Math Real Analysis

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Alannah Leslie Owen
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1 Math 43 - Real Aalysis Solutios to Homework due Septemer 7 I lass, we leared of the famous Cauhy-Shwarz Iequality. Give two -vetors x, y R, the Cauhy-Shwarz iequality relates the dot produt with the orms of the idividual vetors: Writte ompoet-wise with the Cauhy-Shwarz iequality is (x y x y. x = (x, x,..., x ad y = (y, y,, y, ( ( x k y k k= k= x k ( Questio. May times, the Cauhy-Shwarz Iequality a e used to otai some iterestig iequalities y simply hoosig a appropriate vetor x ad y. k= y k. (a Let a,, R. Show that ( Let a,, R. Show that (a 3(a. ( (a a 9. ( Let a, a, a R. Show the Sum of Squares iequality: ( a k a k. k= k= Solutio. (a Cosider the vetors x = (a,, ad y = (,,. Usig the Cauhy-Shwarz Iequality, we get that (a = ( a ( (a = 3(a. ( Cosider the vetors ( a, x =, ad y = ( a,,. Usig the Cauhy-Shwarz Iequality, we get that 9 = ( = ( a ( ( a ( ( a a ( ( = (a a. ( Let x = (/, /,... / ad y = (a, a,... a. Usig the Cauhy-Shwarz iequality, we get that ( ( ( a k = a ( ( k a k = a k. k= k= k= k= k=

2 Questio. It is ofte easier to prove that a give set S is ot ope. To do so, oe eeds to fid a poit x S suh that for o r > 0, B(x; r S. I other words, oe eeds to fid a x S suh that for all r > 0, there exists some y B(x; r suh that y B(x; r ut y S. Show that the followig susets S R are ot ope. (a {a} R ( {(x, 0 R x R} R ( {(x, y R x 0 ad y 0} R Solutio. (a We will show that a is ot a iterior poit. Let ε > 0. We will show that B(a; ε {a}. Notie that a ε/ B(a; ε sie a ε/ a = ε/ < ε. Sie {a} otais oly the elemet a, the a ε/ {a}. Thus, B(a; ε {a} ad therefore a is ot a iterior poit. So, our set is ot ope. ( Let S = {(x, 0 R x R} ad osider (0, 0 S, whih we will show is ot iterior. Let ε > 0; we will show that B((0, 0; ε S. Notie that (0, ε/ B((0, 0; ε sie (0, ε/ (0, 0 = ε/ < ε. However, sie ε/ 0, the (0, 0 S. Thus, B((0, 0; ε S. So, (0, 0 is o-iterior ad S is ot ope. [Note: atually, ay poit i S is o-iterior.] ( Let T = {(x, y R x 0 ad y 0}. We will show that (0, 0 is o-iterior. Let ε > 0; we will show that B((0, 0; ε T. Notie that (0, ε/ T eause (0, ε/ (0, 0 = ε/ < ε. However, sie ε/ < 0, (0, ε/ T. Thus, (0, 0 is ot a iterior poit ad T is ot ope. Questio 3. I what follows, we will demostrate a importat topologial property of Q R. (a Let a Q. Show that a is irratioal for all Z. ( Use (a to show that Q is ot a ope suset of R. Solutio 3. (a Assume, to the otrary, that a / Q. The, sie a Q, the a Q. So, / = a / a Q. Furthermore, sie Q, the = Q. However, it is kow that Q, whih is a otraditio. Thus, a / Q.

3 ( We will show that 0 is ot a iterior poit of Q (i fat, ay a Q will ot e iterior. Let ε > 0. We will show that B(0, ε Q. Sie ε > 0, y the Arhimedea Priiple, there exists a Z suh that < ε. Thus, 0 < < ε. By (a, we kow that / Q. However, / B(0; ε. Thus, B(0; ε Q. So, 0 is ot a iterior poit ad Q is ot ope i R. Give a set S R, a poit x S is alled a isolated poit of S if there exists a ε > 0 suh that B(x; ε S = {x}. I other words, x is isolated i S if there is a small eough ε > 0 suh that B(x; ε itersets S oly at x itself. A set is alled disrete if every poit i S is isolated. Questio 4. Show that the followig sets are or are ot disrete. (a Show that Z is a disrete suset of R ( Show that ever fiite suset of R is a disrete suset of R. } Z is a disrete suset of R ( Show that S = { (d Show that T = { Z } {0} is ot a disrete suset of R. Solutio 4. (a Let Z. We will show that is a isolated poit of Z. Let ε = /. The, B(; / Z = {}. Thus, is isolated ad Z is disrete. ( Let V = {a, a,..., a } e a fiite suset of R. Cosider all the possile distaes etwee every pair of poits i V ad take the miimum. I other words, let ε = mi{ a i a j a i a j V }. Sie there are oly fiitely may pairs of a i s, a miimum exists. Furthermore, sie eah a i a j > 0, ε > 0. We will ow show that a i is a isolated poit i V. Cosider B(a i ; ε. Sie ε a i a j for every a j a i, the the oly elemet of V that is distae less tha ε from a i is a i itself. Therefore, B(a i ; ε V = {a i }. Therefore, a i is a isolated poit. Thus, every poit is isolated ad V is disrete. ( We will show that every / S is a isolated poit. Let ε = = (. Sie <, we kow that ε > 0. We will show that B(/; ε S = {/}. Let m. We will show that /m B(/; ε. First, we take the ase that m >. Thus, m. So, m ad thus m = m = ε. Thus, /m B(/; ε. Next, osider the ase where m <. Thus, m ad so 0 < m. 3

