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 Rudin 【Principle of Mathematical Analysis】Notes & Solutions
04-25-2012, 11:18 AM (This post was last modified: 11-26-2018 04:11 PM by elim.)
Post: #11
 elim Moderator     Posts: 581 Joined: Feb 2010 Reputation: 0
1.17-18 Ordered Fields-- Rudin [Principle of Mathematical Analysis] Notes
1.17 Definition An ordered field is a field $F$ which is also ordered:
$\quad$(i) $\ \forall x,y,z\in F\quad (x < y)\implies (x+z < y+z)$
$\quad$(ii) $\forall x,y\in F\quad\quad\; (x>0)\wedge (y>0)\implies (xy > 0)$
If $x>0$, we call $x$ possitive; if $x < 0$, we call $x$ negative.

$\mathbb{Q}$ is an example of ordered field.

1.18 Proposition Any ordered Field has the following properties:
(a) $(x > 0) \Leftrightarrow (-x < 0)$
(b) $(x > 0)\wedge (y < z)\implies (xy < xz)$
(c) $(x < 0)\wedge (y < z)\implies (xy > xz)$
(d) $(x\ne 0)\implies (x^2 > 0)$. In particular, $1 > 0$
(e) $(0 < x < y)\implies (0 < 1/y < 1/x)$
Proof
(a) $(x > 0)\implies (0 = -x + x > -x + 0 = -x)$, inversely,
$\quad\;\; (x < 0)\implies (0 = -x+x < -x+0)$
(b) $(x>0)\wedge (z > y)\implies (x>0)\wedge (z-y > y-y = 0)$
$\quad\;\;$thus$\; xy + 0 < xy + x(z-y) = xz$
(c) By (a),(b) & Proposition 1.16c, $-(x(z-y)) = (-x)(z-y) > 0$
$\quad\;\;$so $x(z-y)< 0$ or $xz < xy$
(d) If $x > 0$ then Def. 1.17(ii) gives $x^2 > 0$,
$\quad\;\;$if $x < 0,\;$then $-x > 0,\;\; x^2 = (-x)^2 > 0$ by (a) & 1.16(d)
(e) If $((x>0)\wedge (1/x\le 0))\,$then$\; 1 = x(1/x) \le 0$ contradict to (d).
$\quad\;\;$So $1/x > 0\,($likewise $1/y > 0)\,$thus$\,((1/x)(1/y) > 0) \wedge (x < y)$
$\qquad\qquad\qquad\qquad \implies 1/y = x(1/x)(1/y) < y(1/x)(1/y) = 1/x.$
04-26-2012, 02:03 PM (This post was last modified: 01-29-2019 04:41 PM by elim.)
Post: #12
 elim Moderator     Posts: 581 Joined: Feb 2010 Reputation: 0
1.19-1.21 The Real Numbers-- Rudin [Principle Math Analysis] Notes
The Real Field
The core of the chapter: the existence theorem (of Real Field)
1.19 Theorem There exists an ordered field $\mathbb{R}$ which has the
least-upper-bound propery. Moreover, $\mathbb{R}$ contains $\mathbb{Q}$ as a subfield.

The 2nd statement means that $\mathbb{Q}\subset \mathbb{R}$ and the addition and
multiplication in $\mathbb{R}$, when applied to members of $\mathbb{Q}$, coincide with
the usual operations on $\mathbb{Q}$; also, the positive rational numbers
are positive elements of $\mathbb{R}$

The members of $\mathbb{R}$ are called real numbers. The proof of the
theorem, actually constructs $\mathbb{R}$ from $\mathbb{Q}$, will presented as an
Appendix to Chap.1.

We'd like to show how the LUB (least-upper-bound) property
implies the following:

1.20 Theorem
(a) $\forall x\in\mathbb{R}^+,\forall y\in\mathbb{R}\; \exists n\in\mathbb{N}\quad (nx > y)$
(b) $(x,y \in\mathbb{R})\wedge (x < y) \implies (x,y)\cap \mathbb{Q} \ne \varnothing$
Proof
$\quad$(a) If no such an $n$, then $A = \{nx | n\in\mathbb{N}\}$ has upper bound $y$
thus $\alpha = \sup A$ exists in $\mathbb{R}$. Since $x>0$, $\alpha - x$ is not an upper
bound of A hence $\alpha -x < mx \le \alpha$ for some $m\in\mathbb{N}$, but then
$\alpha < (m+1)x$ and so $\alpha$ cannot be an upper bound of $A$.
$\quad$(b) Take(by(a)) $n\in\mathbb{N}$ such that $n(y-x) > 1$. Likewise we
have $m_1,m_2\in\mathbb{N}$ such that $m_1>nx,\; m_2 > -nx$ and so
$-m_2 < nx < m_1$. Thus $m-1\le nx < m$ for some integer $m$
with $-m_2 < m \le m_1$. Combine all these we see that
$nx < m\le 1+nx < n(y-x)+nx = ny$. Since $n$ is positive,
we get $x < \frac{m}{n} < y\quad\square$

