# Metric Spaces, Lecture Notes - Mathematics, Study notes for Calculus. University of California (CA) - UCLA

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Metric Spaces

August 2, 2005

1 Metrics: a notion of distance

So far in analysis on R, every definition (convergence, Cauchy, continuity) has included statements like “|x − y| < ”. The idea is that this |x − y| represents a distance from x to y.

A distance should satisfy the properties: i) |x− y| ≥ 0,∀x, y ∈ R ii) |x− y| = 0→ x = y iii) |x− y| = |y − x| iv) “Triangle inequality”: |x− y|+ |y − z| ≥ |x− z|,∀x, y, z

Definition: a metric space is a set of “objects” S together with a function d on the set of pairs (x, y), x, y ∈ S such that

d(x, y) ∈ R,∀x, y ∈ S 1)d(x, y) ≥ 0,∀x, y ∈ S 2)d(x, y) = 0→ x = y 3)d(x, y) = d(y, x),∀x, y ∈ S 4)d(x, y) + d(y, z) ≥ d(x, z),∀x, y, z ∈ S

d is called the metric on S. May write (S, d) to specify the metric. 4) is called the triangle inequality; it is the most important, and it is

usually the hardest to verify.

2 Examples

1) (R, d) is a metric space where d(x, y) = |x− y|

1

2) “Discrete metric space”: S is any set, and define

d(x, y) =

{ 1 x, y ∈ S, x 6= y 0 x ∈ S, x = y

3) “Euclidean k-space”:

S = {(x1, x2, . . . , xk)|xi ∈ R, 1 ≤ i ≤ k}

d(x,y) =

( k∑

i=1

(xi − yi)2 ) 1

2

Properties 1, 2, and 3 are clear. Property 4 takes work. Best way is to consider dot products of vectors.

This is the most important metric space that we will cover. The given metric space is called the standard metric on Rk.

4) “Manhattan metric”:

S = Rk

d(x,y) = k∑

i=1

|xi − yi|

Properties 1, 2, and 3 are easy. For 4, use the triangle inequality on R,∑ |xi − yi|+

∑ |yi − zi| ≥

∑ |xi − zi|

5) C[0, 1] :

S = {f |f is a continous function on [0, 1]} d(f, g) = sup{|f(x)− g(x)||x ∈ [0, 1]}

Note that f, g continuous on [0, 1] → f − g is continuous on [0, 1] → d(f, g) exists.

To verify property 2,

d(f, g) = 0→ sup{|f(x)− g(x)|} = 0→ |f(x)− g(x)| = 0,∀x→ f(x) = g(x),∀x

2

To verify property 4, use the triangle inequality on R,

d(f, g) + d(g, h) = sup{|f(x)− g(x)|}+ sup{|g(x)− h(x)|} ≥ sup{|f(x)− g(x)|+ |g(x)− h(x)|} ≥ sup{|f(x)− h(x)|} = d(f, h)

6) l1(N) :

S = {(an)∞n=1| ∞∑

n=1

|an| converges}

d((an), (bn)) = ∞∑

n=1

|an − bn|

Properties 1, 2, and 3 are easy. For 4, use the triangle inequality on R,

d((an, bn)) + d((bn), (cn)) = ∞∑

n=1

|an − bn|+ ∞∑

n=1

|bn − cn|

= ∞∑

n=1

|an − bn|+ |bn − cn| ≥ ∞∑

n=1

|an − cn| = d((an), (cn))

Idea is to abstract the qualities of R to many situations. The multitude of examples is one of the most useful aspects of metric spaces.

3 Convergence

Definition: A sequence (sn) ∞ n=1 in a metric space (S, d) is said to converge to

s ∈ S if

∀ > 0,∃N ∈ N such that n > N → d(sn, s) < 

Note: sn converges to s if and only if limn→∞ d(sn, s) = 0 in R. Definition: a sequence (sn) is (S, d) is Cauchy if

∀ > 0,∃N ∈ N such that n,m > N → d(sn, sm) < 

Definition: A metric space (S, d) is said to be complete if every Cauchy sequence converges.

Note: the converse statement that convergence implies Cauchy is always true. Emulate the proof from R.

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Examples: 1) (R, d), d(x, y) = |x− y| is complete by the Completeness axiom 2) (Q, d), d(x, y) = |x− y| is not complete 3) C[0, 1] is complete. 4) (Rk, d), d is the Euclidean metric, is complete. To prove it, we need

lemmas Lemma 1: (xn) is a convergent sequence in Rk → (xni )∞n=1 is convergent

in R,∀i, 1 ≤ i ≤ k. Proof: Assume (xn) is convergent and xn converges to x. Let  > 0 be

given. ∃N such that

n > N →

( k∑

i=1

(xni − xnj )2 ) 1

2

< → ∑

(xni − xi)2 < 2

→ (xni − xi)2 < 2,∀i→ |xni − xi| < ,∀i→ lim n→∞

xni = xi,∀i

Now assume that (xn1 ), (x n 2 ), . . . , (x

n k) converge in R. Let

lim n→∞

xn1 = x1, . . . , lim n→∞

xnk = xk

Let  > 0 be given. Then ∃N1, . . . , Nk ∈ N such that

n > N1 → |x1 − xn1 | < √ k , . . . , n > Nk → |xk − xnk | <

√ k

Let N = max{N1, N2, . . . , Nk}. Then Lemma 2: (xn) is Cauchy in Rk if and only if (xni ) is Cauchy in R,∀i Proof: left to the reader Now, if (xn) is Cauchy in Rk(xni ) is Cauchy in R,∀i→ (xni ) is convergent

∀i→ (xn) is convergent in Rk. So Rk is complete. What’s useful are the lemmas: to examine the properties of something in

Rk, look at the components of the vector one at a time. Definition: A set A ⊆ S, (S, d) is a metric space is bounded if ∃x ∈ S and

∃M ∈ R such that d(a, x) ≤M, ∀a ∈ A. Definition: the open ball around x with radius R is BR(x) = {s ∈

S|d(s, x) < R}. The closed ball = B̄R(x) = {s ∈ S|d(s, x) ≤ R}.

A is bounded if A ⊆ B̄R(x) for some x,R.

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Theorem: if (xn) is a bounded sequence in Rk, then ∃ a convergent sub- sequence (xnl)∞l=1.

Proof: by induction on k: base case k = 1 is the Bolzano-Weirstrass theorem on R.

Assume the theorem is true for Rk. Let (xn) be a bounded sequence in Rk+1. Let yn = (xn1 , xn2 , . . . , xnk) ∈ Rk → (yn) is bounded in Rk and therefor has a convergent subsequence (ynl). So (xnlk+1)

∞ l=1 is bounded in R → ∃ a

convergent subsequence (x nlj k+1)

∞ j=1 → (x

nlj k+1, . . . , x

nlj k , x

nlj k+1) is a convergent

subsequence of (xn). Idea is that we use our knowledge of R to bootstrap ourself up to as many

dimensions as we want.

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