is infinite. The converse is proved similarly, and the last statement of the theorem is clear too.

Transcription

1 12. No-stdrd lysis October 2, 2011 I this sectio we give brief itroductio to o-stdrd lysis. This is firly well-developed field of mthemtics bsed o model theory. It dels ot just with the rels, fuctios o them, cotiuity, etc., but lso with proper elemetry extesio of the rels, ifiitesimls, ifiite itegers, etc. The deeper theory of o-stdrd lysis requires some subtle set-theoreticl cosidertios, but for elemetry cosidertios such s those i this sectio, the pprtus of the precedig sectio is sufficiet. For thorough developmet of this elemetry portio of o-stdrd lysis, two books of H. J. Keisler re recommeded: Foudtios of ifiitesiml clculus (211 pp.) d Elemetry clculus (880pp.). The ltter is complete clculus course bsed o o-stdrd lysis. Ifiitesimls. The bsic theory of ifiitesimls c be formulted with respect to y ordered field R which is proper extesio of the field of rel umbers. So for while tht is ll we ssume bout R. We itroduce some of the bsic otios of o-stdrd lysis: A elemet ε of R is ifiitesiml provided tht ε < r for every positive rel umber r. A elemet x R is fiite if x < r for some rel umber r. Fi(R ) is the collectio of ll fiite elemets of R. Of course, if elemet x R is ot fiite, we sy tht it is ifiite. Elemets x, y R re ifiitely close, i symbols x y, if x y is ifiitesiml. Propositio Fi(R ) is subrig of R. Propositio The set I of ll ifiitesimls is idel i Fi(R ). Proof. It is esy eough to check tht I is subset, eve subrig, of Fi(R ). To check the idel property, suppose tht x Fi(R ) d ε is ifiitesiml. The there is rel umber r such tht x < r. Now if s is y positive rel umber, the s r is lso positive rel umber, hece ε < s r d so εx = ε x < s r r = s. Propositio is equivlece reltio o R. Theorem Every fiite x R is ifiitely close to uique rel umber. Proof. Let X = {s R : s < x}. Now X is oempty, sice we c choose rel umber t such tht x < t, so t < x < t, hece t X. With the sme choice of t we see tht X is bouded bove. So, let r be the lest upper boud of X. We clim tht r is s desired. Let u be y positive rel umber. The r u is o loger upper boud for X, so there is s X such tht r u < s, hece r u < x d r x < u. Also, r + u 2 is ot i X, so x r + u < r +u, hece u < r x. So we hve show tht u < r x < u; 2 hece r x < u. This shows tht r is ifiitely close to x. For uiqueess, suppose tht lso r R is ifiitely close to x. By 12.3, r r. Clerly the r = r. 152

2 A ottio coected with 12.4 is useful. If x is fiite elemet of R, the stdrd prt of x, deoted by st(x), is the uique rel umber ifiitely close to x. Theorem There re positive d egtive ifiitesimls. Proof. It suffices to show tht there re positive ifiitesimls. Let R \R. Cse 1. r < for every rel umber r. The if s is y positive rel umber we hve s <, hece < s, showig tht is positive ifiitesiml. Cse 2. is fiite. The is ifiitely close to some rel umber r, d so r is positive ifiitesiml. Theorem If is ozero elemet of R, the is ifiitesiml iff 1 is ifiite. Hece there exist positive d egtive ifiite elemets. Proof. Suppose tht is ifiitesiml. Let r be y positive rel umber. The < r, d so r <. This shows tht is ifiite. The coverse is proved similrly, d the lst sttemet of the theorem is cler too. No-stdrd rithmetic. As first result i o-stdrd mthemtics we prove little fct bout o-stdrd rithmetic. Let N = ω, +,, 0, S, <. This is the usul rithmetic structure, with ssocited lguge L hvig the obvious reltio d opertio symbols. For ech m ω we defie term m of L s follows, by recursio: 0 = 0, m + 1 = Sm. Here, s i much of our expositio of logic, we re usig 0 d S i two differet wys: first s i the set-theory prt (empty set d successor opertio o sets), d secod s symbols of L. The cotext should mke cler which oe we relly me for exmple, i the defiitio of N we me the set-theoretic otios, d i the defiitio of m we me, o the right side of the defiig equlities, the symbols of L. Note tht m N = m for every m ω. Recll tht m N is the vlue of the term m i the structure N; the geerl ottio is m N () for ω ω, but the c be omitted sice m does ot hve y vribles i it. The followig formul ϕ(v 0 ) sys v 0 is prime : 1 < v 0 v 1 v 2 (v 0 = v 1 v 2 v 0 = v 1 v 0 = v 2 ). The twi prime cojecture (still uresolved, s fr s I kow) sys tht for every turl umber m there is prime p > m such tht p + 2 is lso prime. (Cosider 3, 5; 11, 13; 17, 19; etc.) Propositio Suppose tht A = A, +,, 0, S, < is proper elemetry extesio of N ( proper just mes tht A ω). The the followig coditios re equivlet: (i) The twi prime cojecture. (ii) There is A\ω such tht A = (ϕ(v 0 ) ϕ(v 0 + 2))[]. Note tht (ii) c be loosely stted like this: there is ifiite prime such tht lso + 2 is prime. Oly oe is eeded, rther th ifiitely my thigs s i (i). 153

