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 = 0. 4. 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.
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
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 2+ 2. 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
error tolerace ɛ > 0: 2+ 2 ( 4) < ɛ Simplifyig the lefthad side, this becomes: 4+2 + 4 = 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. 4+2 + 4 = 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 2 + + 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 2 0 2 + d 2 0 3 + + d 0, 8ɛ 4
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 3.... Proof. We clearly have d 0 < 0 0, so that a < b for the geometric series: b = + 0 + ( ) 2 ( 0 + + 0). 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
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 2 2 2 3 2 + = 2/3. Hit: Avoid false iequalities like 2? < or? <. You may use that a c > a 2 2 2 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.
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. 0.23. 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 0.7343434 = 0.734. Hit: Write the decimal as 0.7 plus a multiple of a geometric series with x = 00. 0. 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.