Solutions to Problems

1.16. Again, I’ll only do the triangle inequality. Let x = (a1, . . . , an), y = (b1, . . . , bn), z = (c1, . . . , cn) be three points in Rn, then there is some integer j, 1 ≤ j ≤ n where the max occurs, so that d∞(x, y) = |aj − bj |. Hence, d∞(x, y) = |(aj−cj)+(cj−bj)| ≤ |aj−cj |+|cj−bj | But, |aj−cj | ≤ max{|a1− c1|, . . . , |an − cn|} = d∞(x, z), since it is one of the terms occurring in the maximum. Similarly, |cj − bj | ≤ d∞(z, y). Combining these inequalities, yields d∞(x, y) ≤ d∞(x, z) + d∞(z, y). 1.36. Let p = (a1, a2) ∈ O, then a1 + a2 > 0. Set r = (a1 + a2)/2. We claim that B(p; r) ⊆ O, which will prove that O is open. To see the claim, if q = (b1, b2) ∈ B(p; r), then (a1 − b1) + (a2 − b2) < r2. Hence, each term satisfies, (ai − bi) < r2 and so |ai − bi| < r. This implies that a1− r < b1 < a1 + r and a2− r < b2 < a2 + r. Adding together the lefthand side inequalities, yields a1 + a2 − 2r < b1 + b2, but a1 + a2 − 2r = 0, so 0 < b1 + b2 which implies that q ∈ O. Thus, B(p; r) ⊆ O. 1.37 Given p = (a1, a2) ∈ O, let r = a1 > 0. Claim that B(p; r) ⊆ O. To see the claim, let q = (b1, b2) ∈ B(p; r), then |a1−b1| ≤ √ (a1 − b1) + (a2 − b2) < r, and so, 0 = a1 − r < b1 < a1 + r, and we have b1 > 0. This proves that q ∈ O and so B(p; r) ⊆ O. 1.38. Let p = (a1, a2), let B(p; r) denote a disk of radius r centered at p in the Euclidean metric which is an actual disk, and let B∞(p; r) denote