On a typical compact set as the attractor of generalized iterated function systems of infinite order

TL;DR

This paper demonstrates that in Euclidean and polish spaces, typical compact sets can be viewed as attractors of generalized iterated function systems, extending classical fractal concepts to broader contexts.

Contribution

It shows that typical compact sets in Euclidean and polish spaces are attractors of generalized IFS, broadening the understanding of fractal structures in these spaces.

Findings

Typical compact sets in Euclidean spaces are attractors of Secelean's generalized IFS.

In polish spaces, typical compact sets can be perceived as selfsimilar attractors.

Extends classical IFS theory to generalized systems in infinite-dimensional contexts.

Abstract

In 2013 Balka and M\'ath\'e showed that in uncountable polish spaces the typical compact set is not a fractal of any IFS. In 2008 Miculescu and Mihail introduced a concept of a generalized iterated function system (GIFS in short), a particular extension of classical IFS, in which they considered families of mappings defined on finite Cartesian product $X^m$ with values in $X$. Recently, Secelean extended these considerations to mappings defined on the space $\ell_\infty(X)$ of all bounded sequences of elements of $X$ endowed with supremum metric. In the paper we show that in Euclidean spaces a typical compact set is an attractor in sense of Secelean and that in general in the polish spaces it can be perceived as selfsimilar in such sense.

Equations

$$H^d(K,D):=\max \left\{\sup \{\inf \{d(x,y):x\in K\}:y\in D\},\sup \{\inf \{d(x,y):x\in D\}:y\in K\}\right\}.$$

$$U[U_1,...,U_n]:=\left\{K\in \mathcal{K}(X):K\subset U,K\cap U_i\ne \varnothing \right\}$$

$$A_{\mathcal{F}}=\bigcup _{i=1}^n{f}_i(A_{\mathcal{F}}).$$

$$A_{\mathcal{G}}=\bigcup _{i=1}^n{g}_i(A_{\mathcal{G}}\times ...\times A_{\mathcal{G}}).$$

$$A_{\mathcal{S}}=\bigcup _{g\in \mathcal{S}}\overline{g(A_{\mathcal{S}}\times A_{\mathcal{S}}\times ...)}.$$

$$d_q((x_n),(y_n)):=\underset{n\in \mathbb{N}}{\sup }{q}^{n-1}d(x_n,y_n),(x_n),(y_n)\in {\ell }_{\infty }(X).$$

$$\bigcap _{k\in \mathbb{N}}{f}_1\circ ...\circ f_k(X)$$

$$A_{\mathcal{G}}=\bigcup _{i=1}^n{g}_i(A_{\mathcal{G}}\times ...\times A_{\mathcal{G}}).$$

$${\ell }_{\infty }(X):=\{(x_n)\subset X:(x_n)\text{mbox}isbounded\}.$$

$$d(f(x),f(y))\le \mathrm{Lip}(f)d_1(x,y)\le \mathrm{Lip}(f)\mathrm{diam}\left(\bigcup _{k=1}^{\infty }{K}_k\right).$$

$$A_{\mathcal{S}}=\bigcup _{i=1}^n\overline{f_i\left(\prod _{k=1}^{\infty }{A}_{\mathcal{S}}\right)}.$$

$${\mathcal{H}}_{\delta }^h(A):=\inf \left\{\sum _{i\in I}h\left(\mathrm{diam}(U_i)\right):\{U_i{\}}_{i\in I}\text{ is a δ-cover of A}\right\}.$$

$${\mathcal{H}}^h(A):=\underset{\delta \to 0}{\lim }{\mathcal{H}}_{\delta }^h(A).$$

$${\mathcal{I}}_n:=\prod _{i=1}^n\{1,...,a_i\}$$

$$\mathcal{I}:=\bigcup _{n=1}^{\infty }{\mathcal{I}}_n.$$

$$K=\bigcap _{n\in \mathbb{N}}\left(\bigcup _{i_1=1}^{a_1}\dots \bigcup _{i_n=1}^{a_n}{C}_{i_1,...,i_n}\right),$$

$$\mathrm{dist}(C_{i_1,...,i_n,s},C_{i_1,...,i_n,t})>\mathrm{diam}\left(\bigcup _{j_{n+1}=1}^{a_{n+1}}{C}_{j_1,...,j_n,j_{n+1}}\right).$$

$${\mathcal{H}}^h(K\cap f(K))=0.$$

$$0<{\mathcal{H}}^h(K)={\mathcal{H}}^h\left(\bigcup _{i=1}^n{f}_i(K)\right)\le \sum _{i=1}^n{\mathcal{H}}^h\left(K\cap f_i(K)\right)=0$$

$$A=\bigcup _{i=1}^n\overline{f_i(A\times A\times \dots )}.$$

$$K=\left\{x_{(i_1,i_2,...)}:(i_1,i_2,...)\in \prod _{i=1}^{\infty }\{1,...,a_i\}\right\}$$

$$f_i(x_{{\alpha }_1},x_{{\alpha }_2},...):=x_{i({\alpha }_1,{\alpha }_2,...)}$$

$$i({\alpha }_1,{\alpha }_2,...):=(i,{\beta }_1,{\beta }_2,...)$$

$$d\left(f_i(x_{{\alpha }_1},x_{{\alpha }_2},...),f_i(x_{{\beta }_1},x_{{\beta }_2},...)\right)=d\left(x_{i({\alpha }_1,{\alpha }_2,...)},x_{i({\beta }_1,{\beta }_2,...)}\right)=$$

$$=d\left(x_{(i,1+{\sum }_{i=1}^{a_2-1}{\tilde{\alpha }}_i^{(1)},1+{\sum }_{i=1}^{a_3-1}{\tilde{\alpha }}_i^{(2)},...)},x_{(i,1+{\sum }_{i=1}^{a_2-1}{\tilde{\beta }}_i^{(1)},1+{\sum }_{i=1}^{a_3-1}{\tilde{\beta }}_i^{(2)},...)}\right).$$

$$\eta :=\min \left\{k\in \mathbb{N}:\sum _{i=1}^{a_{k+1}-1}{\tilde{\alpha }}_i^{(k)}\ne \sum _{i=1}^{a_{k+1}-1}{\tilde{\beta }}_i^{(k)}\right\},$$

$$d\left(f_i(x_{{\alpha }_1},x_{{\alpha }_2},...),f_i(x_{{\beta }_1},x_{{\beta }_2},...)\right)\le \mathrm{diam}\left(C_{i,1+{\sum }_{i=1}^{a_2-1}{\tilde{\alpha }}_i^{(1)},...,1+{\sum }_{i=1}^{a_{\eta }-1}{\tilde{\alpha }}_i^{(\eta -1)}}\right)\le b_{\eta }.$$

$${\eta }_j:=\min \{k\in \mathbb{N}:{\alpha }_j^{(k)}\ne {\beta }_j^{(k)}\}\text{(assume min∅:=∞)}$$

$$l:=\min \{k\in \mathbb{N}:\min \{{\eta }_1,...,{\eta }_{a_{k+1}-1}\}\le k\},$$

$$h:=\min \{k\in \{1,...,a_{l+1}-1\}:{\eta }_k\le l\}.$$

$$d_1\left((x_{{\alpha }_1},x_{{\alpha }_2},...),(x_{{\beta }_1},x_{{\beta }_2},...)\right)=\underset{n\in \mathbb{N}}{\sup }d(x_{{\alpha }_n},x_{{\beta }_n})\ge d(x_{{\alpha }_h},x_{{\beta }_h})\ge$$

Full text

In 2008 Miculescu and Mihail introduced a concept of a generalized iterated function system (GIFS in short), a particular extension of classical IFS, in which they considered families of mappings defined on finite Cartesian product $X^{m}$ with values in $X$. Recently, Secelean extended these considerations to mappings defined on the space $\ell_{\infty}(X)$ of all bounded sequences of elements of $X$ endowed with supremum metric.

