# Category: Pure Mathematics

# How to sew a two holed torus

# Constructing numbers from sets

Continued from previous blog: A fast review of set theory; same reference: A Second Course in Analysis by M Ram Murty, Hindustan Book Agency.

Mathematicians and philosophers of the nineteenth century pondered deeply into the nature of a number. The question of “what is a number?” is not a simple one. But since mathematicians decided to give foundations of mathematics using the axiomatic method and sets as the basic building blocks, we are led to define numbers using sets. We follow Richard Dedekind (1831-1916) and Giuseppe Peano (1858-1932) in the following construction. It was as late as 1888 and 1889 when this construction was described in two papers written independently by Dedekind and Peano.

We construct a sequence of sets to represent the natural numbers. As noted earlier, zero is represented by the empty set. We have already described the construction of the natural numbers using the empty set. For each natural number n, the successor of n is denoted by (and sometimes as ) and defined as

Thus, each natural number is a set with n elements, namely,

We designate the set of natural numbers by the symbol . (It is a matter of personal convenience whether to include zero as a natural number or not. In this discussion, zero is a natural number. In other settings, it may not be. There is no universal convention regarding this and the student is expected to understand depending on the context. Some authors use the term “whole numbers” to indicate that zero is included in the discussion.)

The arithmetic operations on are now defined recursively. Addition is defined as a function from to :

where is defined recursively by and . A similar definition is given for multiplication x by defining and

We also define as simply mn which is the familiar symbology.

An equivalence relation on a set S is a subset R of satisfying:

- (reflexive axiom) .
- (symmetry axiom) .
- (transitive axiom) and implies .

The notion of an equivalence relation is an abstaction of our concept of equality, or at least what we implicitly expect of the notion of equality. It is more suggestive to write the equivalence relationn, not as a subset of as indicated above, but rather more symbolically as that our axioms become:

- (reflexive) , .
- (symmetry) if and only if .
- (transitive) and implies .

Equivalence relations play a fundamental role in all of mathematics. They allow us to understand aspects of sets by grouping them using certain properties.

To construct negative integers, we define an equivalence relation on . We write

.

Intuitively, we think of as so that it becomes evident that our definition is now in terms of concepts that have been defined earlier. This is very similar to how the ancients worked with negative numberss that appeared in an equation. They usually moved them to the other side so that the equation became an equation of non-negative numbers. However, with our set theoretic definition, we have reached a more fundamental and higher level of abstraction. Thus, with our equivalence relation above on the natural numbers, we define the set of integers as the set of equivalence classes of such ordered pairs. It is now easy to see that the following lemma holds:

Lemma 1.1 If is an ordered pair of non-negative integers, then exactly one of the following statements holds:

(a) is equivalent to for a unique non-negative integer m;

(b) is equivalent to for a unique non-negative integer m;

(c) is equivalent to .

Sometimes, we denote by the equivalence class of . With this lemma in place, we now denote by m the set of pairs of non-negative integers equivalent to ; by -m the set of pairs equivalent to and by 0 the set of pairs equivalent to . We denote these equivalence classes by .This gives us set theoretic construction of the set of integers.

We can define the operations of addition and multiplication by setting:

This latter definition is best understood if we recall that the symbol represents j-k so that the left hand side of the above equation is

One needs to check that these definitions are “well-defined” in the sense that they are independent of the representatives chosen for the equivalence class. This can be done as exercises.

In this way, we have now extended the notion of addition and multiplication from the set of natural numbers to the set of integers. Subtraction of integers can be defined by

where -1 represents the equivalence class . All of these definitions correspond to our usual notion of addition, subtraction and multiplication. Their virtue lies in their pure set-theoretic formulation.

We can also order the set of integers in the usual way. Thus,

and

.

This corresponds to our usual notion of “less than” and “less than or equal to”.

