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7 B.Sc. Part Il © Semester Ill @ MATHEMATICS Paper y, , Prove that union of two disjoint countable sets is countable,, , , , S and, Sol: Let A = (ay, ys yy reerereereee } and ., = b,, 6 }, be two disjoint countable sets,, B= (by, ba, Dyy crrterereeseeeees ‘ a, As there is no common in 4; and b,, we count the elements of, AUB as, a b,, a, by, 35 b,, snatavaaey, The set AUB = {ay By G3» By» G30 O39 ore }, , is countable set., The union of two disjoint countable sets 1s countable., , Theorem 1: If A,, Aj, Ags eee are countable sets then prove, i i i.e. countable union of countable, that U_, A,, is also countable. i.e. ¢, , sets is countable., Proof ; Let us define the sets Aj, Ag, Agy verses as follows, , | | 1, A, = {a!, a2, 43,, , , , n n n, A, =a a3, 43,., , aj is K™ element of set A,, , Let the height of element aj, be / + k., , a} is the only element of height 2, , a} and q?, are the two elements of height 3, a}, a3, a} are, the three elements of height 4 and so on..., , This show, that for positive integer m > 2 there are m— 1 elements, , of height . Therefore we arrange the elements of U?_, An, . . . =, according to their height as, , J 2 | |, As Ws 42, 4, a, a}, , Pictorially we list the elements of U2, A, nal py, , . . as follow by counting, them in the order indicated,, , Scanned with CamScanner

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countab!, , le Sets 75, | | I 1, we va gy:sveceviens, 2 2 2 2., D> G25 (435 WG yrrseerteee, “ rd 3004, Ay 5 AQ, AZ, Az yserseseres, 1,7, 4 4 4, Ay 5 G25 A354 Ag yreerereees, And so on,, , This counting procedure counts all the elements of wey Age, , :. The set is U_, A,, is countable, , Countable, union of countable sets is countable., Theorem 2 : The set of all rational numbers is countable, , Proof; Let E,, be set of all rational numbers with denominator n., , B= {2 hk tS ceccmee |, n n n n n, As f:I1E, given by, 1, filn=-, nel, n, , « fis 1-1 correspondence, Hence E,, and J are equivalent, , sets. But set I of integers is countable set. Therefore set E,, is also, , countable, Let Q be set of all rational numbers, , We can write,, co, Q= Opel E,, , But we know that countable union of countable sets is countable, , Q is countable set., The set of all rational numbers is countable., , Scanned with CamScanner

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B.Sc. Part Il @ Semester ll @ MATHEMATICS pa,, , Theorem 3 : Any subset of a countable set is countable., Proof : Let A be any countable set given by, , A= {ay 42, Azyesssneessf, , Let B be any subset of A as B is subset of A, each element of g 7, , . Let n, be smallest sub, and so on... Then we get, , some a; ccript for which ay, €B, Let),, : ., , be next smallest, , B={ Ans Inds Inge }, , Thus there one-one correspondence f:N—B given by f (hh,, , keN, N and B are equivalent sets., The set B is countable., , Theorem 4 : The closed interval [0, 1] is uncountable., Proof : We shall prove tiis result by method of contradiction,, , Let the set [0, 1] is countable., We can write, [0, 1] = (ay, yy Ags seereeeees }, Such that every number in [0, 1] occurs among x; expanding each, x, in decimals, we have,, , X= 0.4) ay a3, , Xy = 0. ay) Ay) Ay; «., , , , X3 = 0. 43) G9 433 «., , xy, =0-a,, Gnd Angerseereere, , Let b, be any integer from 0 to 8 such that b, # a1, 6, be a"), integer from 0 to 8 such that b, # 4) and so on, Let y = 0. b, b, &; .. Do ee, , wd,, Then for any 7 , the decimal expansion of y differs from decim!, , expansion of x, as b, # a, , , , nn Hence y# x, for every ” and, , 0s y <1. This contradi, 5: <i. icts the assumptio: ume, in [0, 1] occurs among the x _, , i, , Hence our assumption is wrong,, , The closed interval [0, 1] is uncountable, , Scanned with CamScanner

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countable Sets, , Theorem 5 : The set of, Proof: Let R be set of a, , From theorem (4), we h, uncountable,, , Let if possible R be countable. As [0,, subset of countable set is, , (0, 1] must be countable, , Which is a contradiction,, uncountable,, Prove that set of all ratior, , hal numbers in (0, 1] is count, Sol. :, , 77, all real numbers iS uncountable,, I real numbers,, ve proved the closed interval [0, !] is, , I] is subset of R and every, also countable,, , Hence the set R of all real numbers is, , able., , The set of all rational numbers in {0, 1] is subset of set Q of all, , rational numbers. We know that set Q of all rational numbers is, , countable. Also every sub:, , set of countable sets is countable,, , The set of all rational numbers in [0, 1] is countable., , Ex. 6) Prove that Cartesian product Nx N is countable., , Sol.: Here, , N= { 1, 2, 3,4,, , NxN {C1 , 2), (2, D,, , We count the elements of Mx N in the order indicated in the, following array such that sum of element is same., , (1, 1) (1,2) (1, 3),, , , , x, QD; 2:2) ‘Qc, (3, 7 @, 2), (3, 3),., (n, 1), (4,2), (", 3), «, , , , , , This shows that Nx N = Q is countable., , Let p, be the set of Pol, , 1, f(x) = agx” + ax" +, , ynomial Functions J of degree n., , ive i coefficient dy, ys, Where n is a fixed non-negative integer and coeffici or, , Qoy wescesseves, , 2», , a_ are integers. Prove that Pn, om, , is countable., , Scanned with CamScanner

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1S, , Sol. :, , Sol. :, , d.oc. rally Oe, , , , ~+ ay,, , , , ‘ un-l, f(x) = agx" + ax" Foe, , ony A), , , , = (a » Ay, Ayr +, , +, , : ntl ; :, We define a function f+ P,—7 2 5 where z is set of Integers, , 2, ST iPy > Z, , 2=2+z ie. Cartesian product of two countable sets is Countable, z’ is countable, p, is countable, The result is true for n=1,, , Assume that the result is true for n=k |, , +! is one-one correspondence i.e. P, is, , . wk, SUPE 2, countable set, , Now z is countable and z” is also countable., k42_ ok 2 . ‘, z"** =z" x z°~ is Cartesian product of countable sets is, , also countable., , . k+2;, J: Pha > 2°" is one-one correspondence., , The polynomial p,,; is countable., Hence by mathematical induction Polynomial p,, is countable., , Hence the set P, of Polynomial functions of degree v1 is countable:, , Show that Cartesian product of two countable sets is ls, countable,, , Let A and B be two countable sets. Let us take, A= {a, ay, Dy visasereveves }, , Be= {h,, by. Dyce. |, , Scanned with CamScanner