Now the existence of $n$th roots of positive reals

1.21 Theorem $\forall x\in\mathbb{R}^+ \ \forall n\in\mathbb{N}^+\; \exists ! y\in\mathbb{R}^+\; (y^n = x)\quad$
$\qquad$[We write $y = \sqrt[n]{x}$ or $y = x^{1/n}$]
Proof The uniqueness of $n$th root is obvious: $y_1^n < y_2^n\small\,(0< y_1 < y_2 )$
Let $t = x/(1+x)$, then $0 < t < \min(x, 1)$ thus $t^n \le t < x$ and so
$E = \{t \in\mathbb{R}: t > 0, t^n < x \} \ne \varnothing.$
Clearly that $1+x$ is an upper bound of $E$ and so $y = \sup E$ exists.
We need the inequality $b^n - a^n < n(b - a)b^{n - 1}\quad (0 < a < b)$.
If $y^n < x$, choose $h\in (0,1)$ such that $h < \frac{x - y^n}{n(y+1)^{n -1}}$, then for
$a = y, b= y +h$ the inequality becomes
$\quad\quad(y +h)^n - y^n < nh(y +h)^{n -1} < nh(y+1)^{n -1} < x -y^{n -1}$
and so $y$ cannot be $\sup E$
If $y^n > x$, let $k = \frac{y^n -x}{ny^{n -1}}$, then $0< k< y$ and for $t \ge y -k$, we have
$\quad\quad y^n - t^n \le y^n -(y -k)^n < nky^{n -1}=y^n -x$ and so
$(t\ge y -k)\implies (t^n > x))$
$\quad\quad y -k (< y)$ is an upper bound of $E$ thus $y\ne \sup E$
Now we see that $y = x^n\quad\quad\square$

Corollary If $a$ and $b$ are positive real numbers and $n$ is a positive integer,
$\quad$then $\quad(ab)^{1/n} = a^{1/n} b^{1/n}.$
Proof Put $\alpha = a^{1/n}, \beta = b^{1/n}$, since multiplication is commutative, we
$\quad$have $\quad ab = \alpha^n \beta^n = (\alpha\beta)^n.$ Now the uniqueness of $n$th root implies
$\quad$that$\quad(ab)^{1/n} = \alpha\beta = a^{1/n} b^{1/n}$
04-26-2012, 02:03 PM (This post was last modified: 01-29-2019 04:50 PM by elim.)
Post: #13
 elim Moderator     Posts: 581 Joined: Feb 2010 Reputation: 0
1.22 Real Field/Decimals --Rudin [Principle of Math Analysis]
1.22 Decimals Let $0< x\in\mathbb{R}$, let $n_0 = \lfloor x\rfloor$. Notice that the existence of
$n_0$ relies on the archimedean property of $\mathbb{R}$. Having chosen $n_0,\cdots,n_{k-1}$,
let $n_k$ be the largest integer such that $\displaystyle{n_0 + \frac{n_1}{10}+\cdots +\frac{n_k}{10^k}\le x}$ and let
$\displaystyle{E = \{n_0 + \frac{n_1}{10}+\cdots +\frac{n_k}{10^k} \; | \; k\in \mathbb{N}\}}.$ The decimal expansion of x is
$n_0.n_1 n_2 n_3 \cdots.$ (Notice that $0\le n_k < 10\quad(\forall k>0)$)
Conversely, for any infinite decimal, the corresponding $E$ is bounded above
and the infinite decimal is the decimal expansion of $\sup E$.
04-27-2012, 09:35 AM (This post was last modified: 01-30-2019 01:59 PM by elim.)
Post: #14
 elim Moderator     Posts: 581 Joined: Feb 2010 Reputation: 0
The Extended Real Number System -- Rudin [Principle of Math Analysis]
1.23 Definition The extended real number system consists of the
real field $\mathbb{R}$ and two symbols, $+\infty$ and $-\infty$. Rreserving the original
order in $\mathbb{R}$ and define$\;-\infty < x < +\infty\quad (\forall x\in\mathbb{R})$ $\quad\quad$
It's clear that $+\infty$ is an upper bound of every subset of the extended
real number system, and every nonempty subset has a least upper
bound. We have similar remarks for lower bounds in this system.