In the paper we show that in Euclidean spaces a typical compact set is an attractor in sense of Secelean and that in general in the polish spaces it can be perceived as selfsimilar in such sense.

††journal: J Math Anal Appl

1 Introduction

Let $(X,d)$ be a metric space. By $\mathcal{K}(X,d)$ (shortly $\mathcal{K}(X)$) we denote the space of all nonempty and compact subsets of $X$, endowed with the Hausdorff-Pompeiu metric

[TABLE]

If $d$ is fixed, we will write $H:=H^{d}$. It is well known that $H^{d}$ induces the topology consistent with Vietoris topology on $\mathcal{K}(X)$, i.e. the topology which base consists of nonempty sets

[TABLE]

where $n\in\mathbb{N}$ and $U,U_{1},...,U_{n}\subset X$ are open.

Moreover, $\mathcal{K}(X)$ is complete (compact), provided $X$ is complete (compact).

Let $(X,d),(Y,\rho)$ be metric spaces and $f:X\to Y$. By $\operatorname{Lip}(f)$ we denote the Lipschitz constant of $f$. We say that $f$ is a contraction if $\operatorname{Lip}(f)<1$ and that it is a weak contraction if $\rho(f(x),f(y))<d(x,y)$ provided $x\neq y$. We say that a family $\mathcal{F}:=\{f_{1},...,f_{n}\}$ is an iterated function system (IFS in short) if $f_{i}:X\to X,i=1,...,n,$ are contractions. We call $\mathcal{F}$ a weak IFS if $f_{i}:X\to X$ are weak contractions.

The classical Hutchinson-Barnsley theorem ([1],[2]) states that if $X$ is a complete metric space and $\mathcal{F}$ is an IFS then there exists a unique $A_{\mathcal{F}}\in\mathcal{K}(X)$ such that

[TABLE]

The set $A_{\mathcal{F}}$ is called the fractal or the attractor of the IFS $\mathcal{F}$. Sometimes it is also called self-similar set.

In 2008 some interesting variation on the notion of an IFS was considered. More precisely, Mihail and Miculescu (see [3], [4], [5]) introduced the notion of a generalized iterated function systems (GIFSs in short), in which they considered families of contractions defined on a finite Cartesian product $X^{m}$ (equipped with the maximum metric) of a metric space $X$ with its values in $X$. They obtained a counterpart of the H-B theorem which states that if $X$ is complete and $\mathcal{G}:=\{g_{1},...,g_{n}\}$ is a GIFS, then there exists a unique $A_{\mathcal{G}}\in\mathcal{K}(X)$ such that

[TABLE]

$A_{\mathcal{G}}$ is called the fractal or the attractor of a GIFS $\mathcal{G}$. (1) shows that GIFSs’ fractals are also self-similar, but in more complex way.

Another step in an attempt to describe sets in the language of self-similarity was done by Secelean who considered families of mappings defined on the space $\ell_{\infty}(X)$ of all bounded sequences in $X$ (endowed with supremum metric) with values in $X$. A finite family of contractions from $\ell_{\infty}(X)$ to $X$ we will call a generalized iterated function system of order infinity (GIFS∞ in short). With some extra, technical assumption (C1) (which we will discuss later) he obtained some counterpart of the H-B theorem, namely: If $X$ is complete and $\mathcal{S}$ is a GIFS∞ satisfying (C1), then there exists a unique $A_{\mathcal{S}}\in\mathcal{K}(X)$ such that

[TABLE]

We call $A_{\mathcal{S}}$ the fractal or the attractor of GIFS∞ $\mathcal{S}$. The self-similarity of GIFSs∞’ fractals is even more advanced and may not be completely precise (sometimes we can omit taking closures in (2)).

To obtain his result, Secelean used some iteration procedure which is not a very natural counterpart of the one used by Mihail and Miculescu. Trying to receive such a natural counterpart, Strobin and the author considered the waged supremum metric $d_{q}$ on $\ell_{\infty}(X)$ defined for $q\leq 1$ by

[TABLE]

It appears that under some more rigorous conditions, attractors of GIFSs∞’ have better properties (see [6]).

In 2003 Balka and Máthé ([7], [8]) with the use of generalized Hausdorff measure proved that in uncountable polish spaces, the typical compact set is not an attractor of any weak IFS. Also, using different techniques D’Aniello and Steele ([9]) proved the same claim for Euclidean spaces. It is also worth underlying that in Hilbert spaces a typical compact set is not even an attractor of any GIFS (see [10, Theorem 7.1]). We will modify slightly a construction of Balka and Máthé to prove that the typical (in the Baire category sense) compact set can be perceived as self-similar in the sense of Secelean (i.e. it satisfies (2) for some GIFS∞) and that in Euclidean spaces it is the attractor of some GIFS∞.

It is worth noting that the topological properties of fractals are investigated from many points of view. In particular there appears a notion of a topological IFS fractal (see [11], [12]; in [11] it is called a self similar set), i.e. a compact Hausdorff space $X$ such that for some finite family $\mathcal{F}$ of continuous selfmaps of $X$, $X=\bigcup_{f\in\mathcal{F}}f(X)$ and for every sequence $(f_{k})\subset\mathcal{F}$, the set

[TABLE]

(called sometimes a fibre), is singleton. As was proved in [13] and [14], $X$ is a topological IFS fractal iff $X$ is homeomorphic to the attractor of some weak IFS (in particular, it is metrizable).

In the next section we recall the frameworks of generalized IFSs – GIFSs and GIFSs∞. In particular we explain the (C1) condition. We recall counterparts of H-B theorem obtained by the mentioned authors. Then in Section 3 we introduce some basic notions and definitions. Section 4 is devoted to introducing a slight modification of Balka and Máthé construction of so called balanced set. We also gather results obtained by them. Then, in Section 5, we prove the main theorem which states that in the Polish spaces a typical compact set is self-similar in the sense of Secelean. Also we prove that in Euclidean spaces it is GIFS∞’ fractal. Finally, in the Appendix we correct a proof given by Balka and Máthé in [8] (the theorem, see Theorem 4.7, states that in Polish spaces a typical compact set is either finite or it is a union of balanced set and finite set).

