complete metric spaces - · pdf filed. sanjay mishra complete metric spaces. ... d). we show...

IntroductionComplete metric space and its properties

Completeness of product space Rω

Example of Non-Complete SpacesUniform Metric on Y J

Appendix

Complete Metric Spaces

Sanjay Mishra

Department of MathematicsLovely Professional University

Punjab, India

February 26, 2014

Sanjay Mishra Complete Metric Spaces




Appendix

Outline

1 Introduction

2 Complete metric space and its properties

3 Completeness of product space Rω

4 Example of Non-Complete Spaces

5 Uniform Metric on Y J

Summery

6 Appendix





Appendix

Introduction I

The concepts of completeness for a metric space is basic forall aspects of analysis.

It is metric property rather than a topological property.

Here we will study some results which involve complete metricspaces that are topological in character.





Appendix

Complete Metric Space I

Definition (Cauchy sequence)

Let (X, d) be a metric space. A sequence (xn) of points of X issaid to be a Cauchy sequence in (X, d) if it has the property thatgiven ε > 0, there is an integer N such that

d(xn, xm) < ε whenever n,m ≥ N





Appendix

Complete Metric Space II

Remark

A sequence is called a Cauchy sequence if the terms of thesequence eventually all become arbitrarily close to oneanother.

Any convergent sequence in X is necessarily a Cauchysequence. But Completeness requires that the converse holdalso.





Appendix

Complete Metric Space III

Definition (Complete Metric Space)

The metric space (X, d) is said to be complete if every Cauchysequence in X converges.

Intuitively, a space is complete if there are no ”points missing”from it (inside or at the boundary).For instance, the set of rational numbers is not complete, becausee.g.√

2 is ”missing” from it, even though one can construct aCauchy sequence of rational numbers that converges to it.





Appendix

Complete Metric Space IV

Remark (Subset of Complete metric space)

A closed subset A of a complete metric space (X, d) is necessarilycomplete in the restricted metric.BecauseA Cauchy sequence in A is also Cauchy sequence in X, hence itconverges in X. And because A is a closed subset of X, the limitmust lie in A.





Appendix

Complete Metric Space V

Remark

If X is complete under the metric d, then X will also completeunder the standard bounded metric

d(x, y) = min{d(x, y), 1}

corresponding to d.It converse is also true.For a sequence (xn) is a Cauchy sequence under d if and only if itis a Cauchy sequence under d.And a sequence converges under d if and only if it converges underd.





Appendix

Complete Metric Space VI

Example

1 Euclidean space Rk is complete in either of its usual metrics,the Euclidean metric d or the square metric ρ.

2 The set C(X,Y ) of all continuous functions mapping a spaceS into a metric space Y is complete metric space. This sethas a metric called the uniform metric which is analogous tothe uniform metric defined for RJ .

3 If Y is complete metric space, then function space C(X,Y ) iscomplete in the uniform metric.





Appendix

Complete Metric Space VII

Lemma (Condition of Completeness)

A metric space X is complete if every Cauchy sequence in X has aconvergent subsequence.

Proof:Let (xn) be a Cauchy sequence in (X, d). We show that if (xn)has a subsequence xni that converges to a point x, then thesequence (xn) itself converges to x.Given ε > 0, first choose N large enough that

d(xn, xm) < ε2

for all n,m ≥ N .





Appendix

Complete Metric Space VIII

(Using the fact that (xn) is a Cauchy sequence).Then choose an integer i large enough ni ≥ N

d(xni , x) < ε2

(Using the fact that n1 < n2 < . . . is an increasing sequence ofintegers and xni converge to x).Putting these together, we have the desired results that n ≥ N

d(xn, x) ≤ d(xn, xni) + d(xni , x) < ε





Appendix

Complete Metric Space IX

Theorem (Completeness of Euclidean space)

Euclidean space Rn is complete in either of its usual metrics, theEuclidean metric d or the square metric ρ.

