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!
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.