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}.

Answer 1:

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.

Answer 3a: please refer a previous blog.

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.

Answer 5:

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.

Answer 6:

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.

Answer 7:

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.

Symmetric difference is associative

We want to show that : A \triangle (B \triangle C) = (A \triangle B) \triangle C

Part I: TPT: LHS  \subset RHS

Proof of Part I: let x \in LHS

Then, x \in A \triangle (B \triangle C). By definition of symmetric difference,

x \in \{ A - (B \triangle C)\} \bigcup \{(B \triangle C) -A \}

Hence, x \in A, x \notin (B \triangle C), OR x \in B \triangle C, x \in A

That is, x \notin A \bigcap (B \triangle C).

Hence, x \notin A, and x \notin B \triangle C

Hence, x \notin A and x \in B \bigcap C.

Hence, x \in (B \bigcap C)-A.

Hence, x \in B, x \in C, but x \notin A.

Therefore, x \in B, but x \notin A \bigcap C. —- Call this I.

Consider y \in (A \triangle B) \triangle C.

Therefore, y \notin (A \triangle B) \bigcap C.

Therefore, y \notin C, and y \notin A \triangle B

Therefore, y \notin C, and y \in A \bigcap B.

Therefore, y \in (A \bigcap B)-C.

Hence, y \in A, y \in B, but y \notin C.

That is, y \in B, y \notin A \bigcap C. Call this II.

From I and II, LHS \subset RHS.

Part II: TPT: RHS \subset LHS.

Quite simply, reversing the above steps we prove part II.

QED.

Cheers,

Nalin Pithwa

Compilation of elementary results related to permutations and combinations: Pre RMO, RMO, IITJEE math

1. Disjunctive or Sum Rule:

If an event can occur in m ways and another event can occur in n ways and if these two events cannot occur simultaneously, then one of the two events can occur in m+n ways. More generally, if E_{i} (i=1,2,\ldots,k) are k events such that no two of them can occur at the same time, and if E_{i} can occur in n_{i} ways, then one of the k events can occur in n_{1}+n_{2}+\ldots+n_{k} ways.

2. Sequential or Product Rule:

If an event can occur in m ways and a second event can occur in n ways, and if the number of ways the second event occurs does not depend upon how the first event occurs, then the two events can occur simultaneously in mn ways. More generally, $if E_{1} can occur in n_{1}, E_{2} can occur in n_{2} ways (no matter how E_{1} occurs), E_{3} can occur in n_{3} ways (no matter how E_{1} and E_{2} occur), \ldots, E_{k} can occur in n_{k} ways (no matter how the previous k-1 events occur), then the k events can occur simultaneously in n_{1}n_{2}n_{3}\ldots n_{k} ways.

)3. Definitions and some basic relations:

Suppose X is a collection of n distinct objects and r is a nonnegative integer less than or equal to n. An r-permutation of X is a selection of r out of the n objects but the selections are ordered. 

An n-permutation of X is called a simply a permutation of X.

The number of r-permutations of a collection of n distinct objects is denoted by P(n,r); this number is evaluated as follows: A member of X can be chosen to occupy the first of the r positions in n ways. After that, an object from the remaining collections of (n-1) objects can be chosen to occupy the second position in (n-1) ways. Notice that the number of ways of placing the second object does not depend upon how the first object was placed or chosen. Thus, by the product rule, the first two positions can be filled in n(n-1) ways,….and all r positions can be filled in

P(n,r) = n(n-1)\ldots (n-r+1) = \frac{n!}{(n-r)}! ways.

In particular, P(n,n) = n!

Note: An unordered selection of r out of the n elements of X is called an r-combination of X. In other words, any subset of X with r elements is an r-combination of X. The number of r-combinations or r-subsets of a set of n distinct objects is denoted by n \choose r (read as ” n ‘choose’ r). For each r-subset of X there is a unique complementary (n-r)-subset, whence the important relation {n \choose r} = n \choose {n-r}.

To evaluate n \choose r, note that an r-permutation of an n-set X is necessarily a permutation of some r-subset of X. Moreover, distinct r-subsets generate r-permutations each. Hence, by the sum rule:

P(n,r)=P(r,r)+P(r,r)+\ldots + P(r,r)

The number of terms on the right is the number of r-subsets of X. That is, n \choose r. Thus, P(n,r)=P(r,r) \times {n \choose r}=r! \times {n \choose r}.

