Math 299 Supplement: Real Analysis Nov 2013

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1 Math 299 Supplemet: Real Aalysis Nov 203 Algebra Axioms. I Real Aalysis, we work withi the axiomatic system of real umbers: the set R alog with the additio ad multiplicatio operatios +,, ad the iequality relatio <. We do ot eed to list or describe the elemets of R directly; rather, aythig we wat to kow about R will follow from Axioms 0. We start with the axioms of the additio ad multiplicatio operatios, which iclude the commutative group axioms. For ay a, b, c R, we have:. Closure: a + b R. Closure: ab R 2. Associativity: (a + b) + c = a + (b + c) 2. Associativity: (ab)c = a(bc) 3. Idetity elemet: 0, a + 0 = a 3. Idetity elemet: 0, a = a 4. Iverse: a, b, a + b = Iverse: a 0, b, ab = Deote b = a Deote b = a 5. Commutativity: a + b = b + a 5. Commutativity: ab = ba 6. Distributivity: a(b + c) = ab + ac We defie ew operatios i terms of the basic oes: subtractio a b meas a + ( b); ad divisio a/b or a b meas a b. We also have the axioms of iequality. For ay a, b, c, d R, we have: 7. Trichotomy: Exactly oe of the followig is true: a < b, a = b, a > b. 8. Compatibility of < with +: If a < b ad c < d, the a + c < b + d. 9. Compatibility of < with : If a < b ad 0 < c, the ac < bc. We defie a > b to mea b < a, ad a b to mea a < b or a = b. Completeess. The fial axiom gives a precise meaig to the idea that the real umbers have o holes or gaps, but rather form a cotiuum. First, some defiitios. We say a umber b R is a upper boud for a set S R wheever x b for all x S. Furthermore, l is the least upper boud (or supremum) of S meas l b for every upper boud b of S. We deote this as l = lub(s) or sup(s). Ituitively, the least upper boud is the rightmost edge of S o the real umber lie. examples: (i) Let S = N R. The S is a ubouded set havig o upper bouds, ad hece o least upper boud. (ii) Let S = {0.9, 0.99, 0.999,...}. The every umber b is a upper boud of S, ad l = sup(s) = is the least upper boud. It makes o differece whether is i the set or ot: S {} has the same upper bouds as S, ad sup(s {}) =. (iii) Let S = {x R x 2 < 2}. Some upper bouds for S are upper approximatios to 2, like a =.5,.42,.45,.... The least upper boud is sup(s) = 2 itself, which is a way of producig this irratioal umber without assumig it exists. 0. Axiom of Completeess: Ay set S R which has a upper boud, also has a least upper boud i the reals: sup(s) R. Note that this axiom fails for the ratioal umbers Q, ad this is their mai differece from the real umbers. For example, the set i Example (iii) above has upper bouds i R ad i Q, but it has a least upper boud oly i R: i the ratioals Q, there is a hole where sup(s) = 2 would be, sice 2 is ot ratioal.

