Choosing an appropriate sequence of $\{1,2,3\}$

Question

Let $f_1,f_2,f_3$ be the contracting maps $f_i:x\mapsto \frac{1}{2}(x+p_i)$ from $\mathbb{R^2}$ to itself and $p_i\in \mathbb{R}^2$. Denoted by $S$ the attractor Sierpinki gastek of the iterated function system $(f_1,f_2,f_3)$. I want to prove the following

Given a point $a\in \mathbb{R}^2$ and a sequence $k_n$ of elements $\{1,2,3\}$, we define $x_0=a$ and $x_{n+1}=f_{k_n}(x_n)$. Then

(1) Each accumulation point of $\{x_n\}$ belongs to $S$

(2) For each point $x\in S$ there exist a sequence $k_n$ such that $\{x_n\}$ has $x$ as an accumulation point.

(3) There exist a point $a$ and a sequence $k_n$ such that every point in $S$ is an accumulation point of $\{x_n\}$

I have solved just $(1)$. All this properties should be a consequence of the fixed point's Theorem for contractions and the fact that the function $$A\mapsto \bigcup_{i=1}^3 f[A]$$ is a contraction in the hyperspace of $\mathbb{R}^2$ -the space of all non-empty compact subset of $\mathbb{R}^2$ with the Hausdorff metric.

I have solved (1) using this theorem and finding a sequence in $S$ that behaves "identically" to the sequence $\{x_n\}$ -meaning that i find a sequence $y_n\in S$ such that $|y_n-x_n|<\epsilon$ for an appropriate $n$ and every $\epsilon>0$.

I get stuck with the other ones. I have try several ways to solve this others two problem but I do not find how to satisfy the property of $\{x_n\}$. Intuitively, I have to find how the initial point "walks" to the point $x$ (point $(2)$) but I do not find the appropriate sequence. Any hint?

For $(3)$ the situation it is even more estrange because I should find a point and a sequence in such way that the point "walks everywhere" on $S$.I am pretty sure that the initial point is whatever point in $S$. How can I prove it?

These are classic results above self similar sets and they are really important to understand the very nature of a lot examples of fractals. I really appreciate the help!

Answer 1

I will call $E$ the set of nonempty compact subset of $\Bbb R^2$.
If $x \in \Bbb R^2$ and $Y \in E$, we define the distance from $x$ to $Y$ by $d_x(Y) = \min_{y \in Y} ||x-y||$. Since $Y$ is compact this actually is a minimum, and $d_x(Y) = 0 \iff x \in Y$.

Then the Hausdorff metric on $E$ is defined by $d(X,Y) = \max(\max_{x \in X} d_x(Y) ; \max_{y \in Y} d_y(X))$.

Note that forall $x \in \Bbb R^2$, $d_x : E \to \Bbb R$ is continuous (more precisely, $|d_x(Y)-d_x(Z)| \le d(Y,Z)$).

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let $\phi : E \to E$ defined by $\phi(A) = \bigcup f_i[A]$. As you have noted, $\phi$ is a contraction ($d(\phi(X),\phi(Y)) \le \frac 12 d(X,Y)$) and has a unique fixpoint $S \in E$, and forall $A \in E$, $\phi^{\circ n}(A) \to S$

To prove $(1)$, we pick $A = \{a\}$. Then $A_n$ is the finite set of all possible images of $a$ after $n$ applications of some $f_i$, so it's all the possible values of $x_n$.
Suppose $y$ is an accumulation point of such a sequence $(x_n)$. Then there exists a sequence $(n_i)$ such that $||x_{n_i} - y || \to 0$, and since $x_{n_i} \in \phi^{\circ n_i}(A)$, we have $d_y(\phi^{\circ n_i}(A)) \le ||y-x_{n_i}||$, thus $d_y(\phi^{\circ n_i}(A)) \to 0$.
Since $d_y$ is continuous and $\phi^{\circ n_i}(A) \to S$, we get $d_y(S) = 0$, and therefore $y \in S$.

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On to proving $(2)$ and $(3)$ ($a$ can in fact be any point).

First we see that forall $y \in S, \varepsilon > 0, x \in \Bbb R^2$ there exists $n \in \Bbb N$ and $x' \in \phi^{\circ n}(\{x\})$ such that $||x' - y || < \epsilon$ (we just have to pick $n$ such that $d(\phi^{\circ n}(\{x\}),S) \le \epsilon$). Put another way, there always exists a finite sequence of moves that brings $x$ to a point $x'$ within $\epsilon$ of $y$.

To prove $(2)$ we apply this repeatedly with a fixed $y \in S$ and $\epsilon = 1,1/2,1/3,\ldots$. From any $a$ there is a finite sequence that brings $a$ to $a_1$ such that $||a_1-y|| \le 1$, then there is a finite sequence that brings $a_1$ to $a_2$ such that $||a_2-y|| \le 1/2$, and so on.

To prove $(3)$ we do the same thing but instead of focusing on a particular $y$, we try to get within $\epsilon$ of everyone in $S$ before replacing $\epsilon$ with a smaller value.

Since $S$ is compact, forall $\epsilon > 0$ there is a finite subset $S_\epsilon \subset S$ such that $d(S_\epsilon,S) \le \epsilon/2$.
From any point $a \in \Bbb R^2$ we can build a finite sequence of moves from $a$ such that the sequence obtained gets within $\epsilon/2$ of each point of $S_\epsilon$.
This is enough for that finite sequence to get within $\epsilon$ of any point in $S$.

Once again we repeat this for $\epsilon = 1,1/2,1/3,\ldots$ to build an infinite sequence of moves that gets arbitrarily close to any element of $S$.