# Chapter 1: Topology, Hocking and Young: A theorem due G.D. Birkhoff

Prove that the collection of all topologies on a given set S constitutes a lattice under the partial ordering of finer/coarser topology of a set.

Proof: NB. This theorem has already been proven by eminent mathematician G. D. Birkhoff.

PS: I am making my own attempt here. Your comments are most welcome !

Part 1: TST: Coarser/finer is a partial ordering of topologies.

To prove axiom 1: this order is reflexive.

Let $\{ O_{\alpha}\}$ and $\{ R_{\beta} \}$ be two collections of subsets of a set S, both satisfying the three defining axioms $O_{1}$, $O_{2}$ and $O_{3}$ of a topology. Let S have two topologies. We will say that the topology $\mathcal{T}_{1}$ determined by $\{ O_{\alpha} \}$ is a finer topology than the topology $\mathcal{T}_{2}$ determined by $\{ R_{\beta} \}$ if every set $R_{\beta}$ is a union of sets $O_{\alpha}$, that is, each $R_{\beta}$ is open in the $\mathcal{T}_{1}$ topology. We will denote this situation with the symbol $\mathcal{T}_{1} \geq \mathcal{T}_{2}$.

Axiom 1: reflexive clearly holds because $\mathcal{T}_{1} \geq \mathcal{T}_{1}$ as set inclusion/containment is reflexive.

Axiom 2: antisymmetric also holds true because clearly $\mathcal{T}_{1} \geq \mathcal{T}_{2}$ and $\mathcal{T}_{2} \geq \mathcal{T}_{1}$ together imply that $\mathcal{T}_{2} = \mathcal{T}_{1}$. In other words, these two topologies are equivalent, or they give rise to same basis.

Axiom 3: Transitivity holds because set inclusion/containment is transitive. If $\mathcal{T}_{1} \geq \mathcal{T}_{2}$ and $\mathcal{T}_{2} \geq \mathcal{T}_{3}$, then clearly $\mathcal{T}_{1} \geq \mathcal{T}_{2} \geq \mathcal{T}_{3}$. In other words, there can be a chain of finer/coarser topologies for a given set.

Part 2:

TST: Under this definition of partial ordering of topologies, the partial ordering forms a lattice.

From axiom 3 proof, we know that there can exist a chain of finer/coarser topologies. So supremum and infimum can exist in a partial ordering of topologies. Hence, such a partial ordering forms a lattice.

QED.

Regards,

Nalin Pithwa

# Solutions to Chapter 1: Topology, Hocking and Young

Exercise 1-4: Prove that the collection of all open half planes is a subbasis for the Euclidean topology of the plane.

Proof 1-4:

Note: In the Euclidean plane, we can take as a basis the collection of all interiors of squares.

Note also that a subcollection $\mathcal{B}$ of all open sets of a topological space S is a SUBBASIS of S provided that the collection of all finite intersections of elements of $\mathcal{B}$ is a basis for S.

Clearly, from all open half planes ($x > x_{i}$ and $y > y_{i}$), we can create a collection of all interior of squares.

Hence, the collection of all finite intersections of all open-half planes satisfies:

Axiom $\mathcal{O_{2}}$: the intersection of a finite intersection of open half planes is an open set. (interior of a square).

Axiom $\mathcal{O_{1}}$; (also). The union of any number of finite intersections of all open half planes is also open set (interior of a square).

Axiom $O_{3}$: (clearly). $\phi$ and S are open.

QED.

Exercise 1-5:

Let S be any infinite set. Show that requiring every infinite subset of S to be open imposes the discrete topology on S.

Proof 1-5:

Case: S is countable. We neglect the subcase that a selected subcase is finite. (I am using Prof. Rudin’s definition of countable). The other subcase is that there exists a subset $X_{1} \subset S$, where $X_{1}$ is countable. Let $X_{1}$ be open. Hence, $S-X_{1}$ is closed. But $S-X_{1}$ is also countable. Hence, $S-X_{1}$ is also open. Hence, there is no limit point. Hence, the topology is discrete, that is, there is no limit point.

Case: S is uncountable. Consider again a proper subset $X_{1} \subset S$; hence, $X_{1}$ is open by imposition of hypothesis. Hence, $S-X_{1}$ is closed. But, $S-X_{1} \neq \phi$ and not finite also. Hence, $S-X_{1}$ is infinite. Hence, $S-X_{1}$ is open. Hence, there are no limit points. Hence, it is a discrete topology in this case also.

QED.

Regards,

Nalin Pithwa.

# Tutorial Problems I: Topology: Hocking and Young

Reference: Topoology by Hocking and Young, Dover Publications, Inc., NY. Available in Amazon India.

