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I am new to fractals and dimension theory, so please excuse any errors in my understanding.

For a set $F$, let $dim_b (F)$ be the box counting dimension of $F$, and $dim_H (F)$ be the Hausdorff dimension of $F$.

I understand that, assuming the box counting dimension exists, $dim_H (F) \leq dim_b(F)$, and that this inequality is in general strict.

I have now come across the open set condition in Falconer's book, and my question is as follows:

When $F$ satisfies the open set condition,(regardless of any other properties of $F$) is this an example of when $dim_H(F) = dim_b(F)$?

user3450230
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1 Answers1

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Yes, you are correct. However, much more is true. If $F$ is a strictly self-similar set, then $\dim_H(F) = \dim_b(F)$. This is true regardless of whether the open set condition is satisfied. This surprising fact was proven in 1989 by Falconer; it appears as example 2 on page 550 in his paper "Dimensions and measures of quasi self-similar sets".

The importance of the open set condition is that it makes self-similar sets much more easy to analyze. For example, when OSC is satisfied, then the common value $d$ of the Hausdorff and box dimension is given by the well known Moran equation, namely $$r_1^d + r_2^d + \cdots + r_n^d = 1.$$ In the absence of the open set condition, this formula is no longer applicable and, indeed, the exact common value of the dimensions may not be computable via known methods.

Mark McClure
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