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**Media_Man** Hello all. I am looking at branching binary trees in $\displaystyle \mathbb{R}^n$. Consider such a tree in which each branch is exactly twice the length of its two "child" branches. This can be defined iteratively as a self-similar fractal. I am trying to apply Hausdorff's rule of dimension represented by $\displaystyle N=L^D$, where N is the number of "balls" of a given size 1/L that can cover the figure at a given generation.

stage 1: N=2, L=2

stage 2: N=8, L=4

stage 3: N=32, L=8

stage n: $\displaystyle N=\frac12*4^n$, $\displaystyle L=2^n$

$\displaystyle D=\lim_{n\to\infty} \frac{\log(4^n/2)}{\log(2^n)}=\lim_{n\to\infty} \frac{n\log(4)-\log(2)}{n\log(2)}=\lim_{n\to\infty}\frac{\log(4)} {\log(2)}-\frac1n=2$

This seems a bit counterintuitive. It is easy to show that the linear length of the tree is infinite and the area containing it is finite, but I thought that since it appeared more complex than a simple line it would have dimension slightly greater than one. To think of this as a perfectly 2-dimensional object seems absurd.

If each left branch is assigned 0, and each right branch assigned 1, then the tree is topologically equivalent to the set of all possible infinite strings of 0 and 1, making it slightly larger than [0,1], furthering my suspicion that its dimension was slightly greater than 1.

Can someone verify this result?