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rudin数学分析原理讲义970923

2019-07-01 18:09 网络整理 教案网

外汇评论分析_rudin数学分析原理评论_150107基金评论分析

rudin数学分析原理讲义 台湾国立交通大学白启光教授

rudin数学分析原理讲义970923

9月23日課堂摘要

Recall

An order set S is said to have the least-upper-bound property if the following is true: If E S, E is not empty, and E is bounded above, then sup E exists in S.

Question

外汇评论分析_150107基金评论分析_rudin数学分析原理评论

Does every ordered set with the least-upper-bound property also have the

greatest-lower-bound property ?

Theorem Suppose S is an ordered set with the least-upper-bound property, φ≠B S, and B is bounded below, then inf B exists in S.

Let L be the set of all lower bounds of B.

(1) x∈B, x is an upper bound of L.

(2) α = sup L exists in S. (Why ?)

rudin数学分析原理评论_150107基金评论分析_外汇评论分析

(3) Claim: α = inf B.

(i) α∈L, i.e., if x∈B, then α≤x.

We shall show that if α>γ, then γ B.

Given α>γ,since α = sup Lrudin数学分析原理评论,we get that γ is not an upper bounded of

L. By (1)rudin数学分析原理评论, we see that γ B.

(ii) If α<β, then β is not a lower bounded of B, i.e.,β L.

rudin数学分析原理评论_外汇评论分析_150107基金评论分析

If α<β, then y≤α<β for all y∈L.

Definition A field is a set F with two operations, called addition and multiplication, which satisfy the following so-called “field axioms” (A), (M), and (D):

(A) Axioms for addition

(A1) If x∈F and y∈F, then x+y is in F.

(A2) Addition is commutative: x+y=y+x for all x, y∈F.

(A3) Addition is associative: (x+y)+z=x+(y+z) for all x, y, z∈F.

150107基金评论分析_rudin数学分析原理评论_外汇评论分析

(A4) F contains an element 0 such that 0+x=x for every x∈F.

(A5) For all x∈F there exists an element x∈F such that

x+( x)=0.

(M) Axioms for multiplication

(M1) If x∈F and y∈F, then xy is in F.

(M2) Multiplication is commutative: xy=yx for all x, y∈F.