Question

(1.3) Proposition: Let {x} be a sequence in a metric space (X; d). Then
(x) converges to y in Xiff for every open set U containing y, there exists a
positive integer N such that for every integer ,n>=N,xn belongs to U​

Answer 1

Answer:

Definitions. A metric on a set M is a function d : M × M → R

such that for all x, y, z ∈ M,

• d(x, y) ≥ 0; and d(x, y) = 0 if and only if x = y (d is positive)

• d(x, y) = d(y, x) (d is symmetric)

• d(x, z) ≤ d(x, y) + d(y, z) (d satisfies the triangle inequality)

The pair (M, d) is called a metric space.

If there is no danger of confusion we speak about the metric space

M and, if necessary, denote the distance by, for example, dM .

The open ball centred at a ∈ M with radius r is the set

B(a,r) = {x ∈ M : d(x, a) < r}

the closed ball centred at a ∈ M with radius r is

{x ∈ M : d(x, a) ≤ r}.

A subset S of a metric space M is bounded if there are a ∈ M and

r ∈ (0,∞) so that S ⊂ B(a,r).

MA222 – 2008/2009 – page 1.1

Normed linear spaces

Definition. A norm on a linear (vector) space V (over real or

complex numbers) is a function ( · ( : V → R such that for all

x, y ∈ V,

• (x( ≥ 0; and (x( = 0 if and only if x = 0 (positive)

• (cx( = |c|(x( for every c ∈ R (or c ∈ C) (homogeneous)

• (x + y(≤(x( + (y( (satisfies the triangle inequality)

The pair (V, ( · () is called a normed linear (or vector) space.

Fact 1.1. If ( · ( is a norm on V then d(x, y) = (x − y( is a metric

on V.

Proof. Only the triangle inequality needs an argument:

d(x, z) = (x − z( = ((x − y)+(y − z)(

≤ (x − y( + (y − z( = d(x, y) + d(y, z)

MA222 – 2008/2009 – page 1.2

Examples

Example (Euclidean n spaces). Rn (or Cn) with the norm

(x( =

!""#$n

i=1

|xi|

2 so with metric d(x, y) =

!""#$n

i=1

|xi − yi|

2

Example (n spaces with !p norm, p ≥ 1). Rn (or Cn) with the

norm

(x(p =

%$n

i=1

|xi|

p

&1

p

so with metric dp(x, y) = %$n

i=1

|xi − yi|

p

&1

p

Example (n spaces with max, sup or !∞ metric). Rn (or Cn)

with the norm

(x(∞ = max n

i=1|xi| so with metric d∞(x, y) = max n

i=1|xi − yi|

MA222 – 2008/2009 – page 1.3

Balls in !p norms

Balls in R2 with the !1, ! 3

2

, !2, !4 and !∞ norms.

MA222 – 2008/2009 – page 1.4

Convexity of !p balls

We show that the unit ball (and so all balls) in !p norm are convex.

(This is an important fact, although for us it is only a tool for proving

the triangle inequality for the !p norms.) So we wish to prove:

If (x(p, (y(p ≤ 1, α, β ≥ 0 and α + β = 1 then (αx + βy(p ≤ 1.

Proof for p < ∞; for p = ∞ it is left as an exercise.

Since the function |t|

p is convex (here we use that p ≥ 1!),

|αxi + βyi|

p ≤ α|xi|

p + β|yi|

p.

Summing gives

$n

i=1

|αxi + βyi|

p ≤ α

$n

i=1

|xi|

p + β

$n

i=1

|yi|

p ≤ α + β = 1.

So (αx + βy(p ≤ 1, as required.

MA222 – 2008/2009 – page 1.5

Proof of triangle inequality for !p norms

Proof. (In this proof we write ( · ( instead of ( · (p.)

The triangle inequality (x + y(≤(x( + (y( is obvious if x = 0 or

y = 0, so assume x, y *= 0. Let

xˆ = x

(x(

, yˆ = y

(y(

, λ = 1

(x( + (y(

, α = λ(x( and β = λ(y(.

Then

(xˆ( = 1, (yˆ( = 1, α, β, λ > 0, α + β = 1

and

λ(x + y) = αxˆ + βyˆ.

Since (αxˆ + βyˆ( ≤ 1 by convexity of the unit ball,

(x + y( = ((x( + (y()(λ(x + y)(

= ((x( + (y()(αxˆ + βyˆ(≤(x( + (y(.

MA222 – 2008/2009 – page 1.6

Some exotic metric spaces

Example (Discrete spaces). Any set M with the metric

d(x, y) = '

0 if x = y

1 if x *= y

Example (Sunflower or French railways metric in R2).

d(x, y) = '

(x − y( if x, y lie on the same line passing through origin

(x( + (y( otherwise

Example (Jungle river metric in R2).

d(x, y) = '

|y1 − y2| if x1 = x2

|y1| + |x1 − x2| + |y2| otherwise

MA222 – 2008/2009 – page 1.7

Balls in sunflower metric

d(x, y) = '

(x − y( x, y, 0 colinear

(x( + (y( otherwise

centre (4, 3), radius 6

MA222 – 2008/2009 – page 1.8

Subspaces, product spaces

Subspaces. If M is a metric space and H ⊂ M, we may consider

H as a metric space in its own right by defining dH(x, y) = dM (x, y)

for x, y ∈ H. We call (H, dH) a (metric) subspace of M.

Agreement. If we refer to M ⊂ Rn as a metric space, we have in

mind the Euclidean metric, unless another metric is specified.

Warning. When subspaces are around, confusion easily arises.

For example, in R, the ball B(0, 1) is the interval (−1, 1) while in the

metric space [0, 2], the ball B(0, 1) is the interval [0, 1).

Products. If Mi are metric spaces, the product M1 × ··· × Mn

becomes a metric space with any of the metrics

d(x, y) = %$n

i=1

(di(xi, yi))p

&1

p

or max n

i=1di(xi, yi)

where 1 ≤ p < ∞.

MA222 – 2008/2009 – page 1.9

Function & Sequence Spaces

C([a, b]) with maximum norm. The set C([a, b]) of continuous

functions on [a, b] with the norm

(f ( = sup

x∈[a,b]

|f(x)|

(

= max

x∈[a,b]

|f(x)|

)

C([a, b]) with Lp norm. Very different norms on C([a, b]) are defined

for p ≥ 1 by

(f (p =

%* b

a

|f(x)|

p dx&1

p

Spaces !p. For p ≥ 1, the set of real (or complex) sequences such

that +∞

i=1 |xi|