The $(X,d)$ a complete metric space and $m\geq 1 $ an interger. Consider in $X^{m}$ the distance:
$d_{m}(x,y):=\max\{d(x_{1},y_{1}),\ldots,d(x_{m},y_{m})\}$,
for every $x:=(x_{1},\ldots, x_{m}),y:=(x_{1},\ldots, x_{m})\in X^{m}$. Also, let $r_{n}$ be a sequence in $(0,1)$ with $\sup\{r_{n}:n\geq 1\}<1$ and $f_{n}:X^{m}\longrightarrow X$ a $r_{n}$-contraction, that is,
$f_{n}(x,y)\leq r_{n} d_{m}(x,y)$, for every $x,y\in X^{m}$.
Put $K(X^{m})$ and $K(X)$ the spaces of the compact subsets of $X^{m}$ and $X$, and $d_{H}$ the Hausdorff metric on these spaces. The Hutchinson-Barnsley operator $F:K(X^{m})\longrightarrow K(X)$ is defined then as
$F(K_{1},\ldots, K_{m}):=\overline{\cup_{n\geq 1}f_{n}(K_{1}\times \ldots \times K_{n})}$,
for each compacts $K_{1},\ldots, K_{m}\subset X^{m}$. It is inmediate to check that $F$ is a $r$-contraction for the Hausdorff metric in the space of the compact subsets of $X^{m}$, i.e.
$d_{H}F(A,B)\leq r \max\{d_{H}(A_{1},B_{1}),\ldots, d_{H}(A_{m},B_{m})\}$.
Then, tehre exists a unique set $A\in K(X)$, called the attractor set of $F$, such that $F(A\times\ldots\times A)=A$.
I know, although I have not seen the proof, that for the case $m=1$ the sequence of the attractor sets asociated to $F_{n}:=\cup_{i=1}^{n} f_{i}$ converges (in the Hausdorff metric) to the attractor of $F$. Remains true this assert for $m>1$?
Many thanks for your comments and suggestions.