2 Generalizations of notion of iterated function systems

2.1 Generalized iterated function systems of finite orders

Let $m\in\mathbb{N}$ and $X^{m}$ be the Cartesian product of $m$ copies of $X$, endowed with the maximum metric. A finite family $\mathcal{G}=\{g_{1},...,g_{n}\}$ of contractions $g_{i}:X^{m}\to X$ is called a generalized iterated function systems of order $m$ (GIFS in short). Miculescu and Mihail in [3] and [5] (also [4] for the case of compact $X$) proved the following version of the H-B theorem:

Theorem 2.1**.**

Let $X$ be a complete metric space, $m\in\mathbb{N}$ and $\mathcal{G}=\{g_{1},...,g_{n}\}$ be a GIFS of order $m$. Then there is a unique set $A_{\mathcal{G}}\in\mathcal{K}(X)$ such that

[TABLE]

The set $A_{\mathcal{G}}$ is called the fractal or the attractor of GIFS $\mathcal{G}$.

This result was extended by Strobin and Swaczyna in [15] to mappings which satisfy weaker contractive conditions. The results from articles [5, Example 4.3] and [16] (a construction of a Cantor-type GIFS’ fractal which is not a fractal of any IFS) point out that class of GIFSs’ fractals is essentially wider than class of IFSs’ fractals. Moreover results from [10] combined with [17] shows that there are some GIFSs’ fractals which are not fractals of any weak IFS.

2.2 Generalized iterated function systems defined on the $\ell_{\infty}$-sum – Secelean’s approach

Secelean in [18] considered mappings defined on $\ell_{\infty}$-sum of a space $X$ (mainly with supremum metric) with values in $X$. We will restrict our attention to contractions (although Secelean dealt with more general contractive conditions). Given a metric space $(X,d)$, let $\ell_{\infty}(X)$ be the $\ell_{\infty}$-sum of $X$, i.e.,

[TABLE]

We endow $\ell_{\infty}(X)$ with supremum metric $d_{1}$.

Remark 2.2**.**

Clearly, if $X$ is bounded then $\ell_{\infty}(X)$ is just the Cartesian product.

---

Following [18], we say that $f:\ell_{\infty}(X)\to X$ satisfies the condition (C1) if

$(C1)\;\;\;$ for every $(K_{k})\in\ell_{\infty}(\mathcal{K}(X))$, the closure of the image of the product $\overline{f(\prod_{k=1}^{\infty}K_{k})}\in\mathcal{K}(X)$,

and condition (C2), if

$(C2)\;\;\;$ for every $(K_{k})\in\ell_{\infty}(\mathcal{K}(X))$, the image of the product ${f(\prod_{k=1}^{\infty}K_{k})}\in\mathcal{K}(X).\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;$

Remark 2.3**.**

Observe that since $(K_{k})\in\ell_{\infty}(\mathcal{K}(X))$ the sum $\bigcup_{k=1}^{\infty}K_{k}\subset X$ is bounded and hence $f\left(\prod_{k=1}^{\infty}K_{k}\right)$ is well defined. Notice also that the product $\prod_{k=1}^{\infty}K_{k}$ may not be compact and hence, even if $f$ is continuous, $\overline{f\left(\prod_{k=1}^{\infty}K_{k}\right)}$ may not be compact too.

---

Eventually, if $\mathcal{S}=\{f_{1},...,f_{n}\}$ is a family of contractions $f_{i}:\ell_{\infty}(X)\to X$ which satisfy (at least) (C1) condition, then we say that $\mathcal{S}$ is a generalized iterated function system of infinite order (GIFS∞ in short).

Remark 2.4**.**

Notice that in the case when $X$ is an Euclidean space, any GIFS∞ fulfills (C1). Let $f:\ell_{\infty}(\mathbb{R}^{n})\to\mathbb{R}^{n}$ be a contraction. If $(K_{k})\in\ell_{\infty}(\mathcal{K}(\mathbb{R}^{n}))$ then $\bigcup_{k=1}^{\infty}K_{k}\subset\mathbb{R}^{n}$ is bounded and for any $x,y\in\prod_{k=1}^{\infty}K_{k}$ we have*

[TABLE]

*Hence $f\left(K_{1}\times K_{2}\times...\right)$ is bounded in $\mathbb{R}^{n}$ and $\overline{f\left(K_{1}\times K_{2}\times...\right)}\in\mathcal{K}(\mathbb{R}^{n})$.

---

Secelean obtained the following counterpart of H-B Theorem:

Theorem 2.5**.**

Let $X$ be a complete metric space and $\mathcal{S}=\{f_{1},...,f_{n}\}$ be a GIFS∞. Then there is a unique $A_{\mathcal{S}}\in\mathcal{K}(X)$ such that

[TABLE]

The set $A_{\mathcal{S}}$ is called the fractal or the attractor of GIFS∞ $\mathcal{S}$.

3 Basic definitions

Let $(X,d)$ be a metric space and $A,B\subset X$. By $\mathcal{K}(A)$ we denote the subspace of $\mathcal{K}(X)$ which consists of nonempty, compact subsets of $A$. By $\operatorname{Int}A$ and $\overline{A}$ we denote accordingly the interior and the closure of $A$. By $\operatorname{diam}(A)$ we denote the diameter of $A$ (we set $\operatorname{diam}(\emptyset)=0$). For $r>0$ we define $r$-neighbourhood of $A$ by $B(A,r):=\left\{x\in X:\exists_{a\in A}\ d(a,x)<r\right\}$. The distance between $A$ and $B$ is defined by $\operatorname{dist}(A,B):=\inf\{d(a,b):a\in A,b\in B\}$. We say that $A$ is nowhere dense if $\operatorname{Int}\overline{A}=\emptyset$ and that it is meager if it is a countable union of nowhere dense sets. A set is co-meager if its complement is meager. Note that a set is co-meager iff it contains a dense $G_{\delta}$ set (i.e. a dense set which is a countable intersection of open sets). We call $A$ perfect if it is closed and does not contain isolated points. We say that $X$ is a Polish metric space if it is separable and complete. Cantor-Bendixson Theorem ([19, Theorem 6.4]) states that a Polish metric space $X$ is a disjoint union of perfect set $P$ and countable open set $U$. When $X$ is complete, we say that a typical element of $X$ has property $\mathcal{P}$ if the set $\{x\in X:x\textrm{ has property }\mathcal{P}\}$ is co-meager.

Let $\delta>0$ and $A\subset X$. We say that a family $\{U_{i}\}_{i\in I}$ of subsets of $X$ is a $\delta$-cover of $A$ if $I$ is countable, $A\subset\bigcup_{i\in I}U_{i}$ and $\operatorname{diam}(U_{i})\leq\delta$ for $i\in I$. We call a function $h:[0,\infty)\to[0,\infty)$ a gauge function if it is nondecreasing, right-continuous and $h(x)=0$ iff $x=0$. If $h$ is a gauge function then we set

[TABLE]

Then we put

[TABLE]

$\mathcal{H}^{h}(A)$ is called $h$-Hausdorff measure of $A$. If $0<\mathcal{H}^{h}(A)<\infty$ then we say that $A$ is $h$-visible.

Let us denote by $\mathbb{N}^{\leq n}$ a family of finite sequences of natural numbers of length at most $n$, by $\mathbb{N}^{<\omega}$ a family of all finite sequences of natural numbers. Also by $2\mathbb{N}+1$ denote the set of odd numbers.