Finally, we can define the absolute value on the set of integers by setting

, if

, if

, if

We can now construct the rational numbers from the set of integers. We do this by defining an equivalence relation on the set by stating that two pairs and are equivalent if and only if . Intuitively, we think of as representing the “fraction” and examining what we would mean by by reducing it to notions already defined. The set of rational numbers is then defined as the set of such equivalence classes.

The expected operations of addition and multiplication are now evident:

Again, these definitions are easily verified to be well-defined. Finally, we can now define “division”.If with , we define:

These operations satisfy the familiar laws of associativity, commutativity and distributivity. Subtraction of rational numbers then can be written as :

The ordering of rational numbers can also be written as:

.

These definitions agree with out usual notions of ordering of the rational numbers.

Finally, the definition of absolute value can be extended as:

if

if

.

Again, our familiar properties of the absolute value of rational numbers hold. With this foundational construction in place, we can conveniently represent the equivalence class of as simply the fraction j/k and continue to work with these numbers as we were hopefully taught from childhood.

In the next sections/blogs we construct the real numbers from this axiomatic framework.

Exercises:

Hint (generic): keep the meaning of the symbols in mind and meaning of equivalence relations and equivalence classes. Also note that our basic object is a class and a set is a member of a class.

- let be two elements of . Show that the addition:

is well-defined. That is, prove that for any and , we have that is equivalent to .

2. For , prove the distributive law: .

3. Show that the relations < and on have the following properties:

(a) for all

(b) for all

(c) if and only if

(d) for all

(e) if and only if .

(f) for all .

(g) for all

(h) for if and only if

(i) for all

(j) where if and only if .

Cheers,

Nalin Pithwa.

# Boarbarktree

# Cauchy’s Mean Value Theorem and the Stronger Form of l’Hopital’s Rule

**Reference: G B Thomas, Calculus and Analytic Geometry, 9th Indian Edition. **

The stronger form of l’Hopital’s rule is as follows:

Suppose that and the functions f and g are both differentiable on an open interval that contains the point . Suppose also that at every point in except possibly . Then,

…call this I, provided the limit on the right exists.

Remarks:

The proof of the stronger from of l’Hopital’s rule in based on Cauchy’s mean value theorem, a mean value theorem that involves two functions instead of one. We prove Cauchy’s theorem first and then show how it leads to l’Hopital’s rule.

**Cauchy’s Mean Value Theorem:**

Suppose the functions f and g are continuous on and differentiable through out and suppose also that through out . Then, there exists a number c in at which

.

(Note this becomes the ordinary mean value theorem when ).

**Proof of Cauchy’s Mean Value theorem:**

We apply the ordinary mean value theorem twice. First, we use it to show that . Because if , then the ordinary Mean Value theorem says that

for some c between a and b. This cannot happen because in .

We next apply the Mean Value Theorem to the function

This function is continuous and differentiable where f and g are, and note that . Therefore, by the ordinary mean value theorem, there is a number c between a and b for which . In terms of f and g, this says

or which is equation II above.

* Proof of the stronger form of L’Hopital’s Rule:*We first establish equation I for the case . The method needs almost no change to apply to the case , and the combination of these two cases establishes the result.

Suppose that x lies to the right of . Then, and we can apply Cauchy’s Mean Value theorem to the closed interval from to x. This produces a number c between x and such that

But, so

That

As x approaches , c approaches as it lies between x and . Therefore,

.

This establishes l’Hopital’s Rule for the case where approaches from right. The case where x approaches from the left is proved by applying Cauchy’s Mean Value Theorem to the closed interval when .

QED.

Regards,

Nalin Pithwa

# VII. Complete Metric Spaces

**Reference: Introductory Real Analysis by Kolmogorov and Fomin. Translated by Richard A. Silverman. Dover Publications. **

Available on Amazon India and Amazon USA. This text book can be studied in parallel with Analysis of Walter Rudin.