Exercises 1-1:

Show that if S is a set with the discrete topology and $f: S \rightarrow T$ is any transformation of S into a topologized set T, then f is continuous.

Solution 1-1:

Definition A: The set S has a topology (or is topologized) provided that, for every point p in S and every subset X of S, the question : “is p a limit point of X?” can be answered.

Definition B: A topology is said to be a discrete topology when we assume that for no point p in S, and every subset X of S: the answer to the question: “is p a limit point of X?” is NO.

Definition C: A transformation $f: S \rightarrow T$ is continuous provided that if p is a limit point of a subset X of S, then f(p) is a limit point or a point of f(X).

So, the claim is vacuously true. QED.

Exercises 1-2:

A real-valued function $y=f(x)$ defined on an interval [a,b] is continuous provided that if $a \leq x_{0} \leq b$ and $\epsilon >0$, then there is a number $\delta>0$ such that if $|x-x_{0}|<\delta$, where $x \in [a,b]$, then $|f(x)-f(x_{0})|<\epsilon$. Show that this is equivalent to our definition, using definition 1-1.

Solution 1-2:

Definition 1-1: The real number p is a limit point of a set X of real numbers provided that for every positive number $\epsilon$, there is an element x of the set X such that $0<|p-x|<\epsilon$.

Definition C: A transformation $f: S \rightarrow T$ is continuous provided that if p is a limit point of a subset X of S, then f(p) is a limit point or a point of f(X).

Part 1: Let us assume that given function f is continuous as per definition given just above.

Then, as p is a limit point of X: it means: For any $\delta>0$, there exists a real number p such that there is an element $x \in X$ such that $|p-x|<\delta$..

So, also, by definition C, f(p) is a limit point or a point of f(X); this means the following: if f(p) is a point of f(X), there exists some $x_{0} \in X$ such that $f(x_{0} \in f(X)$, and so quite clearly in this case $p=x_{0}$ so that $|p-x_{0}|=|x_{0}-x_{0}|<\delta$, as $\delta$ is positive.

On the other hand, if f(p) is a limit point of f(X), as per the above definition of continuity, then also for any $\epsilon>0$, there exists a point $y \in f(X)$ such that $|y-f(p)|<\epsilon$. So, in this case also the claim is true.

We have proved Part 1. QED.

Now, part II: We assume the definition of continuity given in the problem statement is true. From here, we got to prove definition C as the basic definition given by the authors.

But this is quite obvious as in this case $p=x_{0}$.

We have proved Part II. QED.

Thus, the two definitions are equivalent.

Cheers,

Nalin Pithwa

# Metric space question and solution

Reference: I had blogged this example earlier, but I myself could not fill in the missing gaps at that time. I am trying again with the help of MathWorld Wolfram and of course, the classic, Introductory Real Analysis by Kolmogorov and Fomin, from which it is picked up for my study.

Consider the set $C_{[a,b]}$ of all continuous functions defined on the closed interval $[a,b]$. Let the distance function (or metric) be defined by the formula:

$\rho(x,y) = (\int_{a}^{b}[x(t)-y(t)]^{2}dt)^{1/2}$ ——– relation I

The resulting metric space will be denoted by $C_{[a,b]}^{2}$.

The first two properties of a metric space are clearly satisfied by the above function. We need to check for the triangle inequality:

Now I satisfies the triangle inequality because of the following Schwarz’s inequality:

$(\int_{a}^{b}x(t)y(t)dt)^{2} \leq \int_{a}^{b}x^{2}(t)dt \int_{a}^{b}y(t)dt$ —— relation II

In order to get to the above relation II, we need to prove the following:

Prove: $(\int_{a}^{b} x(t)y(t)dt)^{2} = \int_{a}^{b}x^{2}(t)dt \int_{a}^{b} y^{2}(t)dt - \frac{1}{2}\int_{a}^{b} \int_{a}^{b}[x(s)y(t)-x(t)y(s)]^{2}dsdt$.

From the above, we can deduce Schwarz’s inequality (relation II here in this blog article).

(My own attempts failed to crack it so I had to look at the internet for help. Fortunately, MathWorld Wolfram has given a crisp clear proof…but some parts of the proof are still out of my reach…nevertheless, I am reproducing the proof here for the sake of completeness of my notes…for whatever understanding I can derive at this stage from the proof…):

CITE THIS AS:

Weisstein, Eric W. “Schwarz’s Inequality.” From MathWorld–A Wolfram Web Resource. https://mathworld.wolfram.com/SchwarzsInequality.html

Schwarz’s Inequality:

Let $\Psi_{1}(x), \Psi_{2}(x)$ be any two real integrable functions in $[a,b]$, then Schwarz’s inequality is given by :

$|< \Psi_{1}, \Psi_{2}>|^{2} \leq < \Psi_{1}|\Psi_{2}> <\Psi_{2}|Psi_{1}>$

Written out explicity,

$|\int_{a}^{b} \Psi_{1}(x), \Psi_{2}(x)|^{2} \leq \int_{a}^{b}[\Psi_{1}(x)]^{2}dx \int_{a}^{b}[\Psi_{2}(x)]^{2}dx$

with equality if and only if $\Psi_{1}(x) = \alpha \Psi_{2}(x)$ with $\alpha$ a constant. Schwarz’s inequality is sometimes also called the Cauchy-Schwarz inequality or Buniakowsky’s inequality.