The extended real number system does not form a field. But we have
the following conventions:
(a) $\displaystyle{x + \infty = +\infty,\quad x-\infty = -\infty,\quad \frac{x}{+\infty} = \frac{x}{-\infty} = 0\quad (\forall x\in\mathbb{R})}$
(b) $x\cdot (+\infty) = +\infty,\quad x\cdot (-\infty) = -\infty\quad (0 < x\in\mathbb{R})$
(c) $x\cdot (+\infty) = - \infty,\quad x\cdot (-\infty) = +\infty\quad (0 > x\in\mathbb{R})$

When it is desired to make the distinction between real numbers on
the one hand and the symbols $\pm \infty$ on the other quite explicit, the
former are called finite.
04-30-2012, 01:08 PM (This post was last modified: 01-29-2019 06:21 PM by elim.)
Post: #15
 elim Moderator     Posts: 581 Joined: Feb 2010 Reputation: 0
1.24~35 Complex Field -- Rudin [Principle of Math Analysis]
1.24 Definition A complex number is a pair $(a,b)$ of real numbers.
"Ordered" means that $(a,b)$ and $(b,a)$ are regarded as distinct if $a\ne b$
$\quad$ Let $x=(a,b),\; y=(c,d)$ be two complex numbers. We write $x=y$
iff $a=c$ and $b=d$. (This is not entirely superfluous: think of equality of
rational numbers, represented as quotients of integers.) We define
$\underset{\,}{\qquad}\qquad x+y=(a+c,b+d),\; xy = (ac-bd,ad+bc)$
1.25 Theorem These definition of addition and multiplication turn the set
of all complex numbers $\mathbb{C}$ into a field, with $(0,0)$ and $(1,0)$ in the rule of
$0$ and $1$.
Proof We simply verify the field axioms in Definition 1.12 using the field
structure of $\mathbb{R}$ of course. Let $x = (a,b),\; y=(c,d),\; z=(e,f)$, for
example, if $x \ne 0 = (0,0)$, then $a^2+b^2 > 0$ by 1.18d, and we can define
$\small\displaystyle{\frac{1}{x} = \left(\frac{a}{a^2+b^2},\frac{-b}{a^2+b^2} \right )}$ since $\small\displaystyle{(a,b)\left(\frac{a}{a^2+b^2},\frac{-b}{a^2+b^2}\right ) = (1,0)}$. That's $(M_5)$

For verify (D) in Definition 1.12, we have
\qquad\begin{align} x(y+z) & = (a,b)(c+e,d+f) \\ & = (ac+ae-bd-bf,ad+af+bc+be) \\ & = (ac-bd,ad+bc)+(ae-bf,af+be) \\ & = xy + xz. \end{align}
1.26 Theorem $(a,0)+(b,0) = (a+b,0),\; (a,0)(b,0) = (ab,0).\quad\square$

Theorem 1.26 shows that the complex numbers of the form $(t,0)$ have
the same arithmetic properties as the corresponding real numbers $t$. We
can therefore identify $(t,0)$ with $t$. This identification gives us the real field
as a subfield of the complex field.

1.27 Definition $i = (0,1)$

1.28 Theorem $i^2 = -1$
Proof $i^2 = (0,1)(0,1) = (-1,0) = -1.\quad\quad\square$

1.29 Theorem $(a,b) = a + bi\quad (\forall a,b\in\mathbb{R})$
Proof $a+bi = (a,0)+(b,0)(0,1)=(a,0)+(0,b)=(a,b).\quad\square$

1.30 Definition If $a,b\in\mathbb{R}$ and $z = a+bi$, then $\overline{Z} = a - bi$ is called
the conjugate of $z$. The numbers $a$ and $b$ are the real part and the
imaginary part of $z$, respectively.
$\quad$ We shall occasionally write $a = \operatorname{Re} (z),\quad b = \operatorname{Im}(z).$

1.31 Theorem If $z$ and $w$ are complex, then
$\quad\quad$(a) $\overline{z+w} = \overline{z} + \overline{w}$
$\quad\quad$(b) $\overline{zw} = \overline{z} \overline{w}$
$\quad\quad$(c) $z + \overline{z} = 2\operatorname{Re}(z),\; z -\overline{z} = 2\operatorname{Im}(z)$
$\quad\quad$(d) $z\overline{z}$ is real and positive (except when $z = 0$)$\quad\square$

