Basic algebra facts of HCF and LCM

1. If x and y are two natural numbers, HCF(x,y) x LCM (x,y) = xy
2. If a, b, c are three natural numbers, then the product abc=HCF(a,b,c) x LCM(ab,bc,ca) and also abc=HCF(ab, bc,ca) x LCM(a,b,c).

Bill Casselman’s Euclid: thanks to ClayMath

The purpose is only to share and spread the awareness of availability of this second master piece on Euclid. Thanks to Clay Math Organization for serving students world wide, and thanks to the generous Mr and Mrs Clayton. I hope my math olympiad students will enjoy this and enrich themselves mathematically.

http://www.math.ubc.ca/~cass/euclid/

Notes I: Sets and Functions: Solutions

Problem 1:

If $\{ A_{i}\}$ and $\{ B_{j}\}$ are two classes of sets such that $\{ A_{i}\} \subseteq \{ B_{j}\}$, prove that $\bigcup_{i}A_{i} \subseteq \bigcup_{i}B_{i}$ and $\bigcap_{j}B_{j} \subseteq \bigcap_{i}A_{i}$.

The first part is clear as the hypothesis says the set of sets $\{ A_{i}\}$ is a subset of the set of sets $\{ B_{j}\}$. So, quite clearly $\bigcup_{i}A_{i} \subseteq \bigcup_{j}A_{j}$. For the second part, we can use the generalized De Morgan’s laws. (Note: if you try to derive it straight, you will prove exactly what is opposite of the required result :-)). For the second part note that $X \subseteq Y \Longleftrightarrow Y^{'} \bigcap X^{'}$. Using the result in the first part, that is, consider $\bigcup_{i}A_{i} \subseteq \bigcup_{j}A_{j}$ and now take the complement of both sides so we get: $\bigcup_{i}A_{i} \subseteq \bigcup_{j}B_{j} \Longleftrightarrow (\bigcup_{j}B_{j})^{'} \subseteq (\bigcup_{i}A_{i})^{'}$, that is, $\bigcap_{j}B_{j} \subseteq \bigcap_{i}A_{i}$. QED.

Problem 2:

The difference between two sets A and B, denoted by $A-B$, is the set of all elements in A and not in B, thus $A-B = A \bigcap B^{'}$. Show the following:

(a)$A-B = A - (A \bigcap B) = (A \bigcup B) -B$.

Answer (a): $A-B$ is the set of elements $x \in A$, but not in B. Hence, $A-B = A-(A \bigcap B)$, simply by definition. Again, if we consider elements in A or B or both, and then take away elements of B, we are just left with elements of A only, but not of B. That is, $A-B = A-(A \bigcap B) = (A \bigcup B) - B$. QED. ps: A venn diagram may help visualize v well.

(b) TPT: $(A-B) - C = A - (B \bigcup C)$ (PS: a Venn diagram can help visualize and perhaps, guide the writing of the proof also).

Let $x \in A-B$, but $x \notin C$. Then, $x \in A$, but $x \notin B$, or $x \notin C$, or not in both B and C. That is, $x \in A-(B \bigcup C)$. By reversing the argument, we get the other subset relationship. Hence, $(A-B)-C = A- (B \bigcap C)$. QED.

(c) TPT: $A - (B-C) = (A-B) \bigcup (A \bigcap C)$.

Answer (c): Let $x \in A-(B-C)$. That is, $x \in A$ and $x \notin (B-C)$. That is, $x \in A, x \notin B, x \in C$. That is, $x \in A, x \notin B$, or x could be both in A and C. That is, $x \in (A-B) \bigcup (A \bigcap C)$. Simply reversing the arguments, we get the reverse subset relation. Hence, the two sets are equal. Hence, $A-(B-C) = (A-B) \bigcup (A \bigcap C)$. QED.

(d) TPT: $(A \bigcup B) - C = (A-C) \bigcup (B-C)$.

answer (d): let $x \in A\bigcup B$, but $x \notin C$. Then, $x \in A$, or $x \in B$, or x is an element of both A and B, but x is not an element of C. Which clearly means, x is an element of A but not C, OR x is an element of B but not C. That is, $x \in (A-C) \bigcup (B-C)$ (upon reversing the arguments, we get the other subset relations. Hence, the two sets are equal). QED.

problem (e): TPT: $A-(B \bigcup C) = (A-B) \bigcap (A-C)$ (once again, note that for set theoretic relations with up to three sets, venn diagrams are helpful to visualize and guide the proofs..)

answer (c): let $x \in A$, but not in $B \bigcup C$. That is, x is an element of A, but not of B, not of C, or not both of B and C. That is, x is an element of A not of B, and x is an element of A not of C. That is, $x \in (A-B) \bigcap (A-C)$. Reversing the arguments, we get the other subset relationship. Hence, the two sets are equal. Hence, $A - (B \bigcup C) = (A-B) \bigcap (B-C)$. QED.

Problem 3:

The symmetric difference of two sets A and B, denoted by $A \triangle B$, is defined by $A \triangle B = (A-B) \bigcup (B-A)$; it is the union of the difference of two sets in opposite orders. Prove the following:

3(a) Symmetric difference is associative: $A \triangle (B \triangle C) = (A \triangle B) \triangle C$.

3(b): TPT: $A \triangle \phi = A$; and $A \triangle A = \phi$. (this is some sort of ‘existence of additive inverse of ‘in a symmetric relation’).

Answer 3b: both results are obvious from definition.

3(c): TPT: Symmetric difference is commutative operation: $A \triangle B = B\triangle A$.

answer 3c: By definition, LHS is $(A-B)\bigcup (B-A) = (B-A) \bigcup (A-B)$, which is RHS. Hence, done.

3(d): TPT: Some sort of distributive law: $A \bigcap (B \triangle C) = (A \bigcap B) \triangle (A \bigcap C)$.

answer 3d: Let $x \in A$, and also simultaneously $x \in (B \triangle C)$. That is, $x \in A$, and $x \notin (B \bigcap C)$. That is, $x \in A, x \notin (A \bigcap B)\bigcap (B \bigcap C)$. That is, $x \in RHS$. Reversing the arguments, we get the other subset relation. Hence, the two sets are equal. Hence, we can say “intersection distributes over symmetric difference.” QED.

