# Introductory real analysis: exercise 3:

Based on previous two blogs. (Reference: Introductory Real Analysis by Kolmogorov and Fomin, Dover Publishers):

Problem 1:Exhibit both a partial ordering and a simple ordering of the set of all complex numbers.

Problem 2:What is the minimal element of the set of all subsets of a given set X, partially ordered by set inclusion. What is the maximal element?

Problem 3: A partially ordered set M is said to be a directed set if, given any two elements $a, b \in M$, there is an element $c \in M$ such that $a < c, b. Are the partially ordered sets in the previous blog(s) Section 3.1 all directed sets?

Problem 4: By the greatest lower bound of two elements a and b of a partially ordered set M, we mean an element $c \in M$ such that $c \leq a, c \leq b$ and there is no element $d \in M$ such that $a < d \leq a, d \leq b$. Similarly, by the least upper bound of a and b, we mean an element $c \in M$ such that $a \leq c, b \leq c$ and there is no element $d \in M$ such that $a \leq d . By a lattice is meant a partially ordered set any two elements of which have both a greatest lower bound and a least upper bound. Prove that the set of all subsets of a given set X, partially ordered by set inclusion, is a lattice. What is the set theoretic meaning of the greatest lower bound and least upper bound of two elements of this set?

Problem 5: Prove that an order preserving mapping of one ordered set onto another is automatically an isomorphism.

Problem 6: Prove that ordered sums and products of ordered sets are associative, that is, prove that if $M_{1}, M_{2}, M_{3}$ are ordered sets, then

$(M_{1}+M_{2})+M_{3}=M_{1}+(M_{2}+M_{3})$,

$(M_{1}.M_{2}).M_{3}=M_{1}.(M_{2}.M_{3})$ where the operations + and . are the same as defined in previous blog(s).

Comment: This allows us to drop the parentheses in writing ordered sums and products.

Problem 7:

Construct well-ordered sets with ordinals

$\omega + n$, $\omega + \omega$, $\omega + \omega + n$, $\omega + \omega + \omega$, $\ldots$.

Show that the sets are all countable.

Problem 8:

Construct well-ordered sets with ordinals

$\omega . n$, $\omega^{2}$, $\omega^{2}.n$, $\omega^{3}$, $\ldots$.

Show that the sets are all countable.

Problem 9:

Show that $\omega + \omega = \omega. 2$, $\omega + \omega + \omega = \omega. 3$, $\ldots$

Problem 10:

Prove that the set $W(\alpha)$ of all ordinals less than a given ordinal $\alpha$ is well-ordered.

Problem 11:

Prove that any non-empty set of ordinals is well-ordered.

Problem 12:

Prove that the set M of all ordinals corresponding to a countable set is itself uncountable.

Problem 13:

Let $\aleph_{1}$ be the power of the set M in the previous problem. Prove that there is no power m such that $\aleph_{0} .

More later,

Nalin Pithwa.

# Introductory Real Analysis: The well-ordering theorem, the axiom of choice,equivalent assertions

Reference: Introductory Real Analysis by Kolmogorov and Fomin, Dover Publications.

Section 3.7 The well-ordering theorem, the axiom of choice, and equivalent assertions:

Theorem 4 of previous blog shows that the powers of two well-ordered sets are always comparable. In 1904, Zermelo succeeded in proving the

Well-ordering theorem: Every set can be well-ordered.

It follows from the well-ordering theorem and Theorem 5 (of previous blog) that the powers of two arbitrary sets are always comparable, a fact already used earlier. Zermelo’s proof, which will not be given here, rests on the following basic:

AXIOM OF CHOICE:

Given any set M, there is a “choice function” f such that $f(A)$ is an element of A for every non-empty subset $A \subset M$.

We will assume the validity of the axiom of choice without further ado. In fact, without the axiom of choice we would be severely hampered in making set-theoretic constructions. However, it should be noted that from the standpoint of the foundations of set theory, there are still deep and controversial problems associated with the use of axiom of choice.

There are a number of assertions equivalent to the axiom of choice, that is, assertions each of which both implies and is implied by the axiom of choice. One of these is the well-ordering theorem, which obviously implies the axiom of choice. In fact, if an arbitrary set M can be well-ordered, then, by merely choosing the “first” element in each subset $A \subset M$, we get the function $f(A)$ figuring in the statement of the axiom of choice. On the other hand, the axiom of choice implies the well-ordering theorem, as already noted without proof.

To state further assertions equivalent to the axiom of choice, we need some more terminology:

DEFINITION 3:

Let M be a partially ordered set, and let A be any subset of M such that a and b are comparable for every $a, b \in A$. Then, A is called a chain (in M). A chain C is said to be maximal if there is no other chain C in M containing C on a proper subset.

DEFINITION 4:

An element a of a partially ordered set M is called an upper bound of a subset $M^{'} \subset M$, if $a^{'} < a$ for every $a^{'} \in M^{'}$.

We now have the vocabulary needed to state two other assertions equivalent to the axiom of choice:

Hausdorff’s Maximal Principle: Every chain in a partially ordered set M is contained in a maximal chain in M.

Zorn’s lemma:  If every chain in a partially ordered set M has an upper bound, then M contains a maximal element.

For the proof of the equivalence of the axiom of choice, the well-ordering theorem, Hausdorff’s maximal principle and Zorn’s lemma, we refer the reader elsewhere. Of these various equivalent assertions, Zorn’s lemma is perhaps the most useful.

Section 3.8 Transfinite Induction:

Mathematical propositions are very often proved by using the following familiar:

THEOREM 4: MATHEMATICAL INDUCTION:

Given a proposition $P(n)$ formulated for every positive integer n, suppose that :

i) P(1) is true.

ii) The validity of P(k) for all $k implies the validity of $P(n+1)$. Then, P(n) is true for all $n=1,2,\ldots$.

PROOF 4:

Suppose P(n) fails to be true for all $n=1,2,\ldots$ and let n, be the smallest integer for which $P(n)$ is false (the existence of such $n_{1}$ follows from the well-ordering of the positive integers). Clearly, $n_{1}>1$, so that $n_{1}-1$ is a positive integer. Therefore, $P(n)$ is valid for all $k \leq n_{1}-1$ but not for $n_{1}$. Contradiction!!

QED.

Replacing the set of all positive integers by an arbitrary well-ordered set, we get

THEOREM 4 (Transfinite induction):

Given a well-ordered set A, let $P(a)$ be a proposition formulated for every element $a \in A$. Suppose that

i) $P(a)$ is true for the smallest element of A

ii) The validity of $P(a)$ for all $a < a^{*}$ implies the validity of $P(a^{*})$. Then, $P(a)$ is true for all $a \in A$.

PROOF 4:

Suppose $P(a)$ fails to be true for all $a \in A$. Then, $P(a)$ is false for all a in some non empty subset $A^{*} \subset A$. By the well-ordering, $A^{*}$ has a smallest element $a^{*}$. Therefore, $P(a)$ is valid for all $a< a^{*}$ but not for $a^{*}$. Contradiction !!