4 Typical compact set in polish spaces

In [7], [8] Balka and Máthé introduced a construction of ”balanced sets”. We will present this construction with a slight modification which will be crucial for concluding about self-similarity of typical compact sets. During this section we will comment if any significant change in the original proofs should be concerned. We begin with a definition of a $q$-balanced set.

Definition 4.1**.**

Let $(a_{n})\subset\mathbb{N}$. For any $n\in\mathbb{N}$, define*

[TABLE]

(i.e. $\mathcal{I}_{n}$ is a set of sequences of length $n$ with $i$-th element from $\{1,...,a_{i}\}$) and put

[TABLE]

*We say that $\Phi:2\mathbb{N}+1\to\mathcal{I}$ is an indexing function according to sequence $(a_{n})$ if $\Phi$ is surjective and $\Phi(n)\in\bigcup_{i=1}^{n}\mathcal{I}_{i}$ for any odd $n$.

---

Definition 4.2**.**

Let $X$ be a complete metric space and $q\geq 2$. We will say that a compact set $K\subset X$ is $q$-balanced if*

[TABLE]

where $(a_{n})\subset\mathbb{N}$ is such that:

• (i)

$a_{1}\geq 2$ and $a_{n+1}\geq na_{1}...a_{n}$,

$C_{i_{1},...,i_{n}}$* are nonempty, closed and satisfy*

• (ii)

$C_{i_{1},...,i_{n},i_{n+1}}\subset C_{i_{1},...,i_{n}}$ for any $(i_{1},...,i_{n})\in\mathcal{I}_{n}$ and $i_{n+1}\in\{1,...,a_{n+1}\}$,

and there exist a sequence $(b_{n})\subset[0,\infty)$ and an indexing function $\Phi:2\mathbb{N}+1\to\mathcal{I}$ according to sequence $(a_{n})$ such that for any $n\in\mathbb{N}$ and $(i_{1},...,i_{n}),(j_{1},...,j_{n})\in\mathcal{I}_{n}$:

• (iii)

$\operatorname{diam}\left(C_{i_{1},...,i_{n}}\right)\leq b_{n}$;

• (iv)

$\operatorname{dist}(C_{i_{1},...,i_{n}},C_{j_{1},...,j_{n}})>qb_{n}$ whenever $(i_{1},...,i_{n})\neq(j_{1},...,j_{n})$;

• (v)

If $n$ is odd and $C_{i_{1},...,i_{n}}\subset C_{\Phi(n)}$, and $C_{j_{1},...,j_{n}}\not\subset C_{\Phi(n)}$ then for any different $s,t\in\{1,...,a_{n+1}\}$

[TABLE]

Remark 4.3**.**

The mentioned modification bases on concerning $q\geq 2$ factor in Definition 4.2(iv). Since this change does not affect the idea of proofs from [7], [8], we will just comment where they can be precisely found.

---

Remark 4.4**.**

Notice that in countable complete metric space there is no $q$-balanced set (since any such set has $\mathfrak{c}$ cardinality).

---

Theorem 4.5**.**

Let $X$ be an uncountable Polish metric space and $q\geq 2$. There exists $q$-balanced set $K\subset X$.

Proof.

[7, Theorem 3.5] ∎

Theorem 4.6**.**

Let $X$ be an uncountable Polish metric space and let $K$ be a $q$-balanced set. Then there exists continuous gauge function $h$ such that $K$ is $h$-visible and for any weak contraction $f:K\to X$

[TABLE]

Proof.

[7, Theorem 5.1] ∎

Theorem 4.7**.**

*Let $X$ be a Polish metric space. Typical compact set $K\subset X$ is either finite or it is a union of finite set and a $q$-balanced set (for any $q\geq 2$).

Additionally, if $X$ is perfect then the typical compact set is a $q$-balanced set (for any $q\geq 2$).*

Remark 4.8**.**

In the original proof of this theorem ([8, Theorem 4.5]) there is a flaw. In Section 6 we will first present a counterexample for the incorrect statement in the proof and then we will correct the flaw.

---

As a consequence the following holds

Corollary 4.9**.**

Let $X$ be Polish metric space. Typical compact set $K\subset X$ is either finite, or there exists continuous gauge function $h$ such that $K$ is $h$-visible and for any weak contraction $f:K\to X$, $\mathcal{H}^{h}(K\cap f(K))=0$.

5 Balanced sets as attractors of GIFSs∞

We will now study mentioned notions of self-similarity of typical compact sets in Polish metric spaces. Directly from the results of Balka and Máthé we get

Theorem 5.1**.**

Let $X$ be Polish metric space. Then a typical compact set is either finite or it is not a fractal of any weak IFS.

Proof.

If $X$ is countable, then a typical compact set is finite. Assume that $X$ is uncountable. Then by Corollary 4.9 a typical compact set $K$ is either finite or there exists continuous gauge function $h$ such that $K$ is $h$-visible and for any weak contraction $f:K\to X$, $\mathcal{H}^{h}(K\cap f(K))=0$. Suppose that $K$ is infinite and take $h$ from the second part of statement. Then $\mathcal{H}^{h}(K)>0$. Suppose that there exists a weak IFS $\mathcal{F}=\{f_{1},...,f_{n}\}$ such that $K$ is its fractal, i.e. $K=\bigcup_{i=1}^{n}f_{i}(K)$. We have

[TABLE]

which is a contradiction. Hence $K$ is not a fractal of any weak IFS. ∎

Remark 5.2**.**

Note that any $q$-balanced set is a Cantor set and hence it is homeomorphic to ternary Cantor set which is clearly an IFS fractal. Moreover a union of topological fractal and a finite set is a topological fractal. Therefore in Polish metric spaces a typical compact set is a topological fractal.

---

Definition 5.3**.**

We say that $A\in\mathcal{K}(X)$ is a generalized fractal if there exists GIFS∞ $\mathcal{F}:=\{f_{1},...,f_{n}\}$ defined on $\ell_{\infty}(A)$ such that $f_{i}:\ell_{\infty}(A)\to X$ are contractions and*

[TABLE]

*We will say that $\mathcal{F}$ witnesses that $A$ is a generalized fractal (shortly that $\mathcal{F}$ witnesses).

---

Remark 5.4**.**

Observe that if $A\in\mathcal{K}(X)$ and there exist continuous mappings $f_{i}:\ell_{\infty}(A)\to A$ such that $A=\bigcup_{i=1}^{n}f_{i}(A\times A\times\ldots)$, then each $f_{i}$ fulfills (C1).

(For any $(K_{k})\in\ell_{\infty}(\mathcal{K}(A))$, $f_{i}(K_{1}\times K_{2}\times...)\subset A$ and so $\overline{f_{i}(K_{1}\times K_{2}\times...)}\in\mathcal{K}(X)$.)

---

We will now show that a typical compact set in Polish metric spaces is a generalized fractal. In correspondence with Theorem 5.1 this result shows that by considering GIFSs∞ we can describe significantly more sets than using only IFS theory (unless we decide to change metric – compare Remark 5.2 and notes on topological fractals in the Introduction).

