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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0I might be missing something, but isn't your set, for each $x$, finite? – 2012-04-05
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0@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. – 2012-04-06
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0Yesterday 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...? – 2012-04-06
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0The 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... – 2012-04-06
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0definition of a continuous transformation? – 2012-11-29