**7.1. Definition and examples:**

*The reader is presumably already familiar with the notion of completeness of the real line. (One good simple reference for this could be: Calculus and analytic geometry by G B Thomas. You can also use, alternatively, Advanced Calculus by Buck and Buck.)*The real line is, of course, a simple example of a metric space. We now make the natural generalisation of the notion of completeness to the case of an arbitrary metric space.

**DEFINITION 1:**

A sequence of points in a metric space R with metric is said to satisfy the * Cauchy criterion *if given any , there is an integer such that for all .

**DEFINITION 2:**

A subsequence of points in a metric space R is called a **Cauchy sequence** (or a **fundamental sequence** ) if it satisfies the Cauchy criterion.

**THEOREM 1:**

Every convergent sequence is fundamental.

**Proof 1:**

If converges to a limit x, then, given any , there is an integer such that

for all .

But, then

for all . QED.

**DEFINITION 3:**

A metric space R is said to be * complete *if every Cauchy sequence in R converges to an element of R. Otherwise, R is said to be

**incomplete.****Example 1:**

Let R be the “space of isolated points” (discrete metric space) defined as follows: Define , if ; let , when . Then, the Cauchy sequence in R are just the “stationary sequences,” that is, the sequences all of whose terms are the same starting from some index n. Every such sequence is obviously convergent to an element of R. Hence, R is complete.

**Example 2:**

The completeness of the real line R is familiar from elementary analysis:

**Example 3:**

The completeness of the Euclidean n-space follows from that of . In fact, let

where

be fundamental sequence of points in . Then, given , there exists an such that

for all . It follows that

for for all , that is, each is a fundamental sequence in .

Let where

Then, obviously .

This proves the completeness of . The completeness of the spaces and introduced in earlier examples/blogs is proved in almost the same way. (HW: supply the details). QED.

**Example 4:**

Let be a Cauchy sequence in the function space introduced earlier. Then, given any , there is an such that

….I

for all and all . It follows that the sequence is uniformly convergent. But the limit of a uniformly convergent sequence of continuous functions is itself a continuous function (see Problem 1 following this Section). Taking the limit as in I, we find that

for all and all , that is, converges in the metric space to a function . Hence, is a complete metric space.

**Example 5:**

Next, let be a sequence in the space so that

, where

Suppose further that is a Cauchy sequence. Then, given any there is a such that

…let us call this II.

if .

It follows that (for )

That is, for every k the sequence is fundamental and hence, convergent. Let

,

Then, as we now show, x is itself a point of and moreover, converges to x in the metric space, so that is a complete metric space.

In fact, the Cauchy criterion here implies that for any fixed M. …let us call this III.

Holding n fixed in III, and taking the limit as , we get

….call this IV.

Since IV holds for arbitrary M, we can in turn take the limit of IV as , obtaining

.

But, as we have learnt earlier in this series of blogs, the convergence of the two series and implies that of the series .

This proves that . Moreover, since is arbitrarily small, III implies that

That is, converges to x in the metric space, as asserted.

**QED.**

*Example 6. *

Consider the space . To recap: consider the set of all functions continuous on the closed interval with the distance metric defined by: .

It is easy to show that the space is incomplete. If

, if

, if

, if

then is a fundamental sequence in since

However, cannot converge to a function in . In fact, consider the discontinuous function

, when

, when .

Then, given any function , it follows from Schwarz’s inequality (obviously still valid for piecewise continuous functions) that

But the integral on the left is non-zero, by the continuity of f, and moreover, it is clear that

Therefore, cannot converge to zero as .

**QED.**

**7.2 The nested sphere theorem. **

A sequence of closed spheres

in a metric space R is said to be * nested *(or decreasing) if

*Using this concept, we can prove a simple criterion for the completeness of R:*

**THEOREM 2: The Nested Sphere Theorem:**

A metric space R is complete if and only if every nested sequence of closed spheres in R such that as has a non empty intersection

.

**Proof of the nested theorem:**

Part I: Assume that R is complete and that if is any nested sequence of closed spheres in R such that as , then the sequence of centres of the spheres is fundamental because

for and as . Therefore, has a limit. Let

.