Problem 4:

A ring of sets is a non-empty class A of sets such that if A and B are in A then $A \triangle B$ is in A and $A \bigcap B$ is also in A. Show that A must contain the empty set, $A \bigcup B$ and $A-B$. Also, show that if a non empty class of sets contains the union and difference of any pair of the sets, then it is a ring of sets. Also, prove that a Boolean algebra of sets is a ring of sets.

Answer 4(i): TPT: A must also contain the empty set, the set $A \bigcup B$ and the set $A-B$. As $A \triangle B$ is in A, so also $A \triangle A = \phi \in A$; now, it is already given that the ring of sets is non-empty, so it contains a non empty universal set also, call it U. We know from problem 2 in this blog that $A - B = A \bigcap B^{'}$. As U exists, so does $B^{'}$ by definition and so $A-B$ is in A. Also, as $A-B$ exists, $A-B^{'}$ exists in A, but $A-B^{'}=A\bigcap B$ also exists in A. QED.

Answer 4(ii): TPT: If a non empty class of sets contains the union and difference of any pair of the sets, then it is a ring of sets.

Let A be any non-empty class of sets. Clearly, it therefore has a non empty universal set. As $\phi \in U$, so also $\phi$ is contained in A. Hence, the complement of a set A is also in A. Now, $A \bigcup B \in A$, so $U - (A \bigcup B)$ is in A, that is, by De Morgan law, $B^{'} \bigcap A^{'}$ is in A. That is, again, $A \bigcap B$ is in A. Now, since $A-B$ is in A, hence, $B-A$ is in A, and since $A \bigcup B$ is in A, then $(A-B) \bigcup (B-A)$ is in A also. Hence, A is a ring of sets.

4(iii) TPT: Show that a boolean algebra of sets is a ring of sets.

Answer 4(iii); Firstly, as a boolean algebra of sets is non-empty, and it also contains the empty set and the Universal set. As A is in A implies that $A^{'}$ is in A, so $A \bigcup B^{'}$ is also in A (by definition of Boolean algebra), but this is precisely that $A-B$ is in A. Now, as $A \triangle B = (A-B) \bigcup (B-A)$ clearly is also in A. Hence, a Boolean algebra of sets is a ring of sets.

Problem 5:

Show that the class of all finite subsets (including the empty set) of an infinite set is a ring of sets but is not a Boolean algebra.

In problem 4 above, we showed that if a non empty class of sets contains the union and difference of any pair of the sets, then it is a ring of sets. The given class is a subclass of all an infinite set such that it contains finite subsets only; that is, again, we know that finite unions of finite subsets and finite intersections of finite subsets also lie in that subclass, but the complements might be infinite sets so that it need not always be a Boolean algebra of sets.

Problem 6:

Show that the class of all finite unions of closed-open intervals on the real line is a ring of sets but is not a Boolean algebra of sets.

Same reasoning as above tells us that it is indeed a ring of sets; but consider for example the closed-open interval $[a,b)$, the complement of this is not in that class as it is an infinite interval so that it is not a Boolean algebra of sets.

Problem 7:

Assuming that the Universal set U is non empty, show that Boolean algebras of sets can be described as rings of sets which contain U.

As $U \neq \phi$, by definition of Boolean algebras, $U^{'}$ is in this class of sets, so that the empty set is also a part of this class of sets. Again, by definition of Boolean algebras, the union of two sets of this class is also in this set so also the intersection of any two sets of this class is also in this class. So, for any two sets A and B, $A-B$ is also in this set (as complement of B exists), and so also, $A \triangle B = (A-B) \bigcup (B-A)$ is also in this set. In other words, this is also a ring of sets.

Regards,

Nalin Pithwa

Set theory: more basic problems to solve and clear and apply

I am producing the list of questions first so that the motivated reader can first read and try them …and can compare with my answers by scrolling down only much later; here we go:

1. If $\{ A_{i}\}$ and $\{ B_{j}\}$ are two classes of sets such that $\{ A_{i}\} \subseteq \{ B_{j}\}$, then prove that $\bigcup_{i}A_{i} \subseteq \{ B_{j}\}$ and $\bigcap_{j} B_{j} \subseteq \bigcap_{i}A_{i}$.
2. The difference between two sets A and B, denoted by $A-B$, is the set of elements in A and not in B; thus, $A - B = A \bigcap B^{'}$. Prove the following simple properties: (a) $A-B = A-(A \bigcap B) = (A \bigcup B)-B$; (b) $(A-B)-C = A-(B \bigcup C)$; (c) $A - (B-C) = (A-B) \bigcup (A \bigcap C)$; (d) $(A \bigcup B) - C = (A-C) \bigcup (B-C)$; (e) $A - (B \bigcup C) = (A-B) \bigcap (A-C)$
3. The symmetric difference of two sets A and B, denoted by $A \triangle B$, is defined by $A \triangle B = (A-B) \bigcup (B-A)$; it is thus the union of their differences in opposite orders. Prove the following : (a) Symmetric difference is associative : $A \triangle (B \triangle C) = (A \triangle B) \triangle C$ (b) $A \triangle \phi=A$ (c) $A \triangle A = \phi$ (c) Symmetric difference is commutative: $A \triangle B = B \triangle A$ (d) Some sort of distributive rule also holds: $A \bigcap (B \triangle C) = (A \bigcap B) \triangle (A \bigcap C)$
4. A ring of sets is a non-empty class A of sets such that if A and B are in A, then $A \triangle B$ and $A \bigcap B$ are also in A. Show that A must also contain the empty set, $A \bigcup B$, and $A-B$. Show that if a non-empty class of sets contains the union and difference of sets any pair of its sets, then it is a ring of sets. Prove that a Boolean algebra of sets is a ring of sets.
5. Show that the class of all finite subsets (including the empty set) of an infinite set is a ring of sets but is not a Boolean algebra of sets.
6. Show that the class of all finite unions of closed-open intervals on the real line is a ring of sets but is not a Boolean algebra of sets.
7. Assuming that the universal set U is non-empty, show that Boolean algebras of sets can be described as a ring of sets which contain U.

I will put up my solutions as soon as I can.

Regards,

Nalin Pithwa.

Number theory: let’s learn it the Nash way !

