I came across this execise in a problem set about Iterated Function System (IFS) and fractals:
"Show that the attractor of an IFS of the form $\{\mathbb{R};~ax+b, cx+d\}$ where $a,b,c,d \in \mathbb{R}$, is either connected or totally disconnected".
All I know about an IFS with contractive mappings $f$ and $g$ is that its attractor $A$ obeys
$f(A)∪g(A)=A$.
I spent quite a long time trying proving it, so I'll be greatful for any help.
Assuming that $|a|<1$ and $|c|<1$ (i.e., the system is contractive), each of the maps has a unique fixed point: call them $x_1,x_2$. The trivial case is when $x_1=x_2$; then $\{x_1\}$ is the invariant set. From now on, assume $x_1\ne x_2$.
Case 1.: $|a|+|c|<1$. This is when the invariant set (call it $K$) is of Cantor type, i.e., totally disconnected. Indeed, being compact, $K$ is contained in some interval $I$. Therefore, $f(I)\cup g(I)$ also contains $K$, etc. But the length of $f(I)\cup g(I)$ is at most $|a|+|c|$ times the length of $I$. Continuing this process, we find that the length of the iterated images of $I$ under $\{f,g\}$ tends to zero. Since each of them contains $K$, it follows that $K$ does not contain any line segment; thus it's totally disconnected.
Case 2.: $|a|+|c|\ge 1$. This is when the invariant set is a line segment $[\alpha,\beta]$. The most direct proof is to find $\alpha,\beta$ explicitly and then verify the invariance of this line segment. It helps to realize that if a map ($f$ or $g$) has positive derivative, its fixed point is an endpoint of the invariant set. For negative derivatives you have to do some linear algebra, which I'll illustrate by an example:
$$f(x)=-x/2,\quad g(x)=(3-x)/2$$ Each map shrinks the line and flips it. Therefore, an invariant segment $[\alpha,\beta] $ must satisfy $$-\beta/2 = \alpha,\quad (3-\alpha)/2 = \beta$$ The solution is $\alpha=-1$, $\beta=2$. It is easy to check that $[-1,2]$ is indeed invariant.