[Analysis] 1. The Real and Complex Number Systems

Section 0. Preview

앞으로 Analysis에서는 PMA 위주로 김김계의 일부분을 섞어 작성할 것이다. 1장에서는 주로 실수가 무엇인지 배운다. 주요 내용은 완비성(completeness, lubp), 유리수로부터 실수를 구축하는 것(데데킨드 컷), 완비순서체는 실수가 유일한 것 정도이다. 복소수도 간략하게 다룬다. 체의 정의와 성질이나 벡터공간에 관한 것은 다른 분야에서 더 formal하게 다룬다고 생각하여 제외했다.

 

Section 1. Ordered Sets

Def 1.1.1 Let $S$ be a set. Order on $S$ is a relation s.t.

(1) $x, y \in S \Rightarrow$ only one of $x<y, x=y, y<x$ holds.

(2) $^\forall x,y,z \in S, x<y, y<z \Rightarrow x<z$

a set with such relation is called ordered set.

 

Def 1.1.2 Let $S$ be an ordered set. $E \subset S$.

if $^\exists \beta \in S$ s.t. $^\forall x \in E, x \leq \beta$, then $E$ is bounded above. $\beta$ is upper bound of $E$.

In same context, define bounded below and lower bound.

 

Def 1.1.3 Let S be an ordered set, $E \subset S$, bdd above. If $^\exists \alpha$ s.t.

(1) $\alpha$, ub of $E$

(2) $r < \alpha \Rightarrow r$, not ub of $E$

Then $\alpha$ is called least ub or supremum. Denote by $\alpha = \sup E$

In same context, define greatest lb = infimum. Denote by $\alpha = \inf E$

 

Def 1.1.4 Let $E \subset S$, $E \neq \phi$, bdd above.

If for every such $E$, $\sup E \in S$, then S has least ub property(lubp).

 

Thm 1.1.1 S, ordered set with lubp. $B \subset S$, $B \neq \phi$, bdd below. $L$, set of lb of $B$.

Then $^\exists \alpha = \sup L \in S$, $\alpha = \inf B$ (i.e. lubp $\Rightarrow$ glbp)

pf. 

$B$, bdd below $\Rightarrow$ $L$, nonempty.

$^\forall y \in L, ^\forall x \in B, y \leq x \Rightarrow x$, ub of $L$

$\therefore L$, bdd above. $^\exists \alpha = \sup L$

$^\forall y \in L, y \leq \alpha, ^\forall x \in B, \alpha \leq x (\because x,$ ub of $ L)$

$\therefore \alpha = \inf B$  $_\blacksquare$

Section 2. Real Fields

Thm 1.2.1 Existance Theorem

Ordered field $\mathbb{R}$ with lubp s.t. $\mathbb{R} \supset \mathbb{Q}$, as a subfield exists.

sketch of pf.

Step 1. Define a 'cut'.

$\alpha \subset \mathbb{Q}$ is a cut if

(1) $\alpha \neq \phi, \mathbb{Q}$

(2) $p \in \alpha, q \in \mathbb{Q}, q < p \Rightarrow q \in \alpha$

(3) $p \in \alpha \Rightarrow ^\exists r \in \alpha$ s.t. $p < r$

 

Step 2. Define $\alpha < \beta$ if $ \alpha \subsetneq \beta$.

Then $<$ is an order, thus a set of cuts $R$ is an ordered set.

 

Step 3. Prove $R$ has lubp.

 

Step 4-7. Prove $R$ is a field.

Define $\alpha + \beta$ as a set of all sums of $r \in \alpha, s \in \beta$.

Define muliplication carefully. (First consider multi between positive, then other cases.)

Let's denote a cut $\{p \in \mathbb{Q}| p < r\}$ as $r^*$.

prove $\langle R, +, \cdot \rangle$ with add/muti id $0^*, 1^*$ is a field.

 

Step 8-9. Define $Q:=\{r^* | r \in \mathbb{Q}\}$. $Q$ is a subfield of $R$ and isomorphic to $\mathbb{Q}$.

Thus there exists some $\mathbb{R}$ satisfies such conditions. $_\blacksquare$

 

Thm 1.2.2 Let S be an ordered set s.t. bdd above. Let $\alpha = \sup S$.

Then $^\forall \epsilon > 0, ^\exists x \in S$ s.t. $\alpha-\epsilon < x \leq \alpha$.

pf.

Spse not. Then $\alpha - \epsilon$ is ub, which is a contradiction. $_\blacksquare$

Similar prop holds for infimum.

 

Thm 1.2.3 Archimedean property.

(1) $x,y \in \mathbb{R}, x>0 \Rightarrow ^\exists n \in \mathbb{N}$ s.t. $nx > y$

(2) $x,y \in \mathbb{R}, x<y \Rightarrow ^\exists p \in \mathbb{Q}$ s.t. $x < p < y$

pf. 

(1) spse not.

Then $S=\{nx| n \in \mathbb{N}\}$ is bdd above. i.e. $^\exists \alpha = \sup S$.

By Thm 2.2, $^\exists m \in \mathbb{N}$ s.t. $\alpha - x < mx \leq \alpha$.

Then $(m+1)x > \alpha$, which is a contradiction.

 

(2) Let $n$ be a positive integer s.t. $n(y-x) > 1$. i.e. $x + \frac{1}{n} < y$.

Let $S=\{\frac{k}{n}| k \in \mathbb{N}, \frac{k}{n} < x\}$. $S$ is bdd above since $x$ is an ub. Let $\alpha=\sup S$.

