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 Rudin 【Principle of Mathematical Analysis】Notes & Solutions
04-25-2012, 11:18 AM (This post was last modified: 10-29-2018 01:19 PM by elim.)
Post: #11
 elim Moderator Posts: 578 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 an ordered set, such that
$\quad$(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
Post: #12
 elim Moderator Posts: 578 Joined: Feb 2010 Reputation: 0
1.19-1.21 The Real Numbers-- Rudin [Principle of Mathematical 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 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
$\quad\quad$(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 $A$.
$\quad\quad$(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 $m_1>nx,\; m_2 > -nx$ i.e. $-m_2 < nx < m_1$. Therefore there is an integer $m$ such that $m-1\le nx < m$ (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 $\forall x\in\mathbb{R}^+ \ \forall n\in\mathbb{N}^+\; \exists ! y\in\mathbb{R}^+\; (y^n = x)\quad$ [We write $y = \sqrt[n]{x}$ or $y = x^{1/n}$]
Proof The uniqueness of $n$th root is obvious since $0< y_1 < y_2 \implies y_1^n < y_2^n$
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 give
$\quad\quad(y +h)^n - y^n < nh(y +h)^{n -1} < nh(y+1)^{n -1} < x -y^{n -1}$ 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 $(t\ge y -k)\implies (t^n > x))$
$\quad\quad y -k ( Now we see that$y = x^n\quad\quad\square$Corollary If$a$and$b$are positive real numbers and$n$is a positive integer, then$(ab)^{1/n} = a^{1/n} b^{1/n}\quad\quad\quad$Proof Put$\alpha = a^{1/n}, \beta = b^{1/n}$, since multiplication is commutative, we have$ab = \alpha^n \beta^n = (\alpha\beta)^n$Now the uniqueness of$n$th root implies that$(ab)^{1/n} = \alpha\beta = a^{1/n} b^{1/n}$04-26-2012, 02:03 PM Post: #13  elim Moderator Posts: 578 Joined: Feb 2010 Reputation: 0 1.22 Real Field/Decimals --Rudin [Principle of Mathematical Analysis] Notes 1.22 Decimals Let$0< x\in\mathbb{R}$, let$n_0$be the largest integer no more than$x$. 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 Post: #14  elim Moderator Posts: 578 Joined: Feb 2010 Reputation: 0 The Extended Real Number System -- Rudin [Principle of Mathematical Analysis] Notes 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 Post: #15  elim Moderator Posts: 578 Joined: Feb 2010 Reputation: 0 1.24~35 Complex Field -- Rudin [Principle of Mathematical Analysis] Notes 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\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 $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$\displaystyle{\frac{1}{x} = \left(\frac{a}{a^2+b^2},\frac{-b}{a^2+b^2} \right )}$since$\displaystyle{(a,b)\left(\frac{a}{a^2+b^2},\frac{-b}{a^2+b^2}\right ) = (1,0) = 1}$. That's$(M_5)For verify (D) in Definition 1.12, we have \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(\forall x,b\in\mathbb{R}) \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\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\quad$The existence (and uniqueness) of$|z|$follows from Theorem 1.21 and Theorem 1.31d.