• 1 Votes - 5 Average
• 1
• 2
• 3
• 4
• 5
 Rudin 【Principle of Mathematical Analysis】Notes & Exercise Solutions
10-04-2010, 05:41 PM (This post was last modified: 12-15-2014 02:52 PM by elim.)
Post: #1
 elim Moderator Posts: 544 Joined: Feb 2010 Reputation: 0
Rudin 【Principle of Mathematical Analysis】Notes & Exercise Solutions

Index
 1. $\mathbb{R}$ and $\mathbb{C}$ [Intro] $\le$ [Fields] [$\mathbb{R}$] [$\overline{\mathbb{R}}$] [$\mathbb{C}$] [$\mathbb{R}^k$] [Appendix] [Ex] 2. Basic Topology $|Set|$ $(M,d)$ [Compact Sets] [Perfect Sets] [Connected Sets] [Ex]
Symbols
 $\mathbb{J}$ $\mathbb{J} = \mathbb{N}^+ = \{n\in\mathbb{N}\mid n > 0\},\;\mathbb{N}$ is the set of natural numbers. $B^A$ $B^A = \{f\mid f:A\to B\}$ whose member is called a function/mapping from $A$ to $B$ $\{a_n\} \in E^{\mathbb{J}}$ means that $\{a_n\}: \mathbb{J} \to E$, call $\{a_n\}$ a sequence in $E$ $\text{In}(\mathscr{F})$ $\text{In}(\mathscr{F}) = \{f\in \mathscr{F}:\;\forall x,y\in A\;((x\ne y)\implies (f(x)\ne f(y)))\}$ (Injections in $\mathscr{F}\subset B^A$) $\mathscr{M}(E,G)$ $\mathscr{M}(E,G) = \{f\in G^E: f(x) < f(y)\implies x < y\}$ ($E,\,G$ are ordered sets) $f\in \mathscr{M}(E,G)$ is said to be monotonically increasing. $\mathscr{M}(E)$ is understood to be $\mathscr{M}(E,G)$ where $G\subset\mathbb{R}$ is assumed in the context. $\{a_{n_j}\}\subset_{seq} \{a_n\}$ $\{a_{n_j}\} = \{a_n\}{\small{\circ}} \{n_j\}\;$ is called a subsequence of $\{a_n\}$ (where $\{n_j\}\in \text{In}(\mathscr{M}(\mathbb{J},\mathbb{J}))$). $\mathscr{C}^n (E, F)$ $\mathscr{C}^n (E, F)=\{f \mid (f\in F^B) \wedge (f^{(m)}\text{ is continuous on }E\;(m < n+1))\}$ ($n$ can be $\infty$ then) If $F$ is understood be given in context, for short write $\mathscr{C}^n(E)$ If both $E,\; F$ are understood be given, it may simply denoted by $\mathscr{C}^n$ We use $f\in \mathscr{C}^n (\{x_1, x_2, \ldots, x_m\})$ as an abbreviation of "$f^{(m)}$ is continuous at $x_1,\ldots, x_m\; (0 \le m < n+1)$ with respect to the domain and range of $f$ understood in the context" $\mathscr{C}[E,G]$ $\mathscr{C}[E,G] = \{f\in \mathscr{C}(E,G):\; \left\|f\right\| (=\sup |f|(E)) < \infty\}$ is a metric space with metric $d(f,g) = \left\| f- g\right\|$ $N_r^{\circledcirc }(p)$ $N_r^{\circledcirc }(p) = N_r(p)\setminus \{p\}\;$ (The $r$ - neighborhood of $p$ without center $p$) $\mathfrak{P}[a,b]$ $\mathfrak{P}[a,b] = \{P\subset [a,b]: a,\, b\in P,\; |P| < \infty\}\;$ (The set of partitions on $[a, b]$) $\mathscr{R}([a,b],\alpha)$ The set $\{f \mid f \text{ is Riemann-Stieltjes integrable with respect to }\alpha \text{ over } [a,b]\}$ If $[a, b]$ is understood to be given in the context, write $\mathscr{R}(\alpha)$ or (when $\alpha(x) = x$) $\mathscr{R}$ for short.
10-04-2010, 05:41 PM
Post: #2
 elim Moderator Posts: 544 Joined: Feb 2010 Reputation: 0
Rudin [Principle of Mathematical Analysis] Notes
1.1 Example We now show that the equation
(1) $p^2 = 2$ is not satisfied by any rational $p$.
$\quad\quad$ If there were such a $p$, we could write $p = m/n$ where $m$ and $n$ are integers that are not both even.
$\quad\quad$ Let us assume this is done. Then (1) implies
(2) $m^2 = 2n^2$. This shows that $m^2$ is even. Hence $m$ is even (if m were odd, m2 would be odd),
$\quad\quad$ and so $m^2$ is divisible by $4$. It follows that the right side of (2) is divisible by $4$, so that $n^2$ is even,
$\quad\quad$ which implies that n is even. The assumption that (1) holds thus leads to the conclusion that
$\quad\quad$ both $m$ and $n$ are even, contrary to our choice of $m$ and $n$. Hence (1) is impossible for rational p.

