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 $T(K)=K$ .