To derive the inequality, let $\Psi(x)$ be a complex function and $\lambda$ a complex constant such that

$\Psi(x) \equiv f(x) + \lambda g(x)$ for some f and g

Since

$\int \overline{\Psi} \Psi dx \geq 0$, where $\overline{z}$ is the complex conjugate.

$\int \overline{\Psi}\Psi dx = \int \overline{f}f dx + \lambda \int \overline{f} g dx + \overline\lambda \int \overline{g} f dx + \lambda \overline{\lambda} \int \overline{g} g dx$

with equality when $\Psi(x) = 0$

Writing this, in compact notation:

$<\overline{f},f> + \lambda <\overline{f},g> + \overline{\lambda} <\overline{g},f> + \lambda \overline{\lambda} <\overline{g},g> \geq 0$….relation A

Now, define $\lambda \equiv - \frac{<\overline{g}, f>}{<\overline{g},g>}$….relation B

and $\overline{\lambda} = - \frac{}{<\overline{g}, g>}$…relation C

Multiply A by $<\overline{g},g>$ and then plug in B and C to obtain:

$<\overline{f}, f><\overline{g}, g> - <\overline{f},g><\overline{g},f> - <\overline{g},f> +<\overline{g}, f> \geq 0$

which simplifies to

$<\overline{g},f><\overline{f},g> \leq <\overline{f},f><\overline{g},g>$

So

$||^{2} \leq $. Bessel’s inequality follows from Schwarz’s inequality. QED.

Regards,

Nalin Pithwa

# VI. Countable sets: My notes.

Reference:

1. Introduction to Topology and Modern Analysis by G. F. Simmons, Tata McGraw Hill Publications, India.
2. Introductory Real Analysis — Kolmogorov and Fomin, Dover Publications, N.Y.(to some extent, I have blogged this topic based on this text earlier. Still, it deserves a mention/revision again).
3. Topology : James Munkres.

The subject of this section and the next — infinite cardinal numbers — lies at the very foundation of modern mathematics. It is a vital instrument in the day to day work of many mathematicians, and we shall make extensive use of it ourselves (in our beginning studying of math ! :-)). This theory, which was created by the German mathematician Georg Cantor, also has great aethetic appeal, for it begins with ideas of extreme simplicity and develops through natural stages into an elaborate and beautiful structure of thought. In the course of our discussion, we shall answer questions which no one before Cantor’s time thought to ask, and we shall ask a question which no one can answer to this day…(perhaps !:-))

Without further ado, we can say that cardinal numbers are those used in counting, such as the positive integers (or natural numbers) 1, 2, 3, …familiar to us all. But, there is much more to the story than this.

The act of counting is undoubtedly one of the oldest of human activities. Men probably learned to count in a crude way at about the same time as they began to develop articulate speech. The earliest men who lived in communities and domesticated animals must have found it necessary to record the number of goats in the village herd by means of a pile of stones or some similar device. If the herd was counted in each night by removing one stone from the pile for each goat accounted for, then stones left over would have indicated strays, and herdsmen would have gone out to search for them. Names for numbers and symbols for them like our 1, 2, 3, …would have been superfluous. The simple yet profound idea of a one-to-one correspondence between the stones and the goats would have fully met the needs of the situation.

In a manner of speaking, we ourselves use the infinite set

$N = \{ 1, 2, 3, \ldots\}$

of all positive integers as “pile of stones.” We carry this set around with us as part of our intellectual equipment. Whenever we want to count a set, say, a stack of dollar bills, we start through the set N and tally off one bill against each positive integer as we come to it. The last number we reach, corresponding to the last bill, is what we call the number of bills in the stack. If this last number happens to be 10, then “10” is our symbol for the number of bills in the stack, as it also is for the number of our fingers, and for the number of our toes, and for the number of elements which can be put into one-to-one correspondence with the finite set $\{ 1,2,3, \ldots, 10\}$. Our procedure is slightly more sophisticated than that of the primitive savage man. We have the symbols 1, 2, 3, …for the numbers which arise in counting; we can record them for future use, and communicate them to other people, and manipulate them by the operations of arithmetic. But the underlying idea, that of the one-to-one correspondence, remains the same for us as it probably was for him.

