微积分讲义Chap 1 Completeness axiom of R

Chapter 1 The real number system

1.3. Completeness axiom of R 1.16 Definition

Let E?R and E??.

(i).The set E is said to be bounded above if there is an M?R s.t a?M for all

a?E.

(ii).A number M is called an upper bound of the set E if a?M for all a?E. (iii).A number S is called a supremum of the set E if S satisfies the following

conditions

(1) if a?S,?a?E,

(2) if M is an upper bound of E then S?M.

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. a?1,?a?E?[0,1]. 2. let M be an upper bound

then a?M for all a?[0,1] ?1?M(?1?[0,1]). 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 a?M for all a?E. Then M is an upper bound.

Let b?0,b?R 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 s1&s2 are the least upper bounds of E. Then s1&s2 are upper bounds of E. ?s1?s2&s2?s1 ?s1?s2.

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] E?R,E??,andsupEexists. Then???0,thereisana?E s.t supE???a?supE.

Proof:. Suppose the conclusion is false. There is an ??0 such that

a?supE??,?a?E..

E?? is an upper bound. ?sup ?supE???supE????0 ??a?E s.t supE???a?supE

1.21. Theorem

If E?N has a supremum, then supE?E

Proof.

Let supE=s.

By Approximation property, there?x0?E s.t s?1?x0?s. If x0?s then supE?E is obvious. If s?1?x0?s, then

?x1?E s.t x0?x1?s?0?x1?x0?s?x0. 1. x1,x0?N?x1?x0?1.

2. s?x1,x0?s?1?x1?x0?s?(s?1)?1. It is a contradiction.

E?E ?sup● [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]

a,b?R,a,b?0??n?N s.t b

1. If b

2. If a

k?b,?k?E?E is bounded above. aBy Completeness of R, supE exists.

?supE?E(byTheorem 1.21) ?sup E?1?E?(suEp?1)a?b take n=supE+1 1.23: Example.

11137 Let A?{1,,,.......} and B?{,,,...}

24248 prove that supA=supB

=1

Proof. 1. A?{1;n?Norn?0} 2n1,n?0,1,. 2,..n2 ?1?x,x? ?1 is an upper bound. Let M be another bound.

1?1 20?supA?1.?M? 2. B?{1? ?1?1?1;n?N} 2n1,?n?N n2 ?1 is an upper bound of B. Let M be an upper bound of B To show M?1.

Suppose not?M?1?1?M?0

By Archimedean principle, there exists n?N such that

??1?1?M for some n?N. n21?1?M, n12n?12n?1?1?(1?M)??M?? M is an upper bound. ?1?n?nn222 ?M?1

B?1 ?sup

● [Well-Order Principle]

E?N,E???E has a least element

(ie.?a?E s.t a?x,?x?E)

1.24. Theorem (Density of rational)

Let a,b?R satisfy a

number c s.t a

1?b?a,n?N(by Archimedean Principle). nk1. If b>0, let E?{k?N;b?}.

nBy Archimedean Principle?E??.

By Well-Order Principle?E has a least element, saysk0.

.?m:?k0?1?m?E(i.e.Let q?m. nm?b) nWe must show that a

q

?a?b?(b?a)??q?a.?a?q?b.k01k0?1???q.nnn

2. If b<0, then

?k?0, k is a natural number s.t b+k>0. ??c?Q s.t a+k

?a?c?k?b ?c?Q ?c?k?Q ie. There is a rational number between a & b. 1.27. Definition. E?R,E??.

1. s is called a lower bound of E if x?s,?x?E. In the case, E is called bounded below 2. t is called the greatest lower bound of E if

1.x?t,?x?E, 2. If M is a lower bound of E then M?t.

3. E is bounded if x?M,?x?E for some M>0. (i.e. E is bounded above and below.)

● Let E be a set of R. We define ?E?{?x;x?E}. 1.28. Theorem E?R,E??.

1. supE exists ?inf(-E) exists

in fact supE= -inf(-E)

2. infE exists?sup(-E) exists

in 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 ?E?s??supE. 1.

E is an upper bound of E ?supE,?x?E??x??supE,?x?E ?x?supE is a lower bound of –E ??sup 2. Suppose that s is a lower bound of -E

Suppose not?s??supE??s?supE on the other hand

?x?s,?x?E

?x??s Hence, -s is a upper bound of E?? By 1.& 2, inf(?E)?&inf(?E)??supE. 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

B?supA,infB?infA. A?B?R,A,B???sup if supBandinfBexist.

Hence, infB?infA?supA?supB Proof: