_1)\setminus S(r_2,s_2)$ for every $s_1\in (1-{(1-r_1)}^2,1)$ and $s_2\in [r_2,1)$. □

Theorem 9.

For every $\beta \in (0,1)$ there exists ${\mathbf{C}}_{\beta }\subset \mathbb{R}$ such that

• $0\in {\mathbf{C}}_{\beta }$ is an accumulation point of ${\mathbf{C}}_{\beta }\cap (0,\infty )$,
• ${\mathbf{C}}_{\beta }\in S(r,s)$ iff $s\ge 1-(1-\beta ){(1-r)}^2$,
• ${\mathbf{C}}_{\beta }\in S^{*}(r,s)$ iff $s\ge 1-(1-\beta ){(1-r)}^2$.

Proof. 

Fix $\beta \in (0,1)$. Inductively, we can find two sequences ${\left(x_n\right)}_{n\ge 1}$ and ${\left(y_n\right)}_{n\ge 1}$, both tending to 0 such that $0<\dots <y_{n+1}<x_n<y_n<\dots <y_1$, ${\displaystyle \frac{x_n-y_{n+1}}{x_n}}={\displaystyle \frac{n-1}{n}}$ and ${\displaystyle \frac{y_n-x_n}{x_n}}=\beta$ for $n\ge 1$. Next, for every $n\ge 1$ let $x_n=t_n^0<t_n^1<\dots <t_n^n=y_n$ satisfy $t_n^{k+1}-t_n^k={\displaystyle \frac{y_n-x_n}{n}}$ for $k=0,1,\dots ,n-1$ (see Figure 2). Finally, let ${\mathbf{C}}_{\beta }=\left\{0\right\}\cup {\bigcup }_{n=1}^{\infty }{\bigcup }_{k=0}^n\left\{t_n^k\right\}$. Since ${\left({\mathbf{C}}_{\beta }\right)}^d=\left\{0\right\}$, we have only to check $[r,s]$-superporosity at 0.

Figure 2. Construction of ${\mathbf{C}}_{\beta }$.

Let $E=\mathbb{R}\setminus {\bigcup }_{n=1}^{\infty }(x_n,y_n)$.

Clearly, $p(E,0)=\beta$ and $p(E\cup {\mathbf{C}}_{\beta },0)=0$. Choose any $s\in (\beta ,1)$. Let $r>0$ be such that $1-s={(1-r)}^2(1-\beta )$. For every $n\ge 1$, define $a_n=x_n(1-r)$ and $b_n={\displaystyle \frac{y_n}{1-r}}$, i.e., ${\displaystyle \frac{x_n-a_n}{x_n}}={\displaystyle \frac{b_n-y_n}{b_n}}=r$. Then, by Corollary 1 $\left(3\right)$, ${\displaystyle \frac{b_n-a_n}{b_n}}=s$. Let $F=\mathbb{R}\setminus \left(\left\{0\right\}\cup {\bigcup }_{n=1}^{\infty }(a_n,b_n)\right)$. Obviously, we can find $n_0\in \mathbb{N}$ such that $(a_{n+1},b_{n+1})\cap (a_n,b_n)=\varnothing$ for $n\ge n_0$. Similarly as in the proof of Theorem 3, we can show $p(F,0)=s$ and $p(F\cup {\mathbf{C}}_{\beta },0)=r$. Therefore, if $s<1-(1-\beta ){(1-r)}^2$, then ${\mathbf{C}}_{\beta }\notin S(r,s)\cup S^{*}(r,s)$.

Choose $r\in (0,1)$, $s\ge 1-(1-\beta ){(1-r)}^2$, and let us take any $F\subset \mathbb{R}$ such that $p(F,x)=s$. If $p(F,x)=p^{-}(F,x)$, then, obviously, $p({\mathbf{C}}_{\beta }\cup F,x)=s>r$. Let $p(F,x)=p^{+}(F,x)=s$. By Theorem 2, there exists a sequence ${\left((a_m,b_m)\right)}_{m\ge 1}$ of open intervals maximal in relation to F such that $0<\dots <b_{m+1}<a_m<\dots <b_1$, ${lim}_{m\to \infty }{a}_m=0$ and $s={lim\; sup}_{m\to \infty }{\displaystyle \frac{b_m-a_m}{b_m}}$. Choose $m\ge 1$.

If ${\mathbf{C}}_{\beta }\cap (a_m,b_m)=\varnothing$, then $(a_m,b_m)\cap ({\mathbf{C}}_{\beta }\cup F)=\varnothing$. If ${\mathbf{C}}_{\beta }\cap (a_m,b_m)$ intersects at least two different intervals $(x_{m_1},y_{m_1})$ and $(x_{m_1+1},y_{m_1+1})$, then $(y_{m_1+1},x_{m_1})\cap ({\mathbf{C}}_{\beta }\cup F)=\varnothing$. Finally, consider the case where ${\mathbf{C}}_{\beta }\cap (a_m,b_m)\subset (x_n,y_n)$ for some n.

By Lemma 4, $(a_m,b_m)$ contains an interval $(c,d)$ disjoint from ${\mathbf{C}}_{\beta }\cup F$ and such that ${\displaystyle \frac{d-c}{d}}\ge 1-\sqrt{{\displaystyle \frac{1-\frac{b_m-a_m}{b_m}}{1-\frac{y_n-x_n}{y_n}}}}=1-\sqrt{{\displaystyle \frac{1-\frac{b_m-a_m}{b_m}}{1-\beta }}}$. Therefore,

$$p({\mathbf{C}}_{\beta }\cup F,0)\ge \underset{k\to \infty }{lim\; sup}\left(1-\sqrt{\frac{1-\frac{b_k-a_k}{b_k}}{1-\beta }}\right)\ge 1-\sqrt{{\displaystyle \frac{1-s}{1-\beta }}}.$$

Hence, if $s\ge 1-(1-\beta ){(1-r)}^2$, then $p({\mathbf{C}}_{\beta }\cup F,x)\ge r$, and if $s>1-(1-\beta ){(1-r)}^2$, then $p({\mathbf{C}}_{\beta }\cup F,x)>r$. The proof is complete. □

Theorem 10.

If $r\in (0,1)$ and $s_2>s_1\ge 1-{(1-r)}^2$, then $S(r,s_1)⫋S(r,s_2)$ and $S^{*}(r,s_1)⫋S^{*}(r,s_2)$.

Proof. 

Inclusions $S(r,s_1)\subset S(r,s_2)$ and $S^{*}(r,s_1)\subset S^{*}(r,s_2)$ are obvious. Take $\beta =1-{\displaystyle \frac{s_1+s_2}{2{(1-r)}^2}}$. Then, $(1-\beta ){(1-r)}^2={\displaystyle \frac{s_1+s_2}{2}}$ and $s_1<(1-\beta ){(1-r)}^2<s_2$. By Theorem 9, ${\mathbf{C}}_{\beta }\in (S(r,s_2)\setminus S(r,s_1))\cap (S^{*}(r,s_2)\setminus S^{*}(r,s_1))$. □

In the next two theorems, we present connections between $[r,s]$-superporosity and the lower porosity.

Theorem 11.

Let $0<r\le s<1$, $E\subset \mathbb{R}$ and $x\in E$. If $\underline{p}(E,x)\ge 1-s+r$, then E is ${[r,s]}^{*}$-superporous at x.

Proof. 

Take any $E,F\subset \mathbb{R}$ such that $\underline{p}(E,x)\ge 1-s+r$ and $p(F,x)>s$. Without loss of generality, we may assume $p(F,x)=p^{+}(F,x)$. We can find decreasing sequence ${\left(x_n\right)}_{n\ge 1}$ tending to x such that ${lim\; sup}_{n\to \infty }{\displaystyle \frac{\Lambda (F,(x,x_n))}{x_n-x}}=p(F,x)$. Since ${\underline{p}}^{+}(E,x)\ge 1-s+r$, ${lim\; inf}_{n\to \infty }{\displaystyle \frac{\Lambda (E,(x,x_n))}{x_n-x}}\ge 1-s+r$. Obviously, $\Lambda (E\cup F,(x,x_n))\ge \Lambda (E,(x,x_n))+\Lambda (F,(x,x_n))-(x_n-x)$. Therefore,

$$\begin{array}{cc}\hfill p(E\cup F,x)\ge & \underset{n\to \infty }{lim\; sup}{\displaystyle \frac{\Lambda (E\cup F,(x,x_n))}{x_n-x}}\ge \underset{n\to \infty }{lim\; inf}{\displaystyle \frac{\Lambda (E,(x,x_n))}{x_n-x}}+\hfill \\ & \underset{n\to \infty }{lim\; sup}{\displaystyle \frac{\Lambda (F,(x,x_n))}{x_n-x}}-1\ge 1-s+r+p(F,x)-1>r.\hfill \end{array}$$

Hence, $p(E\cup F,x)>r$ if $p(F,x)>s$ and E is ${[r,s]}^{*}$-superporous at x. □

Theorem 12.

Let $0<r\le s<1$, $E\subset \mathbb{R}$ and $x\in E$. If E is $[r,s]$-superporous at x, then $\underline{p}(E,x)>0$.

Proof. 

Let $\underline{p}(E,x)=0$ and choose any $0<r\le s<1$. We may assume ${\underline{p}}^{+}(E,x)=0$. Let ${\left(x_n\right)}_{n\ge 1}$ be a decreasing sequence tending to x such that $x_{n+1}-x<{\displaystyle \frac{x_n-x}{n}}$, ${\displaystyle \frac{x_n-x_{n+1}}{x_n-x}}>{\displaystyle \frac{1+s}{2}}$ and ${lim}_{n\to \infty }{\displaystyle \frac{\Lambda (E,(x,x_n))}{x_n-x}}=0$. We may assume ${\displaystyle \frac{\Lambda (E,(x,x_n))}{x_n-x}}<{\displaystyle \frac{r(1-s)}{2-r}}$ for every n. For $n\ge 1$, define $y_n$ such that $y_n<x_n$ and ${\displaystyle \frac{x_n-y_n}{x_n-x}}=s$. Then, $y_n-x=(1-s)(x_n-x)$. Let $F=\mathbb{R}\setminus \left(\left\{x\right\}\cup {\bigcup }_{n=1}^{\infty }(y_n,x_n)\right)$. Obviously, $p(F,x)=s$. Let $(a,b)$ be any open interval disjoint from $E\cup F$. Then, $(a,b)\subset (y_n,x_n)$ for some n. Since ${\displaystyle \frac{\Lambda (E,(x,x_n))}{x_n-x}}<{\displaystyle \frac{r(1-s)}{2-r}}$, $b-a<{\displaystyle \frac{r(1-s)}{2-r}}(x_n-x)$ and

$${\displaystyle \frac{b-a}{b-x}}<{\displaystyle \frac{\frac{r(1-s)}{2-r}(x_n-x)}{\frac{r(1-s)}{2-r}(x_n-x)+(1-s)(x_n-x)}}={\displaystyle \frac{\frac{r}{2-r}}{\frac{r}{2-r}+1}}={\displaystyle \frac{r}{2}}.$$