Theorem 5.5**.**

Let $X$ be a Polish metric space and let $q\geq 2$. Each $q$-balanced set $K\subset X$ is a generalized fractal and a GIFS∞ $\mathcal{F}$ witnessing this is such that $\operatorname{Lip}(f)\leq\frac{1}{q}$ for $f\in\mathcal{F}$.

Proof.

Let $K$ be $q$-balanced set and $(a_{n}),(b_{n})$ be proper sequences and $C_{i_{1},...,i_{n}}$ proper sets from the definition of $K$ (Definition 4.2). We will define mappings from $\ell_{\infty}(K)$ to $X$. Since $K$ is bounded, $\ell_{\infty}(K)=\prod_{i=1}^{\infty}K$. $K$ can be written as

[TABLE]

where for any $(i_{1},i_{2},...)\in\prod_{i=1}^{\infty}\{1,...,a_{i}\}$, $x_{(i_{1},i_{2},...)}$ is the unique element of $\bigcap_{n\in\mathbb{N}}C_{i_{1},...,i_{n}}$.

Define $\mathcal{A}:=\prod_{i=1}^{\infty}\{1,...,a_{i}\}$. For any $i\in\{1,...,a_{1}\}$ define mapping $f_{i}:\ell_{\infty}(K)\to K$ by

[TABLE]

where $x_{\alpha_{1}},x_{\alpha_{2}},...\in K$ and $\alpha_{j}=(\alpha_{j}^{(1)},\alpha_{j}^{(2)},...)\in\mathcal{A}$ for any $j\in\mathbb{N}$ and

[TABLE]

where $\beta_{j}:=1+\sum_{i=1}^{a_{j+1}-1}\tilde{\alpha}_{i}^{(j)}$ for $j\geq 1$ and $\tilde{\alpha}_{i}^{(j)}:=\alpha_{i}^{(j)}\operatorname{mod}2$ for $i,j\in\mathbb{N}$.

Fix $i\in\{1,...,a_{1}\}$. We now show that $\operatorname{Lip}_{d}(f_{i})\leq\frac{1}{q}$. Let $(x_{\alpha_{1}},x_{\alpha_{2}},...)$, $(x_{\beta_{1}},x_{\beta_{2}},...)\in\prod_{i=1}^{\infty}K$. Assume that $f_{i}(x_{\alpha_{1}},x_{\alpha_{2}},...)\neq f_{i}(x_{\beta_{1}},x_{\beta_{2}},...)$.

[TABLE]

Let

[TABLE]

i.e. $\eta$ indicates the first coordinate on which $i(\alpha_{1},\alpha_{2},...)$ and $i(\beta_{1},\beta_{2},...)$ differs.

If $\eta=1$ we have $d\left(f_{i}(x_{\alpha_{1}},x_{\alpha_{2}},...),f_{i}(x_{\beta_{1}},x_{\beta_{2}},...)\right)\leq\operatorname{diam}\left(C_{i}\right)\leq b_{1}=b_{\eta}$. Otherwise

[TABLE]

We now estimate $d_{1}\left((x_{\alpha_{1}},x_{\alpha_{2}},...),(x_{\beta_{1}},x_{\beta_{2}},...)\right)$. For any $j\in\mathbb{N}$ define

[TABLE]

i.e. for each $j$, $\eta_{j}$ is the first coordinate on which $\alpha_{j}$ and $\beta_{j}$ differs. Denote also

[TABLE]

i.e. $l$ is the first coordinate which can impact on the difference between images of $(x_{\alpha_{1}},x_{\alpha_{2}},...)$ and $(x_{\beta_{1}},x_{\beta_{2}},...)$. Notice that $l\leq\eta$. (Since $\sum_{i=1}^{a_{\eta+1}-1}\tilde{\alpha}_{i}^{(\eta)}\neq\sum_{i=1}^{a_{\eta+1}-1}\tilde{\beta}_{i}^{(\eta)}$, there exists some $1\leq j\leq a_{\eta+1}-1$ such that $\alpha_{j}^{(\eta)}\neq\beta_{j}^{(\eta)}$. This implies $\eta_{j}\leq\eta$ and therefore $l\leq\eta$.) Additionally put

[TABLE]

Clearly $\eta_{h}\leq l$. We get

[TABLE]

Hence

[TABLE]

Therefore $\operatorname{Lip}(f_{i})\leq\frac{1}{q}$.

It is easy to see that $K=\bigcup_{i=1}^{a_{1}}f_{i}(K\times K\times...)$. Indeed, if $x_{(i_{1},i_{2},...)}\in K$ for some $(i_{1},i_{2},...)\in\mathcal{A}=\prod_{i=1}^{\infty}\{1,...,a_{i}\}$ then we can take sequence $\alpha=(\alpha_{1},\alpha_{2},...)\in\ell_{\infty}(\mathcal{A})$ such that for any $k\in\mathbb{N}$:

[TABLE]

and $f_{i_{1}}(x_{\alpha_{1}},x_{\alpha_{2}},...)=x_{(i_{1},i_{2},...)}$. By Remark 5.4, $\mathcal{F}:=\{f_{1},...,f_{a_{1}}\}$ fulfills (C1). ∎

Remark 5.6**.**

We can obtain somewhat stronger result – namely for any $r>0$, there exists a witnessing GIFS∞ $\mathcal{F}$ such that $\operatorname{Lip}(f)\leq r$ for $f\in\mathcal{F}$. We choose $p$ so that $q^{-p}<r$ and consider GIFS∞ $\left\{f_{(i_{1},...,i_{p})}:(i_{1},...,i_{p})\in\prod_{i=1}^{p}\{1,...,a_{i}\}\right\}$ given by (we preserve former notation)*

[TABLE]

*where $\beta_{j}:=1+\sum_{i=1}^{a_{j+p}-1}\tilde{\alpha}_{i}^{(j)}$ for $j\in\mathbb{N}$. Then $\operatorname{Lip}(f)\leq q^{-p}<r$ for any $f\in\mathcal{F}$.

---

Remark 5.7**.**

GIFS $\mathcal{F}$ defined in the proof of Theorem 5.5 (and also in Remark 5.6) is such that any $f_{i}\in\mathcal{F}$ fulfills (C2), i.e. for any $(K_{k})\in\ell_{\infty}(\mathcal{K}(K))$, $f_{i}(K_{1}\times K_{2}\times...)\in\mathcal{K}(K)$. It is enough to prove that for any $(K_{k})\in\ell_{\infty}(\mathcal{K}(K))$ the image $f_{i}(K_{1}\times K_{2}\times...)\subset K$ is closed.

Observe that $\left(x_{n}\right)=\left(x_{(i_{1}^{(n)},i_{2}^{(n)},...)}\right)_{n\in\mathbb{N}}\subset K$ converges to some $x_{(i_{1},i_{2},...)}\in K$ iff for any $k\in\mathbb{N}$ the sequence $(i_{k}^{(n)})_{n}\subset\{1,...,a_{k}\}$ is eventually constant (this follows from Definition 4.2(iv)).

