# On the Continuity of the Hutchinson Operator

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## Abstract

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## 1. Introduction

## 2. Hyperspaces, Multifunctions, Iterated Function Systems

**Proposition 1.**Let $\phi :X\to \mathcal{K}\left(X\right)$ be an upper semicontinuous multifunction with compact values. Then the induced Hutchinson operator $F:\mathcal{P}\left(X\right)\to \mathcal{P}\left(X\right)$ transforms compacta into compacta. In particular the restriction $F:\mathcal{K}\left(X\right)\to \mathcal{K}\left(X\right)$ is well-defined and

**Proof.**It is well known that under our assumptions the image of a compact set is again compact ([11], (Proposition 6.2.11, p. 196)). ☐

## 3. Continuity on $\mathcal{K}\left(\mathbf{X}\right)$

**Proposition 2.**Let $K\subset X$. If $\phi :K\to \mathcal{K}\left(X\right)$ is uniformly continuous, then $F:\mathcal{K}\left(K\right)\to \mathcal{K}\left(X\right)$ is uniformly continuous too.

**Proof.**Fix $\epsilon >0$. Find $\delta >0$ such that for every pair ${x}_{1},{x}_{2}\in X$, $d({x}_{1},{x}_{2})<\delta $ implies ${d}_{H}(\phi \left({x}_{1}\right),\phi \left({x}_{2}\right))<\epsilon $.

**Theorem 1.**Let $\phi :X\to \mathcal{K}\left(X\right)$ be a continuous multifunction with compact values. Then the Hutchinson operator $F:\mathcal{K}\left(X\right)\to \mathcal{K}\left(X\right)$ induced by φ is continuous.

**Proof.**Let ${S}_{n},S\in \mathcal{K}\left(X\right)$, ${S}_{n}\to S$ with respect to ${d}_{H}$. Put $K:={\bigcup}_{n=1}^{\infty}{S}_{n}\cup S=\overline{{\bigcup}_{n=1}^{\infty}{S}_{n}}$. Of course $K\in \mathcal{K}\left(X\right)$. Since φ is continuous, it is uniformly continuous on K. Hence $F:\mathcal{K}\left(K\right)\to \mathcal{K}\left(X\right)$ is uniformly continuous by Proposition 2. This yields $F\left({S}_{n}\right)\to F\left(S\right)$. ☐

## 4. Lack of Continuity on ${\mathcal{F}}_{\mathbf{b}}\left(\mathbf{X}\right)$

**Theorem 2.**(Criterion of continuity of F). Let $f:X\to X$ map bounded sets onto bounded sets. Let $F:{\mathcal{F}}_{b}\left(X\right)\to {\mathcal{F}}_{b}\left(X\right)$ be the associated Hutchinson operator. The following are equivalent:

- (i)
- F is continuous,
- (ii)
- f is boundedly uniformly continuous.

**Proof.**The implication (ii) ⇒ (i) follows at once from Theorem 1. We shall prove (i) ⇒ (ii).

**Example 1.**Let X be an infinite dimensional normed space. Let $r>0$. Let ${x}_{n}$ be an r-separated sequence, i.e., $d({x}_{n},{x}_{m})\ge r$ for $n\ne m$, which is bounded. Moreover, let ${y}_{n}$ be a sequence disjoint from ${x}_{n}$ with the property that $d({x}_{n},{y}_{n})<r/3n$ for all n. Such sequences ${x}_{n}$, ${y}_{n}$ always exist. The sets $S:={\left\{{x}_{n}\right\}}_{n=1}^{\infty}$, $Y:={\left\{{y}_{n}\right\}}_{n=1}^{\infty}$ are discrete closed subsets of X. Put $f\left(x\right):={x}_{1}$ for $x\in S$, $f\left(y\right):={y}_{1}$ for $y\in Y$. Since $S\cup Y$ is discrete, the function $f:S\cup Y\to X$ is continuous. Since $S\cup Y$ is closed, the Dugundji-Tietze theorem ([14], (1.3, p. 2)) yields an extension of f on the whole X. Recapitulating: $d({x}_{n},{y}_{n})\to 0$, $d(f\left({x}_{n}\right),f\left({y}_{n}\right))=d({x}_{1},{y}_{1})>0$. We see that f is not boundedly uniformly continuous. Consequently, the associated operator F cannot be continuous on ${\mathcal{F}}_{b}\left(X\right)$.

## 5. Attractors and Continuity

**Proposition 3.**A strict attractor A of the IFS given by an upper semicontinuous multifunction $\phi :X\to \mathcal{P}\left(X\right)$ is invariant, i.e., $F\left(A\right)=A$, where $F:\mathcal{P}\left(X\right)\to \mathcal{P}\left(X\right)$ is the Hutchinson operator generated by φ.