We now examine this situation a little more closely. Let $A$ be the set of all positive rationals $p$ such that $p^2 < 2$ and let $B$ consist of all positive rationals $p$ 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 $p < q$, and for every $p$ in $B$ we can find a rational $q$ in $B$ such that $q < p$. Then (4) To do this, we associate with each rational p > 0 the number

(3) $\displaystyle{q = p-\frac{p^2-2}{p+2} = \frac{2p+2}{p+2}}$ Then
(4) $\displaystyle{q^2- 2 =2(p^2-2)/(p+2)^2}$ .

If $p$ is in $A$ then $p^2 -2 < 0$, (3) shows that $q > p$, and (4) shows that $q^2 < 2$. Thus $q$ is in $A$. If $p$ is in $B$ then $p^2 -2> 0$, (3) shows that $0 < q < p$, and (4) shows that $q^2 > 2$. Thus $q$ is in $B$.

There is no problem to understand the text, but how the expression $q = p - \frac{p^2-2}{p+2}=\frac{2p+2}{p+2}$ been constructed? What's the thoughts behind?
12-15-2010, 02:32 PM
Post: #3
 elim Moderator Posts: 544 Joined: Feb 2010 Reputation: 0
1.1 Example The idea of $q = p - (p^2-2)/(p+2) = (2p+2)/(p+2)$
We'd like to define, for each rational $p>0$, a rational $q > 0$ such that $\displaystyle{0<\frac{q-\sqrt{2}}{p-\sqrt{2}}<1}$
This ensures that $p^2>2 \Leftrightarrow q^2>2$ and $|q-\sqrt{2}|<|p-\sqrt{2}|$
Let $q = p -R(p)$, then $\displaystyle{\frac{q-\sqrt{2}}{p-\sqrt{2}}=1-\frac{R(p)}{p-\sqrt{2}}}$ and so $\displaystyle{0<\frac{R(p)}{p-\sqrt{2}}<1}$. $R(p)$ must be rational with the same sign as $p^2-2$, and so $\displaystyle{R(p)=\frac{p^2-2}{Q(p)}}$, $\displaystyle{\frac{R(p)}{p-\sqrt{2}}=\frac{p+\sqrt{2}}{Q(p)}}$ with $Q(p)$ a rational function of $p$ and to meet the condition $\displaystyle{0<\frac{p+\sqrt{2}}{Q(p)}<1}$, the simplest $Q(p)$ is $p+2$. Finally, this also ensures $\displaystyle{q=p-\frac{p^2-2}{p+2}=\frac{2p+2}{p+2}>0, \quad (p>0)}$
12-15-2010, 02:32 PM (This post was last modified: 11-30-2012 01:54 PM by elim.)
Post: #4
 danmath Junior Member Posts: 1 Joined: Jun 2012 Reputation: 0
Insert: Rudin [Principle of Mathematical Analysis] Notes
(10-04-2010 05:41 PM)elim Wrote:  1.1 Example....

There is no problem to understand the text, but how the expression $q = p - \frac{p^2-2}{p+2}=\frac{2p+2}{p+2}$ been constructed? What's the thoughts behind?