QED.

Remark:

Since any set can be well-ordered, by the well-ordering theorem, transfinite induction can in principle be applied to any set M whatsoever. In practice, however, Zorn’s lemma is a more useful tool, requiring only that M be partially ordered.

Section 3.9: Historical Remarks:

Set theory as a branch of mathematics in its own right stems from the pioneer work of Georg Cantor (1845-1918). Originally met with disbelief, Cantor’s ideas subsequently became widespread. By now, the set theoretic point of view has become standard in the most diverse fields of mathematics. Basic concepts, like groups, rings, fields, linear spaces etc are habitually defined as sets of elements of an arbitrary kind obeying appropriate axioms.

Further development of set theory led to a number of logical difficulties, which naturally gave rise to attempts to replace “naive” set theory by a more rigorous axiomatic set theory. It turns out that certain set theoretic questions, which would at first seem to have “yes” or “no” answers, are in fact of a different kind. Thus, it was shown by Godel in 1940 that a negative answer to the question :”Is there an uncountable set of power less than that of the continuum” is consistent with set theory (axiomatized in a way we will not discuss here), but it was recently shown by Stephen Cohen that an affirmative answer to the question is also consistent in the same sense !!

We will continue to exercises based on last two blogs now,

Regards,

Nalin Pithwa.

# Ordered Sets and Ordinal Numbers: Introductory Real Analysis: Kolmogorov and Fomin

Section 3: Ordered sets and ordinal numbers:

Section 3.1: Partially ordered sets.

A binary relation R on a set M is said to be a partial ordering (and the set M is said to be partially ordered) if

i) R is reflexive (aRa for every $a \in M$).

ii) R is transitive (aRb and bRc together imply aRc)

iii) R is antisymmetric in the sense that aRb and bRa together imply $a=b$.

For example, if M is the set of all real numbers and aRb means that $a, then R is partial ordering. (please verify!) This suggests writing $a \leq b$ (or equivalently $b \geq a$) instead of aRb whenever R is a partial ordering, and we will do so from now on. Similarly, we write $a < b$ if $a \leq b$, but $a \neq b$; and $b > a$ if $b \geq a$ but $b \neq a$.

The following examples give some idea of the generality of the concept of a partial ordering:

Example 1:

Any set M can be trivially partially ordered by setting $a \leq b$ iff $a=b$.

Example 2:

Let M be the set of all continuous functions $f, g, \ldots$ defined in a closed interval $[\alpha, \beta]$. Then, we get a partial ordering by setting $f \leq g$ iff $f(t) \leq g(t)$ for every $t \in [\alpha, \beta]$.

Example 3:

The set of all subsets $M_{1}, M_{2}, \ldots$ is partially ordered if and only if $M_{1} \leq M_{2}$ means that $M_{1} \subset M_{2}$.

Example 4:

The set of all integers greater than 1 is partially ordered if “$a \leq b$” means that “b is divisible by a.”

Remark: An element of a of a partially ordered set is said to be maximal if $a \leq b$ implies $b=a$ and minimal if $b \leq a$ implies $b=a$. Thus, in Example 4 above, every prime number (greater than 1) is a minimal element. (Of course, 1 is neither prime nor composite.)

Section 3.2: Order-preserving mappings. Isomorphisms:

Let M and $M^{'}$ be any two partially ordered sets, and let f be a one-to-one mapping of M onto $M^{'}$. Then, f is said to be order-preserving if $a \leq b$ (where $a, b \in M$) implies $f(a) \leq f(b)$ (in $M^{'}$). An order preserving mapping f such that $f(a) \leq f(b)$ implies $a \leq b$ is called an isomorphism. In other words, an isomorphism between two partially ordered sets M and $M^{'}$ is a one-to-one mapping of M onto $M^{'}$ such that $f(a) \leq f(b)$ if and only if $a \leq b$. Two partially ordered sets M and $M^{'}$ are said to be isomorphic (to each other) if there exists and isomorphism between them.

Example:

Let M be the set of positive integers greater than 1 partially ordered as in example 4 earlier. Let $M^{'}$ be the same set partially ordered in the natural way, that is, in such a way that $a \leq b$ if and only if $b-a >0$. Then, the mapping of M onto $M^{'}$ carrying every integer n into itself is order-preserving, but NOT an isomorphism.

Some remarks:

Isomorphism between partially ordered sets is an equivalence relation as defined in an earlier blog of this series, being obviously reflexive, symmetric, and transitive. Hence, any given family of partially ordered sets can be partitioned into disjoint classes of isomorphic sets. Clearly, two isomorphic partially ordered sets can be regarded as identical in cases where it is the structure of the partial ordering than the specific nature of the elements of the sets that is of interest.

Some other remarks:

For those of you who are aware, compare now the concepts and some examples of homomorphisms and isomorphisms between two algebraic objects called groups with the above concepts.  Two references for the same are: (i) Topics in Algebra by I N Herstein (2) Abstract Algebra by Dummit and Foote.

Some further remarks:

At this stage, it might even be useful to compare the concepts and examples of a cartesian product of two (non-empty) sets A and B; that of a relation R on set A and set B; that of a binary relation on a set A; that, of an equivalence relation on a set A; and also the basic meaning of a function from set A to set B. You will see that there are slight subtle differences.

Sec 3.3 Ordered Sets. Order Types:

Given two elements a and b of a partially ordered set M, it may turn out that neither of the relations $a \leq b$ or $b \leq a$ holds. In this case, a and b are said to be noncomparable. Thus, in general, the relation $\leq$ is defined only for certain pairs of elements, which is why M is said to be partially ordered. However, if M has no non-comparable elements, then M is said to be simply ordered or linearly ordered or ordered. In other words, a set M is ordered if it is partially ordered and if, given any two distinct elements $a, b \in M$, either $a , or $b. Obviously, any subset of an ordered set is itself ordered.

Each of the sets figuring in Examples 1 to 4 of Section 3.1 in this blog, is partially ordered, but not ordered. Simple examples of ordered sets are the set of all positive integers, the set of all rational numbers, the set of all real numbers in the $[0,1]$ and so on (with the usual relations of “greater than” and “less than”.)

Since an ordered set is a special kind of partially ordered set, the concepts of order-preserving mapping and isomorphism apply equally well to ordered sets. Two isomorphic ordered sets are said to have the same order type. Thus “type” is something shared by all isomorphic ordered sets, just as “power” is something shared by all equivalent sets (considered as ‘plain’ sets, without regard for possible orderings).