3 Proof. We prove some simple fcts first. (1) m A = m for ll m ω. This is esily proved by iductio o m. (2) If m ω, A, d < m, the ω. I fct, N = v 0 ( v 0 < m i<m v 0 = i so this holds i A too. It follows tht N = i<m v 0 = i[], hece there is i < m such tht N = v 0 = i[], so = i by (1). (3) If A\ω, the m < for ll m ω. This is true by (2), sice the setece v 0 v 1 (v 0 < v 1 v 0 = v 1 v 1 < v 0 ) holds i N d hece i A. Now we get dow to the proof of the propositio. (i) (ii): Uder the ssumptio (i), the setece v 0 v 1 (v 0 < v 1 ϕ(v 1 ) ϕ(v 1 + 2)) holds i N, hece i A. Now tke y b A\ω. The becuse this setece holds i A, we c choose A such tht b < d A = (ϕ(v 0 ) ϕ(v 0 + 2))[]. By (2) d b <, b / ω, we get / ω. This proves (ii). (ii) (i): Assume (ii). Now tke y m ω; we wt to fid prime p > m such tht p + 2 is lso prime. By (3) d (ii), d the desired coclusio follows. A = (v 0 < v 1 ϕ(v 1 ) ϕ(v 1 + 2))[m, ], hece A = v 1 (v 0 < v 1 ϕ(v 1 ) ϕ(v 1 + 2))[m], hece N = v 1 (v 0 < v 1 ϕ(v 1 ) ϕ(v 1 + 2))[m], ) The full elemetry extesio. Now we itroduce the full pprtus tht is goig to be used i this sectio. Let F be the collectio of ll fiitry opertios o R, d R the set of ll fiitry reltios o R. Recll tht this mes tht F = R R d R = R. ω, ω\{0} Now we tke lguge L which hs opertio symbol G f d reltio symbol S R correspodig to ech member of f F d R R respectively. Note tht L hs 2 2ω symbols; this is iterestig, but is ot very relevt to the work of this sectio. Now we itroduce some ottio d ssumptios used for the rest of this sectio. Let 154