Let $(K_{k})\in\ell_{\infty}(\mathcal{K}(K))$ and set $f:=f_{i}$. We will show that $f(K_{1}\times K_{2}\times...)\subset K$ is closed. Let $(x_{n})=\left(x_{(i,i_{2}^{(n)},...)}\right)\subset f(K_{1}\times K_{2}\times...)$ be convergent to some $x\in K$. For any $n\in\mathbb{N}$ there exists a sequence $\left(y_{1}^{(n)},y_{2}^{(n)},...\right)\in\prod_{i=1}^{\infty}K_{i}$ such that $f\left(y_{1}^{(n)},y_{2}^{(n)},...\right)=x_{n}$. Consider addresses of $y_{i}^{(n)}s$, i.e. sequences $\left(i_{1}^{[i,(n)]},i_{2}^{[i,(n)]},...\right)\in\prod_{i=1}^{\infty}\{1,...,a_{i}\}$ such that $y_{i}^{(n)}=x_{\left(i_{1}^{[i,(n)]},i_{2}^{[i,(n)]},...\right)}\in K$.

Notice that in a multiset $A_{1}:=\left\langle\left(i_{1}^{[1,(n)]},...,i_{1}^{[a_{2}-1,(n)]}\right):n\in\mathbb{N}\right\rangle$ (multiset of tuples of first coordinates of addresses of $y_{1}^{(n)},...,y_{a_{2}-1}^{(n)}$) some $(a_{2}-1)$-tuple appears infinitely many times (since a tuple $\left(i_{1}^{[1,(n)]},...,i_{1}^{[a_{2}-1,(n)]}\right)$ belongs to finite set $\{1,...,a_{1}\}^{a_{2}-1}$). Denote one of them by $\left(j_{1}^{(1)},...,j_{a_{2}-1}^{(1)}\right)$.

Now consider only these $n\in\mathbb{N}$ for which the sequence $\left(y_{1}^{(n)},y_{2}^{(n)},...\right)$ is such that for any $i=1,...,a_{2}-1$, $i_{1}^{[i,(n)]}=j_{i}^{(1)}$ (i.e. for $i=1,...,a_{2}-1$ the address of $y_{i}^{(n)}$ begins with $j_{i}^{(1)}$). Reenumerate obtained subsequence.

Assume that for some $k\in\mathbb{N}$ we have a sequence*

[TABLE]

and infinite number of sequences $\left(y_{1}^{(n)},y_{2}^{(n)},...\right)$ such that for any $i=1,...,a_{{k+1}-1}$ and any $n\in\mathbb{N}$, $\left(i_{1}^{[i,(n)]},...,i_{k}^{[i,(n)]}\right)=\left(j_{i}^{(1)},...,j_{i}^{(k)}\right)$ (i.e. for any $n\in\mathbb{N}$ and $i=1,...,a_{k+1}-1$, address of $y_{i}^{(n)}$ begins with $\left(j_{i}^{(1)},...,j_{i}^{(k)}\right)$). Consider a multiset

[TABLE]

*One of the $(a_{k+2}-1)$-tuples of $(k+1)$-tuples $(i_{1}^{[\cdot]},...,i_{k+1}^{[\cdot]})$ appears infinitely many times. We denote one of them by $\left((j_{1}^{(1)},...,j_{1}^{(k+1)}),...,(j_{a_{k+2}-1}^{(1)},...,j_{a_{k+2}-1}^{(k+1)})\right)$. From the original sequences we leave only these $\left(y_{1}^{(n)},y_{2}^{(n)},...\right)$ such that for any $i=1,...,a_{k+2}-1$ and any $n\in\mathbb{N}$, $\left(i_{1}^{[i,(n)]},...,i_{k+1}^{[i,(n)]}\right)=\left(j_{i}^{(1)},...,j_{i}^{(k+1)}\right)$. Then we reenumerate obtained subsequence.

Observe that during this procedure, the sequence of bigger and bigger tuples $(j_{i}^{(1)},...,j_{i}^{(k)})_{k}$ is growing consistently – in the sense that $j_{i}^{(1)},...,j_{i}^{(k)}$ are the first $k$-elements of $(j_{i}^{(1)},...,j_{i}^{(k+1)})$ for any $i,k\in\mathbb{N}$.

Eventually, observe that for any $n\in\mathbb{N}$ the set $\left(\bigcap_{i\in\mathbb{N}}C_{(j_{n}^{(1)},...,j_{n}^{(i)})}\right)\cap K_{n}$ is nonempty. Fix $n\in\mathbb{N}$. Clearly $\bigcap_{i\in\mathbb{N}}C_{(j_{n}^{(1)},...,j_{n}^{(i)})}=x_{(j_{n}^{(1)},j_{n}^{(2)},...)}$. Moreover $K_{n}$ contains $x_{(j_{n}^{(1)},j_{n}^{(2)},...)}$. This comes from the fact that among sequence $(y_{i}^{(n)})_{i}\subset K_{n}$ we can find $z_{1}$ whose address begins with $j_{n}^{(1)}$, then $z_{2}$ whose address begins with $(j_{n}^{(1)},j_{n}^{(2)})$ and so on. Hence we obtain sequence of elements $z_{i}\in K_{n}\cap C_{(j_{n}^{(1)},...,j_{n}^{(i)})},i\in\mathbb{N}$, which is convergent (since $\operatorname{diam}\left(C_{(j_{n}^{(1)},...,j_{n}^{(i)})}\right)\stackrel{{\scriptstyle i\to\infty}}{{\to}}0$). Let us denote this limit by $z^{(n)}$. Clearly, $z^{(n)}$ belongs to $C_{(j_{n}^{(1)},j_{n}^{(2)},...)}$ and its address is $(j_{n}^{(1)},j_{n}^{(2)},...)$. Also, since $K_{n}\in\mathcal{K}(K)$ is closed, $z^{(n)}$ belongs to $K_{n}$.

Finally, notice that $f\left(z^{(1)},z^{(2)},...\right)=x$. Let $(i,m_{1},m_{2},...)$ be the address of $x$. We need to ensure that for any $k\in\mathbb{N}$, $\sum_{i=1}^{a_{k+1}-1}j_{i}^{(k)}=m_{k}$. This follows from the fact that since $x_{n}\to x$, there are some moments from which the consecutive coordinates in addresses of $x_{n}$s and $x$ match. This means that sequences $\left(y_{1}^{(n)},y_{2}^{(n)},...\right)$ are such that their addresses fulfill $\sum_{i=1}^{a_{k+1}-1}i_{k}^{[i,(n)]}=m_{k}$ for $n$ big enough. This is exactly what $(j_{1}^{(k)},...,j_{a_{k+1}-1}^{(k)})$ fulfills (otherwise it could not appear infinitely many times during construction).

---

Corollary 5.8**.**

Let $X$ be a perfect Polish metric space. Typical compact set $K\subset X$ is a generalized fractal such that for any $r\in(0,1)$ there exists a GIFS∞ $\mathcal{F}$ witnessing this with $\operatorname{Lip}(f)\leq r$ for $f\in\mathcal{F}$.

Proof.

Combine Theorem 4.7 and Remark 5.6. ∎

In case of Euclidean spaces this result can be easily extended in order to guarantee that a typical compact set in $\mathbb{R}^{n}$ is GIFS∞’ fractal (we consider Euclidean metric on $\mathbb{R}^{n}$).