**Proof.**Fix $\epsilon >0$. By Proposition 2 in [15] we know that for some $\delta >0$

## 6. Vietoris Continuity

**Theorem 3.**Let X be a normal topological space. Let $\phi :X\to \mathcal{F}\left(X\right)$ be a Vietoris continuous multifunction. Then the associated Hutchinson operator $F:\mathcal{F}\left(X\right)\to \mathcal{F}\left(X\right)$ is Vietoris continuous.

**Proof.**Let $V\subset X$ be open and $S\in \mathcal{F}\left(X\right)$ such that $F\left(S\right)\in {V}^{+}$. Let us shrink V to an open set W such that $F\left(S\right)\subset W$, $\overline{W}\subset V$. By Vietoris continuity of φ for each $s\in S$ there exists an open ${U}_{s}\ni s$ such that $\phi \left({U}_{s}\right)\subset W$. Put $U:={\bigcup}_{s\in S}{U}_{s}$. Then $S\in {U}^{+}$ and U is open. We have to check that $F\left(C\right)\in {V}^{+}$ for all $C\in {U}^{+}$. Indeed, if $C\subset U$, then

**Theorem 4**(Kieninger)

**.**Let X be a Hausdorff topological space. Let $\phi :X\to \mathcal{K}\left(X\right)$ be a Vietoris continuous multifunction with compact values. Then the Hutchinson operator $F:\mathcal{K}\left(X\right)\to \mathcal{K}\left(X\right)$ induced by φ is Vietoris continuous.

## 7. Infinite Systems

**Theorem 5.**Let X be a normal topological space and I be compact. Let $\Phi :I\times X\to \mathcal{K}\left(X\right)$ be Vietoris continuous. Then the induced Hutchinson operator $F:\mathcal{K}\left(X\right)\to \mathcal{K}\left(X\right)$ is Vietoris continuous.

**Proof.**Denote ${\phi}_{i}:X\to \mathcal{K}\left(X\right)$, ${\phi}_{i}\left(x\right):=\Phi (i,x)$ for $x\in X$, $i\in I$. We will show that $\phi :X\to \mathcal{K}\left(X\right)$, $\phi \left(x\right):={\bigcup}_{i\in I}{\phi}_{i}\left(x\right)=\Phi (I\times \left\{x\right\})$ for $x\in X$, is Vietoris continuous. Then one calls Theorem 3 to finish the proof. From the continuity of ${\phi}_{i}$ we know that there exists an open $U\ni x$ such that $V\cap {\phi}_{i}\left(u\right)\ne \varnothing $ for all $u\in U$. In particular, $V\cap \phi \left(u\right)\ne \varnothing $ for $u\in U$.

**Theorem 6.**Let X be a normal k-space. Let ${\left\{{\phi}_{i}\right\}}_{i\in I}\subset \mathcal{M}(X,X)$ be a collection of multifunctions such that

- (a)
- each ${\phi}_{i}$ is Vietoris continuous and has compact values,
- (b)
- the entire ${\left\{{\phi}_{i}\right\}}_{i\in I}$ is compact with respect to the compact-open topology.

- (i)
- φ constitutes a Vietoris continuous multifunction with compact values,
- (ii)
- the Hutchinson operator $F:\mathcal{K}\left(X\right)\to \mathcal{K}\left(X\right)$ corresponding to φ is Vietoris continuous.

**Proof.**Let us identify I with ${\left\{{\phi}_{i}\right\}}_{i\in I}$. We topologize I by pulling the compact-open topology from ${\left\{{\phi}_{i}\right\}}_{i\in I}$. Define the evaluation mapping $\Phi :I\times X\to \mathcal{K}\left(X\right)$ by $\Phi (i,x):={\phi}_{i}\left(x\right)$ where ${\phi}_{i}$ corresponds to i via identification. Since I is compact, the evaluation is continuous; cf. [19] (2.6.11, p. 110; 3.4.3, p. 158; 3.4.20, p. 163). Thus, we are in position to apply Theorem 5 and complete the proof. ☐

## Acknowledgments

## Author Contributions

## Conflicts of Interest

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Barnsley, M.F.; Leśniak, K.
On the Continuity of the Hutchinson Operator. *Symmetry* **2015**, *7*, 1831-1840.
https://doi.org/10.3390/sym7041831

**AMA Style**

Barnsley MF, Leśniak K.
On the Continuity of the Hutchinson Operator. *Symmetry*. 2015; 7(4):1831-1840.
https://doi.org/10.3390/sym7041831

**Chicago/Turabian Style**

Barnsley, Michael F., and Krzysztof Leśniak.
2015. "On the Continuity of the Hutchinson Operator" *Symmetry* 7, no. 4: 1831-1840.
https://doi.org/10.3390/sym7041831