# The existence of an attractor for a finite family of continuous transformations

Let $(X, \rho)$ be a complete metric space and let $S_1,...,S_N: X\rightarrow X$ be continous functions.

A nonempty compact set $C$ is said to be an attractor for the family $\{S_1,...,S_n\}$ if $$\bigcup_{i=1}^N S_i(C)=C.$$ It is well known that in the case when all the functions $S_i$ are contractions then there exists an attractor for the family $\{S_1,...,S_N\}$ (it can be easily proved by using the Hausdorff distance).

Assume that $$\tag{1} \rho(S_i(x), S_i(y))\leq r(\rho(x,y))\;\;\; (i=1,..,N,\;x,y\in X),$$ where $r:\left[0,\infty\right )\rightarrow \left[0,\infty\right )$ is a nondecreasing and continuous at $0$ function such that $r(0)=0$ and $r^n(t)\to 0$, for all $t\geq 0$.

My question is the following:

Is there exist an attractor for the family $\{S_1,..,S_N\}$ if condition $(1)$ holds?

Denote $$\Gamma^{\infty}(x)=\{S_{i_n}\circ... S_{i_1}(x):\;i_1,...,i_n\in\{1,..,N\},\;n\in\mathbb{N}\}$$

Observe that if $\Gamma$ stands for the Barnsley operator, i.e $\Gamma(A)=\bigcup_{i=1}^N S_i(A)$, then we may write $\Gamma^{\infty}(x)=\bigcup_{n=1}^{\infty}\Gamma^n(\{x\})$. Furthermore, it is easily seen that if $\text{cl } \Gamma^{\infty}(x)$ is compact then it is an attractor for $\{S_1,...,S_N\}$. I have proved that the following conditions are equivalent:

(i) $\text{cl }\Gamma^\infty (x)$ is compact for some $x\in X$,

(ii) $\text{cl }\Gamma^\infty (x)$ is compact for all $x\in X$,

(iii) There exists an attractor for $\{S_1,...,S_N\}$.

So, my question can also be formulate as follows:

Is the set $\text{cl }\Gamma^\infty (x)$ compact if condition $(1)$ holds?

I will be greatfull for any hints.

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I assume that $r^n(t)$ denotes the $n$-th iterate of $r$? And you really don't want to assume that $r$ is continuous everywhere (in which case the condition on $r$ is equivalent to $r(t)<t$)? –  Lukas Geyer Oct 6 '12 at 22:01
Yes, $r^n(t)$ is the n-th iterate of $r$. I don't assume that $r$ is constinuous everywhere, however if you knew the answer to my question in this case, I would be satisfied. –  dawid Oct 7 '12 at 13:33
I could be wrong, but I worry that this isn't true for $N = 1$. What I have in mind is a noncompact space and a single map $S$ for which $d(S^n x, S^n y) \rightarrow 0$ for all $x,y$, and yet $S^n x$ is pushed out to infinity for any $x$ (so the growth of $S^n x$ is very small). A specific instance of this would be $X = [1,\infty) \subset \mathbb{R}$ and $S x = x + \frac{1}{x}$. Indeed, your condition (1) might be considerably stronger than simply $d(S^n x, S^n y) \rightarrow 0$, but I do not know for certain. –  A Blumenthal Jan 17 '13 at 20:48