4 R = (R,, Ö), where for ech f F we hve G f = f, d for ech R R we hve ÖS R = R. Let R = (R,, ) be proper elemetry extesio of A. Thus i prticulr, R is ordered field which is proper extesio of R, so the iitil termiology d results bout ifiitesimls d such re pplicble. This is the officil ottio, but i prctice we will just use f for G f, R for ÖS R, f for G f, d R for S R ; the cotext should mke cler wht is met. If f is member of F, the we let f be the correspodig member of R ; so f is the restrictio of f to tuples from R. Similrly for reltios. Sometimes we use whe we relly should be usig reltio symbol ottio. Also, isted of usig the [...] ottio i tlkig bout formul beig modeled i structure, we simply substitute the relevt elemets ito the formul. My of the cocepts of lysis re cocered ot with fuctios defied o ll of R, but with fuctios f defied oly o subset D. Sice the fuctios i first-order logic re ssumed to be everywhere defied, we hve to use circumlocutio to tke this kid of thig ito ccout. We c imgie y such fuctio f exteded to ll of R i y wy, sy by settig its vlues equl to 0 outside of D, d work with D d the extesio, both of which re expressed i our lguge. The extesio of f defied this wy will be clled the turl extesio of f to R. Our first result i o-stdrd lysis proper gives ostdrd versio of the defiitio of limit. Propositio Suppose tht f : D R, where D R,, u R, d some deleted ope itervl bout is subset of D. Let g be the turl extesio of f to R. The the followig coditios re equivlet: (i) lim x f(x) = u. (ii) For every y R, if y D, y, d y is ifiitely close to, the g (y) is ifiitely close to u. Proof. We tke g s idicted i the remrk. (i) (ii): Assume (i), let y be s i the hypothesis of (ii), d suppose tht ε is positive rel umber. We wt to show tht g (y) u < ε. Choose δ > 0 i R so tht for ll x D, if x < δ d x, the f(x) u < ε. Thus ( ) R = x(x D x x < δ g(x) u < ε). Sice, δ, u, ε re ll i R, the elemetry extesio defiitio implies, from (*), ( ) R = x(x D x x < δ g (x) u < ε). Now y D, y, d sice y is ifiitely close to, by defiitio of tht otio y < δ. Hece by (**), g (y) u < ε. Sice ε ws rbitrry positive elemet of R, this shows tht g (y) is ifiitely close to u. (ii) (i). Suppose tht (i) fils. Accordigly, choose positive ε i R so tht Agi, this implies R = δ[δ > 0 x(x D x x < δ g(x) u ε)]. ( ) R = δ[δ > 0 x(x D x x < δ g (x) u ε)]. 155

5 Let δ > 0 be positive ifiitesiml (by 12.5). By (*), choose y D such tht y, y < δ, d g (y) u ε. Sice δ is positive ifiitesiml, lso y is ifiitesiml. So y is ifiitely close to, but g (y) is ot ifiitely close to u, so tht (ii) fils. A versio of 12.8 for oe-sided limits is just s esy to prove; we omit the formultio d proof of this, d whe we use 12.8 we might tcitly pply the oe-sided versio. Lemm (i) If u v d u is fiite, the v is fiite. (ii) If u v d u is fiite, the st(u) = st(v). (iii) If, b R d < b, d if u [, b], the st(u) [, b]. (iv) If d b re fiite, the st( + b) = st() + st(b). Proof. (i): sy u < r R. Thus r < u < r. We clim tht r 1 < v < r + 1. For if, for exmple, v r 1, the u v = u v > 1, cotrdictio. So v < r +1 d so v is fiite. (ii): The ssumptio, d (i), imply tht st(u) u v st(v), so st(u) = st(v). (iii): We hve R = x [, b]( x b), so lso R = x [, b] ( x b). Hece u b. Suppose tht st(u) / [, b]. By symmetry, sy st(u) <. Thus 0 < st(u) u st(u) = u st(u), cotrdictio sice u d st(u) re ifiitely close. (iv): Let u = st() d v = st(b). The for y positive rel ε, ( + b) (u + v) = ( u) + (b v) u + b v ε 2 + ε 2 = ε. It follows tht + b u + v. Hece st( + b) = u + v = st() + st(b). Theorem Suppose tht f is defied o closed itervl [, b], with < b. Let g be the turl extesio of f to R. The the followig coditios re equivlet: (i) f is cotiuous o [, b]; (ii) For y u [, b], f(st(u)) = st(g (u)). (iii) For ll u, v [, b], if u v the g (u) g (v). Proof. (i) (ii): Assume (i) d suppose tht u [, b]. We my ssume tht u st(u). Now lim x st(u) f(x) = f(st(u)), u st(u), d u is ifiitely close to st(u), so by 12.8, g (u) is ifiitely close to f(st(u)). Hece f(st(u)) = st(g (u)). (ii) (iii): Assume (ii) d suppose tht u, v [, b] with u v. The st(g (u)) = f(st(u)) = f(st(v)) = st(g (v)). (iii) (i): Assume (iii) d suppose tht c [, b]. Tke y y R such tht y c, y [, b], d y is ifiitely close to c. The g (y) is ifiitely close to f(c) by (iii), s desired. Next we wt to give geuie piece of o-stdrd resoig, by provig the itermedite vlue theorem for cotiuous fuctios usig o-stdrd lysis. To do this the followig simple lemm is useful. Lemm There is ifiite iteger, i.e., there is ifiite elemet of ω. 156