By Thm 2.2, $^\exists m \in \mathbb{N}$ s.t. $x - \frac{1}{n} < \frac{m-1}{n} \leq x$.

Then $x < \frac{m}{n} \leq x + \frac{1}{n} < y$. $_\blacksquare$

 

Thm 1.2.4 Ordered field with lubp is unique.

 

sketch of pf.

Let $F, G$ be ordered fields with lubp.

Define $n \cdot 1_F$ as a $n$-sum of $1_F$, $\frac{m}{n} \cdot 1_F = (m \cdot 1_F) (n \cdot 1_F)^{-1}$ (Similar in $G$)

Let $f:F \rightarrow G, x \mapsto \sup\{p \cdot 1_G | p \in \mathbb{Q}, p \cdot 1_F < x\}$. Define $g: G \rightarrow F$ similarly.

Then $g$ is an inverse of $f$, thus $f$ is bijective.

Prove that $^\forall x, y \in F, f(x+y)=f(x)+f(y), f(xy)=f(x)f(y)$ using Archimedean property.

Prove that $f(P_F)=f(P_G)$ where $P_F$ is a positive set of $F$.

Thus $f$ is an isomorphism. $_\blacksquare$

 

Thm 1.2.5 $^\forall x \in \mathbb{R}, x>0, n \in \mathbb{N}, ^{!\exists} y \in \mathbb{R}$ s.t. $y^n=x$

pf.

Let $A={a \in \mathbb{R}^+ | a^n < x}$.

Let $t=\frac{x}{1+x}$. Then $t^n < t < x, t \in A$. $\therefore A \neq \phi$.

Let $t=1+x$. Then $t^n > t > x$. $\therefore A$, bdd above.

$^{!\exists} y = \sup A, y > 0$.

Note that if $0<a<b, b^n - a^n = (b - a)(b^{n-1}+b^{n-2}a+ \cdots + a^{n-1}) < (b-a)nb^{n-1}$.

(i) Spse $y^n < x$.

Let $h = \min \{1, \frac{x-y^n}{n(y+1)^{n-1}} \}$.

$y^n < (y+h)^n < y^n + hn(y+h)^{n-1} \leq y^n+hn(y+1)^{n-1} \leq x$.

i.e. $y+h \in A$. Contradicts.

(ii)Spse $y^n > x$.

Let $k=\frac{y^n-x}{ny^{n-1}}$. $0<k<y$.

$y^n > (y-k)^n > y^n - kny^{n-1} = x$.

i.e. $y - k$, ub of $A$. Contradicts.

$\therefore y^n=x$ and unique.

 

Note. If you want to prove this, expand $(y+h)^n$ or $(y-k)^n$ and restrict $h$ or $k$ properly to get what you want.

 

Def 1.2.1 The Extended Real Number System

$\mathbb{R} \cup \infty \cup -\infty$ with original order in $\mathbb{R}$ and $^\forall x \in \mathbb{R} -\infty < x < \infty$.

 

Section 3. The Complex Field.

Def 1.3.1 Complex Number

is an ordered pair $(a, b), a, b \in \mathbb{R}$ with

$^\forall x=(a,b), y=(c,d), x+y=(a+c, b+d), xy=(ac-bd, ad+bc)$.

 

Thm 1.3.1 The set of complex numbers is a Field. $\mathbb{C}$.

 

Def 1.3.2 $i=(0, 1)$.

Thm 1.3.2 $i^2=-1, (a, b)=a+bi$

 

Def 1.3.3 $z=a+bi \in \mathbb{C}$. $\overline{z}=a-bi$, conjugate of z. $a=\operatorname{Re}(z), b=\operatorname{Im}(z)$.

 

Thm 1.3.3

(1) $\overline{z+w}=\overline{z}+\overline{w}$

(2) $\overline{zw}=\overline{z} \cdot \overline{w}$

(3) $z + \overline{z} = 2\operatorname{Re}(z), z - \overline{z} = 2\operatorname{Im}(z)$

(4) $z\overline{z} \in \mathbb{R}, z \neq 0 \Rightarrow z\overline{z} > 0$

 

Def 1.3.4 $|z|=\sqrt{z\overline{z}}$

 

Thm 1.3.5

(1) $|z| > 0$ if $z \neq 0, |0|=0$

(2) $|\overline{z}|=|z|$

(3) $|zw|=|z||w|$

(4) $|\operatorname{Re}(z)| \leq |z|$

(5) $|z+w| \leq |z| + |w|$

 

Thm 1.3.6 $|\sum{a_j\overline{b_j}}| \leq \sum{|a_j|^2} \sum{|b_j|^2}$ (CBS ineq in complex field)

pf.

Let $A=\sum{|a_j|^2}, B=\sum{|b_j|^2}, C=\sum{a_j\overline{b_j}}$.

$\begin{align} \sum{|Ba_j-Cb_j|^2} &= \sum{(Ba_j-Cb_j)(\overline{Ba_j-Cb_j})} \\&= \sum{(Ba_j-Cb_j)(B\overline{a_j}-\overline{Cb_j})} \\&= \sum{B^2|a_j|^2+|C|^2|b_j|^2-B\overline{C}a_j\overline{b_j}-BC\overline{a_j}b_j} \\&= B^2A+|C|^2B-B|C|^2-B|C|^2 \\&= B^2A-B|C|^2 \\&\geq 0\end{align}$

$\therefore |C|^2 \leq AB$. $_\blacksquare$

 

References

Rudin, W. (1976). Principles of mathematical analysis (Vol. 3). New York: McGraw-hill.

김성기, 김도한, 계승혁. (1995). 해석개론 (제2개정판). 서울: 서울대학교출판문화원