------------------------------------------------------------------------------------------
Hi, i was wondering the same thing so i construct my solution, its not that elegant but there it is:

we want a rational q such that p p then q= p + r for some r>0.

if $q^2$ < 2 then $p^2 + 2pr + r^2 < 2$

then $r< (2-p^2)/(2p + r)$

we know that r < 2 so we take $r = (2-p^2)/(2p+2)$ and then the previously inequality is true for every p so $q = p + (2-p^2)/(2p+2)$ is on A and is bigger than p.
12-18-2010, 11:41 PM (This post was last modified: 11-30-2012 01:55 PM by elim.)
Post: #5
 elim Moderator Posts: 544 Joined: Feb 2010 Reputation: 0
1.2 Remark on Rationals --Rudin [Principle of Mathematical Analysis] Notes
1.2 Remark 1.1 Example Shows that $\mathbb{Q}$ has certain gaps in sprite off the fact that it's dense. The real number system fills these gaps.
12-18-2010, 11:41 PM (This post was last modified: 11-30-2012 01:55 PM by elim.)
Post: #6
 elim Moderator Posts: 544 Joined: Feb 2010 Reputation: 0
1.3~1.4 Basic Set and Notations --Rudin [Principle of Mathematical Analysis] Notes
1.3 Definitions
$\quad\quad$ Use $x \in A$ to indicate that $x$ is a member(an element) of $A$; Otherwise $x \not\in A$
$\quad\quad$ The set $\varnothing$ that contains no element is called the empty set; $A \ne \varnothing$ is called nonempty.
$\quad\quad$ Write $A \subset B$ or $B \supset A$ when $\forall x\in A \; (x\in B)$; We call $A$ is a subset of $B$ when $A \subset B$
$\quad\quad$ Write $A=B$ if $(A\subset B)\wedge (B\subset A)$. Otherwise $A \ne B$
$\quad\quad$ Write $A \subsetneq B$ to indecate that $(A \subset B) \wedge (A \ne B)$ and call $A$ a proper subset of $B$

1.4 Definition Throughout Chap. I, the set of all rational numbers is denoted by $\mathbb{Q}$
12-27-2010, 01:38 PM (This post was last modified: 12-12-2012 04:38 AM by elim.)
Post: #7
 elim Moderator Posts: 544 Joined: Feb 2010 Reputation: 0
1.5 - 1.10 Ordered set: Definitions and Examples. Rudin [Principle of Math Analysis]
ORDERED SETS
1.5 Definition Let $S$ be a set, An order on $S$ is a relation $<$ such that
$\quad\quad$ (i) $\forall x, y\in S$, one and only one of the following is true: $x < y,\; x = y,\; y < x$
$\quad\quad$ (ii) $\forall x, y, z \in S, \quad (x < y) \wedge (y < z) \Rightarrow (x < z)$
We read $x < y$ as "$x$ is less (smaller) than $y$" or "$y > x$ ($y$ is bigger (greater) than $x$)" or "$x$ precedes $y$"
We can write $x \le y$ for $(x < y) \vee (x = y)$ i.e. $\lnot (y < x)$ i.e. $\lnot (x > y)$

1.6 Definition. An ordered set is a set in which an order is defined.
$\quad\quad \mathbb{Q}$ is an ordered set where $r < s$ means $s - r \in \mathbb{Q}$ is positive.

1.7 Definition. If $S$ is an ordered set, $E\subset S$ and $\exists \beta \in S, \forall x \in E (x \le \beta)$, then $E$ is called bounded above, $\beta$ is called an upper bound of $E$
$\quad\quad$ Lower bounds are defined similarly with $\ge$ in place of $\le$

1.8 Definition. If $E \subset S$ is bounded above in ordered set $S$, and $\alpha \in S$ is an upper bound of $E$ such that $(\gamma < \alpha) \Rightarrow (\gamma$ is not an upper bound of $E)$, then $\alpha$ is called the least upper bound of $E$, or the supremum of $E$, and we write $\alpha = \sup E$
$\quad\quad$ Similarly we define The greatest lower bound, or infimum of a set $E$. Thus $\alpha = \inf E$ means that $\alpha$ is a lower bound of $E$ and $E$ has no bigger lower bound than $\alpha$.

1.9 Examples.
(a) In Example 1.1, $B$ ($A$) is exactly the set of upper (lower) bounds of $A$ ($B$) and $A$ ($B$) has no supremum (infimum) in $\mathbb{Q}$.
(b) Even if $\alpha = \sup E$ exists, $\alpha$ may or may not be a member of $E$.
(c) Let $E = \{1/n \mid n \in \mathbb{N}\}$, then $\sup E = 1 \in E, \quad \inf E = 0 \notin E$

