# C0 - C0 = [-1, 1]

The Cantor set C0, while full of holes, has a remarkable property that for any real number in the interval [0, 1] there exists a pair of numbers from the Cantor set whose difference is exactly that number. Taking the difference in the reverse order we see that any number in the interval [-1, 1] is also representable as the difference of two terms of the Cantor set.

For any two sets A and B, the difference set A - B (see also Addition of Shapes) is defined as

A - B = {a - b : a∈A and b∈B}

The aforementioned property of the Cantor set is then expressed concisely:

C0 - C0 = [-1, 1]

In geometric terms, any segment not longer than 1 (unit) can be positioned on the number line such that both its endpoints will rest on the Cantor set. In two dimensions, two such segments can be laid independently - one on the x-axis, the other on y-axis. In other words, any rectangle with side lengths not exceeding 1 can be positioned on the plane so that each coordinate of each of its corners belong to the Cantor set. The same is true for parallelepiped in three dimensions and their generalizations in higher dimensions.

Back to the plane, the set Sx = C0×[0, 1] consists of vertical segments with feet at the Cantor set. (A×B is the direct product of the sets A and B.) Similarly, Sy = [0, 1]× C0 is a collection of horizontal segments growing rightward from a Cantor set on the y-axis. The union of the two sets, S = Sx ∪ Sy is a net-like set with holes all over. Nonetheless, set S will cover any (wire frame) rectangle with side lengths not exceeding 1. A similar construction works in 3D for wire frame parallelepipeds. Now let's prove the statement C0 - C0 = [-1, 1].

### Proof #1

I'll use the ternary characterization of the Cantor set: a real number from [0, 1] belongs to the Canter set iff it has a ternary expansion that contains only digits 0 and 2. The proof consists of two steps. First, we'll see that any number with a finite ternary expansion can be written as the difference of two elements from the Cantor set. Next, we'll invoke a compactness argument to cover numbers with infinite ternary expansion.

The first step is inductive on the length of the ternary expansion. There are just 3 fractions of length 1:

0.0 = 0.0 - 0.0, 0.1 = 0.2 - 0.0222..., 0.2 = 0.2 - 0.0

Expansions of length 2 are obtained by appending a digit to expansions of length 1. The differences are modified easily if the additional digit is 0 or 2. For example, 0.1 = 0.2 - 0.0222... becomes 0.12 = 0.22 - 0.0222... and 0.0 = 0.0 - 0.0 becomes 0.02 = 0.02 - 0.0. With the additional digit equal to 1, we have to consider two cases:

1. 0.0 = 0.0 - 0.0 and 0.2 = 0.2 - 0.0 lead to 0.01 = 0.02 - 0.0022... and 0.21 = 0.22 - 0.0022..., respectively.

2. 0.1 = 0.2 - 0.0222... leads to 0.11 = 0.21 - 0.0222... = 0.2 - 0.02.

More generally, let a be a number from [0, 1] with a finite ternary expansion of length n: a = 0.a1a2a3...an and an≠0. We'll construct two elements b and c of the Cantor set such that a = b - c. Additionally,

1. the expansion of b is not longer than that of a, and
2. the expansion of c is either infinite or finite. In the latter case, it's no longer than a. In the former case, all digits ci starting with i = n+1 equal 2.

We may assume that the two numbers with the required properties have already been found for 0.a1a2...an-1:

 (1) 0.a1a2...an-1 = 0.b1b2...bn-1 - 0.c1c2...cn-1, cn-1≠0 (but bn-1 may be 0), or (2) 0.a1a2...an-1 = 0.b1b2...bn-1 - 0.c1c2...222..., where ci = 2, for i = n+1, n+2, ...

If an is 0, no modifications in either (1) or (2) are necessary. If an = 2, just set bn = 2. Now, assume an = 1. In case (1), set

0.a1a2...an-11 = 0.b1b2...bn-12 - 0.c1c2...cn-10222...

