Self similar set which does not fulfill the open set condition

Question

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?

See these notes for some discussion of this. – Jim Belk Jan 23 '14 at 03:49 — Jim Belk, Jan 23 '14 at 03:49
The Cantor set isn't open... – Martín-Blas Pérez Pinilla Feb 01 '14 at 09:11 — Martín-Blas Pérez Pinilla, Feb 01 '14 at 09:11

score 3 · Accepted Answer · answered Feb 01 '14 at 09:07

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.

Self similar set which does not fulfill the open set condition

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