The simplest example of an ordered set is the set of all positive integers $1,2,3,\ldots$ arranged in increasing order, with the usual meaning of the symbol <. The order type of this set is denoted by the symbol $\omega$. Two isomorphic ordered sets obviously have the same power (because an isomorphism is a one-to-one correspondence). Thus, it makes sense to talk about the power corresponding to a given order type. The converse is not true, since a set of a given power can in general be ordered in many different ways. It is only in the finite case that the number of elements in a set uniquely determines its type, designated by the same symbol n as the number of elements in the set. For example, besides the “natural” order type $\omega$ of the set of positive integers, there is another order type corresponding to the sequence: $1,3,5, \ldots, 2,4,6, \ldots$ where odd and even numbers are separately arranged in increasing order but any odd number precedes any even number. It can be shown that the number of distinct order types of a set of power $\aleph_{0}$ is infinite and in fact, uncountable.

Section 3.4 Ordered sets and products of ordered sets.

Let $M_{1}$ and $M_{2}$ be two ordered sets of types $\theta_{1}$ and $\theta_{2}$ respectively. Then, we can introduce an ordering in the union $M_{1} \bigcup M_{2}$ of the two sets by assuming that :

(i) a and b have the same ordering as in $M_{1}$ if $a,b \in M_{1}$.

(ii) a and b have the same ordering as in $M_{2}$ if $a,b \in M_{2}$

(iii) $a if $a \in M_{1}$ and $b \in M_{2}$.

(Verify that this is actually an ordering of $M_{1} \bigcup M_{2}$). The set $M_{1} \bigcup M_{2}$ ordered in this way is called the ordered sum of $M_{1}$ and $M_{2}$ denoted by $M_{1}+M_{2}$. Note that the order of terms matters here, that is, in general $M_{2}+M_{1}$ is not isomorphic to $M_{1}+M_{2}$. More generally, we can define the ordered sum of any finite number of ordered sets by writing (this will be an exercise problem after this blogged section) :

$M_{1}+M_{2}+M_{3}=(M_{1}+M_{2})+M_{3}$

$M_{1}+M_{2}+M_{3}+M_{4}=(M_{1}+M_{2}+M_{3})+M_{4}$,

and so on.

By the ordered sum of the types $\theta_{1}+\theta_{2}$, denoted by $\theta_{1}+\theta_{2}$ we mean the order type of the set $M_{1}+M_{2}$.

Example:

Consider the order types $\omega$ and $n$. It is easy to see that $n + \omega = \omega$. In fact, if finitely many terms are written to the left of the sequence $1,2,3,\ldots, k, \ldots$, we again get a set of the same type (Quiz: why?). On the other hand, the order type $\omega + n$, that is the order type of the set (***) :

$\{ 1,2,3, \ldots, k, \ldots, a_{1}, a_{2}, \ldots, a_{n}\}$ is obviously not equal to $\omega$.

(Note: ***: In the above listing of elements of the set, we have used the usual curly braces notation, but with the emphasis on the order as shown.)

Again, let $M_{1}$ and $M_{2}$ be two ordered sets of types $\theta_{1}$ and $\theta_{2}$ respectively, Suppose we replace each element of $M_{2}$ by a “replica” of the set $M_{1}$. Then, the resulting set, denoted  by $M_{1}.M_{2}$ is the set of all pairs $(a,b)$ where $a \in M_{1}$ and $a \in M_{2}$ ordered in such a way that:

(i) $(a_{1},b_{1})<(a_{2},b_{2})$ if $b_{1} for arbitrary $a_{1},a_{2}$

(ii) $(a_{1},b)<(a_{2},b)$ if $a_{1}.

Note that the order of factors matters here, that is, in general $M_{2}.M_{1}$ is not isomorphic to $M_{1}.M_{2}$. The ordered product of any finite number of ordered sets can be defined by writing (again a problem in the exercise later):

$M_{1}.M_{2}.M_{3}=(M_{1}.M_{2}).M_{3}$

$M_{1}.M_{2}.M_{3}.M_{4}=(M_{1}.M_{2}.M_{3}).M_{4}$

and so on. By the ordered product of the types $\theta_{1}$ and $\theta_{2}$, denoted by $\theta_{1}.\theta_{2}$, we mean the order type of the set $M_{1}.M_{2}$.

Section 3.5. Well-ordered sets. Ordinal numbers.

A key concept in the theory of ordered sets is given by:

DEFINITION 1: An ordered set M is said to be well-ordered if every non-empty subset A of M has a smallest (or, “first“) element, that is, an element $\mu$ such that $\mu  for every $a \in A$.

Example 1: Every finite ordered set is obviously well-ordered.

Example 2: Every non-empty subset of a well-ordered set is itself well-ordered.

Example 3: The set M of numbers in the interval $[0,1]$ is ordered but not well-ordered. It is true that M has a smallest element, namely, the number 0 but the subset of M consisting of all positive rational numbers has no smallest element.

DEFINITION 2: The order type of a well-ordered set is called an ordinal number or simply, an ordinal. If the set is infinite, the ordinal is said to be transfinite. (NB: This is a good place to point out that the terms “cardinal number” and “power” of a set are synonymous).

Example 4: The set of positive integers $1,2,\ldots, k, \ldots$ arranged in increasing order is well-ordered, and hence, its order type $\omega$ is a transfinite ordinal. The order type $\omega + n$ of the set $\{1,2,\ldots, k, \ldots, a_{1}, a_{2}, \ldots, a_{n} \}$ is also an ordinal.

Example 5: The set $\{ \ldots, -k, \ldots, -3,-2,-1\}$ is ordered but not well-ordered. It is true that any non empty subset A of this set has a largest element (that is, an element $\mathscr{v}$ such that $a ) for every $a \in A$), but in general set A will not have a smallest element. In fact, the given set itself has no smallest element. Hence, the order type of given set is denoted by $\omega^{*}$ is not an ordinal number.

THEOREM 1The ordered sum of a finite number of well-ordered sets $M_{1}, M_{2}, \ldots, M_{n}$ is itself a well-ordered set.

PROOF 1: Let M be an arbitrary subset of the ordered sum $M_{1}+M_{2}+\ldots+M_{n}$ and let $M_{k}$ be the first of the sets $M_{1}, M_{2}, \ldots, M_{n}$, (namely, the set with the smallest index) containing elements of M. Then, $M \bigcap M_{k}$ is a subset of the well-ordered set $M_{k}$ and as such has a smallest element $\mu$. Clearly, $\mu$ is the smallest element of M itself. QED.

COROLLARYThe ordered sum of a finite number of ordinal numbers is itself an ordinal number.

Thus, new ordinal numbers can be constructed from any given set of ordinal numbers. For example, starting from the positive integers, (that is, the finite ordinal numbers) and the ordinal number $\omega$, we can construct the new ordinal numbers :

$\omega + n$, $\omega + \omega$, $\omega + \omega + n$, $\omega + \omega + \omega$, and so on.

THEOREM 2The ordered product of two well-ordered sets $M_{1}, M_{2}$ is itself a well-ordered set.

