Cantor Set A Collection of Some Facts

Geometrically the Cantor set E is defined as follows: Let E_{1} be obtained from [0, 1] by removing the middle third, i.e.

E_{1} = [0, \frac{1}{3}] ∪ [\frac{2}{3}, \frac{3}{3}]

Define E_{2} by removing the middle thirds of the two disjoint subintervals of E_{1}, i.e.

E_{2} = [0, \frac{1}{9}] ∪ [\frac{2}{9}, \frac{3}{9}] ∪ [\frac{6}{9}, \frac{7}{9}] ∪ [\frac{8}{9}, \frac{9}{9}]

Continuing this way define inductively E_{k} by removing the middle thirds of all the components of E_{k−1}. Each E_{k} consists of 2^{k} disjoint subintervals of length 1/3^{k} each. The length of E_{k} is then (2/3)^{k} and hence converges to zero as k goes to infinity. Finally define the Cantor set as

E := ⋂_{k=1}^{∞} E_{k}

Note that E_{1} ⊃ E_{2} ⊃ ⋯ ⊃ E_{k} ⊃ ⋯, so E ⊂ E_{k} for any k and hence |E| < |E_{k}| = (2/3)^{k}. This means that E is a set of measure zero. Moreover by Cantor’s Intersection theorem E is nonempty. Since E is the intersection of closed sets, it is closed and certainly bounded, so compact.

Every number in [0, 1] can be expressed as a ternary expansion, x = ∑_{k=1}^{∞} \frac{b_{k}}{3^{k}}

where each b_{k} is 0, 1 or 2. We can denote this sum by x = (b_{1}, b_{2}, ...)_{3}. This expansion is not unique. For example 1/3 = (1, 0, 0, ..., 0, ...)_{3} = (0, 2, 2, ..., 2, ...)_{3}. We can agree to choose the repeating 2’s expansion whenever possible and this makes the expansion of each number unique. Similarly for binary expansion.

With this convention if x ∈ E_{1}, then b_{1} of the ternary expansion of x is not 1, otherwise x would be in the removed middle third. If x ∈ E_{2}, then similarly b_{1} ≠ 1. If b_{k} = 0, then x is in [0, 1/3], and b_{2} decides whether x is in [0, 1/9] or in [2/9, 1/3], and hence b_{2} ≠ 2. By induction we see that x ∈ E_{n} if and only if b_{k} ≠ 1 for k = 1, ..., n. Hence x ∈ E if and only if it has a ternary expansion without 1’s.

Define a function f : E → [0, 1] as follows. Let x = (b_{1}, b_{2}, ...)_{3}. Let f(x) = (b_{1}/2, b_{2}/2, ..., b_{k}/2, ...)_{2}. Since every x ∈ [0, 1] has a binary expansion, this map is onto. Since E can be mapped onto an uncountable set, E is uncountable. This is an example of an uncountable set of measure zero.

From the definition it is clear that f is increasing on E, i.e. x, y ∈ E and x < y implies that f(x) < f(y).

Recall that E_{k} consists of 2^{k} subintervals of length (1/3)^{k} each. It can be proven by induction that each of these intervals can be written as [(2a)/3^{k}, (2a+ 1)/3^{k}] for some nonnegative integer a. By subdividing this interval we obtain [(6a)/3^{k+1}, (6a + 1)/3^{k+1}] ∪ [(6a + 2)/3^{k+1}, (6a + 3)/3^{k+1}] ⊂ E_{k+1}.

If x ∈ [0, 1] is not in E then there is a k ≥ 1 such that x ∈ E_{k−1} but x ∉ E_{k}, where we take E_{0} = [0, 1]. Then (2a+1)/3^{k} < x < (2a+2)/3^{k} for some integer a. It is straightforeward to show that f((2a + 1)/3^{k}) = f((2a + 2)/3^{k}). Set f(x) to this common value. This extends f to [0, 1]. Using a straightforward ϵ-δ argument and noticing that E_{k} contains no interval of length greater than 1/3^{k} it can be shown that f is continuous.