A zero-dimension set and self-referencial equation 
Let $K$ be a compact set in $\mathbb{R}^2$. Let $f_1,..., f_n$ be contracting similarities of $\mathbb{R}^2$ to itself. Suposse $K$ satisfies the self-referencial equation
  $$K=\bigcup_{i=1}^{n}f_i[K]$$
  and $f_i[K]\cap f_j[K]=\emptyset$. Prove that $K$ is zero-dimensional.

Since $f_i[K]\cap f_j[K]=\emptyset$ I know that the distance between both set is positive, because those sets are compact. Then each $f_i[K]$ is a clopen set of $K$. I don't know how to use that $f_1,..., f_n$ are contraction. Any hint?
Thanks!
 A: I will used all the time the notation given in the problem. Then the function $f_i$ will always mean that it supposed to mean in the hypothesis. 
To prove the theorem we will find a clopen basis. For simplicity define
$$\bigcirc_{k=1}^sf_k =f_s\circ f_{s-1}\circ\cdots\circ f_{1}$$
That is a successive composed functions. Now take a sequence $\{i_k\}_{n=1}^\infty$  such that all its member are integers between $1$ and $n$ ($1\leq i_k\leq n$ for every $k$).
The letters $r_1,..., r_n$ will denote the constant of contraction (that is the real number such that $|f_i(x)-f_i(y)|=r_i|x-y|$ for every $x,y$ in the domain).  
Since $K$ is compact then $A=\text{diam } K$ is finite. 
First claim: Each $f_{i}[K]$ is a clopen set. 
proof: Each $f_{i}[K]$ is compact, since $K$ is compact. By the condition $f_{i}[K]\cap f_j[K]=\emptyset$ the distance between each $f_i[K]$ is positive furthermore this implies too that $f_i[K]$ is closed being the image of a compact set in a eucliden space. Its complement is 
$$\bigcup_{j\not= i}f_j[K] $$
and this set is closed, being the finite union of closed sets, then $f_i[K]$ is open therefore clopen. 
Second claim: For every sequence $f_{i_k}$ where $\{i_k\}_{n=1}^\infty$ is a sequence defined as above the following is true:
$$\lim_{s\rightarrow \infty}\text{diam }\left[\bigcirc_{k=1}^sf_k(K)\right]=0 $$
proof: Defined $ g_s=\bigcirc_{k=1}^sf_{i_k}(K)$. For each $x,y\in f_{i_k}(K)$, the equality $|f_{i_k}(x)-f_{i_k}(y)|=r_{i_k}|x-y|\leq r_{i_k}A $ is the basic step on the proof by induction of 
$$|g_{s+1}(x)-g_{s+1}(x)|=r_{i_s}r_{i_{s-1}}\cdots r_{i_1}|x-y|\leq r_{i_s}r_{i_{s-1}}\cdots r_{i_1} A $$. Which implies 
$$\text{diam }g_s[K]\leq r_{i_s}r_{i_{s-1}}\cdots r_{i_1} A $$
and the sequence $r_{i_s}r_{i_{s-1}}\cdots r_{i_1} A$ is bounded by $AR^s$ where $R=\max\{r_1,...,r_n\}<1$ then $AR^n\rightarrow 0$ when $n\rightarrow\infty$. This proves the second claim. 
Third claim: Taking indices $i_1,..., i_n$ varying between $1$ to $n$. We will have that 
$$K=\bigcup_{i_n,...,i_1=1}^n \bigcirc_{k=1}^nf_{i_k}[K] $$
and that $f_{i_n}\circ\cdots\circ f_{i_1}[K]\cap f_{i_n'}\circ\cdots\circ f_{i_1}[K]=\emptyset$ if $i_n\not= i_n'$. Which implies that this set are clopen. 
As ilustration i will show the case $n=2$. We have that
$$K=\bigcup_{i=1}^nf_i[K]$$
Then $f_jK=\bigcup_{i=1}^nf_jf_i[K]$, consequently
$$K=\bigcup_{j=1}^n\bigcup_{i=1}^nf_jf_i[K]$$
And, since $f_i[K]\subset K$ then $f_jf_i[K]\subset f_j[K]$  then $f_jf_i[K]\cap f_{j'}f_i[K]=\emptyset$ if $j\not=j'$. And it is clear that the set $f_jf_i[K]$ are clopen (the reason is analagous to the proof in the first claim).
This three claims proof the theorem, since for each neighborhood of a pint in $K$ I can take a sufficiently small  (by the firts claim) set of the form $f_{i_k}\circ\cdots\circ f_{i_1}[K]$ wich is clopen by the second claim.
Then the family of sets $f_{i_k}\circ\cdots\circ f_{i_1}[K]$ form a clopen basis, then $K$ is zero dimensional. 
A: Hint: Prove that $K$ is a Cantor set (of Hausdorff dimension $< 2$).