Proof 2:

Let M be an arbitrary subset of $M_{1}\dot M_{2}$ so that M is a set of ordered pairs $(a,b)$ with $a \in M_{1}, b\in M_{2}$. The set of all second elements b of pairs in M is a subset of $M_{2}$, and as such has a smallest element since $M_{2}$ is well-ordered. Let $b_{1}$ denote this smallest element, and consider all pairs of the form $(a,b_{1})$ contained in M. The set of all first elements a of these pairs is a subset of $M_{1}$, and as such has a smallest element since $M_{1}$ is well-ordered. Let $a_{1}$ denote this smallest element. Then, the pair $(a,b)$ is clearly the smallest element of M. QED.

COROLLARY 1:

The ordered product of a finite number of well-ordered sets is itself a well-ordered set.

COROLLARY 2:

The ordered product of a finite number of ordinal numbers is itself an ordinal number.

Thus it makes sense to talk about ordinal numbers $\omega . n, \omega^{2}, \omega^{2} . n, \omega^{3}, \ldots$ and so on. It is also possible to define such ordinal numbers as $\omega^{\omega}, \omega^{\omega^{\omega}}, \ldots$.

Section 3.6 Comparison of ordinal numbers:

If $n_{1}$ and $n_{2}$ are two ordinal numbers, then they either coincide or else one is larger than the other. As we now show, the same is true for transfinite ordinal numbers. We begin by observing that every element $a$ of a well-ordered set M determines an (initial section) P, the set of all $x \in M$ such that $x, and a remainder Q, the set of all $x \in M$ such that $x \geq a$. Given any two ordinal numbers $\alpha$ and $\beta$, Let M and N be well-ordered sets of order type $\alpha$ and $\beta$. Then, we say that

i) $\alpha=\beta$ if M and N are isomorphic;

ii) $\alpha < \beta$ if M is isomorphic to some section of N

iii) $\alpha > \beta$ if N is isomorphic to some section of M.

(note that this definition makes sense for finite $\alpha$ and $\beta$).

LEMMA:

Let f be an isomorphism of a well-ordered set A onto some subset $B \subset A$. Then, $f(a) \geq a$ for all $a \in A$.

Proof of lemma:

If there are elements $a \in A$ such that $f(a), then there is a least such element since A is well-ordered. Let $a_{0}$ be this element, and let $b_{0}=f(a_{0})$. Then $b_{0}, and hence, $f(b_{0}), since f is an isomorphism. But, then $a_{0}$ is not the smallest element such that $f(a). Contradiction! QED.

It follows from the lemma that a well-ordered set A cannot be isomorphic to any of its sections, since if A were isomorphic to the section determined by a, then clearly $f(a). In other words, the two relations

$\alpha=\beta$, $\alpha<\beta$ are incompatible, and so are $\alpha=\beta$, $\alpha>\beta$. Moreover, the two relations $\alpha<\beta$, $\alpha>\beta$ are incompatible, since otherwise we could use the transitivity to deduce $\alpha < \alpha$, which is impossible the lemma. Therefore, if one of the three relations

$\alpha<\beta$, $\alpha=\beta$, $\alpha>\beta$….II

holds, the other two are automatically excluded. We must still show that one of the relations in II above always holds, thereby proving that any two ordinal numbers are always comparable.

THEOREM 3:

Two given ordinal numbers $\alpha$ and $\beta$ always satisfy one and only of the relations: $\alpha=\beta$, $\alpha < \beta$ and $\beta<\alpha$.

Proof 3:

Let $W(\alpha)$ be the set of all ordinals $<\alpha$. Any two numbers $\gamma$ and $\gamma^{'}$ in $W(\alpha)$ are comparable (Recall the meaning of $\gamma < \alpha, \gamma^{'}<\alpha$ and use the fact that a section of a section of a well-ordered set is itself a section of a well-ordered set) and the corresponding ordering of $W(\alpha)$ makes it a well-ordered set of type $\alpha$. In fact, if a set $A = \{ \ldots, a, \ldots, b, \ldots\}$ is of type $\alpha$, then by definition, the ordinals less than $\alpha$ are the types of well-ordered sets isomorphic to sections of A. Hence, the ordinals themselves are in one-one correspondence with the elements of A. In other words, the elements of a set of type $\alpha$ can be numbered by using the ordinals less than $\alpha$: $A = \{ a_{1}, a_{2}, \ldots, a_{n}, \ldots\}$.

Now, let $\alpha$ and $\beta$ be two ordinals. Then, $W(\alpha)$ and $W(\beta)$ are well-ordered sets of types $\alpha$ and $\beta$, respectively. Moreover, let $C=A \bigcap B$ be the intersection of the sets A and B, that is, the set of all ordinals less than both $\alpha$ and $\beta$. Then, C is well-ordered, of type $\gamma$, say. We now show that $\gamma \leq \alpha$. If $C=A$, then obviously $\gamma=\alpha$. On the other hand, if $C \neq A$, then C is a section of A and hence, $\gamma < \alpha$. In fact, let $\xi \in C$, $\eta \in A - C$. Then, $\xi$ and $\eta$ are comparable, that is, either $\xi < \eta$ or $\xi > \eta$. But, $\eta < \xi < \alpha$ is impossible, since then $\eta \in C$. Therefore, $\xi < \eta$ and hence, C is a section of A, which implies $\gamma < \alpha$. Moreover, $\gamma$ is the first element of the set $A-C$. Thus, $\gamma < alpha$, as asserted, and similarly, $\gamma < \beta$. The case $\gamma < \alpha$, $\gamma < \beta$ is impossible, since then $\gamma \in A-C$ and $\gamma \in B-C$. But then $\gamma \notin C$ on the one hand and $\gamma \in A \bigcap B = C$ on the other hand. It follows that there are only three possibilities:

$\gamma=\alpha$ and $\gamma = \beta$ and $\alpha=\beta$

$\gamma=\alpha$ and $\gamma< \beta$ and $\alpha < \beta$

$\gamma < \alpha$ and $\gamma = \beta$ and $\alpha > \beta$,

that is, $\alpha$ and $\beta$ are comparable.

QED.

THEOREM 4:

Let A and B be well-ordered sets. Then, either A is equivalent to B or one of the sets is of greater power than the other, that is, the powers of A and B are comparable.

PROOF 4:

There is a definite power corresponding to each ordinal. But we have just seen that ordinals are comparable, and so are the corresponding powers.

[[recall the definition of inequality of powers given in Sec 2.5 in the blog series of Kolmogorov and Fomin). (Reproduced here for the sake of helping the memory: Given any two sets M and N, suppose M and N are equivalent. Then, M and N are said to have the same power. Roughly speaking, “power” is something shared by equivalent sets. If M and N are finite, then M and N have the same number of elements, and the concept of the power of a set reduces to the usual notion of the number of elements in a set. The power of the set $\mathcal{Z_{+}}$ is $\aleph_{0}$ and sets which are equivalent to the set $[0,1]$ are said to have the power of the continuum, c. For the powers of finite sets, that is, for the set of positive integers, we have the notions of “greater than” or “less than”, as well as the notion of equality. These concepts can be extended to infinite sets. If A and B are two sets with powers m(A) and m(B) respectively, either $m(A)=m(B)$, or A is equivalent to a subset of B such that no subset of B is equivalent to A, then by analogy with the final case, m(A) is less than m(B).]]