Finally, $p(E\cup F,x)\le {\displaystyle \frac{r}{2}}<r$ and E is not $[r,s]$-superporous at x. The proof is complete. □

In the last part of this section, we discuss relationships between $S(r,s)$, $S^{*}(r,s)$, superporous sets and strongly superporous sets. Fix any $0<r\le s<1$, where $s\ge 1-{(1-r)}^2$. For any $\alpha \in (r,1)$, the set ${\mathbf{B}}_{\alpha }$ belongs to $S(r,s)\cap S^{*}(r,s)$ and is not strongly superporous. For any $\alpha \in (0,r)$, the set ${\mathbf{B}}_{\alpha }$ is superporous and belongs neither to $S(r,s)$ nor $S^{*}(r,s)$. For any $\beta \in (s,1)$, the set ${\mathbf{C}}_{\beta }$ is strongly superporous and does not belong to $S(r,s)\cup S^{*}(r,s)$. For any $\beta >0$, the set ${\mathbf{C}}_{\beta }$ belongs to $S(r,s)\cap S^{*}(r,s)$ (for sufficiently large s) and is not superporous.

5. Applications

We now apply $[r,s]$-superporosity to determine multipliers and adders of porouscontinuous functions. In the theory of real functions, families of functions that are not closed under addition or multiplication are often considered. Therefore, it seems natural for any class of functions X to search for functions g with the property that for each $f\in X$, the product $f·g$ (the sum $f+g$) belongs to X as well. Such a family of functions is called a maximal multiplicative (additive) class. The concept of maximal multiplicative class has been generalized in many papers as follows. Given two function classes X and Y, we can identify all functions g such that the product $f·g$ belongs to X whenever f belongs to Y. Such a family of functions $X/Y$ is said to be a multiplier of the set X over the set Y. There are many papers devoted to multipliers, see for example [11,12,13,14,15]. Similarly, we can generalize the concept of maximal additive class. Given two families of real function X and Y, we can consider all functions $g:\mathbb{R}\to \mathbb{R}$ such that the sum $f+g$ belongs to X whenever f belongs to Y. Such a function g is said to be an adder of the set X over the set Y, and it is denoted by $\mathcal{A}(X/Y)$.

We now recall the notion of porouscontinuity [9]. Let $r\in [0,1)$ and $A\subset \mathbb{R}$. A point $x\in \mathbb{R}$ will be called a point of ${\pi }_r$-density of the set A if $p(\mathbb{R}\setminus A,x)>r$. Let $r\in (0,1]$ and $A\subset \mathbb{R}$. A point $x\in \mathbb{R}$ will be called a point of ${\mu }_r$-density of the set A if $p(\mathbb{R}\setminus A,x)\ge r$. Let $r\in [0,1)$. The function $f:\mathbb{R}\to \mathbb{R}$ will be called

• ${\mathcal{P}}_r$-continuous at x if there exists $A\subset \mathbb{R}$ such that $x\in A$, x is a point of ${\pi }_r$-density of A and $f_{\upharpoonright A}$ is continuous at x and
• ${\mathcal{S}}_r$-continuous at x if for each $\epsilon >0$, there exists a set $A\subset \mathbb{R}$ such that $x\in A$, x is a point of ${\pi }_r$-density of A and $f\left(A\right)\subset \left(f\right(x)-\epsilon ,f(x)+\epsilon )$.

Let $r\in (0,1]$. The function $f:\mathbb{R}\to \mathbb{R}$ will be called

• ${\mathcal{M}}_r$-continuous at x if there exists $A\subset \mathbb{R}$ such that $x\in A$, x is a point of ${\mu }_r$-density of A and $f_{\upharpoonright A}$ is continuous at x, and
• ${\mathcal{N}}_r$-continuous at x if for each $\epsilon >0$ there exists a set $A\subset \mathbb{R}$ such that $x\in A$, x is a point of ${\mu }_r$-density of A and $f\left(A\right)\subset \left(f\right(x)-\epsilon ,f(x)+\epsilon )$.

All of these functions are called porouscontinuous functions.

We denote the class of ${\mathcal{P}}_r$-continuous, ${\mathcal{S}}_r$-continuous, ${\mathcal{M}}_r$-continuous, ${\mathcal{N}}_r$-continuous functions by ${\mathcal{P}}_r$, ${\mathcal{S}}_r$, ${\mathcal{M}}_r$, ${\mathcal{N}}_r$, respectively. Symbols ${\mathcal{P}}_r\left(f\right)$, ${\mathcal{S}}_r\left(f\right)$, ${\mathcal{M}}_r\left(f\right)$ and ${\mathcal{N}}_r\left(f\right)$ denote the set of all points at which f is ${\mathcal{P}}_r$-continuous, ${\mathcal{S}}_r$-continuous, ${\mathcal{M}}_r$-continuous and ${\mathcal{N}}_r$-continuous, respectively.