The following is our summary:

  1. P(n,r) = \frac{n!}{(n-r)!}
  2. {n \choose r}=\frac{P(n,r)}{r!}=\frac{n!}{r! (n-r)!}=n \choose {n-r}

4. The Pigeonhole Principle: Basic Version:

If n pigeonholes (or mailboxes) shelter n+1 or more pigeons (or letters), at least 1 pigeonhole (or mailbox) shelters at least 2 pigeons (or letters).

5. The number of ways in which m+n things can be divided into two groups containing m and n equal things respectively is given by : \frac{(m+n)!}{m!n!}

Note: If m=n, the groups are equal (and hence, indistinguishable), and in this case the number of different ways of subdivision is \frac{(2m)!}{2!m!m!}

6. The number of ways in which m+n+p things can be divided into three groups containing m, n, p things severally is given by: \frac{(m+n+p)!}{m!n!p!}

Note: If we put m=n=p, we obtain \frac{(3m)!}{m!m!m !} but this formula regards as different all the possible orders in which the three groups can occur in any one mode of subdivision. And, since there are 3! such orders corresponding to each mode of subdivision, the number of different ways in which subdivision into three equal groups can be made in \frac{(3m)!}{m!m!m!3!} ways.

7. The number of ways in which n things can be arranged amongst themselves, taking them all at a time, when p of the things are exactly alike of one kind, q of them are exactly alike of a another kind, r of them are exactly alike of a third kind, and the rest are all different is as follows: \frac{n!}{p!q!r!}

8. The number of permutations of n things r at a time, when such things may be repeated once, twice, thrice…up to r times in any arrangement is given by: n^{r}. Cute quiz: In how many ways, can 5 prizes be given away to 4 boys, when each boy is eligible for all the prizes? (Compare your answers with your friends’ answers :-))

9. The total number of ways in which it is possible to make a selection by taking some or all of n things is given by : 2^{n}-1

10. The total number of ways in which it is possible to make a selection by taking some or all out of p+q+r+\ldots things, whereof p are alike of one kind, q alike of a second kind, r alike of a third kind, and so on is given by : (p+1)(q+1)(r+1)\ldots-1.

Regards,

Nalin Pithwa.

Solutions to “next number in sequence”: preRMO, pRMO and RMO

What is the next number in sequence?

A) 15, 20, 20, 6, 6, 19, 19, 5, 14, 20, 5, ?

Solution to A:

Ans is 20. The sequence is the position in the letter of the alphabet of the first letter in the numbers 1 to 12, when given in full. e.g. ONE: O=15.

B) 1, 8, 11, 18, 80, ?

Ans is 81. The sequence comprises whole numbers beginning with a vowel.

C) 1, 2, 4, 14, 21, 22, 24, 31, ?

Ans is 32, The sequence comprises whole numbers containing the letter O.

D) 4, 1, 3, 1, 2, 4, 3, ?

Ans. is 2. The sequence is as follows: there is one number between the two I’s, two numbers between the two 2’s, three numbers between the two 3’s and four numbers between the two 4’s.

E) 1, 2, 4, 7, 28, 33, 198, ?

Ans is 205. 1 + 1 \times 2 + 3 \times 4 + 5 \times 6 + 7

F) 17, 8, 16, 23, 28, 38, 49, 62, ?

Answer is 70. Sum of digits in all previous numbers in the sequence.

G) 27, 216, 279, 300, ?

Ans is 307. Difference divided by 3 and added to the last number.

H) 9,7,17,79,545, ?

Answer is 4895. Each number is multiplied by its rank in the sequence, and the next number is subtracted.

9 \times 1 - 2 = 7 \times 3 -4 = 17 \times 5 - 6 = 79 \times 7 - 8 = 545 \times 9 - 10 = 4895

I) 2,3,10,12,13, 20,?

Answer is 21. They all begin with the letter T.

J) 34, 58, 56, 60, 42, ?

Answer is 52. The numbers are the totals of the letters in the words ONE, TWO, THREE, FOUR, FIVE, SIX when A=1, B=2, C=3, etc.

