The word “dust” has appeared in the title of every post in this series, and we will finally begin to see why. Many counterintuitive marvels remain to be unmasked in the counterintuitive worlds of cardinality and measure theory. Georg Ferdinand Ludwig Philipp Cantor, the mathematician who laid the foundations for the modern understanding of infinity, was responsible for the discovery of some of these truly unbelievable marvels.
There seems to be a whole sub-genre of math blog posts that focus on defining or presenting the Cantor set at different levels of rigor! For our purposes, I will offer two descriptions (without proof).
- The Cantor set can be described quite “simply” as the set of all real numbers between 0 and 1 whose ternary representation contains only 0 and 2. (We use the decimal place value system in everyday life, while computers use the binary, using only 0 and 1.) The ternary place value system uses 0, 1, and 2 as the only digits. So the number 0.1 in ternary represents the fraction \( \frac{1}{3} \), and 0.22 the fraction \( \frac{8}{9} \), and so on. It is also worth noting that just as we take the decimal 0.999… to be equal to the number 1, we must also take the ternary number 0.0222… to be equal to the ternary number 0.1, and so on.
- The Cantor set can also be seen as the limit of an infinite process of cutting up the closed interval \( [0, 1] \). The rule is simple: at every stage, delete the middle third (excluding the endpoints) of every remaining interval. Repeat ad infinitum. The first few stages of such a process are shown in this picture.
- Its counting measure is infinite. This is quite easy to see with the ternary representations: there is a very obvious bijective correspondence between \( \mathbb{N} \) and subset \( {0.2, 0.22, 0.222, ...} \) (in ternary representation). Since this subset is infinite in terms of counting measure, the whole Cantor set itself must also be infinite by this measure.
- In terms of cardinality, it is uncountable. This is not obvious at all. However, it can be proven by constructing a surjective map from the Cantor set to \( [0, 1] \). (There are clever ways to do this, but setting one up and showing that it is in fact surjective is probably a little too detailed for now.) This will show that the Cantor set has at least the same cardinality as \( [0, 1] \). Since it’s also a proper subset of \( [0, 1] \), it cannot have a cardinality greater than \( [0, 1] \). Therefore, it has the same cardinality as \( [0, 1] \); i.e., it is uncountable.
- Its Lebesgue measure is zero. To see this, we begin by noting that the Lebesgue measure of \( [0, 1] \) is 1. We then add up the Lebesgue measure of each of the deleted subintervals: \( \frac{1}{3} + 2 \frac{1}{3^2} + 2^2 \frac{1}{3^3} + ... \). (It is not immediately obvious that adding up an infinite sum of Lebesgue measures is a legitimate procedure: infinities can be a nasty thing. However, because none of the deleted intervals overlap in any way, the addition process ends up being legitimate.) This infinite sum is the geometric series \( \frac{1}{3} \Sigma_{i=0}^{\infty} \left( \frac{2}{3} \right)^i \), which evaluates to exactly 1. Thus, the Cantor set has Lebesgue measure zero!
- It is nowhere dense in \( [0, 1] \). How does that make sense?
- Well, for one, the Cantor set is a closed set in \( [0, 1] \). One way to see this is to “intuit” (or to formally prove) that every sequence of numbers in the Cantor set converges to some point in the Cantor set. (It’s easier to “see” this if you think about our description of the Cantor set in terms of numbers with ternary representations containing only 0 and 2.) Another way is to remember that the complement in \( [0, 1] \) of the Cantor set is a countable union of open intervals. Since this complement is open, the Cantor set itself must be closed in \( [0, 1] \). (This is of course false for \( \mathbb{Q} \).)
- But like \( \mathbb{Q} \), it’s also true that for any number in the Cantor set, there will always be some number from \( [0, 1] \) arbitrarily close to it that does not belong to it. Thus, the interior of the Cantor set is empty. By definition, a nowhere dense set is one the interior of whose closure is empty; and so we see that the Cantor set is nowhere dense in \( [0, 1] \).
- And as a cross-check, the complement of the Cantor set in \( [0, 1] \) is both open (since it’s a countable union of open intervals) and dense in \( [0, 1] \) (because, well, it includes almost all of \( [0, 1] \) anyway, and for those points that it doesn’t contain (i.e., for the points belonging to the Cantor set), it’s possible to get arbitrarily close to them!
Thus, the Cantor set is a remarkable beast: uncountable, nowhere dense, and Lebesgue measure zero. But there are even cooler sets, known as the “fat” Cantor sets, or more formally as the Smith-Volterra-Cantor sets. Like the Cantor set, these are uncountable and nowhere dense; unlike the Cantor set, they have positive Lebesgue measure. Indeed, it’s possible to define fat Cantor sets with Lebesgue measure as close to 1 as we want!
When we constructed the Cantor set by removing one-third of each interval at each step, we were in a sense removing as much as we could if we wanted to keep our intervals equal in length at each step. We could, however, remove less than one-third. If, for instance, we removed the central one-sixth of each interval at each step, we would end up removing a total length of \( \frac{1}{6} + 2 \frac{1}{6^2} + 4 \frac{1}{6^3} + ... = \frac{1}{4} \). This gives us a set of measure \( \frac{3}{4} \).
In general, if we remove \( x^n \) from each interval on the \( n^{\text{th}} \) step, we will end up with a fat Cantor set of Lebesgue measure \( \frac{1-3x}{1-2x} \). When \( x = \frac{1}{3} \), we get the “regular” Cantor set of measure zero; and clearly \( \lim_{x \to \infty} \frac{1-3x}{1-2x} = 1 \). Ta-da!
Why “dust”?
I’ll wrap this all up with one quick remark: why “dust”? The answer is: there ain’t no good answer. There are higher-dimension analogs of the Cantor set which retain its strange properties of nowhere denseness and Lebesgue measure zero. Approximate visual representations of these beautiful fractals resemble stationary clouds of dust just hanging out there in a cube, spread out around it but taking up no space at all. Stunningly beautiful.
When we constructed the Cantor set by removing one-third of each interval at each step, we were in a sense removing as much as we could if we wanted to keep our intervals equal in length at each step. We could, however, remove less than one-third. If, for instance, we removed the central one-sixth of each interval at each step, we would end up removing a total length of \( \frac{1}{6} + 2 \frac{1}{6^2} + 4 \frac{1}{6^3} + ... = \frac{1}{4} \). This gives us a set of measure \( \frac{3}{4} \).
In general, if we remove \( x^n \) from each interval on the \( n^{\text{th}} \) step, we will end up with a fat Cantor set of Lebesgue measure \( \frac{1-3x}{1-2x} \). When \( x = \frac{1}{3} \), we get the “regular” Cantor set of measure zero; and clearly \( \lim_{x \to \infty} \frac{1-3x}{1-2x} = 1 \). Ta-da!
Why “dust”?
I’ll wrap this all up with one quick remark: why “dust”? The answer is: there ain’t no good answer. There are higher-dimension analogs of the Cantor set which retain its strange properties of nowhere denseness and Lebesgue measure zero. Approximate visual representations of these beautiful fractals resemble stationary clouds of dust just hanging out there in a cube, spread out around it but taking up no space at all. Stunningly beautiful.
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