1.32 Definition The absolute value $|z|\; (z\in\mathbb{C})$ is a non-negative
square root: $|z|=(z\overline{z})^{1/2}$
$\quad$ The existence (and uniqueness) of $|z|$ follows from Th. 1.21&1.31d.
$\quad$ Note that when $x \in \mathbb{R}\subset \mathbb{C}$, then $\overline{x} = x$, hence
$\qquad\qquad |x| = \sqrt{x^2} = x$ if $x\ge 0$, $|x| = -x$ if $x < 0$

1.33 Theorem Let $z$ and $w$ be complex numbers. Then
$\quad\quad$ (a) $|z| > 0$ unless $z = 0,\; |0| = 0$
$\quad\quad$ (b) $|\overline{z}| = |z|$
$\quad\quad$ (c) $|zw| = |z||w|$
$\quad\quad$ (d) $|\operatorname{Re} z| \le |z|$
$\quad\quad$ (e) $|z+w| < |z|+|w|$
Proof (a) and (b) are trivial. Put $z = a+bi,\; w = c+di,\;(a,b,c,d\in\mathbb{R})$.
$\quad\quad$Then$\,|zw|^2 = (ac-bd)^2+(ad+bc)^2 = (a^2+b^2)(c^2+d^2)$
$\quad\quad=|z|^2|w|^2 = (|z||w|)^2$. Now (c) follows from Theorem 1.21.
$\quad\quad$ For (d) we have $(a^2 \le a^2+b^2) \implies (|a|\le \sqrt{a^2+b^2})$
$\quad\quad$ For (e) we have
\qquad\begin{align}|z+w|^2 &=(z+w)(\overline{z}+\overline{w})\\ &= z\overline{z}+z\overline{w}+\overline{z}w+w\overline{w}\\ &= |z|^2 + 2\operatorname{Re}(z\overline{w}) + |w|^2\\ &\le |z|^2 + 2|z\overline{w}| + |w|^2\\ &= |z|^2 + 2|z||w| + |w|^2 = (|z|+|w|)^2. \end{align}
1.34 Notation If $x_1,\cdots,x_n$ are complex numbers, we write
$\qquad\qquad x_1+x_2+\cdots +x_n = {\displaystyle\sum_{j=1}^n} x_j$
We conclude the section with an important inequality, usually known
as the Schwarz inequality.

1.35 Theorem If $a_1,\cdots,a_n,\; b_1,\cdots,b_n \in \mathbb{C}$ then
$\qquad\qquad\qquad\displaystyle \bigg|\sum_{j=1}^n a_j \overline{b}_j\bigg|^2 \le \sum_{j=1}^n |a_j|^2\sum_{j=1}^n |b_j|^2$
Proof$\quad$ Put $A = \sum |a_j|^2,\; B = \sum |b_j|^2,\; C = \sum a_j\overline{b}_j$ (for this proof,
$\quad j = \overline{1,n}$ in all sums). If $B = 0$, then $b_j = 0\quad (j=\overline{1,n})$ and the
$\quad$conclusion is trivial. Assume $B > 0$. By Theorem 1.31 we then have
$\quad\sum |Ba_j - Cb_j|^2 = \sum (Ba_j - Cb_j)(B\overline{a}_j - \overline{C b_j})$
$\qquad = B^2 \sum |a_j|^2 - B\overline{C}\sum a_j\overline{b}_j - BC\sum \overline{a}_j b_j +|C|^2 \sum |b_j|^2$
$\qquad = B^2 A - B|C|^2 = B(BA - |C|^2)$
$\quad$ Since $B > 0$, we get $AB - |C|^2 \ge 0$. this completes the proof. $\;\;\square$
04-30-2012, 06:03 PM (This post was last modified: 01-30-2019 02:27 PM by elim.)
Post: #16
 elim Moderator     Posts: 581 Joined: Feb 2010 Reputation: 0
1.36~1.37 EUCLIDEAN SPACES --Rudin [Principle of Analysis]
1.36 Definitions For each positive integer $k$, let $\mathbb{R}^k$ be the set of
all ordered $k$-tuples $\mathbf{x} = (x_1,x_2,\cdots,x_k)$ Where $x_1,\cdots,x_k\in\mathbb{R}$,
called the coordinates of $\mathbf{x}$. The elements of $\mathbb{R}^k$ are called points,
or vectors, especially when $k>1$. We shall denote vectors by
boldfaced letters. If $\mathbf{y} = (y_1,\cdots,y_k)$ and $\alpha \in\mathbb{R}$, put
$\qquad\mathbf{x} + \mathbf{y} = (x_1+y_1,\cdots,x_k+y_k), \quad \alpha \mathbf{x} = (\alpha x_1,\cdots,\alpha x_k)$
So $\mathbf{x}+\mathbf{y},\; \alpha \mathbf{x} \in\mathbb{R}^k$. This defines addition of vectors, as well as
multiplication of a vector by a real number (a scalar). These two
operations satisfy the commutative, associative and distributive
laws (trivial) and make $\mathbb{R}^k$ into a vector space over the real field.
The zero element of $\mathbb{R}^k$ (sometimes called the origin or the null
vector) is the point $\mathbf{0}$ with all the coordinates are $0\underset{\,}{.}$
We shall define the so called "inner product" (or scalar product)
of $\mathbf{x}$ and $\mathbf{y}$ by$\;\;\mathbf{x}\cdot \mathbf{y}={\small\displaystyle\sum_{i=1}^k} x_i y_i$ and the norm of $\mathbf{x}$ by
$\qquad\qquad\qquad\quad |\mathbf{x}| = (\mathbf{x}\cdot \mathbf{x})^{1/2} = \bigg({\small\displaystyle\sum_{i=1}^k} x_i^2\bigg)$
The structure now defined (the vector space $\mathbb{R}^k$ with the above
inner product and norm) is called euclidean $k$-space.