In case (2), set

0.a1a2...an-11 = 0.b1b2...bn-1 - 0.c1c2...cn-1cn

As an example, for a = 0.1202112, we have successively

 0.1 = 0.2 - 0.0222... 0.12 = 0.22 - 0.0222... 0.12 = 0.220 - 0.0222... 0.1202 = 0.2202 - 0.0222... 0.12021 = 0.2202 - 0.02222 0.120211 = 0.220202 - 0.022220222... 0.120211 = 0.2202022 - 0.022220222...

This part of the proof is entirely constructive: for any finite ternary fraction a, we have an algorithm that finds two elements - b and c - from the Cantor set such that a = b - c. Now lets turn to the infinite fractions.

Let A be a real number from [0, 1] with an infinite ternary expansion. A = 0.a1a2a3... is a limit of its partial expansions Ak = 0.a1a2...ak. From the foregoing discussion, there exist members of the Cantor set Bk and Ck such that Ak = Bk - Ck.

The Cantor set is a compact subset of the number line. On the number line (and, more generally, in metric spaces), compactness is equivalent to the following property (known as sequential compactness):

A set S is sequentially compact if every sequence {si}⊂S contains a subsequence {sij} convergent in S.

Consider the sequence {Bi}. It contains a convergent sequence {Bij}. The sequence {Cij}, in turn, contains a convergent subsequence {Cijr}. Since any subsequence of a convergent sequence is also convergent (and to the same limit), the subsequence {Bijr} is also convergent. Denote their limits B and respectively C. B and C are both elements of the Cantor set and, additionally, A = B - C, which completes the proof. ### Proof #2

We'll show that every line y = x + a, where a belongs to [-1, 1], intersects the product C0×C0. If (c, b) is the point of intersection, we automatically have b = c + a or a = b - c, where both b and c belong to the Cantor set.

The product C0×C0 can be constructed similarly to the Cantor set itself. Draw a 1×1 square, and divide it into 9 squares of side 1/3. Remove 5 "central" squares. Repeat the process in each of the remaining four squares. That is, divide each of them into 9 equal squares of size 1/9 and remove 5 "central" squares. And so on. It's not hard to convince oneself that the line y = x + a intersects at least one of the 4 "corner" squares with side 1/3. Denote any of these as S1. By the same token, any line with slope 1 that intersects the square S1 intersects at least one of the corner squares of side length 1/9 that are contained in S1. Denote one of them S2. S2⊂S1. In this manner we obtain a decreasing sequence {Si} of (closed) squares. Since the sequence is decreasing, the intersection of any finite subsequence is not empty. (It's simply the smallest square in the subsequence.) This is another facet of compactness. A sequence of closed sets is said to possess a finite intersection property, if any finite subsequence has a non-empty intersection. A topological space is compact iff the intersection of every sequence of closed sets with the finite intersection property is not empty.

The original unit square being closed and bounded is compact. Therefore, the squares {Si} have a common point. This point belongs to both C0×C0 and the line y = x + a.

### Proof #3

This proof by Stuart Anderson appears on a separate page.

### Remark

Following an observation by Paolo Di Muccio, Stuart Anderson also showed that the two sequences {Bn} and {Cn} derived algorithmically in Proof #1, are both Cauchy and hence convergent. Since the Cantor set is closed, it contains both limits. The property of compactness of the set appears unnecessary for the proof. In Stuart's words:

The algorithm you use to generate the digit sequences never changes a digit once it has been set. Hence if one wants to think of the finite digit sequences as partial sums of the finished infinite sequence, one may do so.

Then for any N > 0, and for any m and k both > N, the partial sums 0.b0b1 ... bk and 0.b0b1 ... bm agree to at least N places, and so differ by at most 3-N. But this clearly shows that the sequence is a Cauchy sequence (for any ε > 0, simply choose N so 3-N < ε), hence it converges independent of any compactness assumption. Of course the same applies to {cn} well. ## References

1. B. R. Gelbaum and J. M. H. Olmsted, Counterexamples in Analysis, Dover, 2003
2. I. Stewart, Game, Set and Math, Penguin Books, 1989 ### Cantor Set 