Reference: A Beautiful Mind by Sylvia Nasar.

Comment: This is approach is quite similar to what Prof. Joseph Silverman explains in his text, “A Friendly Introduction to Number Theory.”

Peter Sarnak, a brash thirty-five-year-old number theorist whose primary interest is the Riemann Hypothesis, joined the Princeton faculty in the fall of 1990. He had just given a seminar. The tall, thin, white-haired man who had been sitting in the back asked for a copy of Sarnak’s paper after the crowd had dispersed.

Sarnak, who had been a student of Paul Cohen’s at Stanford, knew Nash by reputation as well as by sight, naturally. Having been told many times Nash was completely mad, he wanted to be kind. He promised to send Nash the paper. A few days later, at tea-time, Nash approached him again. He had a few questions, he said, avoiding looking Sarnak in the face. At first, Sarnak just listened politely. But within a few minutes, Sarnak found himself having to concentrate quite hard. Later, as he turned the conversation over in his mind, he felt rather astonished. Nash had spotted a real problem in one of Sarnak’s arguments. What’s more, he also suggested a way around it. “The way he views things is very different from other people,” Sarnak said later. ‘He comes up with instant insights I don’t know I would ever get to. Very, very outstanding insights. Very unusual insights.”

They talked from time to time. After each conversation, Nash would disappear for a few days and then return with a sheaf of computer printouts. Nash was obviously very, very good with the computer. He would think up some miniature problem, usually very ingeniously, and then play with it. If something worked on a small scale, in his head, Sarnak realized, Nash would go to the computer to try to find out if it was “also true the next few hundred thousand times.”

{What really bowled Sarnak over, though, was that Nash seemed perfectly rational, a far cry from the supposedly demented man he had heard other mathematicians describe. Sarnak was more than a little outraged. Here was this giant and he had been all but forgotten by the mathematics profession. And the justification for the neglect was obviously no longer valid, if it had ever been.}

Cheers,

Nalin Pithwa

PS: For RMO and INMO (of Homi Bhabha Science Foundation/TIFR), it helps a lot to use the following: (it can be used with the above mentioned text of Joseph Silverman also): TI nSpire CAS CX graphing calculator.

https://www.amazon.in/INSTRUMENTS-TI-Nspire-CX-II-CAS/dp/B07XCM6SZ3/ref=sr_1_1?crid=3RNR2QRV1PEPH&keywords=ti+nspire+cx+cas&qid=1585782633&s=electronics&sprefix=TI+n%2Caps%2C253&sr=8-1

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https://www.amazon.in/INSTRUMENTS-TI-Nspire-CX-II-CAS/dp/B07XCM6SZ3/ref=sr_1_1?crid=3RNR2QRV1PEPH&keywords=ti+nspire+cx+cas&qid=1585782633&s=electronics&sprefix=TI+n%2Caps%2C253&sr=8-1

Towards Baby Analysis: Part I: INMO, IMO and CMI Entrance

$\bf{Reference: \hspace{0.1in}Introductory \hspace{0.1in} Real Analysis: \hspace{0.1in} Kolmogorov \hspace{0.1in} and \hspace{0.1in} Fomin; \hspace{0.1in}Dover \hspace{0.1in }Publications}$

$\bf{Equivalence \hspace{0.1in} of \hspace{0.1in} Sets \hspace{0.1in} The \hspace{0.1in}Power \hspace{0.1in }of \hspace{0.1in }a \hspace{0.1in}Set}$

$\bf{Section 1}$:

$\bf{Finite \hspace{0.1in} and \hspace{0.1in} infinite \hspace{0.1in} sets}$

The set of all vertices of a given polyhedron, the set of all prime numbers less than a given number, and the set of all residents of NYC (at a given time) have a certain property in common, namely, each set has a definite number of elements which can be found in principle, if not in practice. Accordingly, these sets are all said to be $\it{finite}$.$\it{Clearly \hspace{0.1in} we \hspace{0.1in}can \hspace{0.1in} be \hspace{0.1in} sure \hspace{0.1in} that \hspace{0.1in} a \hspace{0.1in} set \hspace{0.1in}is \hspace{0.1in}finite \hspace{0.1in} without \hspace{0.1in} knowing \hspace{0.1in} the \hspace{0.1in} number \hspace{0.1in} of elements \hspace{0.1in}in \hspace{0.1in}it.}$

On the other hand, the set of all positive integers, the set of all points on the line, the set of all circles in the plane, and the set of all polynomials with rational coefficients have a different property in common, namely, $\it{if \hspace{0.1in } we \hspace{0.1in}remove \hspace{0.1in} one \hspace{0.1in} element \hspace{0.1in}from \hspace{0.1in}each \hspace{0.1in}set, \hspace{0.1in}then \hspace{0.1in}remove \hspace{0.1in}two \hspace{0.1in}elements, \hspace{0.1in}three \hspace{0.1in}elements, \hspace{0.1in}and \hspace{0.1in}so \hspace{0.1in}on, \hspace{0.1in}there \hspace{0.1in}will \hspace{0.1in}still \hspace{0.1in}be \hspace{0.1in}elements \hspace{0.1in}left \hspace{0.1in}in \hspace{0.1in}the \hspace{0.1in}set \hspace{0.1in}in \hspace{0.1in}each \hspace{0.1in}stage}$. Accordingly, sets of these kind are called $\it{infinite}$ sets.

Given two finite sets, we can always decide whether or not they have the same number of elements, and if not, we can always determine which set has more elements than the other. It is natural to ask whether the same is true of infinite sets. In other words, does it make sense to ask, for example, whether there are more circles in the plane than rational points on the line, or more functions defined in the interval [0,1] than lines in space? As will soon be apparent, questions of this kind can indeed be answered.

To compare two finite sets A and B, we can count the number of elements in each set and then compare the two numbers, but alternatively, we can try to establish a $\it{one-\hspace{0.1in}to-\hspace{0.1in}one \hspace{0.1in}correspondence}$ between the elements of set A and set B, that is, a correspondence such that each element in A corresponds to one and only element in B, and vice-versa. It is clear that a one-to-one correspondence between two finite sets can be set up if and only if the two sets have the same number of elements. For example, to ascertain if or not the number of students in an assembly is the same as the number of seats in the auditorium, there is no need to count the number of students and the number of seats. We need merely observe whether or not there are empty seats or students with no place to sit down. If the students can all be seated with no empty seats left, that is, if there is a one-to-one correspondence between the set of students and the set of seats, then these two sets obviously have the same number of elements. The important point here is that the first method(counting elements) works only for finite sets, while the second method(setting up a one-to-one correspondence) works for infinite sets as well as for finite sets.