Then,

Not only that, we can in fact say that contains every point of the sequence except possibly the points and hence, x is a limit point of every sphere . But, is closed, and hence, for all n.

* Conversely, * suppose every nested sequence of closed spheres in R with radii converging to zero has a non empty intersection, and let be any fundamental sequence in R. Then, x has a limit point in R. To see this, use the fact that is fundamental to choose a term of the sequence such that

for all , and let be the closed sphere of radius 1 with centre . Then, choose a term of such that and for all , and let be the closed sphere of radius with centre .

Continue this construction indefinitely, that is, once having chosen terms (where ), choose a term such that and

for all .

Let be the closed sphere of radius with centre , and so on. This gives a nested sequence of closed spheres with radii converging to zero. By hypothesis, these spheres have a non empty intersection, that is, there is a point x in all the spheres. This point is obviously the limit of the sequence . But, if a fundamental sequence contains a subsequence converging to x, then the sequence itself must converge to x (HW quiz). That is,

.

**QED. **

**7.3 Baire’s theorem:**

We know that a subset A of a metric space R is said to be **nowhere dense** in R if it is dense in no (open) sphere at all, or *equivalently*, if every sphere contains another sphere such that . (Quiz: check the equivalence).

*This concept plays an important role in the following:*

**THEOREM 3: Baire’s Theorem:**

A complete metric space R cannot be represented as the union of a countable number of nowhere dense sets.

Proof of Theorem 3: Baire’s Theorem:

Suppose the contrary. Let ….call this VI.

where every set is nowhere dense in R. Let be a closed sphere of radius 1. Since is nowhere dense in , being nowhere dense in R, there is a closed sphere of radius less than such that and . Since is nowhere dense in , being nowhere dense in , there is a closed sphere of radius less than such that and , and so on. In this way, we get a nested sequence of closed spheres with radii converging to zero such that

, where

By the nested sphere theorem, the intersection contains a point x. By construction, x cannot belong to any of the sets , that is,

It follows that contrary to VI.

Hence, the representation VI is impossible.

**QED.**

**COROLLARY TO Baire’s theorem: **

A complete metric space R without isolated points is uncountable.

**Proof:**

Every single element set is nowhere dense in R.

**QED.**

**7.4 Completion of a metric space:**

As we now show, an incomplete metric space can always be enlarged (in an essentially unique way) to give a complete metric space.

**DEFINITION 4: **Completion of a metric space:

Given a metric space R with closure , a complete metric space is called a completion of R if and , that is, if R is a subset of everywhere dense in .

*Example 1. *

Clearly, if R is already complete. (Quiz: homework).

*Example 2:*

The space of all real numbers is the completion of the space of all rational numbers.

**THEOREM 4:**

Every metric space R has a completion. This completion is unique to within an isometric mapping carrying every point into itself.

**Proof of Theorem 4:**

*(The proof is somewhat lengthy but quite straight forward).*

First , we prove the uniqueness showing that if and are two completions of R, then there is a one-to-one mapping onto such that for all and

….call this VII.

(), where is the distance metric in and the distance metric in . The required mapping is constructed as follows: Let be an arbitrary point of . Then, by the definition of a completion, there is a sequence of points of R converging to . The points of the sequence also belong to , where they form a fundamental sequence (quiz: why?). Therefore, converges to a point since is complete. It is clear that is independent of the choice of the sequence converging to the point (homework quiz: why?). If we set , then is the required mapping. In fact, for all , since if , then obviously , . Moreover, suppose , in , while , . Then, if is the distance in R,

…call this VIII.

While at the same time, ….call this VIII-A. But VIII and VIII-A imply VII.