2 Algebra Propositios. All the usual facts of algebra (icludig iequalities) ca be deduced from Axioms 9. Throughout, we implicitly use Axioms 2, 2 to write expressios like a + b + c istead of (a + b) + c, ad abc istead of (ab)c. propositio (Mulitplicatio by zero): 0a = 0. Proof. By Axioms 3 ad 6, we have: 0a = (0 + 0)a = 0a + 0a. Addig 0a to the left ad right sides of this equality, we get: 0a 0a = 0a + 0a 0a, which we ca simplify by Axioms 4 ad 3 to 0 = 0a as desired. propositio 2 (Multiplicatio with sigs): (i) ( a) = a, (ii) ( a)b = (ab), (iii) ( a)( b) = ab. Proof. (i) By Axiom 4, 0 = ( a) ( a). Addig a to both sides gives: a = a + ( a) ( a) = 0 ( a) = ( a). (ii) We have: ( a)b + ab = ( a + a)b = 0b = 0 by Prop.. Switchig the sides: 0 = ( a)b+ab, ad addig (ab) to both sides: (ab) = ( a)b+ab (ab) = ( a)b+0 = ( a)b. (iii) Applyig (ii) twice, we have: ( a)( b) = (a( b)) = (( b)a) = ( (ba)) = ( (ab)). Thus ( a)( b) = ( (ab)) = ab by (i). propositio 3. If a < b, the b < a. Proof. Let a < b. Usig Axiom 8, we add a b to both sides, gettig: a + ( a b) < b + ( a b). Simplifyig the left ad right sides by Axioms 2, 3, 4, 5 gives b < a. propositio 4. 0 <. Proof. Surprisigly, this is ot immediate. Suppose for a cotradictio that 0. By Axiom 7, this meas 0, but Axiom 3 says 0. Thus 0 >, ad by Prop. 3, 0 <, so by Axiom 9, 0( ) < ( )( ). By Prop., 0( ) = 0, ad by Prop. 2, ( )( ) = ()() =, which meas 0 <. But we already saw 0 >, so this cotradicts the uiqueess part of Axiom 7. This cotradictio shows 0 <. propositio 5 (Trasitivity of <): If a < b ad b < c, the a < c. Proof. Suppose a < b < c. By Axiom 8, we ca subtract a from the first iequality to get 0 < b a, ad subtract b from the secod iequality to get 0 < c b. Addig these two iequalities: 0 < (b a) + (c b) = c a. Addig a to ths iequality gives a < (c a) + a = c. defiitio: x = x if x 0, ad x = x if x < 0. propositio 6: For a, b R: (i) ab = a b. (ii) a + b a + b ; (iii) (Triagle iequality) For x, y, z R, x z x y + y z. Proof. (i) If a, b 0, the ab > 0 by Axiom 9, ad by defiitio ab = ab = a b. If b < 0 a, the a, b 0 ad (ab) = a( b) > 0, so ab < 0; thus ab = ab = a( b) = a b by Prop. 2. Similarly for a < 0 b. If a, b < 0, the ab = ( a)( b) > 0 ad ab = ( a)( b) = a b. Axiom 7 guaratees that we have cosidered all possible cases. (ii) We have x = max{x, x}, so a + b = max{a + b, a b}, whereas we easily see: a + b = max{a + b, a b, a + b, a b}. The larger set clearly has a larger maximum, so a + b a + b. (iii) This follows from (ii) takig a = x y, b = y z, so that a + b = x z. 2

3 Limits. Cosider a ifiite sequece (a ) = = (a, a 2, a 3,...) with a i R. defiitio: We say (a ) coverges to the umber L, writte lim a = L, meaig that for ay error boud ɛ > 0, there exists a threshold N N (depedig o ɛ) such that N forces a ito the error iterval L ɛ < a < L + ɛ. I symbols: ɛ > 0, N N, > N a L ɛ. prop 7: If lim = L, lim = M, the (i) lim +b = L+M; (ii) lim b = LM. Proof. By the defiitio of the limits i the hypothesis, for ay ɛ > 0, there is N such that N a L < ɛ ; ad for ay ɛ 2 > 0, there is N 2 such that N 2 b M < ɛ 2. (i) Now let ɛ > 0 ad take ɛ = ɛ 2 = 2 ɛ. The takig N = max(n, N 2 ) above, for ay N we have: (a + b ) (L + M) = (a L) + (b M) a L + b M by Prop 6, Triagle Iequality < ɛ + ɛ 2 = ɛ sice N N, N 2. (ii) Now let ɛ > 0, ad take ɛ = ɛ 2 L + ad ɛ 2 = ɛ 2 M +. Also take ɛ = 2, so that ), for ay N a L < 2, ad a < L + 2. The takig N = max(n, N 2, N N we have: a b LM = a b a M + a M LM = a (b M) + (a L)M a b M + a L M by Prop 6(ii) < a ɛ 2 + ɛ M sice N N, N 2 < ( L + 2 ) ɛ 2 L + + ɛ 2 M + M sice N < ɛ 2 + ɛ 2 = ɛ sice x+ 2 2x+ = 2, x 2x+ < 2 if x 0. Ifiite Approximatios. We use limits to hadle real umbers which we caot defie directly, but oly through approximatios, such as a derivative, a ifiite series, or a ifiite decimal. propositio 8: The derivative of the fuctio f(x) = x at x = 2, amely the taget slope of y = x at (x, y) = (2, 2 ), is: f (2) = dy dx x=2 = 4. Rough Draft of Proof. The taget lie is very difficult to costruct, because it is defied as touchig the curve at oly the oe poit (2, 2 ), ad oe poit does ot defie a lie. However, it is easy to costruct secat lies, which cut the curve at two earby poits, (2, 2 ) ad (2+, ). The great idea of differetial calculus is that the taget slope is 2+ the limit of secat slopes: f f(2+ (2) = lim ) f(2) = lim We wat to show this limit coverges to 4, ad we work backwards from the desired coclusio, that the distace betwee the sequece ad the limit is withi ay desired 3

