# When the set $\{(S_{i_1}\circ\cdots\circ S_{i_n})(x): n\in \mathbb{N},\;\; i_1,\ldots,i_n\in I\}$ is relatively compact?

Let $(X,\rho)$ be a metric space and let $S_1,\ldots,S_N:X\rightarrow X$ be continuous transformations. Denote $I=\{1,\ldots,N\}$. Is it possible to find some minimal assumptions on $S_i$ which would ensure relative compactness of the set $\{(S_{i_1}\circ\cdots\circ S_{i_n})(x): n\in \mathbb{N},\;\; i_1,\ldots,i_n\in I\}$ at any point $x\in X$? I know it is a general question and perhaps it is known. It is important for me, so I will be grateful for any propositions.

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I might be missing something, but isn't your set, for each $x$, finite? – Martin Argerami Apr 5 '12 at 22:57
@MartinArgerami: The set contains elements of the form $S_1^{\circ n}(x)$, so it may not be finite. The orbit of each element of the set by one of the $S_i$, $i\in I$ has to be relatively compact. Maybe dawid has more precise ideas about the assumptions he needs. – Davide Giraudo Apr 6 '12 at 8:06
Yesterday I came up with something like this: Assume that $\rho(S_i(x),S_i(y))\leqslant w(\rho(x,y))$, for all $i$, where $w$ is an increasing function such that $w^n(t)\to 0$ ($t\geq 0$, as $n\to \infty$) and $S_1,...,S_N$ admits a common fixed point. Then our set is relatively compact. Am I right...? – dawid Apr 6 '12 at 14:28
The conditions I have given above holds true when for example $S_1,...S_N$ are contractions with a common fixed point, I have tried this assumptions and it seems to be ok, but I am not sure... – dawid Apr 6 '12 at 14:32
definition of a continuous transformation? – T. Eskin Nov 29 '12 at 21:51

I am not sure about the exact answer to this. However, it seems to me that the requirement that the maps $S_i$ have a common fixed point is too much. Assuming that the $S_i$ are contractions, and defining the sets $$K_n={S_{i_1}\circ \cdots \circ S_{i_n}(x): i_1,\ldots, i_n\in I},$$ then $K_n$ converges (regardless of the initial point $x$) in the Hausdorff distance to the unique compact fixed point of the map $$F(K)=\bigcup_{i\in I}S_i(K)$$ In fact, $F$ is a contraction in the Hausdorff distance and the fixed point $F(K^*)=K^*$ is the selfsimilar set defined by contractions $\{S_i\}$.
Yes, I think it is known fact, the set $K^*$ is called the attractor for the family $\{S_i\}$ and from the existence of the attractor we get the compactness of the set $\bigcup_n F^n(K^*)=K^*$, but I need the compactness of the set $\bigcup_n F^n(\{x\})$, for all $x$ in this kind of notation. – dawid Apr 14 '12 at 14:57