1.37 Theorem Let $\mathbf{x,y,z} \in \mathbb{R}^k$ and $\alpha \in\mathbb{R}$, Then
$\quad\quad$ (a) $|\mathbf{x}| \ge 0$
$\quad\quad$ (b) $|\mathbf{x}| = 0 \Longleftrightarrow \mathbf{x} = \mathbf{0}$
$\quad\quad$ (c) $|\alpha \mathbf{x}| = |\alpha||\mathbf{x}|$
$\quad\quad$ (d) $|\mathbf{x}\cdot \mathbf{y}| \le |\mathbf{x}||\mathbf{y}|$
$\quad\quad$ (e) $|\mathbf{x}+\mathbf{y}| \le |\mathbf{x}|+|\mathbf{y}|$
$\quad\quad$ (f) $|\mathbf{x}-\mathbf{z}| \le |\mathbf{x}-\mathbf{y}|+|\mathbf{y}-\mathbf{z}|$
Proof (a),(b) and (c) are obvious and (d) is an immediate
$\quad$consequence of the Schwarz inequality. Then by (d) we have
\qquad\begin{align} |\mathbf{x}+\mathbf{y}|^2 & = (\mathbf{x}+\mathbf{y})\cdot (\mathbf{x}+\mathbf{y})\\ & =\mathbf{x\cdot x} + 2\mathbf{x}\cdot \mathbf{y}+\mathbf{y}\cdot\mathbf{y}\\ & \le |\mathbf{x}|^2 +2|\mathbf{x}||\mathbf{y}|+|\mathbf{y}|^2\\ & = (|\mathbf{x}|+|\mathbf{y}|)^2 \end{align}
$\quad$and so (e) is proved. Finally, (f) follows from (e) when $\mathbf{x,\;y}$
$\quad$are replaced by $\mathbf{x-y,\; y-z}$ respectively.
04-30-2012, 06:03 PM
Post: #17
 elim Moderator     Posts: 581 Joined: Feb 2010 Reputation: 0
1.38 Remark(EUCLIDEAN SPACES) --Rudin [Principle of Mathematical Analysis] Notes
1.38 Remarks Theorem 1.37(a),(b) and (f) will allow us (see Chap.2) to regard $\mathbb{R}^k$ as a metric space.
$\quad\quad \mathbb{R}^1$ (the set of all real numbers) is usually called the line, or the real line. Likewise $\mathbb{R}^2$ is called the plane, or the complex plane (compare Definitions 1.24 and 1.36). In these two cases the norm is just the absolute value of the corresponding real or complex number.
05-01-2012, 01:25 PM (This post was last modified: 01-30-2019 05:47 PM by elim.)
Post: #18
 elim Moderator     Posts: 581 Joined: Feb 2010 Reputation: 0
Appendix Ch1 Construction of Real Field --Rudin [Principle of Analysis]
APPENDIX The proof of Theorem 1.19, i.e. constructing $\mathbb{R}$ from $\mathbb{Q}$ will
presented in several steps.