$\bf{Section 2}$:

$\bf{Countable \hspace{0.1in} Sets}$.

The simplest infinite set is the set $\mathscr{Z^{+}}$ of all positive integers. An infinite set is called $\bf{countable}$ if its elements can be put into one-to-one correspondence with those of $\mathscr{Z^{+}}$. In other words, a countable set is a set whose elements can be numbered $a_{1}, a_{2}, a_{3}, \ldots a_{n}, \ldots$. By an $\bf{uncountable}$ set we mean, of course, an infinite set which is not countable.

We now give some examples of countable sets:

$\bf{Example 1}$:

The set $\mathscr{Z}$ of all integers, positive, negative, or zero is countable. In fact, we can set up the following one-to-one correspondence between $\mathscr{Z}$ and $\mathscr{Z^{+}}$ of all positive integers: (0,1), (-1,2), (1,3), (-2,4), (2,5), and so on. More explicitly, we associate the non-negative integer $n \geq 0$ with the odd number $2n+1$, and the negative integer $n<0$ with the even number $2|n|$, that is,

$n \leftrightarrow (2n+1)$, if $n \geq 0$, and $n \in \mathscr{Z}$
$n \leftrightarrow 2|n|$, if $n<0$, and $n \in \mathscr{Z}$

$\bf{Example 2}$:

The set of all positive even numbers is countable, as shown by the obvious correspondence $n \leftrightarrow 2n$.

$\bf{Example 3}$:

The set 2,4,8,$\ldots 2^{n}$ is countable as shown by the obvious correspondence $n \leftrightarrow 2^{n}$.