Proof:Let x = (x1, . . . , xn) and y = (y1, . . . , yn) be arbitrary elements ofRn, then d is defined as

d(x, y) =

[n∑r=1

(xr − yr)2]1/2





Appendix

Complete Metric Space X

An element (x1, . . . , xn) ∈ Rn can be regarded as a real function definedon {1, . . . , n}. Thus for the sake of convenience we write

fx = (x1, . . . , xn), ∀x ∈ Rn

Let fn be a Cauchy sequence in Rn so that given ε > 0, there existsn0 ∈ N such that

p, q ≥ n0 ⇒ d(fp, fq) < ε

⇒ d2(fp, fq) < ε2

⇒

[n∑r=1

(fp(r)− fp(r))2]< ε2





Appendix

Complete Metric Space XI

From this we can deduce that

|fp(r)− fp(r)| < ε, where r = 1, 2, . . . , n





Appendix

Complete Metric Space XII

This show that (fp(r)) is a Cauchy sequence.

AlsoR is complete⇒ (fp(r)) converge point-wise to a limit function,

say f(r) ∈ R⇒ lim

p→∞fp(r) = f(r) for r = 1, 2, . . . , n

Since {1, 2, . . . n}is finite and hence this convergence in uniform

⇒ |fp(r)− f(r)| < ε√n∀ p ≥ n0

squaring and adding

⇒

[n∑r=1

(fp(r)− fr)2]1/2

< ( ε2

n · n)1/2 = ε

⇒ d(fp, f) < ε∀p ≥ n0

This show that the Cauchy sequence (fn) in Rn converges to f ∈ Rn.

Consequently (Rn, d) is complete.Sanjay Mishra Complete Metric Spaces




Appendix

Completeness of product space Rω I

Before discussing the completeness of product space Rω, first wewill discuss about the sequences in a product space.

Lemma (Sequences in a product space)

Let X be the product space X =∏Xα and xα be a sequence of

points of X, then xn → x if and only if πα(xn)→ πα(x) for eachα.





Appendix

Completeness of product space Rω II

Theorem

There is a metric for the product space Rω relative to which Rω iscomplete.

Proof:Let d(a, b) = min{|a− b|, 1} be the standard bounded metric on RLet D be the metric on Rω defined by

D(x, y) = sup{d(xi, yi)

i}

Then by the theorem Go D deduces the product topology on Rωis complete under D.





Appendix

Completeness of product space Rω III

Let xn be a Cauchy sequence in (Rω, D). Now because

d(πi(x), πi(y)) ≤ iD(x, y)

we can say that for fixed i the sequence πi(xn) is a Cauchysequence in R, so it converges, say to ai.Then the sequence xn converges to the point a = (a1, a2, . . . ofRω.Finally we can say that Rω is complete.





Appendix

Non-Complete Spaces I

Example (Q as Non-Complete Space)

The space Q of rational numbers with the usual metricd(x, y) = |x− y| is non-complete space.Because the sequence

1.4, 1.41, 1.414, 1.4142, .141421, . . .

of fine decimals converging (in R) to√

2 is a Cauchy sequence inQ that does not converge (in Q).

Question: Find out the condition when Q will be complete space ?





Appendix

Non-Complete Spaces II

Example (Open Interval of R as Non-Complete Space)

The open interval (−1, 1) in R with the usual metricd(x, y) = |x− y| is non-complete space.Because in this space the sequence (xn) defined by

(xn) = 1− 1

n

is a Cauchy sequence but not converge.

Question: Find out the condition when (−1, 1) will be completespace ?





Appendix

Non-Complete Spaces III

What we learn from the last two examples Go and Go .

Completeness is not a topological property. Or we can say that

It is not preserved by homeomorphism.Because (−1, 1) is homeomorphism to the real line R and R iscomplete in its usual metric.





Appendix

Summery

Completeness of Y J I

As we know that product spaces Rn and Rω have metrics relativeto which they are complete.Now very important question aries thatQuestion: Can we prove the same result as like Rn and Rω for theproduct space RJ in general is complete.Because as we know that RJ is not metrizable if J is uncountable.Here we will discuss thatThere is another topology on the set RJ , which is given by theuniform metric will be complete. But before this we will learnabout uniform metric in general case.