Proposition 5.9** ([20, Lemma 1.1]).**

Let $X$ be a metric space, $A\subset X$ and $f:A\to\mathbb{R}$ be such that $\operatorname{Lip}(f)<\infty$. Then there exists $\tilde{f}:X\to\mathbb{R}$ such that $\tilde{f}|_{A}=f$ and $\operatorname{Lip}(\tilde{f})=\operatorname{Lip}(f)$.

Corollary 5.10**.**

Let $n\in\mathbb{N}$, $X$ be a metric space, $A\subset X$ and $f:A\to\mathbb{R}^{n}$ be such that $\operatorname{Lip}(f)<\infty$. Then there exists $\tilde{f}:X\to\mathbb{R}^{n}$ such that $\tilde{f}|_{A}=f$ and $\operatorname{Lip}(\tilde{f})\leq\sqrt{n}\operatorname{Lip}(f)$.

Theorem 5.11**.**

Let $n\in\mathbb{N}$, $q\geq 2$ and $K\subset\mathbb{R}^{n}$ be a $q$-balanced set. For any $r\in(0,1)$, $K$ is a fractal of some GIFS∞ $\mathcal{F}$ such that mappings $f\in\mathcal{F}$ fulfill $\operatorname{Lip}(f)\leq r$.

Proof.

Let $r\in(0,1)$. By Remark 5.6, there exists GIFS∞ $\mathcal{F}:=\{f_{1},...,f_{p}\}$ defined on $\ell_{\infty}(K)$ witnessing that $K$ is a generalized fractal and such that $\operatorname{Lip}(f)\leq\frac{1}{\sqrt{n}}r$ for $i=1,...,p$.

From Corollary 5.10 for each $f_{i}$ there exists extension $\tilde{f}_{i}:\ell_{\infty}(\mathbb{R}^{n})\to\mathbb{R}^{n}$ with $\operatorname{Lip}(\tilde{f}_{i})\leq\sqrt{n}\operatorname{Lip}(f_{i})\leq r$. By Remark 2.4 each $\tilde{f}_{i}$ satisfies (C1). Hence $K$ is a fractal of the GIFS∞ $\{\tilde{f}_{1},...,\tilde{f}_{p}\}$. ∎

Corollary 5.12**.**

Typical compact set $K\subset\mathbb{R}^{n}$ is such that for any $r\in(0,1)$ it is a fractal of some GIFS∞ $\mathcal{F}$ with $\operatorname{Lip}(f)\leq r$ for $f\in\mathcal{F}$.

We now generalize Theorem 5.5 to cover case of sets which are union of balanced set and a finite set.

Theorem 5.13**.**

Let $X$ be a Polish metric space and $q\geq 2$. Let $R=K\cup P$ be a union of $q$-balanced set $K$ and finite set $P$ with $K\cap P=\emptyset$. Then $R$ is a generalized fractal.

Proof.

Let $\eta:=\operatorname{dist}(K,P)>0$ and $r\in(0,1)$. By Remark 5.6 there exists GIFS∞ $\mathcal{F}:=\{f_{1},...,f_{p}\}$ witnessing that $K$ is a generalized fractal (in particular $f_{i}$ are defined on $\ell_{\infty}(K)$) with $\operatorname{Lip}(f_{i})\leq r\cdot\min\left\{1,\frac{\eta}{\operatorname{diam}(K)}\right\}$. Fix some $x\in K$. For any $i=1,...,p$ define $\tilde{f}_{i}:\ell_{\infty}(R)\to R$ by

[TABLE]

Then $\operatorname{Lip}(\tilde{f}_{i})\leq\operatorname{Lip}{f_{i}}\cdot\max\left\{1,\frac{\operatorname{diam}(K)}{\eta}\right\}\leq r$. Indeed, let $(x_{1},x_{2},...),(y_{1},y_{2},...)\in\ell_{\infty}(R)$. Consider cases:

$(x_{1},x_{2},...),(y_{1},y_{2},...)\in\ell_{\infty}(K)$. Then

[TABLE]

• 2)

$(x_{1},x_{2},...),(y_{1},y_{2},...)\notin\ell_{\infty}(K)$. Then $d\left(\tilde{f}_{i}(x_{1},x_{2},...),\tilde{f}_{i}(y_{1},y_{2},...)\right)=0$.

• 3)

$(x_{1},x_{2},...)\in\ell_{\infty}(K)$ and $(y_{1},y_{2},...)\notin\ell_{\infty}(K)$. Since $(y_{1},y_{2},...)\notin\ell_{\infty}(K)$ there exists $k\in\mathbb{N}$ such that $y_{k}\in P$. Then

[TABLE]

Additionally, for any $x\in P$ define constant mappings $h_{x}:\ell_{\infty}(R)\to R$ by $h_{x}\equiv x$. Then $\tilde{F}:=\mathcal{F}\cup\{h_{x}:x\in P\}$ is a GIFS∞ defined on $\ell_{\infty}(R)$ and its mappings are such that $\operatorname{Lip}(g)\leq r$ for any $g\in\tilde{F}$. Moreover

[TABLE]

Therefore $R$ is a generalized fractal and $\tilde{F}$ witnesses this. ∎

Corollary 5.14**.**

Let $X$ be a Polish metric space. Typical compact set $K\subset X$ is a generalized fractal such that for any $r\in(0,1)$ there exists a GIFS∞ $\mathcal{F}$ witnessing this with $\operatorname{Lip}(f)\leq r$ for $f\in\mathcal{F}$.

6 Appendix: correction of the proof of Theorem 4.7 ([8, Theorem 4.5])

We will correct proof of the following theorem

See 4.7

First we shall see where is the flaw. Let $X^{*}\subset X$ be a perfect subset of $X$ such that $X\setminus X^{*}$ is countable open ($X^{*}$ exists by Cantor-Bendixon Theorem). From the first part of proof (when considering perfect space) one get a family $\mathcal{F}^{*}$ which is dense $G_{\delta}$ in $\mathcal{K}(X^{*})$ and which consists of $q$-balanced sets. In order to show that the typical compact set is a union of finite set and $q$-balanced set the authors consider the mapping $R:\mathcal{K}(X)\to\mathcal{K}(X^{*})\cup\{\emptyset\}$ given by

[TABLE]

(put $H^{d}(\emptyset,A)=1$ whenever $A\in\mathcal{K}(X^{*})$; then $\emptyset$ is an isolated point in $\mathcal{K}(X^{*})\cup\{\emptyset\}$).

They show that $R$ is open and claim that $\tilde{\mathcal{F}}:=R^{-1}(\mathcal{F}^{*}\cup\{\emptyset\})$ is dense $G_{\delta}$ in $\mathcal{K}(X)$. However, it may not be the case since $R$ is not continuous and $\tilde{\mathcal{F}}$ may not be $G_{\delta}$. Consider following example. (We will repair the flaw afterward).