The positive integers are adequate for our purpose of counting any non-empty finite set, and since outside of mathematics all sets appear to of this kind, they suffice for all non-mathematical counting. But in the world of mathematics, we are obliged to consider many infinite sets, such as the set of positive integers itself, the set of all integers, the set of all rational numbers, the set of all real numbers, the set of all points in a plane, and so on. It is often important to be able to count such sets, and it was Cantor’s idea to do this, and to develop a theory of infinite cardinal numbers, by means of one-to-one correspondence.

In comparing the sizes of two sets, the basic concept is that of numerical equivalence as defined in the previous section. We recall that two non-empty sets are numerically equivalent if there exists a one-to-one mapping of a set onto the other, or — and this amounts to the same thing — if there can be found a one-to-one correspondence between them. To say that two non-empty finite sets are numerically equivalent is of course to say that they have the same number of elements in the ordinary sense. If we count one of them, we simply establish a one-to-one correspondence between its elements and a set of positive integers of the form $\{1,2,3, \ldots, n \}$ and we then say that n is the number of elements possessed by both, or the cardinal number of both. The positive integers are the finite cardinal numbers. We encounter many surprises as we follow Cantor and consider numerical equivalences for infinite sets.

The set $N = \{ 1,2,3, \ldots\}$ of all positive integers is obviously “larger” than the set $\{ 2,4,6, \ldots\}$ of all even positive integers, for it contains this set as a proper subset. It appears on the surface that N has “more” elements. But it is very important to avoid jumping to conclusions when dealing with infinite sets, and we must remember that our criterion in these matters is whether there exists a one-to-one correspondence between the sets (not whether one set is a proper subset or not of the other) . As a matter of fact, consider the “pairing”

$1,2,3, \ldots, n, \ldots$

$2,4,6, \ldots, 2n, \ldots$

serves to establish a one-to-one correspondence between these sets, in which each positive integer in the upper row is matched with the even positive integer (its double) directly below it, and these two sets must therefore be regarded as having the same number of elements. This is a very remarkable circumstance, for it seems to contradict our intuition and yet is based only on solid common sense. We shall see below, in Problems 6 and 7-4, that every infinite set is numerically equivalent to a proper subset of itself. Since this property is clearly not possessed by any finite set, some writers even use it as the definition of an infinite set.

In much the same way as above, we can show that N is numerically equivalent to the set of all even integers:

$1, 2, 3,4, 5,6, 7, \ldots$

$0,2,-2,4,-4,4,6,-6, \ldots$

Here, our device is start with zero and follow each even positive integer as we come to it by its negative. Similarly, N is numerically equivalent to the set of all integers:

$1,2,3,4,5,6,7, \ldots$

$0,1,-1, 2, -2, 3, -3, \ldots$

It is of considerable interest historical interest to note that Galileo had observed in the early seventeenth century that there are precisely as many perfect squares (1,4,9,16,25, etc.) among the positive integers as there are positive integers altogether. This is clear from the “pairing”:

$1,2,3,4,5, \ldots$

$1^{2}, 2^{2}, 3^{2}, 4^{2}, 5^{2}, \ldots$

It struck him as very strange that this should be true, considering how sparsely strewn the squares are among all the positive integers. But, the time appears not to have been ripe for the exploration of this phenomenon, or perhaps he had other things on his mind; in any case, he did not follow up his idea.

These examples should make it clear that all that is really necessary in showing that an infinite set X is numerically equivalent to N is that we be able to list the elements of X, with a first, a second, a third, and so on, in such a way that it is completed exhausted by this counting off of its elements. It is for this reason that any infinite set which is numerically equivalent to N is said to be countably infinite. (Prof. Walter Rudin also, in his classic on mathematical analysis, considers a countable set to be either finite or countably infinite. ) We say that a set is countable it is non-empty and finite (in which case it can obviously be counted) or if it is countably infinite.

One of Cantor’s earliest discoveries in his study of infinite sets was that the set of all positive rational numbers (which is very large : it contains N and a great many other numbers besides) is actually countable. We cannot list the positive rational numbers in order of size, as we can the positive integers, beginning with the smallest, then the next smallest, and so on, for there is no smallest, and between any two there are infinitely many others. We must find some other way of counting them, and following Cantor, we arrange them not not in order of size, but according to the size of the sum of numerator and denominator. We begin with all positive rationals whose numerator and denominator sum add up to 2: there is only one $\frac{1}{1}=1$. Next, we list (with increasing numerators) all those for which this sum is 3: $\frac{1}{2}, \frac{2}{1}=2$. Next, all those for which the sum is 4: $\frac{1}{3}, \frac{2}{2}=1, \frac{3}{1}=3$. Next, all those for which this sum is 5: $\frac{1}{4}, \frac{2}{3}, \frac{3}{2}, \frac{4}{1}=4$. Next, all those for which this sum is 6: $\frac{1}{5}, \frac{2}{4}, \frac{3}{3}=1, \frac{4}{2}=2, \frac{5}{1}=5$. And, so on. If we now list all these together from the beginning, omitting those already listed when we come to them, we get a sequence like:

