Metric Spaces 2000, Lecture 03

Then dp is a metric. To prove this one must check the axioms. First, since |xk − yk| = |yk − xk| ≥ 0, it is obvious that dp(x, y) = dp(y, x) ≥ 0 for all x and y. Furthermore, since ∑ k |xk−yk| = 0 if and only if all the terms |xk−yk| are zero, we see that dp(x, y) = 0 if and only if x = y. To verify the remaining axiom we use Minkowski’s Inequality. Let x, y, z ∈ C, and define ak = yk − xk and bk = xk − zk, where xk, yk and zk are the k-th coordinates of x, y and z respectively. Then ak + bk = yk − zk, and so