In [9], the equality ${\mathcal{M}}_r\left(f\right)={\mathcal{N}}_r\left(f\right)$ for every f and every $r\in (0,1]$ was proved.

Observe that if f is left-handed or right-handed continuous at some x, then f is porouscontinuous (in each sense) at x.

Let $r\in [0,1)$, $x\in \mathbb{R}$ and $f:\mathbb{R}\to \mathbb{R}$. Then, $x\in {\mathcal{P}}_r\left(f\right)$ if and only if $\underset{\epsilon \to 0^{+}}{lim}p(\mathbb{R}\setminus \{t:|f\left(x\right)-f\left(t\right)|<\epsilon \},x)>r$.

Applying $[r,s]$-superporosity, we describe multipliers and adders for porouscontinuous functions.

Definition 4.

Let $0<r\le s<1$. We say that $f:\mathbb{R}\to \mathbb{R}$ is $S_{r,s}$-continuous at $x\in \mathbb{R}$ if for every $\epsilon >0$, the set $\mathbb{R}\setminus \{y\in \mathbb{R}:|f\left(x\right)-f\left(y\right)|<\epsilon \}$ is $[r,s]$-superporous at x. The set of points at which f is $S_{r,s}$-continuous is denoted by $S_{r,s}\left(f\right)$ and ${\mathcal{S}}_{r,s}$ denotes the family of $f:\mathbb{R}\to \mathbb{R}$ which are $S_{r,s}$-continuous at every $x\in \mathbb{R}$.

Definition 5.

Let $0<r\le s<1$. We say that $f:\mathbb{R}\to \mathbb{R}$ is $S_{r,s}^{*}$-continuous at $x\in \mathbb{R}$ if for every $\epsilon >0$, the set $\mathbb{R}\setminus \{y\in \mathbb{R}:|f\left(x\right)-f\left(y\right)|<\epsilon \}$ is ${[r,s]}^{*}$-superporous at x. The set of points at which f is $S_{r,s}$-continuous is denoted by $S_{r,s}^{*}\left(f\right)$, and ${\mathcal{S}}_{r,s}^{*}$ denotes the family of $f:\mathbb{R}\to \mathbb{R}$ which are $S_{r,s}^{*}$-continuous at every $x\in \mathbb{R}$.

Definition 6.

Let $0<r\le s<1$. We say that $f:\mathbb{R}\to \mathbb{R}$ is $S_{r,s}^p$-continuous at $x\in \mathbb{R}$ if there exists $E\subset \mathbb{R}$ such that $x\in E$, $\mathbb{R}\setminus E$ is ${[r,s]}^{*}$-superporous at x and f restricted to E is continuous at x. The set of points at which f is $S_{r,s}^p$-continuous is denoted by $S_{r,s}^p\left(f\right)$ and ${\mathcal{S}}_{r,s}^p$ denotes the family of $f:\mathbb{R}\to \mathbb{R}$ which are $S_{r,s}^p$-continuous at every $x\in \mathbb{R}$.

Theorem 13.

${\mathcal{S}}_{r,s}^{*}=\mathcal{A}({\mathcal{S}}_r/{\mathcal{S}}_s)$ for every $0<r\le s<1$. Moreover, $\mathcal{A}({\mathcal{P}}_r/{\mathcal{P}}_s)\subset {\mathcal{S}}_{r,s}^{*}$ for every $0<r\le s<1$.

Proof. 

Fix $0<r\le s<1$. Take any $f\in {\mathcal{S}}_{r,s}^{*}$, $g\in {\mathcal{S}}_s$, $x\in \mathbb{R}$ and $\epsilon >0$. Obviously,

$$\{t\in \mathbb{R}:|(f+g)\left(t\right)-(f+g)\left(x\right)|<\epsilon \}\supseteq \{t\in \mathbb{R}:|f\left(t\right)-f\left(x\right)|<\frac{\epsilon }{2}\}\cap \{t\in \mathbb{R}:|g\left(t\right)-g\left(x\right)|<\frac{\epsilon }{2}\}.$$

Hence,

$$\begin{array}{cc}\hfill \mathbb{R}\setminus \{t\in \mathbb{R}:|(f+g)\left(t\right)-(f+g)\left(x\right)|<& \epsilon \}\subset (\mathbb{R}\setminus \{t\in \mathbb{R}:|f\left(t\right)-f\left(x\right)|>\frac{\epsilon }{2}\left\}\right)\cup \hfill \\ & (\mathbb{R}\setminus \{t\in \mathbb{R}:|g\left(t\right)-g\left(x\right)|<\frac{\epsilon }{2}\left\}\right).\hfill \end{array}$$

Since $\mathbb{R}\setminus \{t\in \mathbb{R}:|f\left(t\right)-f\left(x\right)|<{\displaystyle \frac{\epsilon }{2}}\}$ is ${[r,s]}^{*}$-superporous and $p(\mathbb{R}\setminus \{t\in \mathbb{R}:|g\left(t\right)-g\left(x\right)|<{\displaystyle \frac{\epsilon }{2}}\},x)>s$, $p(\mathbb{R}\setminus \{t\in \mathbb{R}:\left|\right(f+g\left)\right(t)-(f+g\left)\right(x\left)\right|<\epsilon \},x)>r$. Thus, $f+g\in {\mathcal{S}}_r$ and ${\mathcal{S}}_{r,s}^{*}\subset \mathcal{A}({\mathcal{S}}_r/{\mathcal{S}}_s)$.