$\quad\quad$Note that when$x \in \mathbb{R}\subset \mathbb{C}$, then$\overline{x} = x$, hence$|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$, with$a,b,c,d\in\mathbb{R}$. Then$\quad\quad |zw|^2 = (ac-bd)^2+(ad+bc)^2 = (a^2+b^2)(c^2+d^2)=|z|^2|w|^2 = (|z||w|)^2$. Now (c) follows from the uniqueness assertion of 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\quadFor (e) we have \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 Ifx_1,\cdots,x_n$are complex numbers, we write $x_1+x_2+\cdots +x_n = \sum_{j=1}^n x_j$$\quad\quad$We conclude the section with an important inequality, usually known as the Schwarz inequality. 1.35 Theorem If$a_1,\cdots,a_n,\; b_1,\cdots,b_n \in \mathbb{C}$then $\left|\sum_{j=1}^n a_j \overline{b}_j\right|^2 \le \sum_{j=1}^n |a_j|^2\sum_{j=1}^n |b_j|^2$ Proof$\quad$Put$A = \sum |a_j|^2,\; B = \sum |b_j|^2,\; C = \sum a_j\overline{b}_j$(for this proof,$j = \overline{1,n}$in all sums). If$B = 0$, then$b_j = 0\quad (j=\overline{1,n})$and the conclusion is trivial. Assume$B > 0$. By Theorem 1.31 we then have$\begin{align}\sum |Ba_j - Cb_j|^2 & = \sum (Ba_j - Cb_j)(B\overline{a}_j - \overline{C b_j})\\
& = B^2 \sum |a_j|^2 - B\overline{C}\sum a_j\overline{b}_j - BC\sum \overline{a}_j b_j +|C|^2 \sum |b_j|^2\\
& = B^2 A - B|C|^2 = B(BA - |C|^2) \end{align}\quad\quad$Since$B > 0$, we see that$AB - |C|^2 \ge 0$. this is the desired inequality.$\quad\quad\square$04-30-2012, 06:03 PM Post: #16  elim Moderator Posts: 578 Joined: Feb 2010 Reputation: 0 1.36~1.37 EUCLIDEAN SPACES --Rudin [Principle of Mathematical Analysis] Notes 1.36 Definitions For each positive integer$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 $\mathbf{x} + \mathbf{y} = (x_1+y_1,\cdots,x_k+y_k), \quad \alpha \mathbf{x} = (\alpha x_1,\cdots,\alpha x_k)$ 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$0\quad\quad$We shall define the so called "inner product" (or scalar product) of$\mathbf{x}$and$\mathbf{y}$by $\mathbf{x}\cdot \mathbf{y}=\sum_{i=1}^k x_i y_i$ and the norm of$\mathbf{x}$by $|\mathbf{x}| = (\mathbf{x}\cdot \mathbf{x})^{1/2} = \left(\sum_{i=1}^k x_i^2 \right )^{1/2}$$\quad\quad$The structure now defined (the vector space$\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 consequence of the Schwarz inequality. Then by (d) we have \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} and so (e) is proved. Finally, (f) follows from (e) when\mathbf{x,\;y}$are replaced by$\mathbf{x-y,\; y-z}$respectively. 04-30-2012, 06:03 PM Post: #17  elim Moderator Posts: 578 Joined: Feb 2010 Reputation: 0 1.38 Remark(EUCLIDEAN SPACES) --Rudin [Principle of Mathematical Analysis] Notes 1.38 Remarks Theorem 1.37(a),(b) and (f) will allow us (see Chap.2) to regard$\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$is called the plane, or the complex plane (compare Definitions 1.24 and 1.36). In these two cases the norm is just the absolute value of the corresponding real or complex number. 05-01-2012, 01:25 PM (This post was last modified: 10-29-2018 12:37 PM by elim.) Post: #18  elim Moderator Posts: 578 Joined: Feb 2010 Reputation: 0 Appendix Ch1 Construction of Real Field --Rudin [Principle of Mathematical Analysis] Notes APPENDIX The proof of Theorem 1.19, i.e. constructing$\mathbb{R}$from$\mathbb{Q}$will presented in several steps. Step 1 A set$\alpha$is called a cut of$\mathbb{Q}$if it is contained in$\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$denote cuts.$\quad\quad$(II) implies two facts which will be used freely:$\quad\quad\quad\quad\quad (p\in\alpha)\wedge (q\not\in\alpha)\implies (p < q)\;$, and$(s > r\not\in\alpha) \implies (s\not\in\alpha)\quad\quad$The the axioms (M) and (D) of Definition 1.12 hold with$\mathbb{R}^+$in place of$F$and$1^*$in the role of$1$.