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Let $(X,\rho)$ be a separable complete metric space. We consider an Itereted Function System with place-dependent probabilities. Namely, assume that $I=\{1,\ldots,N\}$, $S_i: X\rightarrow X$ and $p_i: X\rightarrow \left[0,1\right]$, $i\in I$, are continuos functions and $\sum_{i=1}^N p_i(x)=1$, for all $x \in X$. The Markov transition operator for the system is given by $$P\mu (A)=\sum_{i=1}^N\int_{X} 1_A(S_i(x))\,p_i(x)\,\mu(x)\;\;\; (\mu \in \mathcal{M}, A\in \mathcal{B}),$$ where $\mathcal{M}$ stands for the set of all Borel probability measures on $X$. The operator $P$ is said to be asymptotically stable if there exists (a unique) invariant distribution $\pi \in\mathcal{M}$ such that $\mu P^n \stackrel{w}{\rightarrow}\pi$, for any $\mu \in \mathcal{M}$. The following theorem (nontrivial) is taken from [T. Szarek, Invariant measures for Markov operators with applications to function systems, Studia Math. 154 (2003), 207–222.]:

Assume that the following conditions hold: $$\sum_{i=1}^N |p_i(x)-p_i(y)|\leq w(\rho(x,y))\tag{1},$$ where $w: \mathbb{R}^+\rightarrow\mathbb{R}^+$ is nondecreasing, continuous at $0$, $w(0)=0$ and it satisfies the Dini condition, that is $\sum_{n=1}^{\infty} w(s^n t)<\infty$, for $s\in (0,1)$ and $t\geq 0$. $$\tag{2} \sum_{i=1}^N p_i(x) \rho(S_i(x), S_i(y))\leq r\rho(x,y)\;\;\; (x,y\in X),$$ where $r<1$,

there exists an $\eta>0$ such that $$\tag{3} \sum_{i\in J(x,y)} p_i(x)p_i(y)\geq\eta\;\;\; (x,y \in X),$$ where $J(x,y)=\{i\in I: \rho(S_i(x),S_i(y))\leq r\rho(x,y)\}$.

Then, the operator $P$ is asymptotically stable.

Under the assumptions of the above theorem we obtain the following: Let $z \in \text{supp } \pi$ and fix $\epsilon>0$. By the asymptotic stability and the Aleksandrov theorem we obtain $$\liminf_{n\to \infty} P^n\delta_x (B(z,\epsilon))\geq \pi (B(z,\epsilon))>0\;\;\;(x \in X).$$ Choose $n_0\in \mathbb{N}$ such that $$P^n\delta_x (B(z,\epsilon))>\frac{1}{2}\pi (B(z,\epsilon))\;\;\;(n\geq n_0)$$ It is easy to show that for an arbitrary $x \in X$ we have $$P^n\delta_x(A)=\sum_{(i_1,\ldots,i_n)\in I^n}1_A(S_{i_n}\circ\cdots\circ S_{i_1}(x))p_{i_1}(x)p_{i_2}(S_{i_1}(x))\cdots p_{i_n}(S_{i_{n-1}}\circ\cdots\circ S_{i_1}(x)),$$ so in particular there is a sequence $\mathbf{i}^n=(i_1^n,\ldots,i_{k_n}^n)\in I^{k_n}$, $n\in \mathbb{N}$ such that $S_{\mathbf{i}^n}(x)=S_{i_{k_n}^n}\circ\cdots\circ S_{i_1^n}(x) \in B(z,\epsilon)$, for $n\geq n_0$. Finally obtain the following coclusion:

Let $(1)-(3)$ hold. Then, there exists a point $z\in X$ with the following property: for every $\epsilon>0$ there is an $n_0$ such that for each $x\in X$ we can choose a sequence $\mathbf{i}^n=(i_1^n,\ldots,i_{k_n}^n)\in I^{k_n}$ such that $$S_{\mathbf{i}^n}(x) \in B(z,\epsilon)\;\;\; (n\geq n_0)$$

Observe that there are no probabilities in the result above. So my questions is: how to prove the last result without using the Probability theory? Is there a simpler direct proof o this?

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