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.
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?
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real-analysis
functional-analysis
metric-spaces
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0definition of a continuous transformation? – 2012-11-29
1 Answers
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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 \begin{equation} K_n={S_{i_1}\circ \cdots \circ S_{i_n}(x): i_1,\ldots, i_n\in I}, \end{equation} then $K_n$ converges (regardless of the initial point $x$) in the Hausdorff distance to the unique compact fixed point of the map \begin{equation} F(K)=\bigcup_{i\in I}S_i(K) \end{equation} In fact, $F$ is a contraction in the Hausdorff distance and the fixed point $F(K^*)=K^*$ is the selfsimilar set defined by contractions $\{S_i\}$.
A nice reference for this is J. Kigami, "Analysis on Fractals".
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0Yes, 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. – 2012-04-14