Let $f\notin {\mathcal{S}}_{r,s}^{*}$. We can find $x\in \mathbb{R}$ and $\epsilon >0$ such that $\mathbb{R}\setminus \{t\in \mathbb{R}:|f\left(t\right)-f\left(x\right)|<\epsilon \}$ is not ${[r,s]}^{*}$-superporous at x. Hence, there is $F\subset \mathbb{R}$ for which $p(F,x)>s$ and $p\left(\right(\mathbb{R}\setminus \{t\in \mathbb{R}:|f\left(t\right)-f\left(x\right)|<\epsilon \})\cup F,x)\le r$. We may assume $p(F,x)=p^{+}(F,x)$. By Theorem 2, we can find a sequence of open intervals ${\left((a_n,b_n)\right)}_{n\ge 1}$ such that ${\bigcup }_{n=1}^{\infty }(a_n,b_n)\cap F=\varnothing$, $x<\dots <b_{n+1}<a_n<b_n<\dots <b_1$, ${lim}_{n\to \infty }{a}_n=x$ and $p^{+}(F,x)={lim\; sup}_{n\to \infty }{\displaystyle \frac{b_n-a_n}{b_n-x}}$. For every $n\ge 1$, let $b_{n+1}=c_0^n<c_1^n<\dots <c_n^n=a_n$ such that $c_{i+1}^n-c_i^n={\displaystyle \frac{a_n-b_{n+1}}{n}}$. Define $g:\mathbb{R}\to \mathbb{R}$ (see Figure 3.) by

$$g\left(t\right)=\left\{\begin{array}{cc}0\hfill & \mathrm{if}t\in \left\{x\right\}\cup \bigcup _{n=1}^{\infty }[a_n,b_n],\hfill \\ -f\left(t\right)+f\left(x\right)+\epsilon \hfill & \mathrm{if}t\in \bigcup _{n=1}^{\infty }\left(\{x-{\displaystyle \frac{1}{n}}\}\cup \bigcup _{k=1}^{n-1}\left\{c_k^n\right\}\right),\hfill \\ \mathrm{affine}\hfill & \mathrm{otherwise}.\hfill \end{array}\right.$$

Figure 3. Function g.

Clearly, g is continuous at every $t\ne x$. Moreover, $g\left(x\right)=0$, $\{t:g\left(t\right)=0\}\supset {\bigcup }_{n=1}^{\infty }(a_n,b_n)$ and

$$p(\mathbb{R}\setminus \{t:g\left(t\right)=0\},x)\ge p(\mathbb{R}\setminus \bigcup _{n=1}^{\infty }[a_n,b_n],x)\ge \underset{n\to \infty }{lim\; sup}{\displaystyle \frac{b_n-a_n}{b_n-x}}=p(F,x)>s.$$

Hence, $g\in {\mathcal{P}}_s\subset {\mathcal{S}}_s$. On the other hand,

$$\{t:|(f+g)\left(t\right)-(f+g)\left(x\right)|\ge \epsilon \}\supset \bigcup _{n=1}^{\infty }\left(\left\{x-{\displaystyle \frac{1}{n}}\right\}\cup \bigcup _{k=1}^{n-1}\left\{c_k^n\right\}\right).$$

Therefore,

$$\begin{array}{cc}\hfill p(\mathbb{R}\setminus & \{t:|(f+g)\left(t\right)-(f+g)\left(x\right)|<\epsilon \},x)\le \hfill \\ \hfill & p\left(\left(\mathbb{R}\setminus \{t:|(f+g)\left(t\right)-(f+g)\left(x\right)|<\epsilon \}\cap \bigcup _{n=1}^{\infty }[a_n,b_n]\right),x\right)=\hfill \\ \hfill & p\left(\mathbb{R}\setminus \left(\{t:|\left(f\right)\left(t\right)-\left(f\right)\left(x\right)|<\epsilon \}\cap \bigcup _{n=1}^{\infty }[a_n,b_n]\right),x\right)\le \hfill \\ \hfill & p\left(\left(\mathbb{R}\setminus \bigcup _{n=1}^{\infty }[a_n,b_n]\right)\cup \left(\mathbb{R}\setminus \{t:|f\left(t\right)-f\left(x\right)|<\epsilon \}\right),x\right)\le \hfill \\ \hfill & p\left(F\cup \left(\mathbb{R}\setminus \{t:|f\left(t\right)-f\left(x\right)|<\epsilon \}\right),x\right)<r.\hfill \end{array}$$

Finally, $f+g\notin {\mathcal{S}}_s$ and $f\notin \mathcal{A}({\mathcal{S}}_r/{\mathcal{S}}_s)\cap \mathcal{A}({\mathcal{P}}_r/{\mathcal{P}}_s)$. The proof is complete. □

Theorem 14.

${\mathcal{S}}_{r,s}=\mathcal{A}({\mathcal{M}}_r/{\mathcal{M}}_s)$ for every $0<r\le s<1$.

Proof. 

The proof is very similar to the proof of Theorem 13 and we leave it to the reader. □

Theorem 15.

${\mathcal{S}}_{r,s}^p\subset \mathcal{A}({\mathcal{P}}_r/{\mathcal{P}}_s)\subset {\mathcal{S}}_{r,s}^{*}$ for every $0<r\le s<1$.