4 Notie that sie <, the ( < ( ad so Furthermore, Thus, sie we kow that Thus, /m B(/; ε. ε = ε = ( < (. ( =. m, ( = m = m. Thus, the oly elemet of S that is also i B(/; ε is / itself. So, / is isolated. Sie every poit is isolated, S is disrete. (d We will show that 0 is ot a isolated poit. Let ε > 0. By the Arhimedea priiple, there exists a Z suh that < ε ad thus 0 < < ε. Thus, B(0; ε T. So, it is ot true that B(0; ε T = {0}. Thus, 0 is ot isolated ad therefore T is ot disrete. Questio 5. Let U, V R e ope sets. Cosider the produt set U V = {(x, y x U, y V } R. Show that U V is ope y showig that eah (x, y U V is a iterior poit. Solutio 5. Let (x, y U V. Thus, x U ad y V. We will show that (x, y is iterior to U V. Sie U is ope, there exists a ε > 0 suh that B(x; ε U. Similarly, sie V is ope, there exists a ε > 0 suh that B(y; ε V. Sie B(x; ε U ad B(y; ε V, the B(x; ε B(y; ε U V. Cosider ε = mi{ε, ε }; thus ε ε ad ε ε. We will show that (x, y is a iterior poit of U V y showig that B((x, y; ε B(x; ε B(y; ε U V. Sie the seod ilusio is already estalished, we fous o the first ilusio. Let (a, B((x, y; ε. Thus, (a, (x, y = (a x ( y < ε. We wish to show that (a, B(x; ε B(y; ε y showig that a x < ε ad y < ε. Assume, to the otrary, that this is ot true. The, a x ε or y ε. If a x ε, the (a, (x, y = (a x ( y (a x = a x ε ε. This otradits the fat that (a, (x, y < ε. 4

5 A similar omputatio gives the same otraditio for the ase whe y ε. Thus, we olude that a x < ε ad y < ε. Thus, So, (x, y is iterior ad U V is ope. B((x, y; ε B(x; ε B(y; ε U V. Questio 6. Cosider the set T = {x R x < }. Geometrially, this set is just a ope disk of radius aout the origi. Cosider S = {x R x = }. Geometrially, S is the irle of radius aout the origi. We will show that every poit i S is a aumulatio poit of T (ad therefore a adheret poit of T. As a hit, you may wat to follow somethig similar to the elow outlie: Let x S. We will show that for all ε > 0, B(x; ε (T {x}. First, ote that T {x} = T sie x T. Thus, we wish to show that B(x; ( ε T. The, osider the ases: ε > or 0 < ε. I the last ase, it might e wise to osider ε x. Solutio 6. Let x S. Thus, x =. We will show that for all ε > 0, B(x; ε (T {x}. First, ote that T {x} = T sie x T. Thus, we wish to show that B(x; ε T. The, osider the ases: ε > or 0 < ε. If ε >, osider the poit (0, 0 T. Notie that (0, 0 x = x = < ε. Thus, (0, 0 B(x; ε T, whih is thus o-empty. If 0 < ε. Cosider the poit ( ε/x. Sie 0 < ε, the 0 < ε/ / <. Thus, 0 < ε/ <. Note that ( ε/x T sie ( ε/x = ε/ x = ( ε/ = ε/ <. Next, we will show that ( ε/x B(x; ε. To see this, ote that ( ε/x x = ε/x = ε/ x = ε/ < ε. Thus, ( ε/x B(x; ε T, whih is thus o-empty. So, every x S is a aumulatio poit (ad thus a adheret poit. 5

3. Greatest Common Divisor - Least Common Multiple

3. Greatest Common Divisor - Least Common Multiple 3 Greatest Commo Divisor - Least Commo Multiple Defiitio 31: The greatest commo divisor of two atural umbers a ad b is the largest atural umber c which divides both a ad b We deote the greatest commo gcd