$1, 1/2, 2, 1/3, 1/4, 2/3, 3/2, 4, 1/5, 5, \ldots$

which contains each positive rational number once and only once. (There is a nice schematic representation of this : Cantor’s diagonalization process; please google it). In this figure, the first row contains all positive rationals with numerator 1, the second all with numerator 2, and so on. Our listing amounts to traversing this array of numbers in a zig-zag manner (again, please google), where of course, all those numbers already encountered are left out.

It is high time that we christened the infinite cardinal number we have been discussing, and for this purpose, we use the first letter of the Hebrew alphabet, $\bf{aleph_{0}}$. We say that $\aleph_{0}$ is the number of elements in any countably infinite set. Our complete list of cardinal numbers so far is

$1, 2, 3, \ldots, \aleph_{0}$.

We expand this list in the next section.

Suppose now that m and n are two cardinal numbers (finite or infinite). The statement that m is less than n (written m < n) is defined to mean the following: if X and Y are sets with m and n elements, then (i) there exists a one-to-one mapping of X into Y, and (ii) there does not exist a mapping of X onto Y. Using this concept, it is easy to relate our cardinal numbers to one another by means of:

$1<2<3< \ldots < \aleph_{0}$.

With respect to the finite cardinal numbers, this ordering corresponds to their usual ordering as real numbers.

Regards,

Nalin Pithwa

# V. Exercises: Partitions and Equivalence Relations

Reference: Topology and Modern Analysis, G. F. Simmons, Tata McGraw Hill Publications, India.

1. Let $f: X \rightarrow Y$ be an arbitrary mapping. Define a relation in X as follows: $x_{1} \sim x_{2}$ means that $f(x_{1})=f(x_{2})$. Prove that this is an equivalence relation and describe the equivalent sets.

Proof : HW. It is easy. Try it. 🙂

2. In the set $\Re$ of real numbers, let $x \sim y$ means that $x-y$ is an integer. Prove that this is an equivalence relation and describe the equivalence sets.

Proof: HW. It is easy. Try it. 🙂

3. Let I be the set of all integers, and let m be a fixed positive integer. Two integers a and b are said to be congruent modulo m — symbolized by $a \equiv b \pmod {m}$ — if a-b is exactly divisible by m, that is, if $a-b$ is an integral multiple of m. Show that this is an equivalence relation, describe the equivalence sets, and state the number of distinct equivalence sets.

Proof: HW. It is easy. Try it. 🙂

4. Decide which one of the three properties of reflexivity, symmetry and transitivity are true for each of the following relations in the set of all positive integers: $m \leq n$, $m < n$, $m|n$. Are any of these equivalence relations?

Proof. HW. It is easy. Try it. 🙂

5. Give an example of a relation which is (a) reflexive, but not symmetric or transitive. (b) symmetric but not reflexive or transitive. (c) transitive but not reflexive or symmetric (d) reflexive and symmetric but not transitive (e) reflexive and transitive but not symmetric. (f) symmetric and transitive but not reflexive.

Solutions. (i) You can try to Google (ii) Consider complex numbers (iii) there are many examples given in the classic text “Discrete Mathematics” by Rosen.

6) Let X be a non-empty set and $\sim$ a relation in X. The following purports to be a proof of the statement that if this relation is symmetric and transitive, then it is necessarily reflexive: $x \sim y \Longrightarrow y \sim x$ ; $x \sim y$ and $y \sim x \Longrightarrow x$; therefore, $x \sim x$ for every x. In view of problem 5f above, this cannot be a valid proof. What is the flaw in the reasoning? 🙂

7) Let X be a non-empty set. A relation $\sim$ in X is called circular if $x \sim y$ and $y \sim x \Longrightarrow z \sim x$, and triangular if $x \sim y and x \sim z \Longrightarrow y \sim z$. Prove that a relation in X is an equivalence relation if and only if it is reflexive and circular if and only if it is reflexive and triangular.

HW: Try it please. Let me know if you need any help.

Regards,

Nalin Pithwa.

PS: There are more examples on this topic of relations in Abstract Algebra of I. N. Herstein and Discrete Mathematics by Rosen.

