Interesting Math

Who needs fingerprints when you have dimension prints?

I'd like to apologize for the weird typesetting that appeared in my article last issue; it looked fine when we compiled it electronically, but things messed up when we printed. Anyways, I suppose I should start with Hausdorff dimension for this issue. Some of you may already know what this is, for those of you who don't: Suppose U is a non-empty subset of n-dimensional Euclidean space; we define the diameter of U, denoted |U|, by taking the supremum of |x – y| over x, y in U. We say that {U_i} is a δ-cover of a set F if {U_i} is a countable collection of sets of diameter at most δ, whose union contains F. For a set F and a non-negative s, we define

 

H_δ^s(F) = inf{Σ_i |U_i|^s : {U_i} is a δ-cover of F}.

 

That is, we look at all covers of F by sets of diameter at most δ and attempt to minimize the sum of the s^th powers of the diameters. Notice as δ decreases, the set of allowable covers is reduced, so that the infimum increases, and hence approaches a limit as δ goes to 0. We define

 

H^s(F) = lim_{δ to 0} H_δ^s(F).

 

For those of you who know what a measure is, this is indeed a measure, called the s-Hausdorff measure. In addition, the special case of s = 1 gives the Lebesgue measure. It turns out by a relatively easy proof that for a fixed F, there exists a special value s such that H^t(F) = 0 for t > s, and H^t(F) = infinity for t < s. At this special value of s, which we call the Hausdorff dimension, the s-Hausdorff measure of F may be 0, infinity, or some positive finite value.

Hausdorff dimension is extremely useful for examining sets of "weird" size, like the Cantor set. The Lebesgue measure of this uncountable set is 0, which in a way tells us that there are a lot of points, yet not a lot of points. In this case, the Hausdorff dimension corresponds to the "dimension" I described in my article "The (approx.) 1.585^th Dimension", so that the Hausdorff dimension of the Cantor set is log 2/log 3, approximately 0.631. This corresponds to the notion that it's "more than" isolated points, with dimension 0, and "less than" a whole line, with dimension 1.

However, there's a problem with Hausdorff dimension — Two sets may have differing characteristics which we may want to distinguish, even though the Hausdorff dimensions are equal. We thus introduce a variant on this notion of dimension, known as dimension print. We now restrict our δ-covers to be rectangles (sides not necessarily parallel to the axes). For a rectangle U, with side lengths a(U) and b(U). Withough loss of generality, we will require a(U) to be the larger of the two (or the same, if U is a square). For F a set and s, tnon-negative numbers, we define

 

H_δ^s,t(F) = inf{Σ_i a(U)^sb(U)^t : {U_i} is a δ-cover of F}.

 

As before, we can get a measure, call it the s,t-Hausdorff measure by letting δ go to 0:

 

H^s,t(F) = lim_{δ to 0} H_δ^s,t(F).

 

In the special case t = 0, this is just the variant of s-Hausdorff measure where only rectangles are allowed, which turns out to be equal to the usual s-Hausdorff measure. We get a notion of "dimension" the same way we did for s-Hausdorff measures, to get the dimension print, except we get a region of points instead of a single point. You may notice that its intersection with the s-axis yields the Hausdorff dimension.

The Cantor set, which is formed by removing a proportion 1/3 from the middle of each interval on each step, will have Hausdorff dimension log 2/log 3, as mentioned. It is not too difficult to show that the middle-λ Cantor set (obtained by removing a proportion 0 < λ < 1 from the middle of intervals) will have Hausdorff dimension log 2/log(2/(1 – λ)), so that conceivablywe may construct Cantor-like sets of dimension 3/4 and 1/2. You can also show, after some work, that the Hausdorff dimension of the product of two Cantor sets of dimension 3/4 is 3/2 (this set is dust-like), and the Hausdorff dimension of the product of a line segment (dimension 1) and a Cantor set of dimension 1/2 is 3/2 (this set is stratified). (In general, the Hausdorff dimension of a product is at least as large of the sum of the Hausdorff dimensions, but this is not always the case). Now comes the cool part: the dimension prints are different for these sets. I'll invite you to try it out yourself, since a rigorous proof here will waste too much time and space.

Vince's problem of the issue

For a straight line, solid square, circle, dust-like Cantor set, or stratified Cantor set, the dimension prints are convex. Find a set whose dimension print is not convex.

Vince Chan
v2chan@student.math.uwaterloo.ca