1. Let X = 2 and define d2 : 2 × 2 → byd2((x1, y1), (x2 , y2)) = max{| x1 − x2 |, | y1 − y2 |}.(a) Verify that d2 is a metric on 2.(b) Draw the neighborhood N (0; 1) for d2, where 0 is the origin in 2.2. Let X = × × = 3 and define the metric d : 3 × 3 → byd((x1, y1, z1), (x2 , y2 , z2)) = max{| x1 − x2 |, | y1 − y2 |, | z1 − z2 |}.Describe the neighborhood N (0; 1), where 0 is the origin in 3.

1. Let X = 2 and define d2 : 2 × 2 → by d2((x1, y1), (x2 , y2)) = max{| x1 − x2 |, | y1 − y2 |}. (a) Verify that d2 is a metric on 2. (b) Draw the neighborhood N (0; 1) for d2, where 0 is the origin...

1. Let X =

²
and define d₂
:

²
×

²
→

by

d₂((x₁, y₁), (x₂
, y₂)) = max{| x₁
− x₂
|, | y₁
− y₂
|}.

(a) Verify that d₂
is a metric on

².

(b) Draw the neighborhood N (0; 1) for d₂, where 0 is the origin in

².

2. Let X =

×

×

=

³
and define the metric d :

³
×

³
→

by

d((x₁, y₁, z₁), (x₂
, y₂
, z₂)) = max{| x₁
− x₂
|, | y₁
− y₂
|, | z₁
− z₂
|}.

Describe the neighborhood N (0; 1), where 0 is the origin in

³.

May 05, 2022