# proof of existence and unicity of self-similar fractals

We consider the space $\mathcal{K}(X)=\{K\subset X\colon K\mathrm{\ compact\ and\ non\ empty}\}$ endowed with the Hausdorff distance $\delta$. Since Hausdorff metric inherits completeness, being $X$ complete     , $(\mathcal{K}(X),\delta)$ is complete too. We then consider the mapping $T\colon\mathcal{K}(X)\to\mathcal{K}(X)$ defined by

 $T(A)=\bigcup_{i=1}^{N}T_{i}(A).$

We claim that $T$ is a contraction. In fact, recalling that $\delta(A_{1}\cup A_{2},B_{1}\cup B_{2})\leq\max\{\delta(A_{1},B_{1}),\delta(A_% {2},B_{2})\}$ while $\delta(T_{i}(A),T_{i}(B))\leq\lambda_{i}\delta(A,B)$ if $T_{i}$ is $\lambda_{i}$-Lipschitz  , we have

 $\displaystyle\delta(T(A),T(B))$ $\displaystyle=\delta(\bigcup_{i}T_{i}(A),\bigcup_{i}T_{i}(B))\leq\max_{i}% \delta(T_{i}(A),T_{i}(B))$ $\displaystyle\leq\max_{i}\lambda_{i}\delta(A,B)=\lambda\delta(A,B)$

with $\lambda=\max_{i}\lambda_{i}<1$.

So $T$ is a contraction on the complete metric space $\mathcal{K}(X)$ and hence, by Banach Fixed Point Theorem  , there exists one and only one $K\in\mathcal{K}(X)$ such that $T(K)=K$.

Title proof of existence and unicity of self-similar fractals ProofOfExistenceAndUnicityOfSelfsimilarFractals 2013-03-22 16:05:30 2013-03-22 16:05:30 paolini (1187) paolini (1187) 5 paolini (1187) Proof msc 28A80