We must now prove the existence of a completion of R. Given an arbitrary metric space R, we say that two Cauchy sequences and in R are equilvalent and write

if

As anticipated by the notation and terminology, is reflective, symmetric and transitive, that is, is an equivalence relation. Therefore, the set of all Cauchy sequences of points in the space R can be partitioned into classes of equivalent sequences. Let these classes be the points of a new space . Then, we define the distance between two arbitrary points by the formula

….call this IX.

where is any “representative” of (namely, any Cauchy sequence in the class ) and is any representative of .

The next step is to verify that IX is indeed a distance metric. That is, also to check that IX exists, independent of the choice of sequence , , and satisfies the three properties of a distance metric function. Given any , it follows from the triangle inequality in R (this can be proved with a little effort: homework quiz) that

That is,

that is, ….call this X

for all sufficiently large and .

Therefore, the sequence of real numbers is fundamental and hence, has a limit. This limit is independent of the choice of , . In fact, suppose that

,

Then,

by a calculation analogous to X. But,

since and , and hence,

.

As for the three properties of a metric, it is obvious that

, and the fact that

if and only if is an immediate consequence of the definition of equivalent Cauchy sequences.

To verify the triangle inequality in , we start from the triangle inequality:

in the original space R, and then take the limit as , obtaining

That is,

We now come to the crucial step of showing that is a completion of R. Suppose that with every point , we associate the class of all Cauchy sequences converging to x. Let

,

Then, clearly

(the above too can be proven with a slight effort: HW quiz); while, on the other hand,

by definition. Therefore,

and hence, the mapping of R into carrying x into is isometric. Accordingly, we need no longer distinguish between the original space R and its image in , in particular between the two metrics and . In other words, R can be regarded as a subset of . The theorem will be proved once we succeed in showing that

(i) R is everywhere dense in , that is, ;

(2) is complete.

Towards that end, given any point and any , choose a representative of , namely a Cauchy sequence in the class . Let N be such that for all . Then,

if , that is, every neighbourhood of the point contains a point of R. It follows that .

Finally, to show that is complete, we first note that by the very definition of , any Cauchy sequence consisting of points in R converges to some point in , namely to the point defined by . Moreover, since R is dense in , given any Cauchy sequence consisting of points in , we can find an equivalent sequence consisting of points in R. In fact, we need only choose to be any point of R such that . The resulting sequence is fundamental, and, as just shown, converges to a point . But, then the sequence also converges to .

**QED. **

*Example. *

If R is the space of all rational numbers, then is the space of all real numbers, both equipped with the distance . In this way, we can “construct the real number system.” However, there still remains the problem of suitably defining sums and products of real numbers and verifying that the usual axioms of arithmetic are satisfied.

Regards,

Nalin Pithwa.

# Problem Set based on VI. Convergence, open and closed sets.

**Problem 1. **

Give an example of a metric space R and two open spheres , in R such that although .

**Problem 2:**

Prove that every contact point of a set M is either a limit point of M or is an isolated point of M.

Comment. In particular, can only contain points of the following three types:

a) Limit points of M belonging to M.

b) Limit points of M which do not belong to M.

c) Isolated points of M.

Thus, is the union of M and the set of all its limit points.

**Problem 3:**

Prove that if and as then .

Hint : use the following problem: Given a metric space prove that

**Problem 4:**

Let f be a mapping of one metric space X into another metric space Y. Prove that f is continuous at a point if and only if the sequence converges to whenever the sequence converges to .

**Problem 5:**

Prove that :

(a) the closure of any set M is a closed set.

(b) is the smallest closed set containing M.

**Problem 6:**

Is the union of infinitely many closed sets necessarily closed? How about the intersection of infinitely many open sets? Give examples.

**Problem 7:**

Prove directly that the point belongs to the Cantor set F, although it is not the end point of any of the open interval deleted in constructing F. Hint: The point divides the interval in the ratio and so on.

**Problem 8:**

Let F be the Cantor set. Prove that

(a) the points of the first kind form an everywhere dense subset of F.

(b) the numbers of the form where fill the whole interval .

**Problem 9:**

Given a metric space R, let A be a subset of R, and . Then, the number is called the distance between A and x. Prove that

(a) implies but not conversely

(b) is a continuous function of x (for fixed A).