Step 1 A set $\alpha$ is called a cut of $\mathbb{Q}$ if $\alpha\subset\mathbb{Q}$ with the following properties:
\quad\quad \begin{align} \text{(I)} & \varnothing \ne \alpha \ne \mathbb{Q}\\ \text{(II)} & (p\in\alpha)\wedge (p> q\in\mathbb{Q}) \implies (q\in\alpha) \\ \text{(III)} & \forall p\in\alpha \exists r\in\alpha\; (p < r) \end{align}
Let $\mathbb{R} = \{\alpha \mid \alpha \text{ is a cut of } \mathbb{Q}\}$, and assume that in this appendix, the
letters $p,q,r,\cdots$ denote rational numbers and $\alpha,\beta,\gamma,\cdots$ denote cuts.
$\qquad$ (II) implies two facts which will be used freely:
$\qquad\qquad(p\in\alpha)\wedge (q\not\in\alpha)\implies (p < q),\;\;(s > r\not\in\alpha) \implies (s\not\in\alpha)$
The the axioms (M) and (D) of Definition 1.12 hold with $\mathbb{R}^+$ in place of
$F$ and $1^*$ in the role of $1$. The proofs are so similar to the ones given in
detail in Step 4 that we omit them. Note, in particular, the multiplication
property of Definition 1.17 holds: $\alpha,\beta\in 0^* \implies \alpha\beta > 0^*$

Step 2 Define "$\alpha < \beta$" to mean: $\alpha \subsetneq \beta$ ($\alpha$ ia a proper subset of $\beta$)
$\quad\quad$ We need to show this meets the requirements of Definition 1.5.
$\quad\quad$ Clearly $(\alpha < \beta < \gamma \implies \alpha < \gamma)$ and at most one of $3$ relations
$\qquad\;\alpha < \beta, \alpha = \beta, \beta < \alpha$ can hold for any pair of $\alpha,\beta$. To show at
$\qquad\;$least one such relation holds, assume that first 2 relations fail,
$\qquad\;$then $\alpha$ is not a subset of $\beta$ hence there is a $p\in\alpha$ with $p\not\in\beta$.
$\qquad\;$So $q\in\beta \implies q < p \implies q\in\alpha$, i.e. $\beta < \alpha$
$\quad\quad$ Thus $\mathbb{R}$ is now an ordered set.

Step 3 $\mathbb{R}$ has the least-upper-bound property.
$\qquad$ Let $\varnothing \ne A \subset \mathbb{R}$, and $\beta\in\mathbb{R}$ is an upper bound of $A$. Let $\gamma = \bigcup A$,
$\qquad$then $\varnothing \ne \gamma \subset \beta$ hence $\gamma \ne \mathbb{Q}$. It's easy to further prove that $\gamma$ is a cut
$\qquad$(a member of $\mathbb{R}$) and an upper bound of $A$.
$\qquad$Suppose $\delta < \gamma$, then $s\not\in\delta$ for some $s \in\gamma$. So $s\in\alpha$ for some $\alpha\in A$
$\qquad$and $\delta < \alpha$. Therefore $\delta$ is not an upper bound of $A$. We conclude that
$\qquad\gamma = \sup A$.

