Verify that the attractor of an IFS is the graph of an interpolation

Question

I am going through exercises in Fractals Everywhere by Michael Barnsley. I have a confusion about Exercise 2.1 in Chapter 7:

The function $f(x) = 1 + x$ is an interpolation function for the set of data $$\{(0,1), (1,2)\}.$$ Consider the hyperbolic IFS $\{\mathbb{R}^2; \omega_1, \omega_2\}$, where $$ \omega_1 \begin{pmatrix} x \\y \end{pmatrix}= \begin{pmatrix} 0.5 &0 \\ 0 &0.5 \end{pmatrix} \begin{pmatrix} x \\y \end{pmatrix} + \begin{pmatrix} 0 \\ 0.5 \end{pmatrix} $$ $$ \omega_2 \begin{pmatrix} x \\y \end{pmatrix}= \begin{pmatrix} 0.5 &0 \\ 0 &0.5 \end{pmatrix} \begin{pmatrix} x \\y \end{pmatrix} + \begin{pmatrix}0.5 \\ 1 \end{pmatrix} $$ Let $G$ denote the attractor of the IFS. Then it is readily verified that G is the straight line segment that connects the pair of points $(0, 1)$ and $(1, 2)$. In other words, $G$ is the graph of the interpolation function $f(x)$ over the interval $[0, 1]$.

It is intuitive to think that the line segment between $(0,1)$ and $(1,2)$ is the attractor of the IFS, since all the $x \in \mathbb{R}$ will converge to the interval $[0,1]$ under these two operations. However I am struggling to come up with "formal steps" to verify that it's indeed the attractor.

Can someone give some hint?

Answer 1

Use the uniqueness: the attractor of an IFS is the unique non-empty compact set which is fixed by the IFS.

Elaborating a bit more: if you have a set $S$ such that

• $S$ is compact and non-empty,
• $S = \omega_1(S) \cup \omega_2(S)$,

then by uniqueness, $S$ must be the attractor. This is contained in theorem 7.1 of the book you're using.