6 Proof. By 12.6 let be positive ifiite elemet of R. Now R = x y(y ω x < y), so R = x y(y ω x < y). Hece choose ω so tht <. This is s desired. Theorem (Itermedite vlue theorem) Suppose tht f : D R, [, b] D, f is cotiuous o [, b], d f() < d < f(b). The there is elemet c (, b) such tht f(c) = d. Remrk: ote tht this is purely bout R; it is usul formultio of the itermedite vlue theorem. Proof. Let g be the turl extesio of f to R. Note tht ( [ ( R = ozero ω i ω i < f + i b ) ( d < f + (i + 1) b )]). I fct, give ozero ω, ote first tht d < f(b) = f(+ b ), d hece there is j such tht d < f(+j b b ). Tke the lest such j. Now f(+0 ) = f() < d, so j 0. Write j = i + 1. Clerly i is s desired. It follows tht ( ) ( R = ozero ω i ω (i < [g + i b ) d < g ( + (i + 1) b )]). By the lemm, let be ifiite iteger. The by (*) choose i <, i ω, so tht g ( + i b ) d < g ( + (i + 1) b ). Note tht i might be ifiite (i fct, it is esy to see tht it relly is, but we do ot eed tht). The differece betwee + i b d + (i + 1) b b is, which is ifiitesiml; hece by the cotiuity of f d 12.10, lso g ( + i b ) is ifiitely close to g ( + (i + 1) b ); so g ( + i b ) is ifiitely close to d. By the cotiuity gi, f(st( + i b )) = st(g ( + i b )) = d, s desired. We devote the rest of this sectio to the ostdrd defiitio of the Riem itegrl d descriptio of its simplest properties. Suppose tht f : D R,, b R, < b, d [, b] D. We defie the Riem sum fuctio s fb : (0, ) R s follows. Let x be give positive rel umber. Let be the lrgest iteger such tht + x b; mybe = 0. The we defie s fb ( x) = i< f( + i x) x + f( + x)(b x). Theorem Suppose tht f : D R,, b R, < b, [, b] D, d f is cotiuous o [, b]. Suppose tht dx is positive ifiitesiml. The s fb (dx) is fiite. Proof. Let M be greter th the mximum vlue of f o [, b], d let m be smller th the miimum vlue of f o [, b]. The for y positive rel umber x, with s bove, s fb ( x) i< M x + M(b x) = M(b ), 157

7 d similrly m(b ) s fb ( s). So R = v 0 (v 0 > 0 m(b ) s fb (v 0 ) M(b )), d so R = v 0 (v 0 > 0 m(b ) s fb (v 0) M(b )). I prticulr, s fb (dx) is fiite. This justifies the followig defiitio of the defiite itegrl: for y positive ifiitesiml dx, f(x)dx = st(s fb (dx)). Note tht dx here is ot just vcuous expressio s i the stdrd defiitio of itegrl; it is defiite positive ifiitesiml, d o the fce of it we might get differet itegrls by tkig differet ifiitesimls. We prove two elemetry properties of itegrls which will eble us to show tht the defiitio does ot relly deped o the prticulr positive ifiitesiml dx chose. Theorem Suppose tht f(x) = c for ll x [, b], where < b. The for y positive ifiitesiml dx, f(x)dx = c(b ). Proof. For y positive rel umber x we hve, with chose s bove, s fb ( x) = i< c x + c(b x) = c(b ). Thus for ll x > 0 i R we hve s fb ( x) = c(b ). Tht is, R = > 0(s fb ( ) = c(b )). Hece by the defiitio of elemetry extesio it follows tht for every positive d R we hve s fb (d) = c(b ), d the desired coclusio follows. Theorem Suppose tht f d g re both defied d cotiuous o [, b], with < b, d dx is positive ifiitesiml. The (f(x) + g(x))dx = f(x)dx + g(x)dx. Proof. Tke y positive rel umber x d choose s usul. The s f+g,,b ( x) = i<(f( + i x) + g( + i x)) x + (f( + x) + g( + x))(b x) = i< f( + i x) x + f( + x)(b x) + i< g( + i x) x + g( + x)(b x) = s fb ( x) + s gb ( x). 158