(c) if and only if is a contact point of A.

(d) , where M is the set of all points x such that .

**Problem 10:**

Let A and B be two subsets of a metric space R. Then, the number is called the distance between A and B. Show that if , but not conversely.

**Problem 11:**

Let be the set of all functions f in satisfying a *Lipschitz condition,* that is, the set of all f such that for all , where K is a fixed positive number. Prove that:

a) is closed and in fact is the closure of the set of all differentiable functions on such that

(b) the set of all functions satisfying a Lipschitz condition for some K is not closed;

(c) The closure of M is the whole space

**Problem 12:**

An open set G in n-dimensional Euclidean space is said to be connected if any points can be joined by a polygonal line(by a polygonal line we mean a curve obtained by joining a finite number of straight line segments end to end.) lying entirely in G. For example, the open disk is connected, but not the union of the two disks , (even though they share a contact point). An open subset of an open set G is called a component of G if it is connected and is not contained in a larger connected subset of G. Use Zorn’s lemma to prove that every open set G in is the union of no more than countably many pairwise disjoint components.

Comment: In the case , that is, the case on the real line, every connected open set is an open interval, possibility one of the infinite intervals , or . Thus, theorem 6 (namely: Every open set G on the real line is the union of a finite or countable system of pairwise disjoint open intervals) on the structures of open sets on the line is tantamount to two assertions:

(i) Every open set on the line is the union of a finite or countable number of components.

(ii) Every open connected set on the line is an open interval.

The first assertion holds for open sets in (and, in fact, is susceptible to further generalizations), while the second assertion pertains specifically to the real line.

Cheers,

Happy analysis !!

Nalin Pithwa

# II Metric Spaces:

**Reference: Introductory Real Analysis Kolmogorov and Fomin, translated by Richard A Silverman, Dover Publications. **

**Reference: Analysis by Walter Rudin, Third Edition.**

**Reference:** https://en.wikipedia.org/wiki/Maurice_Ren%C3%A9_Fr%C3%A9chet

Chapter II Metric Spaces:

V: Basic Concepts:

**Section 5.1: Definitions and examples:**

One of the most important operations in mathematical analysis is the taking of limits. Here what matters is not so much as the algebraic nature of the real numbers (that is, the fact that real numbers form a field), but rather the fact that distance from one point to another on the real line(or, in two or three dimensional space) is well-defined and has certain properties. Roughly speaking, a metric space is a set equipped with a distance (or, ‘metric’) which has these same properties. More exactly, we have:

**Definition 1: **

By a **metric space ***is meant a pair * consisting of a set X and a **distance **, that is, a single valued, nonnegative, real function defined for all which has the following three properties:

- if and only if
- Symmetry:
- Triangle Inequality:

We will often refer to the set X as a “space” and its elements x, y, …, as “points.” Metric spaces are usually denoted by a single letter, like

or, even by the same letter X as used for the underlying space, in cases where there is no possibility of confusion.

**Example 1:**

Setting , if and , when , where x and y are elements of an arbitrary set X, we obviously get a metric space, which might be called a “discrete space” or a “space of isolated points.”

Check: does this satisfy all the three axioms of a metric space: clearly, the first axiom is true. So, also the second axiom is true because is zero when and is 1 when or .

Now we have to check: . Case I: if , LHS is zero and again RHS could be zero or one depending on y and z. In all cases, the inequality holds. Case II: If , then LHS is 1. Now, , then is 1 and depending on z, is zero or 1. So, we get LHS = RHS = 1 or LHS=1 less than RHS, which is 2.

So, yes indeed this is a well-defined metric function.

Some remarks: To think further: Suppose we are given the following function : when and when . Can what can we say about this function with respect to the above discrete space ? What are the limit points of such a function ? Is such a function continuous (if so, at which points) in this metric space? Is it dense in this metric space?

