# Hausdorff Dimension

We work out the area of a square by multiplying the side length by itself. We think of the area as a measure of the size of the square. This value remains meaningful however we think of the square. Whether as a region of R^2 (the plane), or as a subset of the complex numbers, viewed as the Argand diagram, or as an object sitting in the 3-dimensional real world.

But this then becomes a problem if we want to keep using the word ‘measure’. Because the standard Lebesgue measure on R^2 coincides with this definition of area, but if we are in R^3, then the measure of a square is zero. Of course, we are really articulating the idea that dimension is key. The area of a square may be large or small, but its volume will always be zero. Similarly, if we were to try and measure the ‘length’ of a square, in some sense, the only sensible answer could be that the length is infinite. Why? Well, any measure must satisfy the constraint of being increasing under containment. In this sense, any square contains an arbitrarily large number of disjoint lines of length k/2, where k is the side-length of the square, and so the only possibility for the length measure is infinite.

So we realise that dimension is key to resolving this problem. And we are clear in our minds that a square is two-dimensional, since it can be embedded in R^2 but not in R. But this is all rather vague. We would ideally like to have a definition of dimension that is independent of choice of embedding. Not least, the property used in the case of the square has an algebraic flavour that is clear here, but which will certainly not generalise. In vector space terminology, a square is not a subspace, but it certainly generates a subspace of dimension 2. But we want a definition that exploits topological properties. Not least because, for example, the surface of a sphere sits in R^3 and indeed generates R^3 as a vector space, but is very much two-dimensional under any sensible definition.

Ideally we want a way to find the dimension of a space, given only its metric properties. Heuristically, we feel that a measure is the metric to the power of the dimension, so this would resolve all the problems we have been considering.

The rest of this post will talk about one possible solution, the Hausdorff dimension, which specifies the dimension of a metric space in an embedding-independent manner. The motivation is as follows. We can cover a square with lots of smaller squares. Indeed, a square of side-length 1 can be decomposed into $(1+o(1))r^{-2}$ squares of side-length r, as $r \rightarrow 0$. Although they do not tessellate as nicely, the same result holds for small balls. Such a square can be covered by $\Theta(r^{-2})$ balls of radius r. This is a useful observation, because a ball is specified only by metric properties of the space.

From this idea, we obtain a way of constructing a family of measures on the Borel sets of a metric space, which specify the size of subsets of any dimension. We define the $\alpha$-Hausdorff measure of A by:

$\mathcal{H}^\alpha(A):= \liminf_{\delta\downarrow 0} \left\{\sum_{i=1}^\infty \text{diam}(E_i)^\alpha:(E_i)\text{ a covering of }A, \text{ diam}(E_i)<\delta.\right\}$

In other words, we take a covering of A by small sets of diameter at most $\delta$. These might as well be balls at this point. Then we calculate the infimum of the $\alpha$-dimensional measure of these balls, across all such coverings. Letting $\delta\downarrow 0$ gives a measure of the $\alpha$-dimensional volume required to cover a subset with small $\alpha$-dimensional balls. We expect this to be zero when the set has dimension less than $\alpha$, and infinite when the set has dimension greater than $\alpha$. Naturally then, we define the Hausdorff dimension by

$\text{dim}(E)=\inf\{s\geq 0: \mathcal{H}^s(E)=0\}.$

An inquisitive child might well ask the following question: we are happy with the notion of 2-dimensional space, and 3-dimensional space, but what about other non-integer values? Can we have a 2.5-dimensional space, and what does it mean?

As an example of why the integers might not be enough to specify dimension, it is useful to consider fractals. A fractal is some object that has properties of self-similarity, which might mean that, for example, it is impossible to tell how zoomed in a view is. These are typically constructed as the limit of an iterative procedure where each unit present is broken down into several smaller units in a self-similar way. Examples include the Koch snowflake, where the middle third of a line is replaced the other two sides of the equilateral triangle with base equal to that middle third. The same operation is then successively applied to each smaller line segment and so on. The Sierpinski gasket is constructed by applying a similar operation to the areas of triangles.

In the case of the Koch snowflake, in each operation, the total length of the object increases by a factor of 4/3, and so the length of the limiting object is infinite, even though it is constructed from lines, and the endpoints are still distance 1 apart. So it is not clear what the dimension of this object should be. It doesn’t fill space to quite the same extent as a plane area, but by self-similarity, it fills space a bit more than a line locally to any given point.

It’s perhaps easiest to consider as an example for calculation the Cantor Set. Recall this is constructed by starting with the unit interval [0,1], then removing the middle third to leave [0,1/3] u [2/3,1] and then successively removing the middle third of every remaining interval. What remains are precisely those reals which have no ‘1’ in their ternary expansion. Unsurprisingly, this is nowhere dense, and has zero Lebesgue measure, since the measure after the nth iteration is $(\frac23)^n\rightarrow 0$.

However, we can calculate its Hausdorff dimension, by taking advantage of its self-similarity. We suppose that this dimension is s, and we consider the s-Hausdorff measure. It is clear from the definition that the measure of a disjoint union of sets is equal to the sum of the individual measures. So we apply this to the decomposition of the Cantor set C as:

$C=(C\cap [0,\frac13] )\cup (C\cap [\frac23,1]).$

Note that $C\cap[0,\frac13]$ and $C\cap[\frac23,1]$ are isometric to $\frac{C}{3}$. So a covering of C with diameter less than d induces a covering of C/3 with diameter less than d/3, and the s-dimensional volume of the covering sets will be scaled by a factor of $3^{-s}$. We conclude that

$H^s(C)=2H^s(\frac{C}{3})=\frac{2H^s(C)}{3^s}$

$\Rightarrow H^s(C)=\frac{\log 2}{\log 3},$

giving us an example of a non-integral Hausdorff dimension. It turns out that a space with any real Hausdorff dimension can be constructed in this way by adjusting the constants of self-similarity in the iterative construction.

Unsurprisingly, once we are comfortable with objects which are self-similar, it makes sense to look at the Hausdorff dimension of random objects which are self-similar in distribution. The primary example is Brownian motion. Again, we might assume a priori that since Brownian motion is a function of a one-dimensional variable, typically time, it should have dimension one. In fact, recalling that the variation of BM is infinite, it is clear that we have the same situation as in the Koch snowflake example discussed above. Some of the tools involved to deal with the dimension of BM in the two cases d=1 and d=>2 are interesting and non-trivial, so I will postpone discussion of this until a possible further post.