Example 6.1**.**

Let*

[TABLE]

*Then $X=X^{*}\cup S$ where $X^{*}=\{0\}\times([0,1]\cup[2,3])$ and $S=X\setminus X^{*}$ is a set of isolated points. $R$ is not continuous since the sequence $K_{n}:=\{(0,0)\}\cup\left\{\left(\frac{1}{n},\frac{i}{n}\right):i=0,...,n\right\}$ tends to $K:=\{0\}\times[0,1]$, but $R(K_{n})=\{(0,0)\}$ and $R(K)=K$.

---

Proof.

(of Theorem 4.7): The first assertion of the proof (when perfect space is considered) holds ([8, Theorem 4.5]).

Let now $X$ be a separable, complete metric space, $X^{*}\subset X$ be a perfect subset of $X$ such that $U:=X\setminus X^{*}$ is countable open. By $S$ let us denote the set of isolated points of $X$. Then $S\subset U$, $S$ is open and $U\subset\overline{S}$. Since $S$ is open and dense in $\overline{S}$, $\mathcal{K}(S)$ is open and dense in $\mathcal{K}(\overline{S})$. Moreover, any element of $\mathcal{K}(S)$ is finite.

From the first part of proof there exists a family $\mathcal{F}^{*}$ which is a dense $G_{\delta}$ set in $\mathcal{K}(X^{*})$ and which consists of $q$-balanced sets.

Let

[TABLE]

Then any $K\in\mathcal{F}$ is a union of emptyset or $q$-balanced set and finite set consisting of isolated points. We will show that $\mathcal{F}$ contains a dense $G_{\delta}$ set.

If $X^{*}\setminus\overline{S}=\emptyset$ then $\mathcal{K}(S)$ is open and dense in $\mathcal{K}(\overline{S})=\mathcal{K}(X)$, so $\mathcal{K}(S)$ is comeager in $\mathcal{K}(X)$.

Assume that $X^{*}\setminus\overline{S}\neq\emptyset$. There are dense and open $\mathcal{F}_{n}\subset\mathcal{K}(X^{*})$ such that $\mathcal{F}^{*}=\bigcap_{n\in\mathbb{N}}\mathcal{F}_{n}$. For any $n\in\mathbb{N}$ define

[TABLE]

Let $K\in\bigcap_{n\in\mathbb{N}}\mathcal{F}_{n}^{\prime}$. Then, since $K\in\mathcal{F}_{1}^{\prime}$, $K$ is nonempty and $K=A\cup B$ where $A\in\mathcal{F}_{1}\cap\mathcal{K}(X^{*}\setminus\overline{S}))\cup\{\emptyset\}$ and $B\in\mathcal{K}(S)\cup\{\emptyset\}$. If $A=\emptyset$ then $K=B\in\mathcal{K}(S)\subset\mathcal{F}$. Assume that $A\neq\emptyset$. Then $K\in\bigcap_{n\in\mathbb{N}}\mathcal{F}_{n}^{\prime}$ implies $A\in\mathcal{F}_{n}$ for any $n\in\mathbb{N}$. Therefore $K\in\mathcal{F}$.

We now show that $\mathcal{F}_{n}^{\prime}$ are dense and open in $\mathcal{K}(X)$. Let $n\in\mathbb{N}$. We first show openness. Let $K\in\mathcal{F}_{n}^{\prime}$. Assume that $K=A\cup B$ where $A\in\mathcal{F}_{n}\cap\mathcal{K}(X^{*}\setminus\overline{S})$ and $B\in\mathcal{K}(S)$ (in rest of cases we conduct analogously). Since $A\in\mathcal{K}(X^{*}\setminus\overline{S})$ and $X^{*}\setminus\overline{S}$ is open, there exists $r_{1}>0$ such that $B(A,r_{1})\subset X^{*}\setminus\overline{S}$. Since $\mathcal{F}_{n}$ is open in $\mathcal{K}(X^{*})$, there exists $r_{2}>0$ such that $B_{H}(A,r_{2})\cap\mathcal{K}(X^{*})\subset\mathcal{F}_{n}$. For $B\in\mathcal{K}(S)$, which is a finite union of isolated points in $X$, there exists $r_{3}>0$ such that $B_{H}(B,r_{3})=\{B\}$ (in particular $B(B,r_{3})\subset B$). Then for $r:=\frac{1}{2}\min\{r_{1},r_{2},r_{3}\}$ we have

[TABLE]

Indeed, let $F\in B_{H}(A\cup B,r)$. Notice that $B(A\cup B,r)=B(A,r)\cup B(B,r)$. Hence, since $F\in B_{H}(A\cup B,r)$, we have $F\subset B(A\cup B,r)=B(A,r)\cup B$. Define $C:=F\cap B(A,r)$ and $D:=F\cap B$. Then $C,D$ are disjoint and $C\in\mathcal{K}(X^{*}\setminus\overline{S})$ and $D\in\mathcal{K}(S)\cup\{\emptyset\}$. It remains to show that $C\in\mathcal{F}_{n}$. Since $A\cup B\subset B(F,r)=B(C,r)\cup B(D,r)$ and

[TABLE]

we have $A\subset B(C,r)$. Therefore $H(A,C)<r$, so $C\in B_{H}(A,r)\cap\mathcal{K}(X^{*})\subset\mathcal{F}_{n}$. Eventually $F=C\cup D\in\mathcal{F}_{n}^{\prime}$.

We show denseness. Let $K\in\mathcal{K}(X)$ and $\varepsilon>0$. Since a family of finite sets is dense in $\mathcal{K}(X)$, there exists $L=\{x_{1},...,x_{k}\}\in\mathcal{K}(X)$ such that $H(K,L)<\frac{\varepsilon}{2}$. Consider $L_{1}:=L\cap(X^{*}\setminus\overline{S})$ and $L_{2}:=L\cap\overline{S}$. Assume that $L_{1},L_{2}$ are nonempty (other cases are analogous). There exists $r>0$ such that $B_{H}(L_{1},r)\subset\mathcal{K}(X^{*}\setminus\overline{S})$. Then $U:=B_{H}(L_{1},\min\left\{\frac{\varepsilon}{4},r\right\})\subset\mathcal{K}(X^{*}\setminus\overline{S})\subset\mathcal{K}(X^{*})$ and since $\mathcal{F}_{n}\subset\mathcal{K}(X^{*})$ is dense, therefore there is $A\in\mathcal{F}_{n}\cap U\subset\mathcal{F}_{n}\cap\mathcal{K}(X^{*}\setminus\overline{S})$. In particular $H(A,L_{1})<\frac{\varepsilon}{4}$. On the other hand, $L_{2}\subset\overline{S}$ is finite and for any $x\in L_{2}$ there exists $x^{\prime}\in S$ such that $d(x,x^{\prime})<\frac{\varepsilon}{4}$. Then $B:=\{x^{\prime}:x\in L_{2}\}\in\mathcal{K}(S)$ belongs to $B_{H}(L_{2},\frac{\varepsilon}{4})\subset\mathcal{K}(S)$. We have $A\cup B\in\mathcal{F}_{n}^{\prime}$ and

[TABLE]

∎

Acknowledgments

Research is supported by Narodowe Centrum Nauki, grant number 2017/26/A/ST1/00189.

The author would like to thank Filip Strobin for many discussions and valuable remarks.