Maps on the hyperspace of compact sets

Question

In the theory of fractals via iterated function systems, it is well-known that an IFS $\{f_i\}_{i=1}^n$ (being a finite collection of contractions defined on a metric space $X$) induces a single map $F$ on the hyperspace $\mathbb{H}$ of nonempty compact subsets of $X$, where

$$F(A) = \bigcup_{i=1}^n f_i(A) $$

Furthermore, $F$ will be a contraction on $\mathbb{H}$ as measured by the Hausdorff distance $\rho$.

Now since we can get contractions on $\mathbb{H}$ this way, you could ask if all contractions on $\mathbb{H}$ are the result of some IFS on $X$. The answer is pretty clearly no -- if $X = \mathbb{R}$, then we can define a constant function $F : \mathbb{H} \to \mathbb{H}$ by taking $F(A) = [0, 1]$ for all $A \in \mathbb{H}$. This function can't be the result of an IFS on $X$ because for any singleton set $\{x\}$, $F(\{x\}) = [0, 1]$, whereas functions on $\mathbb{H}$ that come from an IFS will necessarily map finite sets to finite sets.

However this constant map $F$ can be written in terms of a Big Iterated Function System, meaning a potentially infinite collection of contractions. All we have to do is take the family

$$\{ f_r(x) = r \; : \; r \in [0, 1] \}$$

Of course, although an IFS is guaranteed to define a map on $\mathbb{H}$, a BIFS is not. The above example, but parameterized on $r \in (0, 1)$, does not map compact sets to compact sets.

Now here is the question that I haven't been able to answer. Suppose you have a BIFS which does induce a map on $\mathbb{H}$. Will the resulting map necessarily be a contraction?

Answer 1

One of my students gave a good counterexample. If the ambient space is $\mathbb{R}$ under the usual metric, here's a BIFS that that induces a map on $\mathbb{H}$, but where the map is not a contraction.

Define $f_0(x) = 0$, and for any $n \geq 1$, define

$$ f_n(x) = \left\{ \begin{array}{ccc} 0 & & x \leq 0 \\ \\ \frac{1}{2}x & & 0 < x \leq 1 \\ \\ \frac{2}{3}x + c_2 & & 1 < x \leq 2 \\ \\ \cdots & & \cdots \\ \\ \frac{n-1}{n}x + c_{n-1} & & n-2 < x \leq n-1 \\ \\ \frac{n}{n+1}x + c_n & & x > n-1 \end{array} \right. $$

Here the constants $c_n$ are selected to make each $f_n$ continuous. We need to observe three facts:

1) Each $f_n$ is a contraction

2) The collection $\{f_n\}_{n \in \mathbb{N}}$ defines a map $F : \mathbb{H} \to \mathbb{H}$ given by $$F(A) = \bigcup_{n \in \mathbb{N}} f_n(A) $$ This is because any compact set in $\mathbb{R}$ is contained in a compact interval, and over any compact interval the behavior of $F$ is determined by finitely many of the $f_n$'s.

3) $F$ is not a contraction on $\mathbb{H}$, since as $n \to \infty$,

$$ \rho\left( \{n\}, \{n+1\} \right) \to 1 $$

where $\rho$ is the Hausdorff metric on $\mathbb{H}$.

Answer 2

Hutchison's argument corresponds entirely to your first paragraph, so I thought that you knew it.

I looked for a reference that could be readily available and found https://maths-people.anu.edu.au/~john/Assets/Research%20Papers/fractals_self-similarity.pdf.

Have a look at page 12. Oversimplifying the matter, you can first look at the proof of the theorem at the bottom of this page. In the situation that you describe this suggests that the estimate in the proof may not lead to a contraction, unless it can be estimated better.

For an explicit class of counterexamples to what you ask see Theorem 2.10 in http://arxiv.org/pdf/1307.1090.pdf (if the map were a contraction the invariant set would be unique).