Let E1 be the unit interval [0, 1] with its middle third (1/3, 2/3) removed (i.e., E1 = [0, l/3]U[2/3, 1]). Let E2 be E1 with its middle thirds removed; that is,
E2 = [0, 1/9] U [2/9, 1/31 U [2/3, 7/9] U [8/9, 1].
Continuing in this manner, generate nested sets Ek such that each Ek is the union of 2k closed intervals of length 1/3k. The Cantor set is the set

Assume that every point x ˆˆ [0, 1] has a binary expansion and a ternary expansion; that is, there exist ak ˆˆ [0, 1] and bk ˆˆ [0, 1, 2} such that

(e.g., if x = 1 /3, then a2k-1 = 0, a2k = 1 for all k and either b1 = 1, bk = 0 for k > 1 or b1 = 0 and bk = 1 for all k > 1.)
a) Prove that E is a nonempty compact set of measure zero.
b) Show that a point x ˆˆ [0, 1] belongs to E if and only if x has a ternary expansion whose digits satisfy bk ‰  1 for all k ˆˆ N.
c) Define f: E †’ [0, 1] by

Prove that there is a countable subset E0 of E such that f is 1-1 from EE0 onto [0, 1] (i.e., prove that E is uncountable).
d) Extend f from E to [0, 1] by making f constant on the middle thirds Ek-1Ek. Prove that f : [0,1] †’ [0, 1] is continuous and increasing. (The function f is almost everywhere constant on [0, 1]; that is, constant off a set of measure zero. Yet it begins at f(0) = 0 and ends at f(1) = 1.)

k=1 0o 34 ! ! by 3A