Bourgain ’ s Theorem

Exercise 1. Show that distances between sets do not necessarily satisfy the triangle inequality. That is, it is possible that d(S1, S2) + d(S2, S3) > d(S1, S3) for some sets S1, S2 and S3. Exercise 2. Prove that d(x, y) ≥ d(S, x)− d(S, y) and thus d(x, y) ≥ |d(S, x)− d(S, y)|. Proof. Fix ε > 0. Let y′ ∈ S be such that d(y′, y) ≤ d(S, y) + ε (if S is a finite set, there is y′ ∈ S s.t. d(y, y′) = d(S, y)). Then d(x, S) ≤ d(x, y′) ≤ d(x, y) + d(y, y′) ≤ d(x, y) + d(S, y) + ε. We proved that d(x, S) ≤ d(x, y) + d(S, y) + ε for every ε > 0. Therefore, d(x, S) ≤ d(x, y) + d(S, y).