4 error tolerace ɛ > 0: 2+ 2 ( 4) < ɛ Simplifyig the lefthad side, this becomes: = 8+4 < ɛ. Now, 8+4 < 8, so it is eough to show 8 < ɛ. Solvig this last iequality for gives: 8ɛ. Thus, to guaratee the desired coclusio, we oly eed that N, where N is ay iteger greater tha 8ɛ. (If ɛ is a very small error tolerace, this N is very large, but we kow there is a iteger larger tha ay give real umber.) Fial Proof. We wat to compute the derivative of f(x) at x = 2, ad show: f f(2+ (2) = lim ) f(2) = lim 2+ 2 = 4. Give ɛ > 0, take a iteger N > 8ɛ. The for N, we compute: 2+ 2 ( 4) = By defiitio, this proves the covergece of the limit = 8+4 < 8 8N < 8 ( ) = ɛ. theorem 9: Suppose the sequece (a ) is icreasig, with a upper boud b: that is, a a 2 a 3 b. The (a ) coverges to L = sup{a }. Proof. The least upper boud L = sup{a } exists by Axiom 0. The L is a upper boud, so a L < L + ɛ for all ad all ɛ > 0. Sice L is the least upper boud, we kow that L ɛ is ot a upper boud of (a ) for ay ɛ > 0. This ca oly be if L ɛ < a N for some N, ad sice a N a for all N, we have: L ɛ < a < L + ɛ, amely L a < ɛ. I summary, for ay ɛ > 0, there is some N such that N implies L a < ɛ. This is precisely the defiitio of lim a = L. Propositio 0 (Sum of geometric series): For a fixed x R, defie the geometric series (s ) by: s = + x + x x. If x <, the (s ) coverges to L = x. Proof. Recall that we proved (by iductio i HW due 0/23) the formula: s = x+. x I the case that 0 x <, we clearly have s s 2 x, so that (s ) is a icreasig bouded sequece, ad it coverges by the Theorem. We leave the exact value of the limit, ad the case < x < 0, as a exercise. Theorem (Decimal expasios): Let (d ) = (d, d 2,...) be a sequece of digits, with d {0,,..., 9}. Defie a icreasig sequece (a ) by: a = d 0 + d d d 0, 8ɛ 4

5 amely the -digit decimal 0.d d 2... d. The (a ) coverges to a uique real umber, the ifiite decimal 0.d d 2 d Proof. We clearly have d 0 < 0 0, so that a < b for the geometric series: b = ( ) 2 ( ). Now, by the previous propositio, the geometric series (b ) has the upper boud b = x for x = 0, ad we have a < b < b, so that (a ) is a icreasig bouded sequece, ad coverges to some real umber. Propositio : There is positive real umber l with l 2 = 2; that is, 2 R. Proof. Let S = {x R x 2 < 2}. Now, if x S ad < x, the x < x 2 < 2, so clearly b = 2 is a upper boud of S. By the Completeess Axiom, S has a least upper boud l = sup(s) 2. We will show that l 2 = 2 by cotradictio. First, suppose l 2 < 2. The we may choose ɛ with 0 < ɛ < 5 (2 l2 ), ad also ɛ <. We compute: (l + ɛ) 2 = l 2 + (2l+ɛ)ɛ < 2 + (2(2)+)ɛ < 2 + (5)( 5 )(2 l2 ) = 2. By defiitio, this meas l + ɛ S with ɛ > 0; but l is a upper boud of S, so l + ɛ l with ɛ > 0. This cotradictio shows l 2 < 2 is impossible. Next, suppose l 2 > 2. The we may choose ɛ with 0 < ɛ < 4 (l2 2), ad also ɛ <. We compute: (l ɛ) 2 = l 2 2lɛ + ɛ 2 > 2 2(2)ɛ + 0 > 2 (4)( 4 )(l2 2) = 2. Thus, for ay x S, we have x 2 < 2 < (l ɛ) 2, with l ɛ > l 0. Now we apply the exercise that if x 2 < y 2 with y > 0, the x < y: lettig y = l ɛ, this meas x < l ɛ. Thus l ɛ is a upper boud of S, but l is the least upper boud, so l l ɛ with ɛ > 0. This cotradictio shows l 2 > 2 is impossible. 5