To be continued soon,

Regards,

Nalin Pithwa

# Second Problem Set from Kolmogorov and Fomin: Introductory Real Analysis

$bf{Problem 1}$.

Prove that a set with an uncountable subset is itself uncountable.

$\bf{Problem 2}$.

Let M be any infinite set and let set A be any countable set. Prove that $M \sim M\bigcup A$.

$\bf{Problem 3}$.

Prove that each of the following sets is countable:

(a) The set of all numbers with two distinct decimal representations (like 0.50000…and 0.49999…).

(b) The set of all rational points in the plane (that is, points with rational coordinates)

(c) The set of all rational intervals (That is, intervals with rational end points)

(d) The set of all polynomials with rational coefficients.

$\bf{Problem 4}$.

A number $\alpha$ is called algebraic if it is a root of a polynomial equation with rational coefficients. Prove that the set of all algebraic numbers is countable.

$\bf{Problem 5}$

Prove the existence of uncountably many transcendental numbers, that is, numbers which are not algebraic. Hint: Use Theorems 2 and 5 of previous blog(s).

$\bf{Problem 6}$

Prove that the set of all real functions (more generally, functions taking values in a set containing at least two elements) defined on a set M is of power greater than the power of M. In particular, prove that the power of the set of all real functions (continuous and discontinuous) defined in the interval $[0,1]$ is greater than c. Hint: Use the fact that the set of all characteristic functions (that is, functions taking values 0 and 1) on M is equivalent to the set of all subsets of M.

$\bf{Problem7}$

Give an indirect proof of the equivalence of the closed interval $[a,b]$, the open interval $(a,b)$ and the half-open interval $[a,b)$ or $(a,b)$. Hint: Use theorem 7 (the Cantor-Bernstein theorem).

$\bf{Problem8}$

Prove that the union of a finite set of a finite or countable number of sets each of power c is itself of power c.

$\bf{Problem9}$

Prove that the each of the following sets has the power of the continuum:

(a) The set of all infinite sequences of positive integers.

(b) The set of all ordered n-tuples of real numbers.

(c) The set of all infinite sequences of real numbers.

$\bf{Problem10}$

Develop a contradiction inherent in the notion of the “set of all sets which are not members of themselves”. Hint: Is this set a member of itself? Comment: Thus, we will be careful to sets which are “too big,” like the “set of all sets”.

We will go through the solutions of the above soon,

Cheers,

Nalin Pithwa

# Some foundation mathematics

Well-Ordering Principle:

Every non-empty set S of non-negative integers contains a least element; that is, there is some integer a in S such that $a \leq b$ for all b’s belonging to S.

Because this principle plays a role in many proofs related to foundations of mathematics, let us use it to show that the set of positive integers has what is known as the Archimedean property.

Archimedean property:

If a and b are any positive integers, then there exists a positive integer n such that $na \geq b$.

Proof:

Assume that the statement of the theorem is not true so that for some a and b, we have $na for every positive integer n. Then, the set $S = \{ b-na : n \in Z^{+}\}$ consists entirely of positive integers. By the Well-Ordering Principle, S will possess a least element, say, $b-ma$. Notice that $b- (m+1)a$ also lies in S; because S contains all integers of this form. Further, we also have $b-(m+1)a=(b-ma)-a contrary to the choice of $b-ma$ as the smallest integer in S. This contradiction arose out of original assumption that the Archimedean property did not hold; hence, the proof. QED.

First Principle of Finite Induction:

Let S be a set of positive integers with the following properties:

a) the integer 1 belongs to S.

b) Whenever the integer k is in S, the next integer $k+1$ is also in S.

Then, S is the set of all positive integers.

Second Principle of Finite Induction:

Let S be a set of positive integers with the following properties:

a) the integer 1 belongs to S.

b) If k is a positive integer such that $1,2,\ldots k$ belong to S, then $(k+1)$ must also be in S.

Then, S is the set of all positive integers.

So, in lighter vein, we assume a set of positive integers is given just as Kronecker had observed: “God created the natural numbers, all the rest is man-made.”

More later,

Nalin Pithwa.

# Function Algebras — by Walter Rudin

(The following is reproduced from the book “The Way I Remember It” by Walter Rudin. The purpose is just to share the insights of a formidable analyst with the student community.)

When I arrived at MIT in 1950, Banach algebras were one of the hot toppers. Gelfand’s 1941 paper “Normierte Ringe” had apparently only reached the USA in the late forties, and circulated on hard-to-read smudged purple ditto copies. As one application of the general theory presented there, it contained a stunningly short proof of Wiener’s lemma: the Fourier series of the reciprocal of a nowhere vanishing function with absolutely convergent Fourier series also  converges absolutely. Not only was the proof extremely short, it was one of those that are hard to forget. All one needs to remember is that the absolutely convergent Fourier series form a Banach algebra, and that every multiplicative linear functional on this algebra is evaluation at some point of the unit circle.

This may have led some to believe that Banach algebras would now solve all our problems. Of course, they could not, but they did provide the right framework for many questions in analysis (as does most of functional analysis) and conversely, abstract questions about Banach algebras often gave rise to interesting problems in “hard analysis”. (Hard analysis is used here as Hardy and Littlewood used it. For example, you do hard analysis when, in order to estimate some integral, you break it into three pieces and apply different inequalities to each.)

One type of Banach algebras that was soon studied in detail were the so-called function algebras, also known as uniform algebras.

To see what these are, let $C(X)$ be the set of all complex-valued continuous functions on a compact Hausdorff space X. A function algebra on X is a subset A of $C(X)$ such that

(i) If f and g are in A, so are $f+g$, $fg$, and $cf$ for every complex number c (this says that A is an algebra).

(ii) A contains the constant functions.

(iii) A separates points on X (that is, if $p \neq q$, both in X, then $f(p) \neq f(q)$ for some f in A), and

(iv) A is closed, relative to the sup-norm topology of $C(X)$, that is, the topology in which convergence means uniform convergence.

A is said to be self-adjoint if the complex conjugate of every f in A is also in A. The most familiar example of a non-self-adjoint function algebra is the disc algebra $A(U)$ which consists of all f in $C(U)$ that are holomorphic in U. (here, and later, U is the open unit disc in C, the complex plane, and $\overline{U}$ is its closure). I already had an encounter with  $A(U)$, a propos maximum modulus algebras.

One type of question that was asked over and over again was: Suppose that a function algebra on X satisfies … and …and is it C(X)? (In fact, 20 years later a whole book, entitled “Characterizations of C(X) among its Subalgebras”  was published by R. B. Burckel.)  The Stone-Weierstrass Theorem gives the classical answer. Yes, if A is self-adjoint.

