微积分讲义Chap 1 Completeness axiom of R.doc

Chapter 1 The real number system1.3. Completeness axiom of R1.16 Definition Let and . (i).The set E is said to be bounded above if there is an s.t for all . (ii).A number M is called an upper bound of the set E if for all . (iii).A number S is called a supremum of the set E if S satisfies the following conditions (1) if ,(2) if M is an upper bound of E then .Remark The supremum is also called the least upper bound.1.17: Example If E=0,1, prove that 1 is a supremum of E. Proof. 1. . 2. let M be an upper bound then for all . We derive the result. 1.18: Remark If a set has one upper bound, it has infinitely many upper bounds Proof:. Let E be a subset of R. Let for all . Then M is an upper bound. Let then 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 upper bound of E is unique if it exists. Proof. Suppose that are the least upper bounds of E. Then are upper bounds of E. .Notation The supremum is also called least upper bound . We use supE to denote the supremum of nonempty set E.1.20. Theorem Approximation property exists. Then s.t . Proof:. Suppose the conclusion is false. There is an such that . is an upper bound. s.t 1.21. Theorem If has a supremum, then Proof. Let supE=s. By Approximation property, there s.t . If then is obvious. If , then s.t . 1. . 2. . It is a contradiction. 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 s.t bna. Proof:1. If ba, then take n=1.2. If ab , let . . E is bounded above.By Completeness of R, supE exists. take n=supE+11.23: Example. Let and prove that supA=supB=1 Proof. 1. . is an upper bound. Let M be another bound. 2. is an upper bound of B. Let M be an upper bound of B To show . Suppose not By Archimedean principle, there exists such that , for some . M is an upper bound. Well-Order Principle E has a least element (ie. s.t )1.24. Theorem (Density of rational) Let satisfy ab, then there is a rational number c s.t ac0, let By Archimedean Principle.By Well-Order PrincipleE has a least element, says.Let .We must show that aqb.qb is obvious, now we show that aq.2. If b0. s.t a+kc0. (i.e. E is bounded above and below.) Let E be a set of R. We define .1.28. Theorem .1. supE exists inf(-E) existsin fact supE= -inf(-E)2. infE existssup(-E) existsin fact infE= -sup(-E) Proof: 1. supE exists. Now we show that supE=inf(-E). Show that 1.-supE is a lower bound of E.2. if s is a lower bound of . 1. is an upper bound of E is a lower bound of E 2. Suppose that s is a lower bound of -E Suppose not on the other hand Hence, -s is a upper bound of E By 1.& 2, . 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 if exist. Hence, Proof:1. suppose supB exists is bounded above & supB is an upper bound of A By complete axiom of R, 2. S ppose that infB exists. is bounded below & infB is an lower bound of A. By complete axiom of R, .Def: 1.4 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 Definition f is called 1-1 if Def: f is called onto if s.t f(x)=y Definition 1.34: Let E be s a set of R. 1. E is said be finite if or s.t f is 1-1 & onto. 2. E is called countably infinite if 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 count