Appendix

Summery

Completeness of Y J II

Definition (Uniform metric on Y J)

Let (Y, d) ne a metric space.Let d(a, b) = min{d(a, b), 1} be the standard bounded metric onY derived from d.If x = (xα)α∈J and y = (yα)α∈J are points of the cartesianproduct Y J , let

ρ(x, y) = sup{d(xα, yα) : α ∈ J}

Here we can check that ρ is a metric.It is called the uniform metric on Y J corresponding to the metric don Y . And the topology induces by this metric is called uniformtopology on Y J .





Appendix

Summery

Completeness of Y J III

Remark (Notation of Uniform Metric)

As in the definition Go we used he standard ”tuple” notation forthe elements of the Y J . But now we will use functional notationbecause elements of Y J are simply functions from J to Y .In this notation, the definition of the uniform metric takes thefollowing form.If f, g : J → Y , then

ρ(f, g) = sup{d(f(α), g(α)) : α ∈ J}





Appendix

Summery

Completeness of Y J IV

Theorem (Completeness of Y J)

If the space Y is complete in the metric d, then the space Y J iscomplete in the uniform metric ρ corresponding to d.

Proof:As we know that (Y, d) is complete, then (Y, d) will be also.Where d is the bounded metric corresponding to d.Let f1, f2, . . . is a sequence of points of Y J which is Cauchysequence relative to ρ.Given α ∈ J , we can say that

d(fn(α), fm(α)) ≤ ρ(fn, fm), ∀n,m





Appendix

Summery

Completeness of Y J V

⇒ sequence f1(α), f2(α), . . . is a Cauchy sequence in (Y, d).Hence this sequence converges, say to a point yα.Let f : J → Y be the function defined by

f(α) = yα

Now we will show that sequence (fn) converges to f in the metricρ.Given ε > 0, first choose N large enough that

ρ(fn, fm) < ε2 ∀n,m ≥ N

Then, in particular

d(fn(α), fm(α)) < ε2 , ∀n,m ≥ N &α ∈ J





Appendix

Summery

Completeness of Y J VI

Let n and α be fixed, and the m become arbitrarily large, we cansee that

d(fn(α), f(α)) ≤ ε2

This inequality holds for all α ∈ J ,when n ≥ N .Therefore we can say that

ρ(fn, f) ≤ ε2 < ε, ∀n ≥ N

Finally, we can say that the Cauchy sequence (fn) converges to fin the metric ρ. Hence Y J is complete.





Appendix

Summery

Completeness of Y J VII

Now consider Y X instead of Y J , where X is a topological spacerather than merely a set J .Now we will discus two points

1 When considering the set of all functions f : X → Y there isno effect due to the topology of X i.e. in this case topologyon X is irrelevant. But

2 Suppose if we consider subset C(X,Y ) of Y X consisting of allcontinuous functions f : X → Y . Then what happen ?Here in this situation we can see that if Y is complete, thissubset C(X,Y ) of Y X will also complete.And the set B(X,Y ) will also complete. Where B(X,Y ) isset of all bounded functions. Go





Appendix

Summery

Completeness of Y J VIII

Theorem (Completeness of C(X,Y ) and B(X,Y ))

Let X be a topological space and let (Y, d) be a metric space. Theset C(X,Y ) of continuous functions is closed in Y X under theuniform metric. So is the set B(X,Y ) of bounded functions.Therefore, if Y is complete, these spaces are complete in theuniform metric.





Appendix

Summery

Completeness of Y J IX

Before proving this theorem we will discuss two points.

1 Uniform converges sequence Go

2 Uniform limit theorem: Uniformity of convergence dependsnot only the topology but also on its metric. This theoremdeal about uniformly convergent sequence. Go





Appendix

Summery

Completeness of Y J X

Proof:

Given 1 X is topological space and (Y, d) is metric space.

Given 2 Y X be product space which is also metric space withuniform metric.

Given 3 C(X,Y ) be the set of all continuous functions fromX to Y .

Given 4 B(X,Y ) be the set set of all bounded functions

Given 5 The metric space Y is complete.

Aim 1 We will show that space C(X,Y ) is complete andclosed with uniform metric.

Aim 2 We will show that space B(X,Y ) is complete withuniform metric.