Regards,

Nalin Pithwa.

Miscellaneous questions: part II: tutorial practice for preRMO and RMO

Problem 1:

Let a_{1}, a_{2}, \ldots, a_{10} be ten real numbers such that each is greater than 1 and less than 55. Prove that there are three among the given numbers which form the lengths of the sides of a triangle.

Problem 2:

In a collection of 1234 persons, any two persons are mutual friends or enemies. Each person has at most 3 enemies. Prove that it is possible to divide this collection into two parts such that each person has at most 1 enemy in his subcollection.

Problem 3:

A barrel contains 2n balls numbered 1 to 2n. Choose three balls at random, one after the other, and with the balls replaced after each draw. What is the probability that the three element sequence obtained has the properties that the smallest element is odd and that only the smallest element, if any, is repeated?

That’s all, folks !!

You will need to churn a lot…!! In other words, learn to brood now…learn to think for a long time on a single hard problem …

Regards,
Nalin Pithwa

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

A fifth degree equation in two variables: a clever solution

Question:

Verify the identity: (2xy+(x^{2}-2y^{2}))^{5}+(2xy-(x^{2}-2y^{2}))^{5}=(2xy+(x^{2}+2y^{2})i)^{5}+(2xy-(x^{2}+2y^{2})i)^{5}

let us observe first that each of the fifth degree expression is just a quadratic in two variables x and y. Let us say the above identity to be verified is:

P_{1}+P_{2}=P_{3}+P_{4}

Method I:

Use binomial expansion. It is a very longish tedious method.

Method II:

Factorize each of the quadratic expressions P_{1}, P_{2}, P_{3}, P_{4} using quadratic formula method (what is known in India as Sridhar Acharya’s method):

Now fill in the above details.

You will conclude very happily that :

The above identity is transformed to :

P_{1}=(x+y+\sqrt{3}y)^{5}(x+y-\sqrt{3}y)^{5}

P_{2}=(-1)^{5}(x-y-\sqrt{3}y)^{5}(x-y+\sqrt{3}y)^{5}

P_{3}=(i^{2}(x-y-\sqrt{3}y)(x-y+\sqrt{3}y))^{5}

P_{4}=((-i^{2})(x+y+\sqrt{3}y)(x-y-\sqrt{3}y))^{5}

You will find that P_{1}=P_{4} and P_{2}=P_{4}

Hence, it is verified that the given identity P_{1}+P_{2}=P_{3}+P_{4}. QED.

Regards,
Nalin Pithwa.

Set Theory, Relations, Functions: preliminaries: part 10: more tutorial problems for practice

Problem 1:

Prove that a function f is 1-1 iff f^{-1}(f(A))=A for all A \subset X. Given that f: X \longrightarrow Y.

Problem 2:

Prove that a function if is onto iff f(f^{-1}(C))=C for all C \subset Y. Given that f: X \longrightarrow Y.

Problem 3:

(a) How many functions are there from a non-empty set S into \phi\?

(b) How many functions are there from \phi into an arbitrary set S?

(c) Show that the notation \{ X_{i} \}_{i \in I} implicitly involves the notion of a function.

Problem 4:

Let f: X \longrightarrow Y be a function, let A \subset X, B \subset X, C \subset Y and D \subset Y. Prove that

i) f(A \bigcap B) \subset f(A) \bigcap f(B)

ii) f^{-1}(f(A)) \supset A

iii) f(f^{-1}(C)) \subset C

Problem 5:

Let I be a non-empty set and for each i \in I, let X_{i} be a set. Prove that

(a) for any set B, we have B \bigcap \bigcup_{i \in I}X_{i}=\bigcup_{i \in I}(B \bigcap X_{i})

(b) if each X_{i} is a subset of a given set S, then (\bigcup_{i \in I}X_{i})^{'}=\bigcap_{i \in I}(X_{i})^{'} where the prime indicates complement.

Problem 6:

Let A, B, C be subsets of a set S. Prove the following statements:

(i) A- (B-C)=(A-B)\bigcup(A \bigcap B \bigcap C)

(ii) (A-B) \times C=(A \times C)-(B \times C)

🙂 🙂 🙂

Nalin Pithwa