# V. Partitions and Equivalence Relations: My Notes

References:

1. Topology and Modern Analysis, G F Simmons, Tata McGraw Hill Publications, India.
2. Toplcs in Algebra, I N Herstein.
3. Abstract Algebra, Dummit and Foote.
4. Topology by James Munkres.

In the first part of this section, we consider a non-empty set X, and we study decompositions of X into non-empty subsets which fill it out and have no elements in common with one another. We give special attention to the tools (equivalence relation) which are normally used to generate such decompositions.

A partition of X is a disjoint class $\{ X_{i} \}$ of non-empty subsets of X whose union if the full set X itself. The $X_{i}$‘s are called the partition sets. Expressed somewhat differently, a partition of X is the result of splitting it, or subdividing it, into non-empty subsets in such a way that each element of X belongs to one and only one of the given subsets. ]

If X is the set $\{1,2,3,4,5 \}$, then $\{1,3,5 \}$, $\{2,4 \}$ and $\{1,2,3 \}$ and $\{ 4,5\}$ are two different partitions of X. If X is the set $\Re$ of all real numbers, then we can partition $\Re$ into the set of all rationals and the set of all irrationals, or into the infinitely many closed open intervals of the form $[n, n+1)$ where n is an integer. If X is the set of all points in the coordinate plane, then we can partition X in such a way that each partition set consists of all points with the same x coordinate (vertical lines), or so that each partition set consists of all points with the same y coordinate (horizontal lines).

Other partitions of each of these sets will readily occur to the reader. In general, there are many different ways in which any given set can be partitioned. These manufactored examples are admittedly rather uninspiring and serve only to make our ideas more concrete. Later in this section we consider some others which are more germane to our present purposes.

A binary relation in the set X is a mathematical symbol or verbal phrase, which we denote by R in this paragraph, such that for each ordered pair $(x,y)$ of elements of X the statement $x \hspace{0.1in} R \hspace{0.1in} y$ is meaningful, in the sense that it can be classified definitely as true or false. For such a binary relation, $x \hspace{0.1in} R \hspace{0.1in}y$ symbolizes the assertion that x is related by R to y, and $x \not {R} \hspace{0.1in}y$ the negation of this, namely, the assertion that x is not related by R to y. Many examples of binary relations can be given, some familiar and others less so, some mathematical and others not. For instance, if X is the set of all integers and R is interpreted to mean “is less than,” which of course is usually denoted by the symbol <, then we clearly have 6<7 and $5 \not < 2$. We have been speaking of binary relations, which are so named because they apply only to ordered pairs of elements, rather than to ordered triples, etc. In our work, we drop the qualifying adjective and speak simply of a relation in X, since we shall have occasion to consider only relations of this kind. {NB: Some writers prefer to regard a relation R in X as a subset R of $X \times X$. From this point of view, x R y and $x \not {R} y$ are simply equivalent ways of writing $(x,y) \in R$ and $(x,y) \notin R$. This definition has the advantage of being more tangible than our definition, and the disadvantage that few people really think of a relation in this way.” )

We now assume that a partition of our non-empty set X is given, and we associate with this partition a relation on X. This relation is defined to be in the following way: we say that x is equivalent to y and write this as $x \sim y$ (the symbol $\sim$ is pronounced “wiggle”.), if x and y belong to the same partition set. It is obvious that the relation $\sim$ has the following properties:

a) $x \sim x$ for every x (reflexivity)

b) $x \sim y \Longrightarrow y \sim x$ (symmetry)

c) $x \sim y \hspace{0.1in} and \hspace{0.1in} y \sim z \Longrightarrow x \sim z$ (transitivity)

This particular relation in X arose in a special way, in connection with a given partition of X, and its properties are immediate consequences of the definition. Any relation whatever in X which possesses these three properties is called an equivalence relation in X.

We have just seen that each partition of X has associated with it a natural equivalence relation in X. We now reverse the situation and prove that a given equivalence relation in X determines a natural partition of X.

Let $\sim$ be an equivalence relation in X; that is, assume that it is reflexive, symmetric, and transitive in the sense described above. If x is an element of X, the subset of X defined by $[x] = \{ y: y \sim x\}$ is called the equivalence set of x. The equivalence set of x is thus the set of all elements which are equivalent to x. We show that the class of all distinct equivalence sets forms a partition of X. By reflexivity, $x \in [x]$ for each element x in X, so each equivalence set is non-empty and their union is X. It remains to be shown that any two equivalence sets $[x_{1}]$ and $[x_{2}]$ are either disjoint or identical. We prove this by showing that if $[x_{1}]$ and $[x_{2}]$ are not disjoint, then they must be identical. Suppose that $[x_{1}]$ and $[x_{2}]$ are not disjoint, that is, suppose that they have a common element z. Since x belongs to both equivalence sets, $z \sim x_{1}$ and $z \sim x_{2}$, and by symmetry $x_{1} \sim z$. Let y be any element of $x_{1}$, so that $y \sim x_{1}$. Since $y \sim x_{1}$ and $x_{1} \sim z$, transitivity shows that $y \sim z$. By another application of transitivity, $y \sim z$ and $z \sim x_{2}$, imply that $y \sim x_{2}$ so that y is in $[x_{2}]$. Since y was arbitrarily chosen in $[x_{1}]$, we see by this that $[x_{1}] \subseteq [x_{2}]$. The same reasoning shows that $[x_{2}] \subseteq [x_{1}]$ and from this we conclude that $[x_{1}] = [x_{2}]$.