6 Problems Prove the followig statemets usig the above Axioms ad Propositios (sayig which results you use).. If 0 < a < b, the b < a. 2. For y > 0, if x 2 < y 2, the x < y. (+ 3a. Usig the formal defiito, prove: lim ) 2 = 2. Hit: For the rough draft, work backward from the coclusio a L < ɛ. b. What does the above limit mea i calculus? Hit: It cocers the behavior of the fuctio f(x) = x 2 ear x =. c. Prove that lim = 2/3. Hit: Avoid false iequalities like 2? < or? <. You may use that a c > a for a > 0 ad large (how large?). 4a. There is o largest elemet of R. Hit: Cotradictio. b. There is o smallest elemet of the positive reals R >0. c. If x < ɛ for all ɛ > 0, the x = 0. Hit: Cotradictio. 5. The limit lim a coverges to at most oe value. That is, if (a ) coverges to L ad also to L, the L = L. Hit: This is ot just a matter of writig L = lim a = L, sice the whole poit is to prove the limit i the middle is a well-defied, uambiguous value. Rather, write out the defiitio of lim a = L ad lim a = L ad use the Triagle Iequality above to prove that L L < ɛ for every ɛ > 0. Why does this give the coclusio? 6. If (a ) is a coverget sequece, the the sequece is bouded. That is, if lim a = L, the there is some B with a B for all. 7. Cosider the sequece a = 2. a. Prove that lim a =. This meas that for every boud B R, there is a threshold N such that N implies that a > B. That is, a goes above ay boud for large eough values of N. b. Prove that (a ) is diverget; that is, (a ) does ot coverge to ay value L R, meaig lim a = L is false. Hit: This does ot follow immediately from part (a): you must show the defiitio of lim a = implies the egatio of the defiitio of lim a = L, for ay L. This meas that there is a error boud, say ɛ =, such that o matter how large N is, the sequece a is ot forced withi the error boud a L < ɛ.

7 8. For each statemet (a), (b) below, either prove it is true, or fid a couterexample ad prove its properties. Let (a ) ad (b ) be ay two real sequeces, ad defie their sum (c ) by c = a + b. a. If (a ) ad (b ) are coverget, the (c ) is coverget. Hit: We discussed this situatio i class; see also Beck Prop b. If (c ) is coverget, the (a ) ad (b ) are coverget. Hit: Is there ay way (a ) ad (b ) could fluctuate aroud, eve though (c ) is covergig to some L? 9. Use the geometric series formula lim (+x+x 2 + +x ) = x to fid a fractioal form a b for the repeatig decimal = Hit: Write the decimal as 0.7 plus a multiple of a geometric series with x = For bouded subsets S, T R, ad U = {x + y x S, y T }, we have sup(u) = sup(s) + sup(t ).. Defie a lower boud for a set S R to mea some b R with b x for all x S. The greatest lower boud or ifimum of S, deoted glb(s) or if(s), meas a lower boud g with g b for all lower bouds b. Deote S = { x x S}. a. The umber b is a lower boud of S if ad oly if b is a upper boud of S. b. if(s) = sup( S). c. Ay set S havig a lower boud b has a greatest lower boud g = if(s) R.