Can we check whether a Cantor set is self-similar or not?

Question

Given a Cantor set $C$ on the real line, do we have some ways to determine whether it is self-similar or not? In particular, how can we check that $C$ is not self-similar?

Edited:
Definition: Let $\{f_i\}_i$ be a family of contraction maps, i.e. $|f_i(x)−f_i(y)|=r_i|x−y|$ where $0<r_i<1$. $C$ is self-similar if there are such maps so that $E=\bigcup_i f_i(E)$.

To make things concrete, we may consider the following central Cantor set as an example.
Suppose $r_k = \frac{1}{k+2}$ for $k \geq 1$. Let $I_e = [0,1], I_0 = [0, r_1], I_1 = [1-r_1 , 1]$. For each $k \geq 1$, $w \in \{0,1\}^k$, let $I_w$ be a subinterval at level $k$. We take $I_{w1} , I_{w2}$ to be two subintervals of $I_w$ placed on the left and right with lengths $$|I_{w1}| = |I_{w2}| = |I_w| \cdot r_{k+1} . $$ Then $C= \bigcap_{k=1}^\infty \bigcup_{w \in \{0,1 \}^k} I_w$ is a Cantor set. Is it not self-similar?

Answer 1

First of all, self-similar sets are not necessarily topological Cantor sets. For example, closed intervals are self-similar, as it is easy to check. However, if we assume (using your notation) that the pieces $f_i(E)$ are pairwise disjoint, then $E$ is a Cantor set. This is known as the strong separation condition. When dealing with self-similar sets, a weaker separation that is often used and assumed is the open set condition: there exists a nonempty open set $U$ such that $f_i(U)\subset U$ and $f_i(U)$ are pairwise disjoint. Closed intervals are self-similar sets under the open set condition, so the open set condition is not enough to guarantee a self-similar set is a Cantor set.

In general, there isn't a simple way to determine whether a given compact set (Cantor or not) is self-similar. However, self-similar sets are more regular than arbitrary compact sets. For example:

1. Different concepts of fractal dimension such as Hausdorff and box-counting (Minkowski) dimension coincide for arbitrary self-similar sets.

2. If a self-similar set $E$ is not a point, then it has strictly positive Hausdorff dimension. This can be seen as follows: after reordering, let $f_1, f_2$ be two of the maps generating $E$ with distinct fixed points (if all the maps have the same fixed point, then $E$ equals that fixed point). Let $F$ be the self-similar set corresponding to the maps $f_1^n, f_2^n$, where $n$ is large enough that $f_1^n(I)$ and $f_2^n(I)$ are disjoint intervals, where $I$ is the closed interval joining the fixed points of $f_1, f_2$. Then $F$ is a self-similar set with the strong separation condition, for which there is a well-known formula for the Hausdorff dimension which is always strictly positive (this can also be checked directly for $F$), and $F\subset E$.

3. Under the open set condition (but not in general) self-similar sets have positive and finite Hausdorff measure in their dimension.

The particular set $C$ defined in the question has Hausdorff dimension $0$. This can be seen by using the natural covers by sets $I_w, w\in\{0,1\}^k$, and is essentially a consequence of the fact that $r_k\to 0$, so that the relative size of intervals of generation $k+1$ inside intervals of generation $k$ tends to $0$. Since $C$ is not a point, by 2. above, $C$ cannot be self-similar.