$\quad\quad$The proofs are so similar to the ones given in detail in Step 4 that we omit them.$\quad\quad$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$\alpha < \beta, \alpha = \beta, \beta < \alpha$can hold for any pair of$\alpha,\beta$. To show at least one such relation holds, assume that first 2 relations fail, then$\alpha$is not a subset of$\beta$hence there is a$p\in\alpha$with$p\not\in\beta$. 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.$\quad\quad$Let$\varnothing \ne A \subset \mathbb{R}$, and$\beta\in\mathbb{R}$is an upper bound of$A$. Let$\gamma = \bigcup A$, then$\varnothing \ne \gamma \subset \beta$hence$\gamma \ne \mathbb{Q}$. It's easy to further prove that$\gamma$is a cut (a member of$\mathbb{R}$) and an upper bound of$A$.$\quad\quad$Suppose$\delta < \gamma$, then$s\not\in\delta$for some$s \in\gamma$. So$s\in\alpha$for some$\alpha\in A$and$\delta < \alpha$. Therefore$\delta$is not an upper bound of$A$. We conclude that$\gamma = \sup A$. Step 4 Define$\alpha + \beta = \{r+s \mid r\in\alpha,\; s\in\beta \}, \; 0^* = \{r \in\mathbb{Q}: r < 0 \}$. Clearly$0^*$is a cut. We shall verify the axioms for addition (in Definition 1.12) hold in$\mathbb{R}$with$0^*$playing the role of$0$.$\quad\quad (A_1)$Take${r}',{s}' \in\mathbb{Q}\setminus (\alpha \cup \beta)$, then$\forall r\in\alpha \; \forall s\in\beta \;(r+s < {r}'+{s}') \implies ({r}'+{s}' \not\in \alpha+\beta \ne \varnothing)$So$\alpha+\beta$has cut property (I).$\quad\quad$If$p\in\alpha+\beta$then$p = r+s$for some$r\in\alpha,\;s \in\beta$. To see (II) holds, for$q < p$, we have$(q-s < r)\implies q-s\in\alpha \implies q = (q-s)+s \in \alpha +\beta$; Now choose$t\in\alpha$so that$t > r$then$p < t+s \in \alpha+\beta$. Thus (III) holds and so$\alpha + \beta \in\mathbb{R} (\forall \alpha,\beta\in\mathbb{R})\quad\quad (A_2) \quad \alpha +\beta = \{r+s \mid r\in\alpha, s\in\beta\} = \{s+r \mid s\in\beta, r\in\alpha\} = \beta+\alpha\quad\quad (A_3)$This, similar to the above, follows from the associative law in$\mathbb{Q}\quad\quad (A_4)$If$r\in\alpha,\; s\in 0^*$, then$r+s < r$thus$\alpha + 0^*\subset \alpha$. On the other hand, if$p\in\alpha$, then$p < r \in\alpha$for some$r$so$p = r + (p-r)$with$p-r\in 0^*$and so$\alpha \subset \alpha+0^*$. We conclude that$\alpha + 0^* = \alpha\quad\quad (A_5)$Fix$\alpha\in\mathbb{R}$, let$\beta = \{p\in\mathbb{Q}: \exists r > 0\; (-p -r \in\mathbb{Q}\setminus \alpha)\}$. We'll show$\beta\in\mathbb{R}$and$\alpha+\beta = 0^*.\quad$If$s\not\in\alpha$, take$p = -s + 1$then$-p -1 = s\not\in\alpha$hence$p\in\beta$; If$q\in\alpha$then$-q \not\in\beta$so$\varnothing \ne \beta \ne \mathbb{Q}$.$\quad\quad$For$p\in\beta$, pick$r > 0$such that$-p-r \not\in\alpha$. If$q < p$, then$-q -r > -p -r$hence$-q -r \not\in\alpha$and$q\in\beta$. So (II) holds. Take$t = p + (r/2)$, then$-t -(r/2) = -p -r\not\in\alpha$. So$p < t \in\beta$, (III) holds. Therefore$\beta\in\mathbb{R}$.$\quad\quad$Let$r\in\alpha,\; s\in\beta$, then$-s\not\in\alpha \implies r < -s \implies r+s < 0$so$\alpha+\beta \subset 0^*\quad\quad$Conversely, if$v\in 0^*$, let$w = -v/2$then$w > 0$and there is an integer$n$such that$nw\in\alpha$but$(n+1)w \not\in\alpha$ (by Archimedean property of \mathbb{Q}). Put$p = -(n+2)w$then$-p-w = (n+1)w \not\in\alpha \implies p\in\beta$and$v = -2w = nw + p\in\alpha + \beta$thus$0^*\subset \alpha+\beta$.