There are problems even when X is a compact interval I on the real line. For instance, suppose A is a function algebra on I, and to every maximal ideal M of A corresponds a point p in I such that M is the set of all f in A having $f(p)=0$ (In other words, the only maximal ideals of A are the obvious ones). Is $A=C(I)$? This is still unknown, in 1995.

If $f_{1}, f_{2}, \ldots f_{n}$ are in $C(I)$, and the n-tuple $(f_{1}, f_{2}, \ldots, f_{n})$ separates points on I, let $A(f_{1}, \ldots , f_{n})$ be the smallest closed subalgebras of $C(I)$ that contains $f_{1}, f_{2}, \ldots f_{n}$ and I.

When $f_{1}$ is 1-1 on I, it follows from an old theorem of Walsh (Math. Annalen 96, 1926, 437-450) that $A(f_{1})=C(f)$.

Stone-Weierstrass implies that $A(f_{1}, f_{2}, \ldots, f_{n})=C(I)$ if each $f_{i}$ is real-valued.

In the other direction, John Wermer showed in Annals of Math. 62, 1955, 267-270, that $A(f_{1}, f_{2}, f_{3})$ can be a proper subset of $C(I)$!

Here is how he did this:

Let E be an arc in C, of positive two-dimensional measure, and let $A_{E}$ be an algebra of all continuous functions on the Riemann sphere S (the one-point compactification of C). which are holomorphic in the complement of E. He showed that $g(E)=g(S)$ for every g in $A_{E}$, that $A_{E}$ contains a triple $(g_{1}, g_{2}, g_{3})$ that separates points on S and that the restriction of $A_{E}$ to E is closed in $C(E)$. Pick a homeomorphism $\phi$ of I onto E and define $f_{i}=g_{i} \circ \phi$. Then, $A(f_{1}, f_{2}, f_{3}) \neq C(I)$, for if h is in $A(f_{1}, f_{2}, f_{3})$ then $h= g \circ \phi$ for some g in $A_{E}$, so that

$h(I)=g(E)=g(S)$ is the closure of an open subset of $C$ (except when h is constant).

In order to prove the same with two function instead of three I replaced John’s arc E with a Cantor set K, also of positive two-dimensional measure (I use the term “Cantor set” for any totally disconnected compact metric space with no isolated points; these are all homeomorphic to each other.) A small extra twist, applied to John’s argument, with $A_{K}$ in place of $A_{E}$, proved that $A(f_{1}, f_{2})$ can also be smaller than $C(I)$.

I also used $A_{K}$ to show that $C(K)$ contains maximal closed point-separating subalgebras that are not maximal ideals, and that the same is true for $C(X)$ whenever X contains a Cantor set. These ideas were pushed further by Hoffman and Singer in Acta Math. 103, 1960, 217-241.

In the same paper, I showed that $A(f_{1}, f_{2}, \ldots, f_{n})=C(I)$ when $n-1$ of the n given functions are real-valued.

Since Wermer’s paper was being published in the Annals, and mine strengthened his theorem and contained other interesting (at least to me) results, I sent mine there too. It was rejected, almost by return mail, by an anonymous editor, for not being sufficiently interesting. I have had a few others papers rejected over the years, but for better reasons. This one was published in Proc. AMS 7, 1956, 825-830, and is one of six whose Russian transactions were made into a book “Some Questions in Approximation Theory”, the others were three by Bishop and two by Wermer. Good company.

Later, Gabriel Stolzenberg (Acta Math. 115, 1966, 185-198) and Herbert Alexander (Amer. J. Math., 93, 1971, 65-74) went much more deeply into these problems. One of the highlights in Alexander’s paper is:

$A(f_{1}, f_{2}, \ldots f_{n})=C(I)$ if $f_{1}, f_{2}, \ldots, f_{n-1}$ are of bounded variation.

A propos the Annals (published by Princeton University) here is a little Princeton anecdote. During a week that I spent there, in the mid-eighties, the Institute threw a cocktail party. (What I enjoyed best at that affair was being attacked by Armand Borel for having said, in print, that sheaves had vanished into the background.) Next morning I overheard the following conversation in Fine Hall:

Prof. A: That was a nice party yesterday, wasn’t it?

Prof. B: Yes, and wasn’t it nice that they invited the whole department.

Prof. A: Well, only the full professors.

Prof. B: Of course.

The above-mentioned facts about Cantor sets led me to look at the opposite extreme, the so-called scattered spaces. A compact Hausdorff space Q is said to be shattered if Q contains no perfect set, every non-empty compact set F in Q thus contains a point that is not a limit point of F. The principal result proved in Proc. AMS 8, 1957, 39-42 is:

THEOREM: Every closed subalgebra of $C(Q)$ is self-adjoint.

In fact, the scattered spaces are the only ones for which this is true, but I did not state this in that paper.

In 1956, I found a very explicit description of all closed ideals in the disc algebra $A(U)$ (defined at the beginning of this chapter). The description involves inner function. These are the bounded holomorphic functions in U whose radial limits have absolute value 1 at almost every point of the unit circle $\mathcal{T}$. They play a very important role in the study of holomorphic functions in U (see, for instance, Garnett’s book, Bounded Analytic Functions) and their analogues will be mentioned again, on Riemann surfaces, in polydiscs, and in balls in $C^{n}$.

Recall that a point $\zeta$ on $\mathcal{T}$ is called a singular point of a holomorphic  function f in U if f has no analytic continuation to any neighbourhood of $\zeta$. The ideals in question are described in the following:

THEOREM: Let E be a compact subset of $\mathcal{T}$, of Lebesgue measure 0, let u be an inner function all of whose singular points lie in E, and let $J(E,u)$ be the set of all f in $A(U)$ such that

(i) the quotient f/u is bounded in U, and

(ii) $f(\zeta)=0$ at every $\zeta$ in E.

Then, $J(E,u)$ is a closed ideal of A(U), and every closed ideal of $A(U) (\neq \{0\})$ is obtained in this way.

One of several corollaries is that every closed ideal of A(U) is principal, that is, is generated by a single function.

I presented this at the December 1956 AMS meeting in Rochester, and was immediately told by several people that Beurling had proved the same thing, in a course he had given at Harvard, but had not published it. I was also told that Beurling might be quite upset at this, and having Beurling upset at you was not a good thing. Having used this famous paper about the shift operator on a Hilbert space as my guide, I was not surprised that he too had proved this, but I saw no reason to withdraw my already submitted paper. It appeared in Canadian J. Math. 9, 1967, 426-434. The result is now known as Beurling-Rudin theorem. I met him several times later, and he never made a fuss over this.

In the preceding year Lennart Carleson and I, neither of us knowing what the other was doing proved what is now known as Rudin-Carleson interpolation theorem. His paper is in Math. Z. 66, 1957, 447-451, mine in Proc. AMS 7, 1956, 808-811.

THEOREM. If E is a compact subset of $\mathcal{T}$, of Lebesgue measure 0, then every f in C(E) extends to a function F in A(U).

