MA541 : Real Analysis. Tutorial and Practice Problems - 1 Hints and Solutions

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1 MA54 : Real Aalysis Tutorial ad Practice Problems - Hits ad Solutios. Suppose that S is a oempty subset of real umbers that is bouded (i.e. bouded above as well as below). Prove that if S sup S. What ca you say if if S = sup S? Solutio: Secod Part: if S = sup S iff S is a sigleto.. Is the set {x R x < x} bouded above (below)? If so, fid its supremum (ifimum). Solutio: Hit. {x R x < x} = (0, ). 3. Suppose is a positive iteger ad a R, a 0. Show that there is a uique b 0 i R such that b = a. Solutio: Hit. Cosider the set S = {x R : x 0, x a}. Show that S is oempty ad bouded above. The supremum b of S satisfies b = a (show that b < a ad b > a are ot possible). Fially show that that if c 0 ad c = a, the c = b. 4. Show that betwee ay two distict real umbers there is a ratioal umber as well as a irratioal umber. Solutio: (Refer to ay text book.) 5. Prove Well-Orderig Priciple of atural umbers: Every oempty subset of positive itegers has a miimum. Solutio: Let S be a oempty subset of positive itegers. The S is bouded below by 0, ad so has a ifimum, say c. Claim: c is the least elemet of S. Sice c + is ot a lower boud of S, there is m S such that m < c +. Now, for ay positive iteger < m we have m < c ad therefore, / S. Cosequetly, m is the least elemet of S. (Note that this also meas that c = m.) of 5 BK Sarma, 05 Aug 04, 8:4

2 6. Suppose that a real umber a has the property that a for all N. Show that a 0. Solutio: Hit. Follows from Archimedea Priciple. 7. Let S ad T be oempty sets of real umbers such that every real umber is i S or T ad if s S ad t T, the s < t. Prove that there is a uique real umber β such that every real umber less tha β is i S ad every real umber greater tha β is i T. (A decompositio of the reals ito two sets with these properties is a Dedekid cut. This is kow as Dedekid s theorem.) Solutio: Doe i the Tutorial class. 8. Suppose that a sequece (x ) coverges to x ad x 0 for all. Show that x 0. Solutio: Hit. Show that if x < 0, the x < 0 for ifiitely may values of. 9. A subset S of R is closed if every coverget sequece i S has its limit i S. Which of the followig sets are closed: [a, b), [a, b], Q, R \ Q? Justify your aswer. Solutio: [a, b] is closed: If x x ad a x b for each, the a x b. [a, b) is ot closed: If x = b b a, the x [a, b) ad x b / [a, b). Q ad R \ Q are ot closed: Use desity property. 0. Let (x ), (y ), (z ) be sequeces such that x y z for all N. If both (x ) ad (z ) coverges to the same limit c, the show that (y ) coverges to c. [This result is kow as the Sadwich Theorem.] Solutio: Refer to ay text book. ( ) x +. Let (x ) be such that x 0 ad x coverges to L. Show that (x ) (i) coverges to 0, if L <, ad (ii) diverges, if L >. Show that (x ) ca coverge or diverge, if L =. Solutio: Doe i the Tutorial class. of 5 BK Sarma, 05 Aug 04, 8:4