Proof. 

Inclusion $\mathcal{A}({\mathcal{P}}_r/{\mathcal{P}}_s)\subset {\mathcal{S}}_{r,s}^{*}$ was proved in Theorem 13.

Take $f\in {\mathcal{S}}_{r,s}^p$, $g\in {\mathcal{P}}_s$ and $x\in \mathbb{R}$. There exist $E,F\subset \mathbb{R}$ such that $x\in E\cap F$, $E\in S^{*}(r,s)$, $p(\mathbb{R}\setminus F,x)>s$, f restricted to E is continuous at x and g restricted to F is continuous at x. Hence, $p(\mathbb{R}\setminus (E\cap F),x)=p\left(\right(\mathbb{R}\setminus E)\cup (\mathbb{R}\setminus F),x)>r$ and $f+g$ restricted to $E\cap F$ is continuous at x. Therefore, $x\in {\mathcal{P}}_r\left(f\right)$. Since x is arbitrary, $f\in {\mathcal{P}}_r$ and ${\mathcal{S}}_{r,s}^p\subset \mathcal{A}({\mathcal{P}}_r/{\mathcal{P}}_s)$. The proof is completed. □

Question 1.

Is either of the equalities ${\mathcal{S}}_{r,s}^p=\mathcal{A}({\mathcal{P}}_r/{\mathcal{P}}_s)$ or $\mathcal{A}({\mathcal{P}}_r/{\mathcal{P}}_s)={\mathcal{S}}_{r,s}^{*}$ true?

To find multipliers of porouscontinuous functions, we need to introduce the concept of $[r,s]\left(A\right)$-continuity for $A\subset \mathbb{R}$.

Definition 7.

Let $0<r\le s<1$ and $A\subset \mathbb{R}$. We say that $f:\mathbb{R}\to \mathbb{R}$ is $S_{r,s}\left(A\right)$-continuous at $x\in \mathbb{R}$ if, for every $\epsilon >0$, the set $\mathbb{R}\setminus \left(\right\{y\in \mathbb{R}:\left|f\right(x)-f(y\left)\right|<\epsilon \}\cup A)$ is $[r,s]$-superporous at x.

Definition 8.

Let $0<r\le s<1$ and $A\subset \mathbb{R}$. We say that $f:\mathbb{R}\to \mathbb{R}$ is $S_{r,s}^{*}\left(A\right)$-continuous at $x\in \mathbb{R}$ if, for every $\epsilon >0$, the set $\mathbb{R}\setminus \left(\right\{y\in \mathbb{R}:\left|f\right(x)-f(y\left)\right|<\epsilon \}\cup A)$ is ${[r,s]}^{*}$-superporous at x.

Definition 9.

Let $0<r\le s<1$ and $A\subset \mathbb{R}$. We say that $f:\mathbb{R}\to \mathbb{R}$ is $S_{r,s}^p\left(A\right)$-continuous at $x\in \mathbb{R}$ if there exists $E\subset \mathbb{R}$ such that $x\in E$, $\mathbb{R}\setminus (E\cup A)$ is ${[r,s]}^{*}$-superporous and f restricted to E is continuous at x.

Corollary 2.

Let $A\subset \mathbb{R}$ and $x\in \mathbb{R}$. If $\mathbb{R}\setminus A$ is $[r,s]$-superporous (${[r,s]}^{*}$-superporous) at x, then every function $f:\mathbb{R}\to \mathbb{R}$ is $S_{r,s}\left(A\right)$-continuous ($S_{r,s}^{*}\left(A\right)$-continuous) at x.

Theorem 16.

Let $0<r\le s<1$. $f\in {\mathcal{S}}_r/{\mathcal{S}}_s$ if and only if the following two conditions hold:

(i) 

f is $S_{r,s}^{*}$-continuous at every $x\in \mathbb{R}$ at which $f\left(x\right)\ne 0$;

(ii) 

f is $S_{r,s}^{*}\left(N_f\right)$-continuous at every $x\in \mathbb{R}$ at which $f\left(x\right)=0$ and $N_f=\{x\in \mathbb{R}:f\left(x\right)=0\}$.

Moreover, ${\mathcal{P}}_r/{\mathcal{P}}_s\subset {\mathcal{S}}_r/{\mathcal{S}}_s$.

Proof. 

Fix $0<r\le s<1$. Take any $f\in {\mathcal{S}}_{r,s}^{*}$, $g\in {\mathcal{S}}_s$, $x\in \mathbb{R}$ and $\epsilon >0$. First, consider the case where $c=f\left(x\right)\ne 0$. Obviously,

$$\begin{array}{cc}\hfill \{t\in \mathbb{R}:& \left|\right(f·g\left)\right(t)-(f·g\left)\right(x\left)\right|<\epsilon \}\supset \hfill \\ \hfill & \left\{t\in \mathbb{R}:|f\left(t\right)-f\left(x\right)|<\frac{\epsilon }{2\left(\right|g\left(x\right)|+1)}\right\}\cap \{t\in \mathbb{R}:|g\left(t\right)-g\left(x\right)|<\frac{\epsilon }{2c}\}.\hfill \end{array}$$