The above discussion demonstrates that there is no real distinction (other than a difference in language) between partitions of a set and equivalence relation by regarding elements as equivalent if they belong to the same partition set, and if we start with an equivalence relation, we get a partition by grouping together into subsets all elements which are equivalent to one another. We have here a single mathematical idea, which we have been considering from two different points of view, and the approach we choose in any particular application depends entirely on our own convenience. In practice, it is almost invariably the case that we use equivalence relations (which are usually easy to define) to obtain partitions (which are sometimes difficult to describe fully).

We now turn to several of the more important simple examples of equivalence relations.

Let I be the set of integers. If a and b are elements of this set, we write $a = b$ (and say that a equals b) if a and b are the same integer. Thus, $2+3=5$ means that the expression on the left and right are simply different ways of writing the same integer. It is apparent that = used in this sense is an equivalence relation in the set I:

i) a=a for every a

ii) $a=b \Longrightarrow b=a$

iii) $a=b \hspace{0.1in} b=c \Longrightarrow a=c$.

Clearly, each equivalence set consists of precisely one integer.

Another familiar example is this relation of equality commonly used for fractions. We remind the reader that, strictly speaking, a fraction is merely a symbol of the form a/b, where a and b are integers and b is not zero. The fractions 2/3 and 4/6 are obviously not identical, but nevertheless we consider them to be equal. In general, we say that two fractions a/b and c/d are equal, written $\frac{a}{b} = \frac{c}{d}$, if ad and bc are equal as integers in the usual sense (see the paragraph above). (HW quiz: show this is an equivalence relation on the set of fractions). An equivalence set of fractions is what we call a rational number. Every day usage ignores the distinction between fractions and rational numbers, but it is important to recognize that from the strict point of view it is the rational numbers (and not the fractions) which form part of the real number system.

Our final example has a deeper significance, for it provides us with the basic tool for our work of the next two sections.

For the remainder of all this section, we consider a relation between pairs of non-empty sets, and each set mentioned (whether we say so explicitly or not) is assumed to be non-empty. If X and Y are two sets, we say that X is numerically equivalent to Y if there exists a one-to-one correspondence between X and Y, that is, if there exists a one-to-one mapping of X onto Y. This relation is reflexive, since the identity mapping $i_{X}: X \rightarrow X$ is one-to-one onto; it is symmetric since if $f: X \rightarrow Y$ is one-to-one onto, then its inverse mapping $f^{-1}: Y \rightarrow X$ is also one-to-one onto; and it is transitive, since if $f: X \rightarrow Y$ and $g: Y \rightarrow Z$ are one-to-one onto, then $gf: X \rightarrow Z$ is also one-to-one onto. Numerical equivalence has all the properties of an equivalence relation, and if we consider it as an equivalence relation in the class of all non-empty subsets of some universal set U, it groups together into equivalence sets all those subsets of U which have the “same number of elements.” After we state and prove the following very useful but rather technical theorem, we shall continue in Sections 6 and 7 with an exploration of the implications of these ideas.

The theorem we have in mind — the Schroeder-Bernstein theorem: If X and Y are two sets each of which is numerically equivalent to a subset of the other, then all of X is numerically equivalent to all of Y. There are several proofs of this classic theorem, some of which are quite difficult. The very elegant proof we give is essentially due to Birkhoff and MacLane.

Proof:

