The Cantor Set

If we think of a point as having dimension 0, a line or curve having dimension 1, a plane or curved surface having dimension 2, and so on, then a natural question to ask is: do there exist objects having dimensions that are not integers? The surprising answer is yes, as the following example of Georg Cantors` illustrates. We initially start out with a line, L = [0,1], representing the closed unit interval of length 1 containing it`s endpoints. From this we remove the open middle third (1/3, 2/3) subdividing L into two line segments: [0,1/3] U [2/3,1] each a third of the length of the original line. From these we again remove the open middle-thirds of Ll and Lr, these being (1/9, 2/9) and (7/9, 8/9) respectively, giving the third row in the figure having four line segments each 1/3 x 1/3 = 1/9th the length of L. We then continue this procedure over and over again ad infinitum until in the limit what we are left with is the Cantor set in the `final row`. Looking at the structure formed by all of the rows we see this iterative procedure gives us two self-similar left and right copies a third of the size of the whole structure starting at R1, and at R2 we get the same thing happen with two pairs of self-similar left and right copies a ninth in size of the whole structure, one pair below the left of R1, and the other pair below the right of R1. This endless self-similarity shows we are dealing with a fractal set.

________________________________ R0 __________ __________ R1 ____ ____ ____ ____ R2 __ __ __ __ __ __ __ __ R3 . . . . . . . . . . . . . . . . R4 : : : : : :

If we label the rows R0, R1, R2, R3, etc. we can see that for the (k+1)th row Rk, there are 2^k many line segments of length (1/3)^k. So for row Rk the length of all these segments is:

length(Rk) = 2^k x (1/3)^k = (2/3)^k

So R0 has one piece of length (1/3)^0 = 1, and R7, say, has length 2^7 x 1/3 x 1/3 x . . . x 1/3 = (2/3)^7 = 128/2187 ~ 0.05853.

So it would appear at each stage of the iteration the length of the segments is tending towards zero since lim_{k->oo}{(2/3)^k} = 0, and we say the length, or measure, of the remaining set is zero, or analogously that the Cantor set can not contain any interval of non-zero length. We show this by summing all the intervals taken out:

1/3 + 2/9 + 4/27 + 8/81 + . . . = 1/3 x sum_{n=0}^{oo} (2/3)^n = 1/3(1/(1-2/3)) = 1/3 x 3/1 = 1

But this is not to say that this set has no points in it in fact not only does it contain points but it contains uncountably many.

(For those new to this, the set of natural numbers {1, 2, 3, 4,. . . } are said to be countable as they can be put in a bijective (one-to-one) correspondence with themselves: just make the bijection 1<-1, 2<-2, 3<-3, . . . and keep counting! However as Cantor noted by his ingenious diagonal argument the real numbers can not be lined up in so efficient a manner. He showed that any list purporting to contain all the real numbers could be shown to be missing at least one real number by constructing this from the list itself, and so this list, or indeed any list could never list all the reals in such a manner as the diagonal trick could be invoked to procure yet another real number hiding in it. The diagonal argument went like this: If you had a list of all the real numbers

1 <-> 0.12345678. . . 2 <-> 0.13456789. . . 3 <-> 0.14567891. . . 4 <-> 0.15678912. . . 5 <-> 0.16789123. . .

and took the diagonal 0.13579. . . in this list, and then subtracted one from each of the digits to give 0.02468. . . this number will differ from each number in the list in at least one-decimal place and as a consequence will not appear in the list which is then incomplete proving the real numbers can not be put into a one-to-one correspondence with the integers.

So what did all this mean: we can count the natural numbers by our voice, say start at 1 and keep going, oh and never, ever stop, then by doing this you will indeed have listed all the natural numbers, although it would take you longer than any predefined time to count them all! What of the real numbers? They could not be counted in the same way and so were uncountable as they could not be put into a one-to-one correspondence with the natural numbers. This led to the idea of different orders of infinity and brought about the need for the so-called aleph-numbers to indicate the order of infinity such infinite sets possessed. The cardinality (or size) of the natural numbers is aleph-0, while the real numbers have cardinality aleph-1.)

Back to the Cantor set: why then is it not empty when it has zero-measure? Well, when we initially took the middle-third away from our initial line L we removed the open interval (1/3,2/3). Now a look at the diagram should convince you that at each subsequent step the end-points 1/3 and 2/3 from step 1 are never removed, and neither are the end-points 1/9, 2/9, 7/9, 8/9 from step 2 ever removed, and at step k, 2^k new end points are added that are never removed since at each step an internal open set is removed, never touching the end points. So the amount of end points left is given by an infinite sum of ascending powers of 2 with an initial sum of the two endpoints 0 and 1 of [0,1]:

2 + sum_{j=1}^{oo} 2^j = 2 + 2 + 4 + 8 + 16 + 32 + . . . = oo However the summands can be put into a one-to-one correspondence with the natural numbers, and so this infinite sum has cardinality aleph-0. But most members of the Cantor set are not endpoints, and the strange thing is that as many points remain in the set as those taken away, which is all of them when you think about it. Have a think about where the point 1/4 goes is it ever got rid of in one of the steps? When we look at fractions in the ternary system (like binary but with base 3 instead of base 2) employing the digits 0, 1, and 2, i.e., those of the form:

a_1(1/3) + a_2(1/9) + a_3(1/27) + a_4(1/81) + . . . a_i in {0, 1, 2}

we find that the numbers in the interval (1/3, 2/3) are precisely those numbers that in the ternary system have a 1 to the right of the ternary place. (See if you can prove this? If you can extend the result to the other open intervals which are removed. What does this tell you about the Cantor set.) The Cantor set qualifies as a fractal due to its self-similarity and it can be shown to have a Hausdorff fractal dimension given by ln 2/ln 3 = log_{3}(2) = 0.6309, and so is closer to being a 1-dimensional line than a 0-dimensional point.