$\bf{Example 4}: The set$latex \mathscr{Q}$of rational numbers is countable. To see this, we first note that every rational number $\alpha$ can be written as a fraction $\frac{p}{q}$, with $q>0$ with a positive denominator. (Of course, p and q are integers). Call the sum $|p|+q$ as the “height” of the rational number $\alpha$. For example, $\frac{0}{1}=0$ is the only rational number of height zero, $\frac{-1}{1}$, $\frac{1}{1}$ are the only rational numbers of height 2, $\frac{-2}{1}$, $\frac{-1}{2}$, $\frac{1}{2}$, $\frac{2}{1}$ are the only rational numbers of height 3, and so on. We can now arrange all rational numbers in order of increasing “height” (with the numerators increasing in each set of rational numbers of the same height). In other words, we first count the rational numbers of height 1, then those of height 2 (suitably arranged), then those of height 3(suitably arranged), and so on. In this way, we assign every rational number a unique positive integer, that is, we set up a one-to-one correspondence between the set Q of all rational numbers and the set $\mathscr{Z^{+}}$ of all positive integers. $\it{Next \hspace{0.1in}we \hspace{0.1in} prove \hspace{0.1in}some \hspace{0.1in}elementary \hspace{0.1in}theorems \hspace{0.1in}involving \hspace{0.1in}countable \hspace{0.1in}sets}$ $\bf{Theorem1}$. $\bf{Every \hspace{0.1in} subset \hspace{0.1in}of \hspace{0.1in}a \hspace{0.1in}countable \hspace{0.1in}set \hspace{0.1in}is \hspace{0.1in}countable}$. $\bf{Proof}$ Let set A be countable, with elements $a_{1}, a_{2}, a_{3}, \ldots$, and let set B be a subset of A. Among the elements $a_{1}, a_{2}, a_{3}, \ldots$, let $a_{n_{1}}, a_{n_{2}}, a_{n_{3}}, \ldots$ be those in the set B. If the set of numbers $n_{1}, n_{2}, n_{3}, \ldots$ has a largest number, then B is finite. Otherwise, B is countable (consider the one-to-one correspondence $i \leftrightarrow a_{n_{i}}$). $\bf{QED.}$ $\bf{Theorem2}$ $\bf{The \hspace{0.1in}union \hspace{0.1in}of \hspace{0.1in}a \hspace{0.1in}finite \hspace{0.1in}or \hspace{0.1in}countable \hspace{0.1in}number \hspace{0.1in}of \hspace{0.1in}countable \hspace{0.1in}sets \hspace{0.1in}A_{1}, A_{2}, A_{3}, \ldots \hspace{0.1in}is \hspace{0.1in}itself \hspace{0.1in}countable.}$ $\bf{Proof}$ We can assume that no two of the sets $A_{1}, A_{2}, A_{3}, \ldots$ have any elements in common, since otherwise we could consider the sets $A_{1}$, $A_{2}-A_{1}$, $A_{3}-(A_{1}\bigcup A_{2})$, $\ldots$, instead, which are countable by Theorem 1, and have the same union as the original sets. Suppose we write the elements of $A_{1}, A_{2}, A_{3}, \ldots$ in the form of an infinite table $\begin{array}{ccccc} a_{11} & a_{12} & a_{13} & a_{14} &\ldots \\ a_{21} &a_{22} & a_{23} & a_{24} & \ldots \\ a_{31} & a_{32} & a_{33} & a_{34} & \ldots \\ a_{41} & a_{42} & a_{43} & a_{44} & \ldots \\ \ldots & \ldots & \ldots & \ldots & \ldots \end{array}$ where the elements of the set $A_{1}$ appear in the first row, the elements of the set $A_{2}$ appear in the second row, and so on. We now count all the elements in the above array “diagonally”; that is, first we choose $a_{11}$, then $a_{12}$, then move downwards, diagonally to “left”, picking $a_{21}$, then move down vertically picking up $a_{31}$, then move across towards right picking up $a_{22}$, next pick up $a_{13}$ and so on ($a_{14}, a_{23}, a_{32}, a_{41}$)as per the pattern shown: $\begin{array}{cccccccc} a_{11} & \rightarrow & a_{12} &\hspace{0.1in} & a_{13} & \rightarrow a_{14} & \ldots \\ \hspace{0.1in} & \swarrow & \hspace{0.1in} & \nearrow & \hspace{0.01in} & \swarrow & \hspace{0.1in} & \hspace{0.1in}\\ a_{21} & \hspace{0.1in} & a_{22} & \hspace{0.1in} & a_{23} \hspace{0.1in} & a_{24} & \ldots \\ \downarrow & \nearrow & \hspace{0.1in} & \swarrow & \hspace{0.1in} & \hspace{0.1in} & \hspace{0.1in} & \hspace{0.1in}\\ a_{31} & \hspace{0.1in} & a_{32} & \hspace{0.1in} & a_{33} & \hspace{0.1in} & a_{34} & \ldots \\ \hspace{0.1in} & \swarrow & \hspace{0.1in} & \hspace{0.1in} & \hspace{0.1in} & \hspace{0.1in} & \hspace{0.1in} & \hspace{0.1in}\\ a_{41} & \hspace{0.1in} & a_{42} &\hspace{0.1in} & a_{43} &\hspace{0.1in} &a_{44} &\ldots\\ \ldots & \hspace{0.1in} & \ldots & \hspace{0.1in} & \ldots & \hspace{0.1in} & \ldots & \hspace{0.1in} \end{array}$ It is clear that this procedure associates a unique number to each element in each of the sets $A_{1}, A_{2}, \ldots$ thereby establishing a one-to-one correspondence between the union of the sets $A_{1}, A_{2}, \ldots$ and the set $\mathscr{Z^{+}}$ of all positive integers. $\bf{QED.}$ $\bf{Theorem3}$ $\bf{Every \hspace{0.1in}infinite \hspace{0.1in}subset \hspace{0.1in}has \hspace{0.1in}a \hspace{0.1in}countable \hspace{0.1in}subset.}$ $\bf{Proof}$ Let M be an infinite set and $a_{1}$ any element of M. Being infinite, M contains an element $a_{2}$ distinct from $a_{1}$, an element $a_{3}$ distinct from both $a_{2}$ and $a_{1}$, and so on. Continuing this process, (which can never terminate due to “shortage” of elements, since M is infinite), we get a countable subset $A= \{ a_{1}, a_{2}, a_{3}, \ldots, a_{n}, \ldots\}$ of the set $M$. $\bf{QED.}$ $\bf{Remark}$ Theorem 3 shows that countable sets are the “smallest” infinite sets. The question of whether there exist uncountable (infinite) sets will be considered below. $\bf{Section3}$ $\bf{Equivalence \hspace{0.1in} of \hspace{0.1in} sets}$ We arrived at the notion of a countable set M by considering one-to-one correspondences between set M and the set $\mathscr{Z^{+}}$ of all positive integers. More generally, we can consider one-to-one correspondences between any two sets M and N. $\bf{Definition}$ Two sets M and N are said to be $\bf{equivalent}$ (written $M \sim N$) if there is a one-to-one correspondence between the elements of M and the elements of N. The concept of equivalence is applicable both to finite and infinite sets. Two finite sets are equivalent if and only if they have the same number of elements. We can now define a countable set as a set equivalent to the set $\mathscr{Z^{+}}$ of all positive integers. It is clear that two sets are equivalent to a third set are equivalent to each other, and in particular that any two countable sets are equivalent. $\bf{Example1}$ The sets of points in any two closed intervals$[a,b]$and$[c,d]$are equivalent; you can “see’ a one-to-one correspondence by drawing the following diagram: Step 1: draw cd as a base of a triangle. Let the third vertex of the triangle be O. Draw a line segment “ab” above the base of the triangle; where “a” lies on one side of the triangle and “b” lies on the third side of the third triangle. Note that two points p and q correspond to each other if and only if they lie on the same ray emanating from the point O in which the extensions of the line segments ac and bd intersect. $\bf{Example2}$ The set of all points z in the complex plane is equivalent to the set of all points z on a sphere. In fact, a one-to-one correspondence $z \leftrightarrow \alpha$ can be established by using stereographic projection. The origin is the North Pole of the sphere. $\bf{Example3}$ The set of all points x in the open unit interval $(0,1)$ is equivalent to the set of all points y on the whole real line. For example, the formula $y=\frac{1}{\pi}\arctan{x}+\frac{1}{2}$ establishes a one-to-one correspondence between these two sets. $\bf{QED}$. The last example and the examples in Section 2 show that an infinite set is sometimes equivalent to one of its proper subsets. For example, there are “as many” positive integers as integers of arbitrary sign, there are “as many” points in the interval $(0,1)$ as on the whole real line, and so on. This fact is characteristic of all infinite sets (and can be used to define such sets) as shown by: $\bf{Theorem4}$ $\bf{Every \hspace{0.1in} infinite \hspace{0.1in} set \hspace{0.1in}is \hspace{0.1in} equivalent \hspace{0.1in} to \hspace{0.1in}one \hspace{0.1in}of \hspace{0.1in}its \hspace{0.1in}proper \hspace{0.1in}subsets.