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Let $(X,\rho)$ be a Polish space. We are given continuous functions $S_i: X\rightarrow X$ ($i=1,...,N$) and a constant $0<r<1$. We assume the following:

For each $x,y \in X$ there exists $i\leq N$ such that $$\rho(S_i(x),S_i(y))\leq r\rho(x,y).$$

My question is:

Does there exist a point $x_0$ such that for every $x\in X$ there is a sequence $(i_1,i_2,..)$ such that we have $$\lim_{n\to\infty} S_{i_n}\circ...\circ S_{i_1}(x)=x_0\;?$$

I think it is too optymistic, however I can not find a suitable counterexample. I would be grateful for some help.

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I don't quite understand why we cannot simply pick all the $i$'s equal, in which case we know $x_0$ exists since $S_i$ is a strict contraction. –  Christopher A. Wong Sep 28 '12 at 23:24
    
Because none of the functions $S_i$ need not be a contraction. Observe we dont have "there exists $i$ such that for each $x,y$..." but "for each $x,y$ there exists $i$..." –  dawid Sep 28 '12 at 23:33
    
Oh, I see your point. –  Christopher A. Wong Sep 29 '12 at 1:26
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