(It is easy to see that this fails if $m(E)>0$. To say that F is an extension of f means simply that $F(\zeta)=f(\zeta)$ at every $\zeta$ in E.)

Our proofs have some ingredients in common, but they are different, and we each proved more than is stated above. Surprisingly, Carleson, the master of classical hard analysis, used a soft approach, namely duality in Banach spaces, and concluded that F could be so chosen that $||F||_{U} \leq 2||f||_{E}$. (The norms are sup-norms over the sets appearing as subscripts.) In the same paper he used his Banach space argument to prove another interpolation theorem, involving Fourier-Stieltjes transforms.

On the other hand, I did not have functional analysis in mind at all, I did not think of the norms or of Banach spaces, I proved, by a bare-hands construction combined with the Riemann mapping theorem that if $\Omega$ is a closed Jordan domain containing $f(E)$ then f can be chosen so that $F(\overline{U})$ also lies in $\Omega$. If $\Omega$ is a disc, centered at 0, this gives $||F||_{U}=||f||_{E}$, so F is a norm-preserving extension.

What our proofs had in common is that we both used part of the construction that was used in the original proof of the F. and M. Riesz theorem (which says that if a measure $\mu$ on $\mathcal{T}$ gives $\int fd\mu=0$ for every f in $A(U)$ then $\mu$ is absolutely continuous with respect to Lebesgue measure). Carleson showed, in fact, that F. and M. Riesz can be derived quite easily from the interpolation theorem. I tried to prove the implication in the other direction. But that had to wait for Errett Bishop. In Proc. AMS 13, 1962, 140-143, he established this implication in a very general setting which had nothing to do with holomorphic functions or even with algebras, and which, combined with a refinement due to Glicksberg (Trans. AMS 105, 1962, 415-435) makes the interpolation theorem even more precise:

THEOREM: One can choose F in $A(U)$ so that $F(\zeta)=f(\zeta)$ at every $\zeta$ in E, and $|f(z)|<||f||_{E}$ at every z in $\overline{U}\\E$.

This is usually called peak-interpolation.

Several variable analogues of this and related results may be found in Chap. 6 of my Function Theory in Polydiscs and in Chap 10 of my Function Theory in the Unit Ball of $C^{n}$.

The last item in this chapter concerns Riemann surfaces. Some definitions are needed.

A finite Riemann surface is a connected open proper subset R of some compact Riemann surface X, such that the boundary $\partial R$ of R in X is also the boundary of its closure $\overline{R}$ and is the union of finitely many disjoint simple closed analytic curves $\Gamma_{1}, \ldots, \Gamma_{k}$. Shrinking each $\Gamma_{i}$ to a point gives a compact orientable manifold whose genus g is defined to be the genus of R. The numbers g and k determine the topology of R, but not, of course, its conformal structure.

$A(R)$ denotes the algebra of all continuous functions on $\overline{R}$ that are holomorphic in R. If f is in $A(R)$ and $|f(p)|=1$ at every point p in $\partial R$ then, just as in U, f is called inner. A set $S \subset A(R)$ is unramified if every point of $\overline{R}$ has a neighbourhood in which at least one member of S is one-to-one.

I became interested in these algebras when Lee Stout (Math. Z., 92, 1966, 366-379; also 95, 1967, 403-404) showed that every $A(R)$ contains an unramified triple of inner functions that separates points on $\overline{R}$.  He deduced from the resulting embedding of R in $U^{S}$ that $A(R)$ is generated by these 3 functions. Whether every $A(R)$ is generated by some pair of its member is still unknown, but the main result of my paper in Trans. AMS 150, 1969, 423-434 shows that pairs of inner functions won’t always do:

THEOREM: If $A(\overline{R})$ contains a point-separating unramified pair f, g of inner functions, then there exist relatively prime integers s and t such that f is s-to-1 and g is t-to-1 on every $\Gamma_{i}$, and

(*) $(ks-1)(kt-1)=2g+k-1$

For example, when $g=2$ and $k=4$, then (*) holds for no  integers s and t. When $g=23$ and $k=4$, then $s=t=2$ is the only pair that satisfies (*) but it is not relatively prime. Even when $R=U$ the theorem gives some information. In that case, $g=0, k=1$, so (*) becomes $(s-1)(t-1)=0$, which means:

If a pair of finite Blaschke products separates points on $\overline{U}$ and their derivatives have no  common zero in U, then at least one of them is one-to-one (that is, a Mobius transformation).

There are two cases in which (*) is not  only necessary but also sufficient. This happens when $g=0$ and when $g=k=1$.

But there are examples in which the topological condition (*) is satisfied even though the conformal structure of R prevents the existence of a separating unramified pair of inner functions.

This paper is quite different from anything else that I have ever done. As far as I know, no  one has ever referred to it, but I had fun working on it.

*************************************************************************************************

More blogs from Rudin’s autobiography later, till then,

Nalin Pithwa

# Interchanging Limit Processes — by Walter Rudin

As I mentioned earlier, my thesis (Trans. AMS 68, 1950, 278-363) deals with uniqueness questions for series of spherical harmonics, also known as Laplace series. In the more familiar setting of trigonometric series, the first theorem of the kind that I was looking for was proved by Georg Cantor in 1870, based on earlier work of Riemann (1854, published in 1867). Using the notations

$A_{n}(x)=a_{n} \cos{nx}+b_{n}\sin{nx}$,

$s_{p}(x)=A_{0}+A_{1}(x)+ \ldots + A_{p}(x)$, where $a_{n}$ and $b_{n}$ are real numbers. Cantor’s theorem says:

$If \lim_{p \rightarrow \infty}s_{p}(x)=0$ at every real x, then $a_{n}=b_{n}=0$ for every n.

Therefore, two distinct trigonometric series cannot converge to the same sum. This is what is meant by uniqueness.

My aim was to prove this for spherical harmonics and (as had been done for trigonometric series) to whittle away at the hypothesis. Instead of assuming convergence at every point of the sphere, what sort of summability will do? Does one really need convergence (or summability) at every point? If not, what sort of sets can be omitted? Must anything else be assumed at these omitted points? What sort of side conditions, if any, are relevant?

I came up with reasonable answers to these questions, but basically the whole point seemed to be the justification of the interchange of some limit processes. This left me with an uneasy feeling that there ought to be more to Analysis than that. I wanted to do something with more “structure”. I could not  have explained just what I meant by this, but I found it later when I became aware of the close relation between Fourier analysis and group theory, and also in an occasional encounter with number theory and with geometric aspects of several complex variables.

Why was it all an exercise in interchange of limits? Because the “obvious” proof of Cantor’s theorem goes like this: for $p > n$,

$\pi a_{n}= \int_{-\pi}^{\pi}s_{p}(x)\cos{nx}dx = \lim_{p \rightarrow \infty}\int_{-\pi}^{\pi}s_{p}(x)\cos {nx}dx$, which in turn, equals

$\int_{-\pi}^{\pi}(\lim_{p \rightarrow \infty}s_{p}(x))\cos{nx}dx= \int_{-\pi}^{\pi}0 \cos{nx}dx=0$ and similarly, for $b_{n}$. Note that $\lim \int = \int \lim$ was used.