3 . Let a, b R, ad let (x ) be defied as follows: x = a, x = b, ad x = (x + x ) for 3. Prove that (x ) is coverget ad fid its limit. Solutio: For 3, we have x = (x +x ) ad therefore x x = (x +x ), i.e., x x = x + x. sequece ad so coverget. Moreover, Therefore, by the Result give below, (x ) is a Cauchy x x = (x x ) + (x x ) + + (x x ) [ ( = ) ( ] (b a) ) [ = ( ) ] (b a). 3 Takig limits we get lim x a = 3 (b a), i.e., lim x = a + 3 (b a) 3. Let (x ) be a sequece ad y = (x + x + + x ). Show that if (x ) is coverget, the (y ) is coverget. Is the coverse true? Solutio: Suppose x a. Let ϵ > 0 be give. There exists N N such that x a < ϵ/ for all N. Now, for N y a = + + x ) a [ x a + x a + + x a ] = = K N x i a + i= + ( N) ϵ. i=n+ x i a where K = N i= x i a. Choose positive iteger M N such that K/M < ϵ/. The for M we have y a < ϵ. The coverse is ot true. Take for example, x = ( ). 4. Prove or disprove: If (x ) coverges to x, ad (x ) has a subsequece (x k ) such that ( ) k x k 0 for all k, the x = 0. Solutio: Suppose, if possible, x > 0. There exists m N such that x ( x, 3x ) for all m (Note: we took ϵ = x/.) I particular, x > 0 for all m. Now, m, m+ m ad therefore x m, x m+ are positive. This implies that ( ) m x m 0, ( ) m+ x m+ 0 simultaeously caot hold, sice oe of them is egative. Thus, x 0. Similarly, it ca be show that x < 0 is ot possible. 3 of 5 BK Sarma, 05 Aug 04, 8:4

4 5. Examie whether the sequeces (x ) coverge, where x is give below. Fid also the limits, if exist. Here, a, b are real umbers. (i) si3 (ii) a a (iii)! ( (iv) + (vii) (viii) ) (v) (a + b ), where a, b > 0 (vi) ( ) 5 8 (3 ) (ix) ( [a] + [a] + + [a] ) Solutio: (i) si3. Now use Sadwich Theorem. (ii) x = a. If a = 0 or, the clearly (x ) coverges. If a =, the (x ) = (( ) ) diverges. Case: a >. Let a = + h. The a = ( + h) ( ) = + h + h + + h > h. Thus, (x ) is ot bouded, ad so ot coverget. Case: 0 < a <. The /a >. Put /a = + h, i.e., a = +h. Thus, 0 < x = a = ( + h) = + h + + h + h. Use Sadwich Theorem to coclude that x 0. Case: < a < 0. The a 0 ad therefore a 0. (iii) x + = a + 0 <. Thus x 0. x (iv) Show that < x = ( + ) < 3 ad (x ) is mootoically icreasig. Thus the sequece coverges. (lim x is deoted by e.) (v) Assume a b. The b x = (a + b ) (b + b ) = b. Use Sadwich Theorem to coclude x b. Thus, x max{a, b}. (vi) Sice 0 < x = <, the sequece (x ) is bouded. Moreover, x + x = > 0, i.e., (x ) is mootoically icreasig. Hece (x ) coverges. I fact, x = [ + / + + / / ] (vii) x =. We have x > for. Put h = x. The = ( + h ) = + h + ( ) Therefore, for we have 0 h i.e., 0 h get h 0, i.e. x. (viii) Note that x + x = + 3+ /3 <. Hece x 0. 0 dx = l. + x = + h, i.e., h ( ) + > h.. Usig Sadwich Theorem, we (ix) x = ( [a] + [a] + + [a] ). Note that for ay real b, b < [b] b. Thus, ( ) (a ) + (a ) + + (a ) < x ( ) a + a + + a. Simplify this to get + a + < x a +. Use Sadwich Theorem ad get x a. 4 of 5 BK Sarma, 05 Aug 04, 8:4

5 Result. Suppose (x ) be a real sequece such that x + x c x x for all. (0.) If c <, the (x ) is a Cauchy sequece ad therefore coverget. Proof. If x = x, the x = x for all, ad the result follows. Suppose x x. We have for all Moreover, x + x c x x c x x. x +k x x +k x +k + + x x (c k + c + + c + ) x x = ck c x x c c x x. Let ϵ > 0 be give. Sice 0 c <, we have c 0. Choose N N such that c c( c) < ϵ for all x x N. The for ay N, k we have x +k x < ϵ. Thus (x ) is a Cauchy sequece. 5 of 5 BK Sarma, 05 Aug 04, 8:4