So, I have to prove that in the metric space (R^n, d), where d is the standard Euclidean metric, B(x1, r1) = B(x2, r2) <==> x1 = x2 & r1 = r2. I finished the proof, but I'm not sure about one step. Assume B(x1, r1) = B(x2, r2) with x1 = x2. Using the triangle inequality for x1, x and x2, one obtains d(x1, x2) <= d(x1, x) + d(x, x2) < r1 + r2. Now, since this holds for any x1, x and x2, and specially, for any corresponding r1 and r2, i.e. r1 + r2, we conclude that d(x1, x2) = 0, so x1 = x2. Is this correct? If so, then it is easy to prove the rest - assume B(x1, r1) = B(x2, r2) with x1 = x2 and r1 =/ r2, one easily derives a contradiction.