Appendix

Summery

Completeness of Y J XI

First we show that if a sequence of the elements fn of Y X

converges to the elements f of Y X with respect to metric ρ on yX ,then it converges to f uniformly with respect to metric d on Y .As by given (our assumption) given ε > 0, choose integer N suchthat

ρ(f, fn) < ε, ∀n > N

Then for all x ∈ X and all n ≥ N

d(fn(x), f(x)) ≤ ρ(fn, f) < ε

So, by definition of uniform convergence we can say that (fn)converges uniformly to f .





Appendix

Summery

Completeness of Y J XII

Now we show that C(X,Y ) is closed in Y X with respect to ρ.Let f be an element of Y X i.e. it is a limit point of C(X,Y ).Then there is a sequence (fn) of the elements of C(X,Y )converging to f with respect to metric ρ.By the Uniform limit theorem we can say that f is continuous sothat f ∈ C(X,Y ).





Appendix

Summery

Completeness of Y J XIII

Now we will show that B(X,Y ) is closed in Y X .If f is a limit point of B(X,Y ), then there exists a sequence of theelements fn of B(X,Y ) converging to f .Choose N so large that

ρ(fN , f) < 12

Then this show that

d(fN (x), f(x)) < 12 , ∀x ∈ X, (?)

This show thatd(fN (x), f(x)) < 1

2

This show that





Appendix

Summery

Completeness of Y J XIV

If M is the diameter of the set fN (X), then f(X) has diameter atmost M + 1.Hence f ∈ B(X,Y )





Appendix

Summery

Completeness of Y J XV

Now we are going to find out the solution of very importantquestion that”Is it possible every metric space can imbedded isometrically in acomplete metric space”.But before this we will learn some important fact which areprerequisites of this question

1 A new metric called Superim metric on B(X,Y ). Go

2 Isometric Imbedding of one metric space X in another metricspace Y . Go





Appendix

Summery

Completeness of Y J XVI

Theorem (Isometric Imbedding of MS into CMS)

Let (X, d) be a metric space. Then there is an isometricimbedding of X into a complete metric space.

Proof:Let B(X,R) be the set of all bounded functions mapping X into R.Let x0 be a fixed point of X.Given a ∈ X, a map φa : X → R defined by

φa(x) = d(x, a)− d(x, x0) (1)

Here we can see that map φa is bounded map.





Appendix

Summery

Completeness of Y J XVII

For this, consider the inequalities

d(x, a) ≤ d(x, b) + d(a, b) (2)

andd(x, b) ≤ d(x, a) + d(a, b) (3)

Now from inequalities (2) and (3), we can say that

|d(x, a)− d(x, b)| ≤ d(a, b) (4)

Let b = x0 in (1), we get

|φa(x)| ≤ d(a, x0)∀x (5)

Hence by the definition we can say that φa is bounded map.Sanjay Mishra Complete Metric Spaces




Appendix

Summery

Completeness of Y J XVIII

Let us consider new map Φ: X → B(X,R) defined by

Φ(a) = φa (6)

Now we will show that Φ is an isometric imbedding of (X, d) intothe complete metric space (B(X,R), ρ).For this we will show that for every pair of points a, b ∈ X

ρ(φa, φb) = d(a, b) (7)

By the definition of sup metric

ρ(φa, φb) = sup{|φa(x)− φb(x)|;x ∈ X}= sup{|d(x, a)− d(x, b)|;x ∈ X} (8)





Appendix

Summery

Completeness of Y J XIX

We conclude thatρ(φa, φb) ≤ d(a, b) (9)

On other hand, this inequality cannot be strict, for when x = a.

|d(x, a)− d(x, b)| = d(a, b) (10)





Appendix

Summery

Completeness of Y J XX

Definition (Completion of Metric space)

Let X be metric space. If h : X → Y is an isometric imbedding ofX into a complete metric space Y , then the subspace h(X) of Yis a complete metric space.And this isometric imbedding h is called completion of X.





Appendix

Summery

Completeness of Y J XXI

Here we can see that from the definition 14 that completion of Xis uniquely determined up to an isometry.

