微积分讲义Chap1CompletenessaxiomofR

1、chapter 1 the real number system 1.3. completeness axiom of r1.16 definition let reand e. (i).the set e is said to be bounded above if there is an rms.t mafor all ea. (ii).a number m is called an upper bound of the set e if mafor all ea. (iii).a number s is called a supremum of the set e if s satisf

2、ies the following conditions (1) if easa, (2) if m is an upper bound of e then ms. remark the supremum is also called the least upper bound. 1.17: exampleif e=0,1, prove that 1 is a supremum of e. proof. 1. 1 ,0, 1eaa. 2. let m be an upper bound then mafor all 1 , 0a)1 ,01(1m. we derive the result.

3、1.18: remarkif a set has one upper bound, it has infinitely many upper bounds proof:. let e be a subset of r. let mafor all ea. then m is an upper bound. let rbb,0then m+b is also an upper bound. so, e has infinitely many upper bounds. 1.19 . theorem. let e be a nonempty subset of r. then the least

4、upper bound of e is unique if it exists. proof. suppose that 21& ssare the least upper bounds of e.then 21& ssare upper bounds of e. 1221&ssss21ss. notation the supremum is also called least upper bound . we use supe to denote the supremum of nonempty set e. 1.20. theorem approximation p

5、roperty eeres u pa n d,exists. theneaanisthere,0s.t eaesupsup. proof:. suppose the conclusion is false. there is an 0such that .,supeaea. es u pis an upper bound. eesupsup0eas.t eaesupsup1.21. theorem if nehas a supremum, then eesupproof. let supe=s.by approximation property, thereex0s.t sxs01. if s

6、x0then eesupis obvious. if sxs01, then ex1s.t 001100 xsxxsxx. 1. 1,0101xxnxx. 2. 1)1(1,0101ssxxsxxs. it is a contradiction. ees u p complete axiom of r every nonempty subset e of r that is bounded above, then e has the least upper bound. . 1.22 :archimedean principle nnbarba0,s.t bna. proof: 1. if b

7、a, then take n=1. 2. if ab , let ;bkanke. ee,1. ekabk,e is bounded above. by completeness of r, supe exists. baeeeee)1( s u p1su p)21.1t heoremby(suptake n=supe+1 1.23: example. let ,.41,21, 1aand ,.87,43,21bprove that supa=supb =1proof. 1. 1;02nann or n11,0 , 1, 2 , . .2nxxn. 1is an upper bound. le

8、t m be another bound. .1s u p1210am2. ;211nnbnnnn,21111is an upper bound of b.let m be an upper bound of bto show 1m. suppose not011mmby archimedean principle, there exists nnsuch that mn11, mn121for some nn. mmnnnnn212)1(1212211m is an upper bound. 1m1s u p b well-order principle ene,e has a least

9、element (ie.eas.t exxa,) 1.24. theorem (density of rational) let rba,satisfy ab, then there is a rational number c s.t ac0, let .;nkbnkeby archimedean principlee. by well-order principlee has a least element, says0k . .).(1:0bnmeiemkmlet nmq. we must show that aqb. qb is obvious, now we show that aq

10、.11)(00bqaaqqnknnkabba2. if b 0. qcs.t a+kc0. (i.e. e is bounded above and below.) let e be a set of r. we define ;exxe. 1.28. theorem ere,. 1. supe exists inf(-e) exists in fact supe= -inf(-e) 2. infe existssup(-e) exists in fact infe= -sup(-e) proof: 1. supe exists. now we show that supe=inf(-e).s

11、how that 1. -supe is a lower bound of e. 2. if s is a lower bound of esesup. 1. es u pis an upper bound of eexexexex,su p,s u pesu pis a lower bound of e 2. suppose that s is a lower bound of -esuppose notesessupsupon the other hand sxexsx,hence, -s is a upper bound of eby 1.& 2, eeesup)inf(&

12、;)inf(. the proof of converse is similar. remark. the largest lower bound is also called infimum. remark. the completeness axiom of r is equivalent to “ every nonempty, bounded below subset of r has the infimum ”.1.29. theorem .i n fi n f,s u ps u p,ababbarbaif bbinfandsupexist. hence, baabsupsupinf

13、infproof: 1. suppose sup b exists .,su p.,su paxbxbabxbxais bounded above & supb is an upper bound of a by complete axiom of r, .supsup&supbaa2.s ppose thatinfb exists. .,i n f,i n faxbxbabxbxais bounded below & infb is an lower bound of a. by complete axiom of r, baainfinf&:inf. def

14、: s u p, i n f1.4 functions, countability and the algebra of sets. definition let a & b be two sets of r. a function f is a relation between a & b s.t f assigns each element x of a to a unique bydefinition baf :f is called 1-1 if )()(yfxfyxdef: baf :f is called onto if axby,s.tf(x)=y definit

15、ion 1.34: let e be s a set of r. 1. e is said be finite if eor enfnnn.2, 1:&,.3,2, 1s.t f is 1-1 & onto. 2. e is called countably infinite if enf :s.t f is 1-1 & onto. 3. e is called countable if e is finite or countably infinite. 4. e is called uncountable if e is not countable 1.35. theorem . the open interval (0,1) is uncountable pf: suppose (0,1) is countable. then there is a list for (0,1 ) says .0.0.0.0321333231323222121312111nnnnaaaaaaaaaaa