Step 4 Define $\alpha + \beta = \{r+s \mid r\in\alpha,\; s\in\beta \}, \; 0^* = \{r \in\mathbb{Q}: r < 0 \}$.
$\quad$Clearly $0^*$ is a cut. We shall verify the axioms for addition (in Definition
$\quad$1.12) hold in $\mathbb{R}$ with $0^*$ playing the role of $0$.
$\quad(A_1)$ Take ${r}',{s}' \in\mathbb{Q}\setminus (\alpha \cup \beta)$, then $\forall r\in\alpha \; \forall s\in\beta \;(r+s < {r}'+{s}')$
$\qquad\quad\implies ({r}'+{s}' \not\in \alpha+\beta \ne \varnothing)$ So $\alpha+\beta$ has cut property (I).
$\quad$ If $p\in\alpha+\beta$ then $p = r+s$ for some $r\in\alpha,\;s \in\beta$. To see (II) holds,
$\quad$ for $q < p$, we have $(q-s < r)\implies q-s\in\alpha \implies q = (q-s)+s$
$\quad\;\in \alpha +\beta$; Now choose $t\in\alpha$ so that $t > r$ then $p < t+s \in \alpha+\beta$.
$\quad$ Thus (III) holds and so $\alpha + \beta \in\mathbb{R} (\forall \alpha,\beta\in\mathbb{R})$
$\quad(A_2) \quad \alpha +\beta = \{r+s \mid r\in\alpha, s\in\beta\} = \{s+r \mid s\in\beta, r\in\alpha\} = \beta+\alpha$
$\quad(A_3)$ This, similar to the above, follows from the associative law in $\mathbb{Q}$
$\quad(A_4)$ If $r\in\alpha,\; s\in 0^*$, then $r+s < r$ thus $\alpha + 0^*\subset \alpha$. On the other
$\quad$hand, if $p\in\alpha$, then $p < r \in\alpha$ for some $r$ so $p = r + (p-r)$ with $p-r$
$\quad\in 0^*$ and so $\alpha \subset \alpha+0^*$. We conclude that $\alpha + 0^* = \alpha$
$\quad(A_5)$ Fix $\alpha\in\mathbb{R}$, let $\beta = \{p\in\mathbb{Q}: \exists r > 0\; (-p -r \in\mathbb{Q}\setminus \alpha)\}$.
$\quad$We'll show $\beta\in\mathbb{R}$ and $\alpha+\beta = 0^*.\quad$ If $s\not\in\alpha$, take $p = -s + 1$ then
$\quad -p -1 = s\not\in\alpha$ hence $p\in\beta$; If $q\in\alpha$ then $-q \not\in\beta$ so $\varnothing \ne \beta \ne \mathbb{Q}$.
$\quad$ For $p\in\beta$, pick $r > 0$ such that $-p-r \not\in\alpha$. If $q < p$, then $-q -r >$
$\quad -p -r$ hence $-q -r \not\in\alpha$ and $q\in\beta$. So (II) holds. Take $t = p + (r/2)$,
$\quad$then $-t -(r/2) = -p -r\not\in\alpha$. So $p < t \in\beta$, (III) holds. $\therefore\;\beta\in\mathbb{R}$.
$\quad$Let $r\in\alpha,\; s\in\beta$, then$\;-s\not\in\alpha \Rightarrow r < -s \Rightarrow r+s < 0$ so$\small\;\alpha+\beta \subset 0^*$
$\quad$ Conversely, if $v\in 0^*$, let $w = -v/2$ then $w > 0$ and there is an integer
$\quad n$ such that $nw\in\alpha$ but $(n+1)w \not\in\alpha$ (by Archimedean property of $\mathbb{Q}$).
$\quad$Put $p = -(n+2)w$ then $-p-w = (n+1)w \not\in\alpha \implies p\in\beta$ and
$\quad v = -2w = nw + p\in\alpha + \beta$ thus $0^*\subset \alpha+\beta$.
$\quad$We conclude that $\alpha + \beta = 0^*$ and got the reason to denote $\beta$ as $-\alpha$.
$\quad$(see $(A_4)$)

Step 5 Having proved the addition defined in Step 4 satisfied Axioms (A)
$\quad$in Definition 1.12, Proposition 1.14 is thus valid in $\mathbb{R}$ and we can prove
$\quad$the addition property of ordered field (Definition 1.17):
$\quad\quad\quad \forall \alpha,\beta,\gamma \in\mathbb{R}\; (\beta < \gamma)\implies (\alpha + \beta < \alpha + \gamma)$.
$\quad$ Indeed, it's clear in this case that $\alpha+\beta \subset \alpha+\gamma$ by the definition of $+$
$\quad$ in $\mathbb{R}$. If $\alpha+\beta = \alpha+\gamma$, then the cancellation law (Proposition 1.14)
$\quad$ would imply $\beta = \gamma$. It follows that $\alpha > 0^*$ if and only if $-\alpha < 0^*$.

Step 6 Multiplication is a little more bothersome than addition in the
$\quad$present context. since products of negative rationals are positive. For
$\quad$this reason we confine ourselves first to $\mathbb{R}^+ = \{\alpha\in\mathbb{R}: \alpha > 0^*\}$
$\quad$For $\alpha,\beta \in \mathbb{R}^+$ define $\alpha\beta = \{p\in\mathbb{Q}: p < rs, (0< r\in\alpha,\; 0 < s\in\beta)\}$
$\quad$and $1^* = \{q\in\mathbb{Q}: q < 1\}$
$\quad$The this axioms (M) and (D) of Definition 1.12 hold with $\mathbb{R}^+$ in place
$\quad$of $F$ and $1^*$ in the role of $1$.
$\quad$The proofs are so similar to the ones given in detail in Step 4 that we
$\quad$omit them. Note, in particular, the multiplication property of Definition
$\quad$1.17 holds $\alpha,\beta > 0^* \implies \alpha\beta > 0^*$

