Math 650. Homework 2. Solutions

3. Let C be the set of all numbers w on the interval [0,1] which can be written in the form

w = ∞ ∑  k=1  ak 3k

with a_k=0 or 2. This set is called the Cantor set. Show that C can be constructed by the following procedure: from [0,1] remove the middle third (1/3,2/3); from the remainder (that is, the intervals [0,1/3] and [2/3,1]) remove the middle thirds, and so on, ad infinitum. What remains is the Cantor set.

Solution.

Let B₀ be the interval [0,1],  let B₁ be B₂ minus the middle third interval, and so on, ad infinitum. What remains is the intersection

∩ i=k∞ Bk

Let w = .a₁a₂a₃..., with a_i=0,2 be a point in the Cantor set C. To show that w ∈ ∩_k B_k, observe the following. If a₁=0, then w belong to the right interval of B₁, that is, to [0,1/2]. If a₁=2, then w belongs to the left interval of B₁, that is to [2/3,1]. In either case, w ∈ B₁.

Do the same with a₂. The set B₂ is the union of four intervals, namely,

B2=[0,1/9] ∪ [2/9,1/3] ∪ [2/3,7/9] ∪ [8/9,1].

If a₂=0, then w is in [0,1/9] (if a₁=0), or in [2/3,7/9] (if a₁=2). If a₂=2, then w is in [2/9,1/3] (if a₁=0), or in [8/9,1/9] (if a₂=2). In any case, w ∈ B₂. Repeating this reasoning it obtains that w ∈ B_k for every k=1,2,3, ... , that is, w ∈ ∩_kB_k

Conversely, let w ∈ ∩_k B_k, and write w = .a₁a₂a₃..., the ternary expansion with a_k=0,1,2. Because w ∈ B₁, it must be that a₁=0 or 2. In general, because w ∈ B_k, the digit a_k=0 or 2. That is, w is in the Cantor set C.

4. Show that the Cantor set is of Lebesgue measure zero.

Solution. As shown above, the Cantor set C is a countable intersection C= ∩_k B_k, where each B_k is a finite union of intervals. In fact,

Bk=Ik,1∪ ... ∪ Ik,2k

is the union of 2^k intervals I_k,l, each of them of length 2^k-1/3^k. Hence B_k belongs to R_Leb and

m(Bk) = 2k 2k-1 3k = (2/3)k.

Thus, given e > 0, there exists k such that (2/3)^k < e. Since C= ∩_j B_j, the set C ⊂ B_k and the finite collection of intervals I_k,1,..., I_k,2^k is a cover of C with

2k ∑ j=1  m(Ik,j) = (2/3)k < e.