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In Massopoust's Interpolation and approximation with splines and fractals, one can read this theorem:

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

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 $H()$ 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

We know that for $A \in H(X)$, we have that $\mathcal{A} = \lim\limits_{n \to \infty} F^n(A) = \lim\limits_{n \to \infty} \bigcup\limits_{i_1,\ldots,i_n = 1}^n \overline{f}_{i_1,\ldots,i_n}(A)$. So one could understand that restricting to a specific code leads to point in $A$. This is what is done above. But why is the intersection non-empty?

Graphically, one can interpret that we are contracting the set $A$ in successive iterations, so at each step in the sequence we are building a decreasing sequence of compact sets. This decreasing sequence will have a non-empty intersection (provided none of the intermediate steps is compact).

I need to still improve this formalization.

Note: I can clarify any doubt on notation you may have. This lecture notes might be useful anyways.

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The key is that $\gamma$ does not depend on $x \in X$. So we have $\forall x \in A.\gamma(\sigma) = \lim\limits_{n \to \infty} f_{\sigma(n)}(x)$. Here $A \in H(X)$ to ensure the diameter of the set exists (this was required in the proof).

Now, we observe that $\{\overline{f}_{\sigma(n)}(A)\}$ is a decreasing sequence this is because $\overline{f_i}$ are contractions on $H(X)$. We can reason like in the comments to this answer.

The sequence $\{\stackrel{-}{f}_{\sigma(n)}(X)\}$ is a decreasing sequence using that $f:X \to X$, by induction. Furthermore, the diameter of each element of the sequence, reduces strictly. Take a $C$ in the sequence, using The diameter of a compact set., we see that there exists $x_0,y_0 \in X$ such that $diam(B) = d(x_0,y_0)$. We have:

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

By induction, $diam(\stackrel{-}{f}_{\sigma(n)}(X)) \le \lambda^n \cdot diam(X) \to 0$ as $n \to \infty$. Using Diameter of Nested Compact Sets, we conclude that $\cap_{i = 1}^n \{\overline{f}_{\sigma(i)}(X)\}$ contains exactly one element. But the sequence, $\{\overline{f}_{\sigma(n)}(A)\}$ is included in the above and thus reduces to a point or the empty set. We can apply Intersection of compact subsets of metric space / seeking alternative proof to show that there has to be a point.

By the quoted theorem this point needs to be $\gamma(\sigma)$.

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