Let $X \subseteq \mathbb{R}^n$ is closed and $S_N = \{f_i\}$ is a set of $N > 1$ similitudes with $s_i = Lip(S_i)$.
$S_N$ satisfies the open set condition iff there exists a nonempty bounded open set $G \subseteq \mathbb{R}^n$ so that $\stackrel{.}{\cup} f_i(G) \subseteq G$.
Prove that, in the above conditions, $\sum_{i = 1}^N s_i^n < 1$.
Could you verify my approach? Surely there is a simpler proof.
The relevance of the statement
There exist results on the Hausdorff/Box dimension of self-similar fractals. The dimension is computed as the unique solution of $\sum_{i = 1}^N s_i^d = 1$. Then, the above result together with the monotonicity of the sum in $d$ yields that there is indeed a unique solution to the equation.
Proof
$G$ is an open set. Thus, we know that $G = \bigcup\limits_{i \in \mathbb{N}} B(x_i,r_i)$ because balls form a countable basis of the topology.
Case $G = B(x,r)$
We have $\bigcup\limits_{i = 1}^N f_i(B(x,r)) \subseteq B(x,r)$ and then: $$V(\bigcup\limits_{i = 1}^N f_i(B(x,r))) = \sum_{i = 1}^N V(f_i(B(x,r))) = \sum_{i = 1}^N s_i^n r^n \le r^n= V(B(x,r))$$ and thus, we get: $\sum_{i = 1}^N r^ns_i^n \le r^n$ and so $\sum_{i = 1}^N s_i^n \le 1$.
Equality holds iff $G = \bigcup\limits_{i = 1}^N f_i(B(x,r))$. But, since the union is disjoint, this gives a non-trivial partition of the connected ball $B(x,r)$. Contradiction.
Case $G = \bigcup\limits_{j \in \mathbb{N}} B(x_j,r_j)$
We know that $V(G) \le \sum r_j^n \le M^n$ where $B(x,M)$ is the ball bounding the set. Then we consider the mappings $\{\overline{f_i}\}$ acting on each ball $B(x_j,r_j)$. We obtain in this way a set of disjoint balls whose volume is $\sum_{i = 1}^N s_i^n r_j^n$. Summing over all balls we get $(\sum_{j \in \mathbb{N}} r_j^n) (\sum_{i = 1}^N s_i^n) \le \sum r_j^n$, by the open set condition. We get $\sum_{i = 1}^N s_i^n \le 1$.
Equality holds iff $G = \bigcup\limits_{i = 1}^N \overline{f_i}(G)$. But since $f_i$ is a similitude, then (assuming $s > 0$) it is an open mapping and each $\overline{f_i}(G)$ is an open set. This induces an open partition in each connected component of $G$. This forces that each connected component receives one $\overline{f_i}(G)$ so that we don't get a partition of a connected set.
But since $f_i$ is a similitude with $s > 0$, it has an inverse which would have domain a connected component. We would have that a continuous mapping sends a connected set to several connected sets. Contradiction. Thus, the only possibility is that $G$ consists of only one connected component. By connectedness it would also imply that $N = 1$. However, (motivated by the definition of IFS) we imposed that $N > 1$. Contradiction.
References
Massopust: Interpolation and Approximation with Splines and Fractals, page 157.