Hence,

$$\begin{array}{cc}\hfill \mathbb{R}\setminus & \{t\in \mathbb{R}:|(f·g)\left(t\right)-(f·g)\left(x\right)|<\epsilon \}\subset \hfill \\ & (\mathbb{R}\setminus \{t\in \mathbb{R}:|f\left(t\right)-f\left(x\right)|<\frac{\epsilon }{2\left(\right|g\left(x\right)|+1)}\left\}\right)\cup (\mathbb{R}\setminus \{t\in \mathbb{R}:|g\left(t\right)-g\left(x\right)|<\frac{\epsilon }{2c}\left\}\right).\hfill \end{array}$$

Because $\mathbb{R}\setminus \{t\in \mathbb{R}:|f\left(t\right)-f\left(x\right)|<\frac{\epsilon }{2\left(\right|g\left(x\right)|+1)}\}$ is ${[r,s]}^{*}$-superporous at x and $p(\mathbb{R}\setminus \{t\in \mathbb{R}:|f\left(t\right)-g\left(t\right)|<{\displaystyle \frac{\epsilon }{2c}}\},x)>s$, we obtain $p(\mathbb{R}\setminus \{t\in \mathbb{R}:\left|\right(f·g\left)\right(t)-(f·g\left)\right(x\left)\right|<\epsilon \},x)>r$. Therefore, $f·g$ is ${\mathcal{S}}_r$-continuous at x.

Consider the second case where $f\left(x\right)=0$. Obviously,

$$\begin{array}{cc}\hfill \{t\in \mathbb{R}:& \left|\right(f·g\left)\right(t)-(f·g\left)\right(x\left)\right|<\epsilon \}=\{t\in \mathbb{R}:\left|\right(f·g\left)\right(t\left)\right|<\epsilon \}\supset \hfill \\ \hfill & N_f\cup \left(\right\{t\in \mathbb{R}:\left|f\left(t\right)\right|<\frac{\epsilon }{\left|g\right(x\left)\right|+1}\}\cap \{t\in \mathbb{R}:|g\left(t\right)-g\left(x\right)|<1\left\}\right).\hfill \end{array}$$

Hence,

$$\begin{array}{cc}\hfill \mathbb{R}\setminus & \{t\in \mathbb{R}:|(f·g)\left(t\right)-(f·g)\left(x\right)|<\epsilon \}\subset \hfill \\ \hfill & \left(\mathbb{R}\setminus (N_f\cup \left(\right(\mathbb{R}\setminus \{t\in \mathbb{R}:|f\left(t\right)|<\frac{\epsilon }{\left|g\right(x\left)\right|+1}\}\right)\cap (\mathbb{R}\setminus \{t\in \mathbb{R}:|g\left(t\right)-g\left(x\right)|<1\left\}\right)).\hfill \end{array}$$

Since $\mathbb{R}\setminus (N_f\cup \{t\in \mathbb{R}:|f\left(t\right)-f\left(x\right)|<\frac{\epsilon }{\left|g\right(x\left)\right|+1}\left\}\right)$ is ${[r,s]}^{*}$-superporous at x and $p(\mathbb{R}\setminus \{t\in \mathbb{R}:\left|g\right(t)-g(x\left)\right|<1\},x)>s$, we obtain $p(\mathbb{R}\setminus \{t\in \mathbb{R}:\left|\right(f·g\left)\right(t\left)\right|<\epsilon \},x)>r$. Again, $f·g$ is ${\mathcal{S}}_r$-continuous at x. Thus, $f·g\in {\mathcal{S}}_r$ and ${\mathcal{S}}_{r,s}^{*}\subset {\mathcal{S}}_r/{\mathcal{S}}_s$.

Take any $f:\mathbb{R}\to \mathbb{R}$ and $x\in \mathbb{R}$. Assume that f does not satisfy condition $\left(i\right)$, i.e., $f\left(x\right)\ne 0$ and f is not $S_{r,s}^{*}$-continuous at x. Repeating constructions and arguments from the proof of Theorem 13, we can construct $g\in {\mathcal{P}}_s$ such that $f·g$ is not ${\mathcal{S}}_r$-continuous at x. Now, consider the case where f does not satisfy condition $\left(ii\right)$. We can find $\epsilon >0$ such that $\mathbb{R}\setminus (N_f\cup \{t\in \mathbb{R}:|f\left(t\right)-f\left(x\right)|<\epsilon \})$ is not ${[r,s]}^{*}$-superporous at x, and there is $x\in F\subset \mathbb{R}$ for which $p(F,x)>s$ and $p\left(\right(\mathbb{R}\setminus (N_f\cup \{t\in \mathbb{R}:|f\left(t\right)-f\left(x\right)|<\epsilon \}\left)\right)\cup F,x)\le r$. We may assume $p(F,x)=p^{+}(F,x)$. By Theorem 2, we can find a sequence of open intervals ${\left((a_n,b_n)\right)}_{n\ge 1}$ such that ${\bigcup }_{n=1}^{\infty }(a_n,b_n)\cap F=\varnothing$, $x<\dots <b_{n+1}<a_n<b_n<\dots <b_1$, ${lim}_{n\to \infty }{a}_n=x$ and $p^{+}(F,x)={lim\; sup}_{n\to \infty }{\displaystyle \frac{b_n-a_n}{b_n-x}}$.

