Informally, a set is considered self similar if it consists of smaller copies of itself. If this set fulfills the so called open set condition, one can easily calculate the Hausdorff Dimension (see for example Wikipedia).
It is easy to find some proofs on this topic. Most of them state more or less direct that most self similar sets fulfill the open set condition. However, I was unable to find any counter example.
So my question is: Is there any example of a self similar set which does not fulfill the open set condition. If yes: Which set?
Yes. Take, for example, $f_1(x)=rx$ and $f_2(x)=rx+(1-r)$, where $1/2<r<1$, and let your IFS be $\{f_1,f_2\}$. The invariant set of the IFS is then the unit interval $[0,1]$ since $$f_1([0,1]) \cup f_2([0,1]) = [0,r] \cup [1-r,1] = [0,1].$$ Thus, the similarity dimension of $-\log(2)/\log(r)$ is strictly larger than the actual dimension of the attractor so the open set condition cannot be satisfied.
More generally, any IFS with a similarity dimension larger than the ambient space on which the IFS operates cannot satisfy the open set condition.