**Example 2:**

The set of all real numbers with distance is a metric space, which we denote by —- one dimensional real line.

Check: is this a well-defined metric ? Clearly, axiom 1 holds true because and axiom 2 is true because . Now, we need to check: . Here, LHS is where we have used the triangle inequality. So, axiom 3 holds true.

So, this is indeed a well-defined metric.

**Example 3:**

The set of all ordered n-tuples of real numbers with distance

….call this relation I

is a metric space denoted by and called n-dimensional Euclidean space (or, simply Euclidean n-space). The distance (I) obviously satisfies axioms 1 and 2 of definition of a metric. Moreover, it can be seen that (I) satisfies the third axiom also:

In fact, let , , be three points in .

Futher, let and when .

Then, the triangle inequality takes the form:

….let us call this Relation II.

Or equivalently,

….call this as relation II’

It follows from the Cauchy-Schwarz inequality that …call this as relation III.

so that we have now,

.

Taking square roots, we get II’ and hence, II.

QED.

**Example 4:**

Take the same set of ordered tuples as in preceding example , but this time define the distance function by ….call this as relation IV.

It is clear that this is also a well-defined metric function.

Check: Axiom 1 is obvious. So, also axiom 2 because . Axiom 3 holds true because the following general inequality holds true: .

**Example 5:**

Take the same set as in the previous two examples, but now let us define the distance to between two points and to be ….call this V.

This is also a well-defined metric function.

This space, denoted by is often as usual the Euclidean space .

*Remark: The last three examples show that it is sometimes important to use a different notation for a metric space than for the underlying set of points in the space, since the latter can be “metrized” in a variety of different ways.*

**Example 6: **

The set of all continuous functions defined on the closed interval with distance …call this VI. This is a metric space of great importance in analysis.

Let us verify so:

*****

****

***

This metric space and the underlying set of points in the space will both be denoted by . Instead of we just write C. A space like is often called a “function space” to emphasize that its elements are functions.

**Example 7:**

Let be the set of all “infinite” sequences : of real numbers satisfying the convergence condition:

Note: the infinite sequence with the general term can be written as or simply as (this notation is familiar from calculus). It can also be written in “point notation” as , that is, as an “ordered tuple” generalizing the notion of an ordered n-tuple. (In writing ) we have another use of curly brackets, but the context will always prevent any confusion between the *sequence * and the set whose only element is ).

Where distance between points is defined by

…call this VII.

Clearly, VII makes sense for all since it follows from the elementary inequality

that the convergence of the two series and also implies the convergence of the series .

At the same time, we find that if the points and both belong to , then so does the point:

(since the lim of a sum of two sequences is the sum of the individual limits)

The function VII obviously has the first two defining properties of a distance. To verify the triangle inequality, which takes the form:

….call this VIII.

for the metric VII, we first note that all three series converge, for the reason just given. Moreover, the inequality:

…call this IX, holds for all n, (as proved in Example 3 above). Taking the limit as in IX, we get VIII, thereby satisfying the triangle inequality in . Therefore, is a metric space.

**Example 8:**

As in Example 6, consider the set of all functions continuous on the interval , but now let us define the metric by the formula:

….call this X.

instead of VI.

The resulting metric space will be denoted by . The first two axioms of the metric clearly hold, and the fact that X satisfies the triangle inequality is an immediate consequence of the following **Schwarz’s inequality:**

(see Problem 3 in the exercises below), by the continuous analogue of the argument given in example 3 above.

**Example 9:**

Next consider the set of all bounded infinite sequences of real numbers and let

….call this XII.

This gives a metric space which we denote by m. The fact that XII satisfies axioms 1 and 2 of a metric space is obvious by the definition of a supremum.

Axiom 3 can be verified as follows:

***

***

**Example 10:**

As in example 3, consider the set of all ordered n-tuples, , but now let the metric be given by the more general formula as follows:

….call this XIII.

where p is a fixed real number greater than or equal to 1. (Examples 3 and 4 correspond to the cases and , respectively.) This gives a metric space, which we denote by .

