Introduction⚓︎
A satisfied discussion of the main concepts of analysis must be based on an accurately defined number concept. ——WalterRudin
We first assume familiarity with the rational numbers(the numbers of the form \(\frac{m}{n}\),where m and n are integers and \(n\neq 0\))
The rational number system is inadequate for many purposes,both as a field and an ordered set.(These terms will be defined in Secs.1.6 and 1.12)
Motivation Instance:there is no rational \(p\) such that \(p^{2} =2\).
We can use a sequence that is made of infinite decimal expansions to approximate the irrational numbers "actually \(\sqrt{ 2 }\)"
it "tends to \(\sqrt{ 2 }\)" if we know the irrational number.
But without the definition of an irrational number, we can't answer the question: What is it that this sequence "tends to"?
Statement: There is no rational \(p\) satisfying the equation \(p^{2}=2\)
Proof
If there were such a \(p\),we could write \(p= \frac{m}{n}\) where m and n are integers that are not both even.If not, we eliminate the 2.
We can get \(m^{2}=2n^{2}\); this shows that \(m^{2}\) is even and \(m^{2}\) is divisible by 4 (m is divisible by 2). Thus n is also divisible by 2 (n is even). Hence the equation is impossible for rational \(p\).
Now we can use this instance to make two subsets of \(\mathbb{Q+}\) and denote this example 1.1. Let A be the set of all positive rational \(p\) such that \(p^{2}<2\),let B be the set of all positive rational \(p\) such that \(p^{2}>2\) .We shall show that A contains no largest number and B contains no smallest.
More explicitly,for every \(p\) in A we can find a rational \(q\) in A such that \(p<q\) (B is the same)
A quite elegant structure:
If \(p\) is in A then \(p^{2}-2<0\), which shows that \(q>p\) and that \(q^{2}<2\).
Thinking: How can we construct this \(q\)? It looks difficult.
Answer: We can use \(\sqrt{ 2 }\) (or any other symbol) to approximate the number \(q\). To make an appropriate approximation, we should clarify our goal first (what conditions the number q should satisfy).
- it must be bigger than p(only think about the situation \(p^{2}<2\)).
- \(q^{2}<2\),too.
That's a draft and the other situation is the same. \(q>p-(p-\sqrt{ 2 })=p+ \frac{2-p^{2}}{\sqrt{ 2 }+p}\), you will find the numerator is controlled by (2-\(p^{2}\)), and we actually can use any number bigger than \(\sqrt{ 2 }\) to replace the 2 in Rudin's book.
Remark: The rational number system has certain gaps,if \(r<s\) then \(r< \frac{r+s}{2}<s\),the real number system fills these gaps.
In order to elucidate its structure, as well as that of the complex numbers,we start with a discussion of the general concepts of ordered set and field
Some definitions of ordered sets and fields: \(\not\in \emptyset \subset \supset\)
Definition Throughout Chap.1,the set of all rational numbers will be denoted by \(\mathbb{Q}\)
Ordered sets⚓︎
Definition Let \(S\) be a set.An order on \(S\) is a relation,denoted by <,with the following two properties: (1) If \(x\in S\) and \(y \in S\) then only one of the statements is true.
(2) If \(x,y,z\in S\) , if \(x<y\) and \(y<z\) ,then \(x<z\). "\(x<y\)" may be read as "x is less than y" or "x is smaller than y" or "x precedes y". Or we can use y>x to replace the "x<y".The notation \(x\le y\) indicates that "x<y" or "x=y"
Definition An ordered set is a set \(S\) in which an order is defined.
Assume \(S\) is an ordered set and \(E \subset S\).(Used in following definitions)
Definition If there exists a \(\beta \in S\) such that \(x\le \beta\) for every \(x\in E\),we say that \(\mathbb{E}\) is bounded above,and call \(\beta\) an upper bound of \(E\) Lower bounds are defined in the same way.
