Fractals and Dimension - Key Proof

ProofProof of Moran's Formula for Cantor Set Dimension

We prove that the middle-third Cantor set has Hausdorff dimension $d = \log 2/\log 3$ using Moran's formula.

Setup: The Cantor set $C$ is constructed by removing the middle third of $[0,1]$ repeatedly. It's the attractor of the IFS with two similarities:

$S_0(x) = \frac{x}{3}, \quad S_1(x) = \frac{x+2}{3}$

Both have contraction ratio $r = 1/3$ , and $C = S_0(C) \cup S_1(C)$ .

Step 1: Covering argument

At stage $n$ , the Cantor set is covered by $2^n$ intervals, each of length $(1/3)^n$ . Any covering by balls of diameter $\delta \leq (1/3)^n$ must use at least $2^n$ balls to cover all these intervals.

For the $s$ -dimensional Hausdorff measure with $\delta = (1/3)^n$ :

$H^\delta_s(C) \geq 2^n \cdot \left(\frac{1}{3}\right)^{ns} = \left(\frac{2}{3^s}\right)^n$

Step 2: Critical dimension

If $s < \log 2/\log 3$ , then $2/3^s > 1$ , so $(2/3^s)^n \to \infty$ . Thus $H^s(C) = \infty$ for $s < \log 2/\log 3$ .

If $s > \log 2/\log 3$ , then $2/3^s < 1$ . We can cover $C$ at stage $n$ with $2^n$ intervals of length $(1/3)^n$ :

$H^{(1/3)^n}_s(C) \leq 2^n \cdot \left(\frac{1}{3}\right)^{ns} = \left(\frac{2}{3^s}\right)^n \to 0$

Thus $H^s(C) = 0$ for $s > \log 2/\log 3$ .

Step 3: Dimension conclusion

By definition of Hausdorff dimension:

$\dim_H(C) = \inf\{s : H^s(C) = 0\} = \sup\{s : H^s(C) = \infty\} = \frac{\log 2}{\log 3}$

Step 4: Verification of Moran's formula

The dimension satisfies $\sum_{i=1}^2 r_i^d = 1$ :

$2 \cdot \left(\frac{1}{3}\right)^{\log 2/\log 3} = 2 \cdot \frac{1}{3^{\log 2/\log 3}} = 2 \cdot \frac{1}{2} = 1$

using $3^{\log 2/\log 3} = (3^{\log 3})^{\log 2/\log 3 \cdot 1/\log 3} = e^{\log 3 \cdot \log 2/\log 3} = e^{\log 2} = 2$ .

Conclusion: The Cantor set dimension equals the unique solution to $2r^d = 1$ with $r = 1/3$ , confirming Moran's formula.

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This proof demonstrates the self-similar scaling principle: at each level, the set consists of copies scaled by $r = 1/3$ . The dimension $d$ is chosen so that the $d$ -dimensional "volume" (Hausdorff measure) remains finite and positive. Too small $d$ gives infinite measure, too large gives zero measure, and the critical $d$ balances perfectly.

Remark

The proof generalizes to any self-similar set satisfying the open set condition. The key insight is that self-similarity at ratio $r$ with $k$ pieces requires dimension $d$ such that $k r^d = 1$ —the scaling of $d$ -dimensional volume under contraction by $r$ applied $k$ times returns to the original scale. This geometric balance determines dimension uniquely.

ExampleExtending to Sierpinski Gasket

For the Sierpinski gasket, three pieces each scaled by $r = 1/2$ :

$3 \cdot (1/2)^d = 1 \Rightarrow 2^d = 3 \Rightarrow d = \frac{\log 3}{\log 2}$

The same covering argument applies: at stage $n$ , there are $3^n$ triangles of side length $(1/2)^n$ , yielding the critical dimension $\log 3/\log 2$ where Hausdorff measure is finite and positive.

This proof exemplifies how self-similarity leads directly to non-integer dimensions. The geometric iteration creates structure at all scales, and dimension measures how this structure fills space. Understanding this principle is key to intuiting fractal geometry and applying dimensional analysis to chaotic attractors.