It is obvious that if and only if .

It is obvious that .

But, verification of the third axiom of the definition of a metric (XIII) (that is, the triangle inequality) requires a little work as follows:

Let , , be three points in , and let:

, for just as in example 3. Then, the triangle inequality

takes the form of **Minkowski’s inequality:**

Call the above inequality as XIV in the current blog article.

PS: I think the proof of the Minkowski inequality can be found in any standard text on Inequalities by B. J. Venkatachala, for example;; or, in wikipedia.

The above inequality holds true clearly for , and hence, we assume the case .

The proof of XIV in turn again is based on **Holder’s inequality: **

Call the above as inequality XV.

Where the numbers , satisfy the condition:

….call this as XVI.

We begin by observing that the inequality XV is homogeneous, that is, if it holds for any two points and then it holds for the two points and where and are any two real numbers. Therefore, we need only prove XV for the following case:

….call this relation XVII.

Thus, assuming that XVII holds, we now have to prove that: …call this XVIII.

Consider the two areas and , associated with the curve defined in the -plane and given by the equation:

, or equivalently by the equation:

Then, clearly , and

Moreover, it is apparent (if we draw the figure suitably) that for arbitrary positive a and b. It follows that …call this relation (19 or XIX).

Setting , , summing over k from 1 to n, and taking account of (16, or XVI) and (17, or XVII), we get the desired inequality (18, or XVIII). This proves Holder’s inequality (15 or XV). Note that (15 or XV) reduces to Schwarz’s inequality if .

It is now an easy matter to prove Minkowski’s inequality (14 or XIV), starting from the identity

.

In fact, putting , and summing over k from 1 to n, we obtain

Next, we apply Holder’s inequality (15 or XV) to both sums on the right, bearing in mind that :

Dividing both sides of this inequality by

we get

which immediately implies (14 of XIV), thereby proving the triangle inequality in .

QED.

**Example 11:**

Finally, let be the set of all infinite sequences of real numbers satisfying the convergence condition

for some fixed number , where distance between points is defined by

….call this (20 or XX)

(the case has already been considered in Example 7). It follows from Minkowski’s inequality (14 or XIV) that

….call this (21 or XXI) for any n.

Since the series , and converge, by hypothesis, we can take the limit as in (21 or XXI) obtaining

.

This shows that (20 or XX) actually makes sense for arbitrary . At the same time, we have verified that the triangle inequality holds in (the other two properties of a metric space are obviously satisfied). Therefore, is a metric space.

QED.

**Remarks:**

If is a metric space and M is any subset of X, then obviously is again a metric space, called a subspace of the original metric space R. This device gives us infinitely more examples of metric spaces.

**Example 12:**

For , define:

12a)

12b)

12c)

12d)

12e)

Determine for each of these, whether it is a metric or not.

**Solution 12a: **

Axioms 1 and 2 are clearly satisfied. We have to verify if the following holds true:

. Clearly, RHS is whereas LHS is so it may not always be true that LHS is lesser than or equal to RHS.

Hence, this is not a metric function.

**Solution 12b:**

this also satisfies the first two axioms of the definition of a metric.

So, we have to verify if the following is true:

, that is, TPT:

.

Consider the following:

Also,

So the third axiom holds true in this case. So, the given function is a metric.

**Solution 12c:**

Once again, the first two axioms clearly hold.

We have to verify if the following holds true:

, that is, to prove that:

, which is obviously true by triangle inequality of real numbers. So, the given function is a metric.

**Solution 12d:**

Clearly, the first two axioms do not hold. It can easily be checked that the third axiom also does not hold. So, the given function is not a metric.

**Solution 12e:**

The first axiom holds true.

To check the second axiom consider and compare:

whereas clearly again.

To verify the third axiom, we have to check if the following is true:

, that is, to prove that:

. A little algebraic work shows that this is aot always possible. Hence, the given function is not a metric.