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*