Theorem (Uniqueness of the Completion)

Let h : X → Y and h : X → Y ′ be isometric imbeddings of metricspace (X, d) in the complete metric spaces (Y,D) and (Y ′, D′)respectively. Then there is an isometry of (h(X), D) with(h′(X), D′) that equals h′h−1 on the subspace h(X).





Appendix

Summery

Summery I

Finally what we learned from the Complete metric space.

1 When any metric space will be complete ?

2 When Euclidean space Rk is complete ?

3 When Rω and RJand is complete ?

4 When space Y J is complete even Y is complete ?

5 When C(X,Y ) and B(X,Y ) are complete ?

6 How can isometrically imbed a metric space into completemetric space ?





Appendix

Definition (Euclidean Metric)

Given x = (x1, . . . , xn) in Rn, we define the norm of x by theequation

‖x‖ = (x21 + . . .+ x2n)1/2

and we define the Euclidean metric d on Rn by the equation

d(x, y) = ‖x− y‖ = [(x1 − y1)2 + . . .+ (xn − yn)2]1/2





Appendix

Definition (Square Metric)

We define the square metric ρ by the equation

ρ(x, y) = max[|x1 − y1|, . . . |xn − yn|]





Appendix

Definition (Standard Bounded Metric)

Let X be a metric space with metric d. Define d : X ×X → R theequation

d(x, y) = min{d(x, y), 1}

Then d is a metric that induces the same topology as d.And metric d is called the standard bounded metric correspondingto d.





Appendix

Theorem (Product topology on Rω induces by metric)

Let d(a, b) = min{|a− b|, 1} be the standard bounded metric onR. If x and y are two points of Rω, defined by

D(x, y) = sup{d(xi, yi)

i}

Then D is a metric that induces the product topology on Rω.





Appendix

Definition (Bounded function)

A function f : X → Y is said to be bounded if its image f(X) is abounded subset of the metric space (Y, d).





Appendix

Definition (Uniform converges sequence)

Let fn : X → Y be a sequence of functions from the set X tometric space (Y, d). We say that (fn) converges uniformly to thefunction f : X → Y of given ε > 0, there exists an integer N suchthat

d(fn(x), f(x)) < ε

for all n > N and all x ∈ X.





Appendix

Theorem (Uniform limit theorem)

Let fn : X → Y be a sequence of continuous functions from thetopological space X to the metric space Y . If (fn) convergesuniformly to f , then f is continuous.





Appendix

Here we are going to introduce another metric on the set B(X,Y ).

Definition (Superim Metric)

Let (Y, d) is a metric space. Another metric ρ on the set B(X,Y )of bounded functions from X to Y is defined by

ρ(f, g) = sup{d(f(x), g(x)) : x ∈ X}

This metric is called Superim metric on B(X,Y ).





Appendix

Remark (Relation between ρ and ρ)

There is simple relation between sup metric and uniform metric onB(X,Y ).If f, g ∈ B(X,Y ), then

ρ(f, g) = min{ρ(f, g), 1}





Appendix

1 If ρ(f, g) > 1, then d(f(x0), g(x0)) > 1 for at least onex0 ∈ X. And this show that

d(f(x0), g(x0)) = 1 and ρ(f, g) = 1

2 If ρ(f, g) ≤ 1, then

d(f(x), g(x)) = d(f(x), g(x)) ≤ 1, ∀x

andρ(f, g) = ρ(f, g) = 1

Form the above we can conclude that Uniform metric ρ onB(X,Y ) is just the standard bounded metric derived from the supmetric ρ.





Appendix

Remark (Function space with ρ)

1 If X is compact space, then every continuous functionf : X → Y is bounded so we can define sup metric onC(X,Y ).

2 If Y is complete under d, then C(X,Y ) is complete undercorresponding uniform metric ρ, and it will also completeunder the sup metric ρ.





Appendix

Definition (Isometric Imbedding)

let (X, dX) and (Y, dY ) are two metric spaces. Let f : X → Yhave the property that for every pair of points x1, x2 ∈ X

dY (f(x1), f(x2)) = dX(x1, x2)

Then f is an imbedding and called isometric imbedding of X in Y .


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