Prove that (R^n, d'') is a metric space, where the function d'' is defined by d''(x,y)= [i=1]Σ^n lxi - yil

Question

x = (x1, x2, ..., xn), y = (y1, y2, ..., yn) element of R^n.

Answer 1

d''(x, y) = Σ(i=1 to n) |xi - yi| ≥ 0 for all x, y

d''(x, x) = Σ(i=1 to n) |xi - xi| = Σ0 = 0 for all x

d''(x, y) = 0 => Σ(i=1 to n) |xi - yi| = 0
=> |xi - yi| = 0 for all i since |xi - yi| ≥ 0
=> xi = yi for all i
=> x = y

d''(x, y) = Σ(i=1 to n) |xi - yi|
= Σ(i=1 to n) |yi - xi|
= d''(y, x)

d''(x, z) = Σ(i=1 to n) |xi - zi|
≤ Σ(i=1 to n) (|xi - yi| + |yi - zi|) since |x-y| is a metric on R and so satisfies the triangle inequality
= [Σ(i=1 to n) (|xi - yi|)] + [Σ(i=1 to n) (|yi - zi|)]
= d''(x, y) + d''(y, z)

So we have proved:
d''(x, y) ≥ 0 for all x, y and d''(x, y) = 0 <=> x = y;
d''(x, y) = d''(y, x), for all x, y;
d''(x, z) ≤ d''(x, y) + d''(y, z) for all x, y, z.

So (R^n, d'') is a metric space. In fact, d'' is a standard metric known as the 1-norm.

Answer 2

Prove it's not a metric space, bossy.