Step 7 We complete the definition of multiplication by setting
$\qquad\qquad\alpha 0^* = 0^* \alpha = 0^*$, and
$\qquad\alpha\beta =\begin{cases} (-\alpha)(-\beta)& \text{ if } \alpha < 0^*,\;\beta < 0^* \\ -[(-\alpha)\beta]& \text{ if } \alpha < 0^*,\;\beta > 0^* \\ -[\alpha(-\beta)]& \text{ if } \alpha > 0^*,\;\beta < 0^* \end{cases}$
$\quad$The products on the right were defined in Step 6.
$\quad$Having proved (Step 6) that the axioms (M) hold in $\mathbb{R}^+$, it's
$\quad$perfectly simple to prove them in $\mathbb{R}$ by repeated application of the
$\quad$identity $\gamma = -(-\gamma)$ which is part of Proposition 1.14.
$\quad$The distributive law $\alpha(\beta+\gamma) = \alpha \beta + \alpha \gamma$ is proved in several cases.
$\quad$For instance, suppose $\alpha > 0^*,\; \beta < 0^*,\; \beta + \gamma > 0^*$, then $\gamma = (\beta+\gamma)$
$\quad\; + (-\beta)$ and (since we already have the distributive law in
$\quad\mathbb{R}^+$) $\alpha\gamma = \alpha(\beta+\gamma)+\alpha(-\beta)$. But $\alpha (-\beta) = -(\alpha\beta)$,
$\quad$we see that $\alpha(\beta+\gamma) = \alpha \beta + \alpha \gamma$ this case.
$\quad$The other cases are handled in the same way.
$\quad$We have now completed the proof that $\mathbb{R}$ is an ordered field with
$\quad$the least upper-bound property.

Step 8 For $r\in\mathbb{Q}$, let $r^* = \{p\in\mathbb{Q}: p < r\}$. It's clear that $r^*$ is a cut;
$\qquad$ that is, $r^*\in\mathbb{R}$. We have
$\qquad$ (a) $r^* + s^* = (r+s)^*$
$\quad\quad$ (b) $r^* s^* = (rs)^*$
$\quad\quad$ (c) $(r^* < s^*) \Longleftrightarrow (r < s)$

$\quad\quad$ To proof (a), assume $p\in (r^*+s^*)$, then $p = u+v$ for some
$\qquad\; u < r,\; v < s$. So $p < r+s$ thus $p \in (r+s)^*$.
$\qquad$ Conversely, if $p\in (r+s)^*$, then $p < r+s$. Choose $t$ so that
$\qquad\;2t = r+s -p$ and put ${r}' = r-t,\; {s}' = s-t$. Then ${r}'\in r^*,$
$\qquad\;{s}'\in s^*$ and $p = {r}'+{s}'$, so $p\in (r^* + s^*)$.
$\quad\quad$ This proves (a). The proof of (b) is similar.
$\quad\quad$ (c): $r < s\implies (r\in s^*)\wedge (r\not\in r^*)\implies$
$\qquad\qquad r^* < s^* \implies \exists p\in s^* \; (p\not\in r^*)\implies (r\le p < s)\quad\quad\square$

Step 9 We saw in Step 8 that the correspondence "$r \to r^*$" preserves
$\quad$sums, products and order. This can be expressed by saying that the
$\quad$ordered field $\mathbb{Q}$ is isomorphic to the ordered field $\mathbb{Q}^*$ whose elements
$\quad$are the rational cuts.
$\quad$It is this identification of $\mathbb{Q}$ with $\mathbb{Q}^*$ which allows us to regard $\mathbb{Q}$ as a
$\quad$subfield of $\mathbb{R}$. The 2nd part of Theorem 1.19 is to be understood in
$\quad$terms of this identification.
$\quad$It is a fact, which we'll not prove here, that any two ordered fields
$\quad$with the least-upper-bound property are isomorphic. The 1st part of
$\quad$Theorem 1.19 therefore characterizes the real field $\mathbb{R}$ completely.
05-04-2012, 11:17 AM (This post was last modified: 12-03-2019 05:28 PM by elim.)
Post: #19
 elim Moderator     Posts: 581 Joined: Feb 2010 Reputation: 0
Historical Notes of the Number System Theory --Rudin 【PMA】
$\;\;$The book by Landau and Thurston cited in the Bibliography
are entirely devoted to number systems. Chap. 1 of Knopp's
book contains a more leisurely description of how $\mathbb{R}$ can be
obtained from $\mathbb{Q}$. Another construction defines real number
as an equivalence class of Cauchy sequences of rational num-
bers(see Chap.3), is carried out in Section. 5 of the book by
Hewitt and Stromberg.
$\;\;$The cuts in $\mathbb{Q}$ that we used here were invented by Dedekind.
Rational Cauchy sequence equivalence classes' approach of
construction is due to Cantor. Both Cantor and Dedekind pub-
lished their constructions in 1872.
05-04-2012, 11:17 AM
Post: #20
 elim Moderator     Posts: 581 Joined: Feb 2010 Reputation: 0
Chap. I Reader Review --Rudin [Principle of Mathematical Analysis] Notes
Very condensed materials.
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