For every $n\ge 1$, there exists a finite set $B_n\subset [b_{n+1},a_n]\setminus N_f$ such that for every $u\in [b_{n+1},a_n]\setminus N_f$, we can find $v\in B_n$ for which $|u-v|<{\displaystyle \frac{a_n-b_{n+1}}{n}}$. Moreover, we can find a decreasing sequence ${\left(x_n\right)}_{n\ge 1}$ tending to x such that $f\left(x_n\right)\ne 0$ and $p^{-}({\bigcup }_{n\ge 1}\left\{x_n\right\},x)\le r$ (we can do it because $\mathbb{R}\setminus N_f$ is not ${[r,s]}^{*}$-superporous and $p^{-}(\mathbb{R}\setminus N_f,x)\le r$). Define $g:\mathbb{R}\to \mathbb{R}$ by

$$g\left(t\right)=\left\{\begin{array}{cc}1\hfill & \mathrm{if}t\in \left\{x\right\}\cup \bigcup _{n=1}^{\infty }[a_n,b_n],\hfill \\ {\displaystyle \frac{1}{f\left(t\right)}}\hfill & \mathrm{if}t\in \bigcup _{n=1}^{\infty }\left(\left\{x_n\right\}\cup B_n\right),\hfill \\ \mathrm{affine}\hfill & \mathrm{otherwise}.\hfill \end{array}\right.$$

Clearly, g is continuous at every $t\ne x$. Similarly, as in the proof of Theorem 13, we can check that $g\in {\mathcal{P}}_s\subset {\mathcal{S}}_s$.

On the other hand,

$$\{t:|(f·g)\left(t\right)-(f·g)\left(x\right)|\ge \epsilon \}\supset \bigcup _{n=1}^{\infty }\left(\left\{x_n\right\}\cup \bigcup _{n\ge 1}{B}_n\right).$$

Therefore,

$$\begin{array}{cc}\hfill p(\mathbb{R}\setminus & \{t:|(f·g)\left(t\right)-(f·g)\left(x\right)|<\epsilon \},x)\le \hfill \\ \hfill & p\left(\left(\mathbb{R}\setminus \left(N_f\cup \left(\{t:|(f·g)\left(t\right)|<\epsilon \}\cap \bigcup _{n=1}^{\infty }[a_n,b_n]\right)\right)\right),x\right)=\hfill \\ \hfill & p\left(\mathbb{R}\setminus \left(N_f\cup \left(\{t:|\left(f\right)\left(t\right)|<\epsilon \}\cap \bigcup _{n=1}^{\infty }[a_n,b_n]\right)\right),x\right)\le \hfill \\ \hfill & p\left(\left(\mathbb{R}\setminus \bigcup _{n=1}^{\infty }[a_n,b_n]\right)\cup \left(\mathbb{R}\setminus (N_f\cup \{t:|f\left(t\right)-f\left(x\right)|<\epsilon \})\right),x\right)\le \hfill \\ \hfill & p\left(F\cup \left(\mathbb{R}\setminus (N_f\cup \{t:|f\left(t\right)-f\left(x\right)|<\epsilon \})\right),x\right)\le r.\hfill \end{array}$$

Hence, $f·g\notin {\mathcal{S}}_s$ and $f\notin {\mathcal{S}}_r/{\mathcal{S}}_s\cap {\mathcal{P}}_r/{\mathcal{P}}_s$. We have proved that if f does not satisfy $\left(i\right)$ or $\left(ii\right)$ then $f\notin {\mathcal{S}}_r/{\mathcal{S}}_s\cap {\mathcal{P}}_r/{\mathcal{P}}_s$. The proof is complete. □

Theorem 17.

Let $0<r\le s<1$. $f\in {\mathcal{M}}_r/{\mathcal{M}}_s$ if and only if the following two conditions hold

(i) 

f is $S_{r,s}$-continuous at every $x\in \mathbb{R}$ at which $f\left(x\right)\ne 0$;

(ii) 

f is $S_{r,s}\left(N_f\right)$-continuous at every $x\in \mathbb{R}$ at which $f\left(x\right)=0$ and $N_f=\{x\in \mathbb{R}:f\left(x\right)=0\}$.

Proof. 

The proof is very similar to the proof of the previous theorem, and we have omitted it. □

Corollary 3.

${\mathcal{S}}_{r,s}^p\subset {\mathcal{P}}_r/{\mathcal{P}}_s\subset {\mathcal{S}}_r/{\mathcal{S}}_s$ for $0<r\le s<1$.

6. Conclusions

In this paper, we introduced the concept of $[r,s]$ and ${[r,s]}^{*}$-upper superporosity, which proposes a different perspective on the construction of small subsets of $\mathbb{R}$ using the properties of porosity. The motivation for these studies were the results relating superporosity and strong superporosity, obtained by L. Zajíček and V. Kelar. Despite similarities in the definition, $[r,s]$ and ${[r,s]}^{*}$-upper superporosity has diametrically different properties than superporosity and strong superporosity. We focused on relationships between $[r,s]$-upper superporous sets for different indices $[r,s]$. We also showed that there are no inclusions between families of $[r,s]$-superporous sets, superporous sets and strong superporous sets. Moreover, we applied the notion of $[r,s]$-superporosity to find adders of porouscontinuous functions.

It seems to be an interesting question whether similar properties hold for subsets of the plane or metric space. It is known that porosity and strong superporosity were also studied for subsets of metric space. One can also try to define and investigate $[r,s]$ and ${[r,s]}^{*}$-lower superporosity sets, where in the definitions we use lower porosity instead of the usual one. As shown in [16], as a rule, problems in which upper porosity is replaced by lower porosity are much more difficult, but also more interesting.