$\quad\quad$We conclude that$\alpha + \beta = 0^*$and got the reason to denote$\beta$as$-\alpha$. (see$(A_4)$) Step 5 Having proved the addition defined in Step 4 satisfied Axioms (A) in Definition 1.12, Proposition 1.14 is thus valid in$\mathbb{R}$and we can prove the addition property of ordered field (Definition 1.17):$\quad\quad\quad \forall \alpha,\beta,\gamma \in\mathbb{R}\; (\beta < \gamma)\implies (\alpha + \beta < \alpha + \gamma)$.$\quad\quad$Indeed, it's clear in this case that$\alpha+\beta \subset \alpha+\gamma$by the definition of$+$in$\mathbb{R}$. If$\alpha+\beta = \alpha+\gamma$, then the cancellation law (Proposition 1.14) would imply$\beta = \gamma$.$\quad\quad$It follows that$\alpha > 0^*$if and only if$-\alpha < 0^*$. Step 6 Multiplication is a little more bothersome than addition in the present context. since products of negative rationals are positive. For this reason we confine ourselves first to$\mathbb{R}^+ = \{\alpha\in\mathbb{R}: \alpha > 0^*\}\quad\quad$For$\alpha,\beta \in \mathbb{R}^+$define$\alpha\beta = \{p\in\mathbb{Q}: p < rs, (0 $\quad\quad$ The this axioms (M) and (D) of Definition 1.12 hold with $\mathbb{R}^+$ in place of $F$ and $1^*$ in the role of $1$.
$\quad\quad$ The proofs are so similar to the ones given in detail in Step 4 that we omit them.
$\quad\quad$ Note, in particular, the multiplication property of Definition 1.17 holds $\alpha,\beta > 0^* \implies \alpha\beta > 0^*$

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

Step 8 For $r\in\mathbb{Q}$, let $r^* = \{p\in\mathbb{Q}: p$\quad\quad$(a)$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$for some$u < r,\; v < s$. So$p < r+s$thus$p \in (r+s)^*$.$\quad\quad$Conversely, if$p\in (r+s)^*$, then$p < r+s$. Choose$t$so that$2t = r+s -p$ and put${r}' = r-t,\; {s}' = s-t$. Then${r}'\in r^*,\; {s}'\in s^*$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 sums, products and order. This can be expressed by saying that the ordered field$\mathbb{Q}$is isomorphic to the ordered field$\mathbb{Q}^*$whose elements are the rational cuts.$\quad\quad$It is this identification of$\mathbb{Q}$with$\mathbb{Q}^*$which allows us to regard$\mathbb{Q}$as a subfield of$\mathbb{R}$.$\quad\quad$The 2nd part of Theorem 1.19 is to be understood in terms of this identification.$\quad\quad$It is a fact, which we'll not prove here, that any two ordered fields with the least-upper-bound property are isomorphic. The 1st part of Theorem 1.19 therefore characterizes the real field$\mathbb{R}$completely. 05-04-2012, 11:17 AM Post: #19  elim Moderator Posts: 578 Joined: Feb 2010 Reputation: 0 Historical Notes of the Number System Theory --Rudin 【Principle of Mathematical Analysis】Notes$\quad\quad$The book by Landau and Thurston cited inn 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 numbers(see Chap.3), is carried out in Sec. 5 of the book by Hewitt and Stromberg.$\quad\quad$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 published their constructions in 1872.
05-04-2012, 11:17 AM
Post: #20
 elim Moderator Posts: 578 Joined: Feb 2010 Reputation: 0
Chap. I Reader Review --Rudin [Principle of Mathematical Analysis] Notes
Very condensed materials.
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