I'm having problems understanding a proof in the following references:
Massopoust, Interpolation and Approximation with Splines and Fractals, page 159.
Barnsley, Fractals Everywhere, page 124.
The theorem states the relation between the fractal $\mathcal{A}$ generated by an IFS $(X,F_N)$ and the code space $\Sigma_N$:
There is a continuous 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 $X$.
My problem comes with the proof of surjectivity. Massopoust reads as follows:
Could you explain me the surjectivity part of the proof? Why there has to be such a sequence σν? What is the issue here to have such complicated notation σνμ? I feel this is a "typical" subsequence arguments from functional analysis but I'm not really seeing it.
Thanks for your help.
Note: I can clarify any doubt on notation you may have. This lecture notes might be useful anyways.
$(H(X),h)$ is a complete metric space. We have $A = \lim F^n(\{x\})$. This means that $\forall \epsilon > 0. \exists n_0 \in \mathbb{N}. \forall n \ge n_0. h(A_n,A) < \epsilon$ where $A_n = F^n(\{x\})$.
Take $y \in A$. Recall $F^n(A) = \bigcup\limits_{i_1,\ldots,i_n = 1}^N \overline{f}_{i_1,\ldots,i_n}(A)$ so the definition of $A$ is saying that $y = \lim\limits_{n \to \infty} f_{\sigma^{n}(n)}$. Here $\{\sigma^n\} \subseteq \Sigma_N$. Since $\Sigma_N$ is a compact metric space, we know $\{\sigma^n\}$ has a convergent subsequence $\tau^n = \{\sigma^{\tau(n)}\} \to \tau$. This means that $d_F(\tau^n,\tau) \to 0$ where $d_F$ is the Fréchet distance $d_F(\sigma,\tau) = \sum_{n = 1}^{\infty} \frac{|\sigma_n-\tau_n|}{(N+1)^n}$.
Then we write $c(n)$ for the cardinality of the set of indexes $\chi$ such that the $\chi$ prefix of $\tau^n$ equals the $\chi$ prefix of $\tau$. Observe that $c(n) \le n$ and $n \to \infty \implies c(n) \to \infty$ since $d_F(\tau^n,\tau) \to 0$ where $d_F$. This implies that $|f_{\tau^n(n)}(x)-f_{\tau(n)}(x)| \le \lambda^{c(n)} diam(X)$.
Since $\lambda < 1$, when $n \to \infty$, $y = \lim f_{\tau^n(n)}(x) = \lim f_{\tau(n)}(x) = \gamma(\sigma)$.
However to justify the first equality in the last chain I would need $ \lim f_{\tau^n(n)}(x) = \lim f_{\tau^n(\tau(n))}(x)$. We can do that by redifining $\tau^n$ to be $\tau^n = \sigma^{\tau(i)}$ if $n = \tau(i)$ for some $i$. Otherwise, $\tau^n = \sigma^{\tau(i)}$ where $m = \tau(i)$ is the closest natural greater than $n$ and of the form $\tau(i)$.