8 Hece R = > 0(s f+g,,b ( ) = s fb ( )+s gb ( ). It follows tht s f+g,,b (d) = s fb (d)+ s gb (d) for every positive d R, d the desired coclusio follows, usig 12.9(iv). So, ow we c prove the idepedece of the defiitio o the prticulr ifiitesiml chose: Theorem Suppose tht f : D R,, b R, < b, [, b] D, d f is cotiuous o [, b]. Suppose tht dx d du re positive ifiitesimls. The f(x)dx = f(u)du. Proof. Let g be the turl extesio of f to R. This will be fuy proof. We show tht for every positive rel umber r we hve f(x)dx f(u)du + r; sice r is rbitrry, this shows oe iequlity of the desired equlity i the theorem, d the other oe is true by symmetry. Let c = r b, d let h(x) = f(x) + c for ll x [, b]. We clim (*) If x d u re two positive rel umbers d s fb ( x) > s hb ( u), the there re x, u [, b] such tht x u u x + x d f(x) > f(u) + c. To see this, ccordig to the defiitio of s fb d s gb choose itegers, m ω such tht s fb ( x) = i< f( + i x) x + f( + x)(b x) d s hb ( u) = i<m g( + i u) u + g( + m x)(b m u). Now let {c 0,..., c p } = { + i x : i } { + j u : j m} {b}, where c 0 < < c p. The for ech k < p there re uique i k d j k m such tht [c k, c k+1 ] [ + i k x, + (i k + 1) x] [ + j k u, + (j k + 1) u], where if i k = the we replce + (i k + 1) x by b, d similrly if j k = m. Hece s fb ( x) = f( + i k x) c k+1 c k k<p d s hb ( u) = k<p h( + j k u) c k+1 c k. 159

9 It follows tht there is k < p such tht f( + i k x) > h( + j k u), i.e., f( + i k x) > f( + j k u) + c. Let x = + i k x d u = + j k u. So f(x) > f(u) + c. Now d + i k x c k < c k+1 + (i k + 1) x + j k u c k < c k+1 + (j k + 1) u. It follows tht u c k+1 x + x d x < c k+1 u + u, s desired i (*). It follows tht the exteded versio of (*) holds i R : (**) If dx d du re two positive members of R d s fb (dx) > s hb (du), the there re x, u [, b] such tht x du u x + dx d g (x) > g (u) + c. Now we clim: (***) If dx d du re positive ifiitesimls, the s fb (dx) s hb (du). I fct, if ot we c pply (**) to get x, u [, b] such tht x du u x + dx d g (x) > g (u) + c. But the x u d g (x) g (u), cotrdictig the cotiuity of f. This proves (***). Usig (***), the proof is esily fiished: f(x)dx = st(s fb(dx)) st(s hb(du) = = = = (f(u) + c)du f(u)du + cdu f(u)du + c(b ) f(u)du + r. Exercises, sectio 12 For the first three exercises, suppose tht ε d δ re positive ifiitesimls, d M is positive ifiite umber, i proper extesio of R. Determie whether the give expressio is 0, positive ifiitesiml, positive fiite umber ot ifiitesiml, or positive ifiite umber; if more th oe of these is possible, idicte tht, d give resos. E12.1. (ε + δ)/ ε 2 + δ 2. E M + 1 3M

10 E12.3. ε/δ. E12.4. Let M be positive d ifiite. Show tht the followig is fiite, d compute its stdrd prt: 2M 3 + 5M 2 + 3M 17M 3 + M + 1. E12.5. Suppose tht st(b) = 5 d b 5. Compute the stdrd prt of b 2 25 b 5. E12.6. Let A be s i Propositio Prove tht for every m A\ω d ll positive (ordiry) itegers we hve x m + y m z m. E12.7. Let f : [0, 1] R, d let g be its turl extesio to R. Prove tht the followig coditios re equivlet: (i) f is uiformly cotiuous o [, b]. (ii) For ll x, y [0, 1], if x is ifiitely close to y, the g (x) is ifiitely close to g (y). (iii) There is positive δ i R such tht for ll x, y [0, 1], if x y < δ the g (x) is ifiitely close to g (y). 161