Assume that $f: X \rightarrow Y$ is a one-to-one mapping of X into Y, and that $g: Y \rightarrow X$ is a one-to-one mapping of Y into X. We want to produce a mapping $F: X \rightarrow Y$ which is one-to-one onto. We may assume that neither f nor g is onto, since if f is, we can define F to f, and if g is, we can define F to be $g^{-1}$. Since both f and g are one-to-one, it is permissible to use the mappings $f^{-1}$ and $g^{-1}$ as long as we keep in mind that $f^{-1}$ is defined only on f(X) and $g^{-1}$ is defined only on g(Y). We obtain the mapping F by splitting both X and Y into subsets which we characterize in terms of the ancestry of their elements. Let x be an element of X. We apply $g^{-1}$ (if we can) to get the element $g^{-1}(x)$ in Y. If $g^{-1}(x)$ exists, we call it the first ancestor of x. The element x itself we call the zeroth ancestor of x. We now apply $f^{-1}$ to $g^{-1}(x)$ if we can, and if $(f^{-1}g^{-1})(x)$ exists, we call it the second ancestor of x. We now apply $g^{-1}$ to $(f^{-1}g^{-1})(x)$ if we can, and if $(g^{-1}f^{-1}g^{-1})(x)$ exists, we call it the third ancestor of x. As we continue this process of tracing back the ancestry of x, it becomes apparent that there are three possibilities — (1) x has infinitely many ancestors. We denote by $X_{i}$, the subset of X, which consists of all elements with infinitely many ancestors (2) x has an even number of ancestors, this means that x has a last ancestor (that is, one which itself has no first ancestor) in X. We denote by $X_{e}$ the subset of X consisting of all elements with an even number of ancestors. (3) x has an odd number of ancestors; this means that x has a last ancestor in Y. We denote by $X_{o}$ the subset of X which consists of all elements with an odd number of ancestors. The three sets $X_{i}$, $X_{e}$, $X_{o}$ form a disjoint class whose union is X. We decompose Y in just the same way into three subsets $Y_{i}$, $Y_{e}$, $Y_{o}$. It is easy to see that f maps $X_{i}$ onto $Y_{i}$, and $X_{e}$ onto $Y_{e}$, and that $g^{-1}$ maps $X_{o}$ onto $Y_{o}$, and we complete the proof by defining F in the following piecemeal manner:

$F(x) = f(x)$ if $x \in X_{i} \bigcup X_{e}$

and $F(x) = g^{-1}(x)$ if $x \in X_{o}$.

QED.

The Schroeder Bernstein theorem has great theoretical and practical significance. It main value for us lies in its role as a tool by means of which we can prove numerical equivalence with a minimum of effort for many specific sets. We put it to work in Section 7.

Regards,

Nalin Pithwa

# IV. Product of Sets: Exercises

Reference: Topology and Modern Analysis, G F Simmons, Tata McGraw Hill Publications, India.

Problems:

I) The graph of a mapping $f: X \rightarrow Y$ is a subset of the product $X \times Y$. What properties characterize the graphs of mappings among all subsets of $X \times Y$?

Solution I: composition.

II) Let X and Y be non-empty sets. If $A_{1}$ and $A_{2}$ are subsets of X, and $Y_{1}$ and $Y_{2}$ are subsets of Y, then prove the following

(a) $(A_{1} \times B_{1}) \bigcap (A_{2} \times B_{2}) = (A_{1} \bigcap A_{2}) \times (B_{1} \bigcap B_{2})$

(b) $(A_{1} \times B_{1}) - (A_{2} \times B_{2}) = (A_{1}-A_{2}) \times (B_{1}-B_{2}) \bigcup (A_{1} \bigcap A_{2}) \times (B_{1}-B_{2}) \bigcup (A_{1}-A_{2}) \times (B_{1} \bigcap B_{2})$

Solution IIa:

TPT: $(A_{1} \times B_{1}) \times (A_{2} \times B_{2}) = (A_{1} \bigcap A_{2}) \times (B_{1} \bigcap B_{2})$

This is by definition of product and the fact that the co-ordinates are ordered and the fact that $A_{1} \subseteq X$, $A_{2} \subseteq X$, $B_{1} \subseteq Y$, and $B_{2} \subseteq Y$.

Solution IIb:

Let $(a_{i}, b_{j}) \in A_{1} \times B_{1}$, but $(a_{i}, b_{j}) \notin A_{2} \times B_{2}$. So, the element may belong to $(A_{1}-A_{2}) \times (B_{1} - B_{2})$ or it could happen that it belongs to $A_{1} \times A_{1}$, but to $(B_{1}-B_{2})$ (here we need to remember that the element is ordered); so, also it could be the other way: it could belong to $(A_{1}-A_{2})$ but to $(B_{1} \bigcap B_{2})$ also. The same arguments applied in reverse establish the other subset inequality. Hence, done.

III) Let X and Y be non-empty sets, and let A and B be rings of subsets of X and Y, respectively. Show that the class of all finite unions of sets of the form $A \times B$ with $A \in \bf{A}$ and $B \in \bf{B}$ is a ring of subsets of $X \times Y$.

Solution III:

$\bigcup_{i=1}^{n}A_{i} \times B_{i} = \bigcup_{i=1}^{n}A_{i} \times \bigcup_{i=1}^{n}B_{i}$.

From question IIb above, the difference of any two pairs of sets is also in $X \times Y$.

Hence, done.

Regards,

Nalin Pithwa