# Pointless characterization relating between a fractal and its code space

Given an hyperbolic IFS $$(X,\{f_i:i=1,\ldots,N\})$$ and denoting its code space by $$\Sigma_N = \{1,\ldots,N\}^{\mathbb{N}}$$ and the generated fractal set by $$\mathcal{A}$$.

There is a continuous and surjective mapping $$\gamma: \Sigma_N \to \mathcal{A}$$ given by $$\gamma(\sigma) = \lim\limits_{n \to \infty} f_{\sigma(n)}(x)$$ where $$x$$ can be chosen arbitrarily in $$X$$.

Denoting for $$f:X \to X$$ by $$\overline{f}:H(X) \to H(X)$$ the function $$\overline{f}(A) = \{f(a):a \in A\}$$ where $$H(X)$$ is the hyperspace of compact subsets of $$X$$. The author makes the following remark:

Suppose that $$\sigma \in \Sigma_N$$ and let $$A_{\sigma(n)} = \overline{f}_{\sigma(n)}(A)$$ for $$A \in H(X)$$. Then the above theorem states that $$\gamma(\sigma) = \bigcap\limits_{n \in \mathbb{N}} A_{\sigma(n)}$$

How can I show this remark is true?

My try

$$f(X) \subset X$$. This is because $$f:X \to X$$. The strict inequality follows if one takes $$x_0,y_0 \in X$$ such that $$diam(X) = d(x_0,y_0)$$. It is clear that $$x_0 \in f(X)$$ and $$y_0 \in f(X)$$ cannot happen simultaneously, since the diameter decreases strictly:

$$diam(f(X)) = \sup\{d(x,y).x,y \in f(X)\} = \sup\{d(f(x'),f(y')).x',y' \in X\} \le \lambda \cdot \sup\{d(x',y').x',y' \in X\} = \lambda \cdot diam(X) < diam(X)$$

where $$\lambda < 1$$. By monotonicity, $$f(f(X)) \subseteq f(X)$$ and then do induction. For this decreasing sequence, one has $$\cap_{i = 1}^n \{\overline{f}_{\sigma(i)}(X)\} = \overline{f}_{\sigma(n)}(X)$$ so $$\cap_{i = 1}^\infty \{\overline{f}_{\sigma(i)}(X)\} = \lim\limits_{n \to \infty} \overline{f}_{\sigma(n)}(X) = \lim\limits_{n \to \infty} f_{\sigma(n)}(x)$$ for $$x \in X$$.

But this cannot be applied to any $$A \in H(X)$$. It can be applied though to sets such that $$f(S) \subseteq S$$ like the fractal set $$\mathcal{A}$$.

References

These lecture notes.

Massopust's Interpolation and approximation with splines and fractals

$$\newcommand{\de}{\delta} \newcommand{\ga}{\gamma} \newcommand{\si}{\sigma} \newcommand{\ep}{\varepsilon}$$ First here, $$\bigcap\limits_{n \in \mathbb{N}} A_{\sigma(n)}$$ is a subset of $$X$$, whereas $$\gamma(\sigma)$$ is an element of $$X$$. So, in your second highlighted statement $$\gamma(\sigma)$$ should be replaced by $$\{\gamma(\sigma)\}$$. (This is using the notations on pages 25 and and 22 of the lecture notes: $$A_{\sigma(n)}:=f_{\sigma(n)}(A),\quad f_{\sigma(n)}:=f_{\sigma_1}\circ\dots\circ f_{\sigma_n}$$ for $$\sigma=(\sigma_1,\sigma_2,\dots)\in\Sigma_N=\{1,\ldots,N\}^{\mathbb{N}}$$.)
Even after that, your second highlighted statement will be definitely false if e.g. $$A=\emptyset\ne X$$. It will also be false if e.g. $$X=[0,1]$$, $$f_1(x)=x/2$$ for $$x\in[0,1]$$, $$f_2,\dots,f_N$$ are any contraction maps of $$X=[0,1]$$ into itself, $$A$$ is the nonempty compact set $$\{1\}$$, and $$\si=(1,1,\dots)$$.
However, it is easy to see that $$\begin{equation*} \{\gamma(\sigma)\} = \lim_n A_{\si(n)} \end{equation*}$$ for any nonempty bounded $$A\subseteq X$$, in the sense $$\begin{equation*} d(A_{\si(n)},\ga(\si)):=\sup\{d(y,\ga(\si))\colon y\in A_{\si(n)}\}\to0,\tag{1} \end{equation*}$$ where $$d$$ is the distance function on $$X$$. Indeed, let $$M\in[0,\infty)$$ be the diameter of the bounded set $$A$$. Then for some $$r\in[0,1)$$, all $$\si\in\Sigma_N$$, and all natural $$n$$ the diameter of the set $$A_{\si(n)}$$ will be $$\le M r^n$$. So, for any $$x\in A$$ $$\begin{equation*} d(A_{\si(n)},\ga(\si))\le d(A_{\si(n)},f_{\si(n)}(x))+d(f_{\si(n)}(x),\ga(\si)) \le M r^n+d(f_{\si(n)}(x),\ga(\si))\to0 \end{equation*}$$ so that (1) follows.
• the author informed me that $N>1$ is assumed in the book, apparently, that should solve the problem, but how? – Javier Apr 10 at 8:07
• @Javier : Of course, just assuming that $N>1$ will not make that statement in the book true. I have modified, slightly, the second one of my two little counterexamples, to allow $N$ to be $>1$. – Iosif Pinelis Apr 10 at 13:45
• $N$ denotes the number of mappings in the IFS, in your example $N = 1$ right? – Javier Apr 10 at 16:09
• @Javier : No, in that counterexample $N$ can be any natural number. However, because we take $\sigma=(1,1,\dots)$, I did not have to specify $f_2,\dots,f_N$ -- they can be any contraction maps. I have now added this detail. – Iosif Pinelis Apr 10 at 16:24