Definition \(\alpha\) is called the least upper bound (or the supremum) of \(E\) if there exists an \(\alpha\) with the following properties (there is at most one such \(\alpha\), as we can see from (2)): (1) \(\alpha\) is an upper bound of \(E\) (2) If \(\gamma <\alpha\) then \(\gamma\) is not an upper bound of \(E\) and we write \(\alpha=sup\ E\) Similarly, we can define the greatest lower bound, or infimum. We write (for convenience, we use a letter different from \(\alpha\)) \(\beta=inf\ E\).
Examples (a) The set A is bounded above and has no least upper bound in \(\mathbb{Q}\) Back to the example 1.1 and consider A and B as subsets of the ordered set \(\mathbb{Q}\),we may get (a) in the following three conclusions.
(b) If \(\alpha=sup\ E\) exists,then \(\alpha\) may or may not be a member of \(E\). For instance,let \(E_{1}\) be the set of all \(r\in \mathbb{Q}\) with \(r<0\) .Let \(E_{2}\) be the set of all \(r\in \mathbb{Q}\) with \(r\le 0\).Then \(sup\ E_{1}=sup \ E_{2}=0\) and \(0\not\in E_{1},0\in E_{2}\).
(c) \(E\) is an ordered set,\(sup\ E\) and \(inf\ E\) have one in \(E\) Let \(E\) consist of all numbers \(\frac{1}{n}\) ,where \(n=1,2,\dots\) ,Then \(sup\ E=1\) ,which is in \(E\) ,but on the other hand \(inf\ E=0\) ,which is not in \(E\).
Definition An ordered set \(S\) is said to have the least-upper-bound property if the following is true If \(E \subset S\),\(E\) is not empty,and E is bounded above,then \(sup\ E\) exists in \(S\). The example in (a) shows that \(\mathbb{Q}\) does not have the least-upper-bound property.
The close relation between greatest lower bounds and least upper bounds: every ordered set with the least-upper-bound property also has the greatest-lower-bound property. We can summarize it as a theorem.(We should discuss the set with least-upper-bound property)
Theorem Suppose \(S\) is an ordered set with the least-upper-bound property,\(B\subset S\),B is not empty,and B is bounded below.Let L be the set of all lower bounds of B.Then \(\alpha=sup\ L\) exists in \(S\) ,and \(\alpha=inf\ B\).In particular,\(inf\ B\) exists in \(S\).
Proof
Step 1: Since B is bounded below, L is not empty. Since L consits of exactly all of the y which satisfy the inequality \(y\le x\) for every \(x\in B\) ,then every \(x\in B\) is an upper bound of L.(B is not empty) Thus L is bounded above.Our hypothsis about \(S\) implies therefore that L has supremum in \(S\) ; call it \(\alpha\). Step 2: Due to the definition of supremum,think about two situations(left and right): if \(\gamma<\alpha\) then \(\gamma\) is not an upper bound of L,hence \(\gamma \not\in B\)(every element of B is an upper bound of L). It shows that \(\alpha\le x\) for every \(x\in B\) .(Because every \(x\in B\) should be greater than or equal to the \(\alpha\) so we can get the \(\alpha\) is the lower bound of B)Thus \(\alpha\in L\). if \(\alpha<\beta\) then \(\beta \not\in L\) ,since \(\alpha\) is an upper bound of L. In my words, they mean a number that is bigger than \(\alpha\) does not belong to L and thus is not a lower bound of B. From these we can get the conclusion that \(\alpha=inf\ B\) (In definition)
Some words: I think I should not write the definitions in my note or should write them very simply; they waste time! Some more words: the author skips one or two logical explanation sentences in one conclusion.
Fields⚓︎
Q&A Q: Straightly, the first question, what is a field ? A: a set F with two operations, "addition and multiplication" , more precisely, it satisfies the following "field axioms" (A), (M) and (D) explained below. (A) and (M) both have 5 subaxioms 1. 封闭性 2. 交换律 3. 结合律 4. 幺元律 5. 逆元存在 (D) 乘法分配律
Use the field axioms to prove some familiar properties of \(\mathbb{Q}\), so that we won't prove them again for real numbers and complex numbers.
Give one example for each axiom.(Others can be found in the book)
The Real Field⚓︎
Remark: This chapter's notes are too long and like a copy of Rudin's book. I will change the way I write in the next chapters (but on the other hand, is this proof that the book is good?)