}$ $\bf{Proof}$ According to Theorem 3, every infinite set M contains a countable subset. Let this subset be $A=\{a_{1}, a_{2}, a_{3}, \ldots, a_{n}, \ldots \}$ and partition A into two countable subsets $A_{1}=\{a_{1}, a_{3}, a_{5}, \ldots \}$ and $A_{2}=\{a_{2}, a_{4}, a_{6}, \ldots \}$. Obviously, we can establish a one-to-one correspondence between the countable subsets A and $A_{1}$ (merely let $a_{n} \leftrightarrow a_{2n-1}$). This correspondence can be extended to a one-to-one correspondence between the sets $A \bigcup (M-A)=M$ and $A_{1} \bigcup (M-A)=M-A_{2}$ by simply assigning x itself to each element $x \in M-A$. But $M-A_{2}$ is a proper subset of M. $\bf{QED}$. More later, to be continued, Regards, Nalin Pithwa Set theory, relations, functions: preliminaries: part VI Set Theory, Relations, Functions: Preliminaries: Part VIII SETS and FUNCTIONS: Reference: Introductory Real Analysis by A. N. Kolmogorov and S V Fomin, Dover books. Operations on sets: Let A and B be any two sets. Then, by the sum or union of A and B, denoted by $A \bigcup B$, is meant the set consisting of all elements which belong to at least one of the sets A and B. More generally, by the sum or union of an arbitrary number (finite or infinite) of sets $A_{\alpha}$ (indexed by some parameter $\alpha$), we mean the set, denoted by $\bigcup_{\alpha}A_{\alpha}$ of all elements belonging to at least one of the sets $A_{\alpha}$. By the intersection $A \bigcap B$ of two given sets A and B, we mean the set consisting of all elements which belong to both A and B. For example, the intersection of the set of all even numbers and the set of all integers divisible by 3 is the set of all integers divisible by 6. By the intersection of an arbitrary number (finite or infinite) of sets $A_{\alpha}$, we mean the set, denoted by $\bigcap_{\alpha}A_{\alpha}$ of all elements belonging to every one of the sets $A_{\alpha}$. Two sets A and B are said to be disjoint if $A \bigcap B=\phi$, that is, if they have no elements in common. More generally, let $\mathscr{F}$ be a family of sets such that $A \bigcap B=\phi$ for every pair of sets A, B in $\mathscr{F}$. Then, the sets in $\mathscr{F}$ are said to be pairwise disjoint. It is an immediate consequence of the above definitions that the operations $\bigcup$ and $\bigcap$ are commutative and associative, that is, $A \bigcup B=B \bigcup A$ and $(A \bigcup B) \bigcup C= A \bigcup (B \bigcup C)$; $A \bigcap B=B \bigcap A$; and, $(A \bigcap B)\bigcap C = A \bigcap (B \bigcap C)$. Moreover, the operations $\bigcup$ and $\bigcap$ obey the following distributive laws: $(A \bigcup B)\bigcap C = (A \bigcap C) \bigcup (B \bigcap C)$….call this I $(A \bigcap B)\bigcup C = (A \bigcup C) \bigcap (B \bigcup C)$….call this II. For example, suppose that $x \in (A \bigcup B)\bigcap C$, so that x belongs to the left-hand side of I. Then, x belongs to both C and $A \bigcup B$, that is, x belongs to both C and at least one of the sets A and B. But then x belongs to at least one of the sets $A \bigcap C$ and $B \bigcap C$, that is, $x \in (A \bigcap C)\bigcup (B \bigcap C)$, so that x belongs to the right hand side of I. Conversely, suppose that $x \in (A \bigcap C)\bigcup (B \bigcap C)$. Then, x belongs to at least one of the two sets $A \bigcap C$ and $B \bigcap C$. It follows that x belongs to both C and at least one of the two sets A and B, that is, $x \in C$ and $x \in A \bigcup B$, or equivalently $x \in (A \bigcup B)\bigcap C$. This proves I, and II is proved similarly. By the difference of two sets, $A - B$ between two sets A and B (in that order), we mean the set of all elements of A which do not belong to B. Note that it is not assumed that $A \supset B$. It is sometimes convenient (e.g., in measure theory) to consider the symmetric difference of two sets A and B, denoted by $A \delta B$ and defined as the union of the two differences $A-B$ and $B-A$: $A \delta B=(A-B)\bigcup (B-A)$. We will often be concerned later with various sets which are all subsets of some underlying basic set R, for example, various sets of points on the real line. In this case, given a set A, the difference $R-A$ is called the complement of A, denoted by $A^{'}$. An important role is played in set theory and its applications by the following duality principle: $R - \bigcup_{\alpha}A_{\alpha}=\bigcap_{\alpha}(R-A_{\alpha})$…call this III $R - \bigcap_{\alpha}A_{\alpha}=\bigcup_{\alpha}(R-A_{\alpha})$…call this IV. In words, the complement of a union equal the intersection of the complements; and, the complement of an intersection equals the union of the complements. According to the duality principle, any theorem involving a family of subsets of a fixed set R can be converted automatically into another, “dual” theorem by replacing all subsets by their complements, all unions by intersections and all intersections by unions. To prove III, suppose that $x \in R - \bigcup_{\alpha}A_{\alpha}$….call this V. Then, x does not belong to the union $\bigcup_{\alpha}A_{\alpha}$. …call this VI. That is, x does not belong to any of the sets $A_{\alpha}$. It follows that x belongs to each of the complements $R - \bigcup_{\alpha}A_{\alpha}$, and hence, $x \bigcap_{\alpha}(R-A_{\alpha})$….call this VII. Conversely, suppose that VII holds, so that x belongs to every set $R - A_{\alpha}$. Then, x does not belong to any of the sets $A_{\alpha}$, that is, x does not belong to the union VI, or equivalently V holds true. This proves 3. Homework: Prove IV similarly. Remark: The designation “symmetric difference” for the set $A \Delta B$ is not too apt, since $A \Delta B$ has much in common with the sum $A \bigcup B$. In fact, in $A \bigcup B$ the two statements “x belongs to A” and “x belongs to B” are joined by the conjunction “or” used in the “either …or …or both…” sense, while in $A \Delta B$ the same two statements are joined by “or” used in the ordinary “either…or…” sense (as in “to be or not to be”). In other words, x belongs to $A \bigcup B$ if and only if x belongs to either A or B or both, while x belongs to $A \Delta B$ if and only if x belongs to either A or B but not both. The set $A \Delta B$ can be regarded as a kind of “modulo-two sum” of the sets A and B, that is, a sum of the sets A and B in which elements are dropped if they are counted twice (once in A and once in B). Functions and mappings. Images and preimages: (Of course, we have dealt with this topic in quite detail so far in the earlier blog series; but as presented by Kolmgorov and Fomin here, the flavour is different; at any rate, it helps to revise. Revision refines the intellect.) 🙂 🙂 🙂 A rule associating a unique real number $y=f(x)$ with each element of a set real numbers X is said to define a (real) function f on X. The set X is called the domain (of definition) of f, and the set Y of all numbers $f(x)$ such that $x \in X$ is called the range of f. More generally, let M and N be two arbitrary sets. Then a rule associating a unique element $b=f(a) \in N$ with each element $a \in M$ is again said to define a function f on M (or a function f with domain M). In this more general context, f is usually called a mapping of f M into N. By the same token, f is said to map M into N(and a into b). If a is an element of M, the corresponding element $b=f(a)$ is called the image of a (under the mapping f). Every element of M with a given element $b \in N$ as its image is called a preimage of b. Note that in general b may have several pre-images. Moreover, N may contain elements with no pre-images at all. If b has a unique pre-image, we denote this pre-image by $f^{-1}(b)$. If A is a subset of M, the set of all elements $f(a) \in N$ such that $a \in A$ is called the image of A, denoted by $f(A)$. The set of all elements of M whose images belong to a given set $B \subset N$ is called the preimage of B, denoted by $f^{-1}(B)$. If no element of B has a preimage, then $f^{-1}(B)=\phi$. A function f is said to map M into N if $f(M)=N$, as is always the case, and onto N if $f(M)=N$. Thus, every “onto” mapping is an “into” mapping, but converse is not true. Suppose f maps M onto N. Then, f is said to be one-to-one if each element $b \in N$ has a unique preimage $f^{-1}(b)$. In this case, f is said to establish a one-to-one correspondence between M and N, and the mapping $f^{-1}$ associating $f^{-1}(b)$ is called the inverse of f. Theorem I: The preimage of the union of two sets is the union of the preimages of the sets $f^{-1}(A \bigcup B)=f^{-1}(A) \bigcup f^{-1}(B)$. Proof of Theorem I: If $x \in f^{-1}(A \bigcup B)$, then $f(x) \in A \bigcup B$ so that $f(x)$ belongs to at least one of the sets A and B. But, then x belongs to at least one of the sets $f^{-1}(A)$ and $f^{-1}(B)$, that is, $x \in f^{-1}(A) \bigcup f^{-1}(B)$. Conversely, if $x \in f^{-1}(A) \bigcup f^{-1}(B)$, then x belongs to at least one of the sets $f^{-1}(A)$ and $f^{-1}(B)$. Therefore, $f(x)$ belongs to at least one of the sets A and B, that is, $f(x) \in A \bigcup B$. But, then $x \in f^{-1}(A \bigcup B)$. QED. Theorem 2: The preimage of the intersection of two sets is the intersection of the preimages of the sets: $f^{-1}(A \bigcap B)=f^{-1}(A) \bigcap f^{-1}(B)$. Proof of Theorem 2: If $x \in f^{-1}(A \bigcap B)$, then $f(x) \in A \bigcap B$, so that $f(x) \in A$ and (meaning, simultaneously) $f(x) \in B$. But, then $x \in f^{-1}(A)$ and $x \in f^{-1}(B)$, that is, $x \in f^{-1}(A) \bigcap f^{-1}(B)$. Conversely, if $x \in f^{-1}(A) \bigcap f^{-1}(B)$, then $x \in f^{-1}(A)$ and $x \in f^{-1}(B)$. Therefore, $f(x) \in A$ and $f(x) \in B$, that is, $f(x) \in A \bigcap B$. But, then $x \in f^{-1}(A \bigcap B)$. QED. Theorem 3: The image of the union of two sets equals the union of the images of the sets $f(A \bigcup B) = f(A) \bigcup f(B)$. Proof of theorem 3: If $y \in f(A \bigcup B)$, then $y=f(x)$ where x belongs to at least one of the sets A and B. Therefore, $y=f(x)$ belongs to at least one of the sets $f(A)$ and $f(B)$. That is, $x \in f(A) \bigcup f(B)$. Conversely, if $y \in f(A) \bigcup f(B)$, then $y=f(x)$. where x belongs to at least one of the sets A and B, that is, $x \in A \bigcup B$ and hence, $y=f(x) \in f(A \bigcup B)$. QED. Remark I: Surprisingly enough, the image of the intersection of two sets is not so “well-behaved”. The image of the intersection of two sets does not necessarily equal the intersection of the images of the sets. For example, suppose the mapping f projects the xy-plane onto the x-axis, carrying the point $(x,y)$ into the $(x,0)$. Then, the segments $0 \leq x \leq 1$ with $y=0$, and $0 \leq x \leq 1$ with $y=1$ do not intersect, although their images coincide. Remark 2: In the light of above remark: consider the following: If a function is not specified on elements, it is important in general to check that f is well-defined. That is, unambiguously defined. For example, if the set A is the union of two subsets $A_{1}$ and $A_{2}$, and we are considering a function $f: A \longrightarrow B$, then one can try to specify a function from set A to the set B $\{ 0,1\}$ by declaring that f is to map everything in $A_{1}$ to 0 and is to map everything in $A_{2}$ to 1. This unambiguously defines f unless $A_{1}$ and $A_{2}$ have elements in common in which case it is not clear whether these elements should map to 0 or to 1. Checking that this f is well-defined therefore amounts to checking that $A_{1}$ and $A_{2}$ have no intersection. Remark 3: Theorems 1-3 continue to hold for unions and intersections of an arbitrary number (finite or infinite) of sets $A_{\alpha}$: $f^{-1}(\bigcup_{\alpha})A_{\alpha}=\bigcup_{\alpha}f^{-1}(A_{\alpha})$ $f^{-1}(\bigcap_{\alpha}A_{\alpha})=\bigcap_{\alpha}f^{-1}(A_{\alpha})$ $f(\bigcup_{\alpha})A_{\alpha}=\bigcup_{\alpha}f(A_{\alpha})$ Decomposition of a set into classes. Equivalence relations. Decompositions of a given set into pairwise disjoint subsets play an important role in a great variety of problems. For example, the plane (regarded as a point set) can be decomposed into lines parallel to the x-axis, three dimensional space can be decomposed into concentric spheres, the inhabitants of a given city can be decomposed into different age groups, and so on. Any such representation of a given set M as the union of a family of pairwise disjoint subsets of M is called a decomposition or partition of M into classes. A decomposition is usually made on the basis of a certain criterion, allowing us to assign the elements of M to one class or another. For example, the set of all triangles in the plane can be decomposed into classes of congruent triangles, or classes of triangles of equal area, the set of all functions of x can be decomposed into classes of functions all taking the same value at a given point x, and so on. Despite the great variety of such criteria, they are not completely arbitrary. For example, it is obviously impossible to partition all real numbers into classes by assigning the number b to the same class as number a if and only if $b>a$. In fact, if $b>a$, b must be assigned to same class as a but then a cannot be assigned to same class as b, since $a. Moreover, since a is not greater than itself, a cannot be assigned to the class containing itself!! As another example, it is impossible to partition the points of the plane into classes by assigning two points to the same class if and only if the distance between them is less than 1. In fact, if the distance between a and b is less than 1, and if the distance between b and c is less than 1, it does not follow that distance between a and c is less than 1. (Hint: Think of triangle inequality). Thus, by assigning a to the same class as b and b to the same class as c, we may well find that two points fall in the same class even though the distance between them is greater than 1! These examples suggest conditions that which must be satisfied by any criterion it it is to be used as the basis for partitioning a given set into classes. Let M be a set, and let certain ordered pairs (a,b) of elements of M be called “labelled.” If $(a,b)$ is a labelled pair, we say that a is related to b by the binary relation R and we write it symbolically as aRb. For example, if a and b are real numbers, $aRb$ might mean $a, while if a and b are triangles, $aRb$ might mean that a and b have the same area. A relation between elements of M is called a relation on M, if there exists at least one labelled pair $(a,b)$ for every $a \in M$. A relation R on M is called an equivalence relation (on M) if it satisfies the following three conditions: • Reflexivity: $aRa$ for every $a \in M$. • Symmetry: If $aRb$, then $bRa$. • Transititivity: If $aRb$ and $bRc$, then $aRc$. 🙂 🙂 🙂 Nalin Pithwa, more later Set theory, relations, functions: preliminaries: Part VII FINITE, COUNTABLE AND UNCOUNTABLE SETS: (Reference: Principles of Mathematical Analysis, Third Edition, Walter Rudin:) We begin with the function concept. Definition: Consider two sets A and B whose elements may be any objects whatsoever, and suppose that with each element x of A there is associated in some manner, an element of B, which we denote by $f(x)$. Then, f is said to be a function from A to B (or a mapping of A into B). The set A is called the domain of f (we also say f is defined on A), and the elements f(x) are called the values of f. The set of all values of f is called the range of f. Definition: Let A and B be two sets and let f be a mapping of A into B. If $E \subset A$, then $f(E)$ is defined to be the set of all elements f(x), for $x \in E$. We call $f(E)$ the image of E under f. In this notation, $f(A)$ is the range of f. It is clear that $f(A) \subset B$. If $f(A)=B$, we say that f maps A onto B. (Note that, according to this usage, onto is more specific than into). If $E \subset B$, $f^{-1}(E)$ denotes the set of all $x \in A$ such that $f(x) \in E$. We call $f^{-1}(E)$ the inverse image of E under f. If $y \in B$, then $f^{-1}(y)$ is the set of all$latex  \in A\$ such that $f(x)=y$. If, for each $y \in B$, $f^{-1}(y)$ consists of at most one element of A, then f is said to be a 1-1 (one-one) mapping of A into B. This may also be expressed as follows: f is a 1-1 mapping of A into B provided that $f(x_{1}) \neq f(x_{2})$ whenever $x_{1} \neq x_{2}$ and $x_{1} \in A$ and $x_{2} \in A$.

(The notation $x_{1} \neq x_{2}$ , means that $x_{1}$ and $x_{2}$ are distinct elements, otherwise we write $x_{1} = x_{2}$).

Definition:

If there exists a 1-1 mapping A of onto B, we say that A and B can be put in 1-1 correspondence, or that A and B have the same cardinal number, or briefly, that A and B are equivalent, and we write $A \sim B$. This relation clearly has the following properties:

It is reflexive: $A \sim A$

It is symmetric: If $A \sim B$, then $B \sim A$.

It is transitive: If $A \sim B$, and $B \sim C$, then $A \sim C$.

Any relation with these three properties is called an equivalence relation.

Definition:

For any positive integer n, let $J_{n}$ be the set whose elements are the integers $1, 2, \ldots, n$. And, let J be the set consisting of all positive integers. For any set A, we say:

(a) A is finite if $A \sim J_{n}$ for some n (the empty set is also considered to be finite).

(b) A is infinite, if A is not finite.

(c) A is countable if $A \sim J$.

(d) A is uncountable if A is neither finite nor countable.

(e) A is at most countable if A is finite or countable.

Countable sets are sometimes called enumerable, or denumerable.

For two finite sets A and B, we evidently have $A \sim B$ if and only if A and B contain the same number of elements. For infinite sets, however, the idea of “having the same number of elements” becomes quite vague, whereas the notion of 1-1 correspondence retains its clarity.

Example:

Let A be the set of all integers. Then, A is countable. For, consider the following arrangement of the sets A and J:

A: 0,1,-1,2,-2,3,-3,…

J:1,2,3,4,5,6,7,…

We can, in this example, even give an explicit formula for a function f from J to A which sets up a 1-1 correspondence:

When n is even, $f(n)=\frac{n}{2}$

When n is odd, $f(n)=-\frac{n-1}{2}$

Remark:

A finite set cannot be equivalent to one of its proper subsets. That this is, however, possible for infinite sets, is shown by the above example, in which J is a proper subset of A.

In fact, we can define infinite sets as follows: A set A is infinite if A is equivalent to one of its proper subsets.

Definition: SEQUENCES:

By a sequence, we mean a function f defined on the set J of all positive integers. If $f(n)=x_{n}$, for $n \in J_{n}$, it is customary to denote the sequence f by the symbol $\{ x_{n}\}$, or sometimes by $x_{1}, x_{2}, x_{3}, \ldots$. The values of f, that is, the elements $x_{n}$ are called the terms of the sequence. If A is a set and if $x_{n} \in A$, for all $n \in J$, then $\{ x_{n}\}$ is said to be a sequence in A, or a sequence of elements of A.

Note that the terms $x_{1}, x_{2}, x_{3}, \ldots$ of a sequence need not be distinct.

Since every countable set is the range of a 1-1 function defined on J, we may regard every countable set as the range of a sequence of distinct terms. Speaking more loosely, we may say that the elements of any countable set can be “arranged” in a sequence.

Sometimes, it is convenient to replace J in this definition by the set of all non-negative integers, that is, to start with 0 rather than with 1.

Regards,

Nalin Pithwa

PS: the above exposition will be extremely useful to RMO, INMO and IITJEE advanced maths students and students preparing for Chennai Mathematics Institute Entrance Examination.