Answers to Chapter 6

6.1 Answer: for (the most difficult case of) the closed interval [−2, 2], n = 1, 2, …, the partition is given by ∪ j = 0 ∞ I k , 2 j ∪ ∪ j = 0 ∞ I k + 1 , 2 j + 1 , where I _{0, 0} = [−2, 2^{−n + 1}], I _{0, j} = (1 + 2^−j, 1 + 2^−j+1], I _{k, 0} = [2^−k + 2^−k−1, 2^−k+1], I_k,j = [2^−k + 2^−k−j−1, 2^−k + 2^−k−j ), k = 1, …, n − 1, I_n,j = I _0,j, n = 2, 3, …, j = 0, 1, …. For n = ω: I 0 , 2 k − 1 ∪ ∪ j = 0 ∞ I k , 2 j ∪ ∪ j = 0 ∞ I k + 1 , 2 j + 1 , where I _{0, −1} = ∅, I _{0, 0} = [−2, 0].