Suppose you have the unit interval $[0,1]$. For the first iteration you remove the segment $(1/5,3/5)$. So you are left with two intervals of lengths $1/5$ and $2/5$. You now repeat the process on the remaining intervals to get a set $F$. What is the Hausdorff dimension of this set and can you prove this rigorously.

I think you should note that in level $k$ (starting counting at with $[0,1]$) one has 2^k intervals. The minimum length of the interval is $(\frac{1}{5})^k$ and the maximum length fo an interval is $(\frac{2}{5})^k$

I am having problem with this question as the usual method of solving $\sum c_i^s=1$ doesnt work.

This links to my previous question which is on the same topic.

  • $\begingroup$ I would like to this question without using the standard formula in a similiar fashion to Falconer in Example 2.7 of Fractal Geometry $\endgroup$ – Permian Oct 13 '14 at 12:47

You have a set that consists of two copies of itself - one scaled by the factor $1/5$ and one scaled by the factor $2/5$. The similarity dimension is the unique number $s$ satisfying $$(1/5)^s + (2/5)^s = 1.$$ Again, this characterizes the dimension uniquely, though doesn't express it in closed form. In general, an equation of this type can be solved in closed form only when the scaling factors are exponentially commensurable. I provide some examples in my answer to this question.

Of course, we can compute your dimension as closely as we want - it is approximately $0.56389552425993647949$.

  • $\begingroup$ I have been told that in this instance you cannot use the formula $\sum_{i} c_i^s = 1$ as the conditions to use it are not met $\endgroup$ – Permian Oct 14 '14 at 8:58
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    $\begingroup$ @1234 By "repeat the process on the remaining intervals", I assume that you retain the left-most 1/5 of each and the right-most 2/5 of each, including endpoints. If so, the process produces two smaller copies of the whole - one scaled by the factor 1/5 on the left and one scaled by the factor 2/5 on the right. The result is the invariant set of the IFS $\{x/5,2x/5+3/5\}$ which satisfies the open set condition with the open unit interval so the conditions to apply the formula are satisfied. Have I misinterpreted the instruction to "repeat the process on the remaining intervals". $\endgroup$ – Mark McClure Oct 14 '14 at 12:26
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    $\begingroup$ no you have not $\endgroup$ – Permian Nov 4 '14 at 12:03

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