In Riemann’s above mentioned paper, the derives the conclusion of Cantor’s theorem under an additional hypothesis, namely, $a_{n} \rightarrow 0$ and $b_{n} \rightarrow 0$ as $n \rightarrow \infty$. He associates to $\sum {A_{n}(x)}$ the twice integrated series

$F(x)=-\sum_{1}^{\infty}n^{-2}A_{n}(x)$

and then finds it necessary to prove, in some detail, that this series converges and that its sum F is continuous! (Weierstrass had not yet invented uniform convergence.) This is astonishingly different from most of his other publications, such as his paper  on hypergeometric functions in which mind-boggling relations and transformations are merely stated, with only a few hints, or  his  painfully brief paper on the zeta-function.

In Crelle’s J. 73, 1870, 130-138, Cantor showed that Riemann’s additional hypothesis was redundant, by proving that

(*) $\lim_{n \rightarrow \infty}A_{n}(x)=0$ for all x implies $\lim_{n \rightarrow \infty}a_{n}= \lim_{n \rightarrow \infty}b_{n}=0$.

He included the statement: This cannot be proved, as is commonly believed, by term-by-term integration.

Apparently, it took a while before this was generally understood. Ten years later, in Math. America 16, 1880, 113-114, he patiently explains the differenence between pointwise convergence and uniform convergence, in order to refute a “simpler proof” published by Appell. But then, referring to his second (still quite complicated) proof, the one in Math. Annalen 4, 1871, 139-143, he sticks his neck out and writes: ” In my opinion, no further simplification can be achieved, given the nature of ths subject.”

That was a bit reckless. 25 years later, Lebesgue’s dominated convergence theorem became part of every analyst’s tool chest, and since then (*) can be proved in a few lines:

Rewrite $A_{n}(x)$ in the form $A_{n}(x)=c_{n}\cos {(nx+\alpha_{n})}$, where $c_{n}^{2}=a_{n}^{2}+b_{n}^{2}$. Put

$\gamma_{n}==\min \{1, |c_{n}| \}$, $B_{n}(x)=\gamma_{n}\cos{(nx+\alpha_{n})}$.

Then, $B_{n}^{2}(x) \leq 1$, $B_{n}^{2}(x) \rightarrow 0$ at every x, so that the D. C.Th., combined with

$\int_{-\pi}^{\pi}B_{n}^{2}(x)dx=\pi \gamma_{n}^{2}$ shows that $\gamma_{n} \rightarrow 0$. Therefore, $|c_{n}|=\gamma_{n}$ for all large n, and $c_{n} \rightarrow 0$. Done.

The point of all this is that my attitude was probably wrong. Interchanging limit processes occupied some of the best mathematicians for most of the 19th century. Thomas Hawkins’ book “Lebesgue’s Theory” gives an excellent description of the difficulties that they had to overcome. Perhaps, we should not be too surprised that even a hundred years later many students are baffled by uniform convergence, uniform continuity etc., and that some never get it at all.

In Trans. AMS 70, 1961,  387-403, I applied the techniques of my thesis to another problem of this type, with Hermite functions in place of spherical harmonics.

(Note: The above article has been picked from Walter Rudin’t book, “The Way I  Remember It)) — hope it helps advanced graduates in Analysis.

More later,

Nalin Pithwa

# Analysis: Chapter 1: part 11: algebraic operations with real numbers: continued

(iii) Multiplication.

When we come to multiplication, it is most convenient to confine ourselves to positive numbers (among which we may include zero) in the first instance, and to go back for a moment to the sections of positive rational numbers only which we considered in articles 4-7. We may then follow practically the same road as in the case of addition, taking (c) to be (ab) and (O) to be (AB). The argument is the same, except when we are proving that all rational numbers with at most one exception must belong to (c) or (C). This depends, as in the case of addition, on showing that we can choose a, A, b, and B so that C-c is as small as we please. Here we use the identity

$C-c=AB-ab=(A-a)B+a(B-b)$.

Finally, we include negative numbers within the scope of our definition by agreeing that, if $\alpha$ and $\beta$ are positive, then

$(-\alpha)\beta=-\alpha\beta$, $\alpha(-\beta)=-\alpha\beta$, $(-\alpha)(-\beta)=\alpha\beta$.

(iv) Division.

In order to define division, we begin by defining the reciprocal $\frac{1}{\alpha}$ of a number $\alpha$ (other than zero). Confining ourselves in the first instance to positive numbers and sections of positive rational numbers, we define the reciprocal of a positive number $\alpha$ by means of the lower class $(1/A)$ and the upper class $(1/a)$. We then define the reciprocal of a negative number $-\alpha$ by the equation $1/(-\alpha)=-(1/\alpha)$. Finally, we define $\frac{\alpha}{\beta}$ by the equation

$\frac{\alpha}{\beta}=\alpha \times (1/\beta)$.

We are then in a position to apply to all real numbers, rational or  irrational the whole of the ideas and methods of elementary algebra. Naturally, we do not propose to carry out this task in detail. It will be more profitable and more interesting to turn our attention to some special, but particularly important, classes of irrational numbers.

More later,

Nalin Pithwa

# Analysis: Chapter 1: part 10: algebraic operations with real numbers

Algebraic operations with real numbers.

We now proceed to meaning of the elementary algebraic operations such as addition, as applied to real numbers in general.

(i),  Addition. In order to define the sum of two numbers $\alpha$ and $\beta$, we consider the following two classes: (i) the class (c) formed by all sums $c=a+b$, (ii) the class (C) formed by all sums $C=A+B$. Clearly, $c < C$ in all cases.

Again, there cannot be more than one rational number which does not belong either to (c) or to (C). For suppose there were two, say r and s, and let s be the greater. Then, both r and s must be greater than every c and less than every C; and so $C-c$ cannot be less than $s-r$. But,

$C-c=(A-a)+(B-b)$;

and we can choose a, b, A, B so that both $A-a$ and $B-b$ are as small as we like; and this plainly contradicts our hypothesis.

If every rational number belongs to (c) or to (C), the classes (c), (C) form a section of the rational numbers, that is to say, a number $\gamma$. If there is one which does not, we add it to (C). We have now a section or real number $\gamma$, which must clearly be rational, since it corresponds to the least member of (C). In any case we call $\gamma$ the sum of $\alpha$ and $\beta$and write

$\gamma=\alpha + \beta$.

If both $\alpha$ and $\beta$ are rational, they are the least members of the upper classes (A) and (B). In this case it is clear that $\alpha + \beta$ is the least member of (C), so that our definition agrees with our previous ideas of addition.

(ii) Subtraction.

We define $\alpha - \beta$ by the equation $\alpha-\beta=\alpha +(-\beta)$.

The idea of subtraction accordingly presents no fresh difficulties.

More later,

Nalin Pithwa