1.10 Definition. An ordered set $S$ is said to have the least-upper-bound property If $\forall E\subset S \; ((E \ne \varnothing) \wedge (E \text{ is bounded above})) \Rightarrow (\exists \alpha \in S, \; \alpha = \sup E )$
12-28-2010, 04:47 PM
Post: #8
 elim Moderator Posts: 544 Joined: Feb 2010 Reputation: 0
1.11 Theorem [Symmetric Properties between $\inf$ and $\sup$]
1.11 Theorem Suppose ordered set $S$ has least-upper-bound property,
$\quad \varnothing \ne B\subset S,\quad \varnothing \ne L=\{x\in S \mid \forall y \in B, x \le y\}$ Then $\exists \;\alpha \in S, \quad \inf B =\alpha = \sup L$

$\quad$ Proof. Since $L$ is not empty and bounded above, $\exists \;\alpha \in S\quad \alpha = \sup L$
$\quad$ If $\gamma < \alpha$ then $\exists \lambda \in L,\;\forall x \in B \quad \gamma < \lambda \le x$ and so $\gamma \not\in B$, therefore $\forall x \in B, \quad \alpha \le x$
$\quad$ i.e. $\alpha \in L$ or $\alpha$ is a lower bound of $B$.
$\quad$ If $\alpha < \beta$, then $\beta \not\in L$ since $\alpha = \sup L$. So $\alpha$ is the greatest lower bound of $B$, i.e. $\alpha = \inf B \quad\quad \square$
01-10-2011, 10:16 AM
Post: #9
 elim Moderator Posts: 544 Joined: Feb 2010 Reputation: 0
1.12 Fields, 1.13 Remarks Rudin [Principle of Mathematical Analysis] Notes
1.12 Definition A field is a set $F$ with two operations addition and multiplication, satisfying axioms (A),(M) and (D):

$\quad\quad$ (A1) Addition is closed in $F$: $(x, y \in F)\Rightarrow (x+y\in F)$
$\quad\quad$ (A2) Addition is commutative: $x + y = y + x\quad (\forall x,y\in F)$
$\quad\quad$ (A3) Addition is associative: $(x+y)+z = x+(y+z)\quad (\forall x,y,z\in F)$
$\quad\quad$ (A4) $F$ has additive unit: $\exists 0\in F\ \forall x\in F\quad 0 + x = x$
$\quad\quad$ (A5) $F$ has additive inverse:$\forall x\in F\ \exists -x\in F \quad(x+(-x) = 0)$

(M) Axioms for multiplication
$\quad\quad$ (M1) multiplicationis closed in $F$: $(x, y \in F)\Rightarrow (xy\in F)$
$\quad\quad$ (M2) multiplication is commutative: $xy = yx\quad (\forall x,y\in F)$
$\quad\quad$ (M3) multiplication is associative: $(xy)z = x(yz)\quad (\forall x,y,z\in F)$
$\quad\quad$ (M4) $F$ has multiplicative unit: $\exists 1\in F\ \forall x\in F\quad 1x = x$
$\quad\quad$ (M5) $F$ has multiplicative inverse:$\forall x\in F\ \exists 1/x\in F \quad(x(1/x) = 1)$

(D) The distributive law: $x(y+z) = xy+xz\quad (\forall x,y,z\in F)$

1.13 Remarks
(a) In any field one usually write
$\quad\quad \displaystyle{x-y, \frac{x}{y}, x+y+z, xyz, x^2, x^3, 2x, 3x,\cdots}$
$\ \quad$ In place of
$\quad\quad \displaystyle{x+(-y), x\cdot \left(\frac{1}{y} \right),(x+y)+z,(xy)z, xx, xxx,x+x, x+x+x,\cdots}$
(b) Rational numbers $\mathbb{Q}$ with its customary addition and multiplication is a field.
(c) We shall see that some familiar properties of $\mathbb{Q}$ are actually the
$\quad\quad$consequences of field axioms thus hold in any field (complex numbers, etc)
04-23-2012, 11:04 AM
Post: #10
 elim Moderator Posts: 544 Joined: Feb 2010 Reputation: 0
1.14 - 1.16 Propositions from Field Axioms. Rudin 【Principle of Mathematical Analysis】Notes
1.14 Proposition From axioms for addition
(a) $(x+y=x+z) \implies (y = z)$
(b) $(x+y = x) \implies (y = 0)$
(c) $(x+y=0) \implies (y=-x)$
(d) $-(-x) = x$
Proof:
(a) $y\overset{A_4}{=} 0+y \overset{A_5}{=} (-x + x) + y \overset{A_3}{=} -x + (x+y) \overset{former}{=} -x + (x+z) \overset{A_3}{=} (-x+x) + z \overset{A_5}= 0+z \overset{A_4}{=}z$
(b) This is the case $z = 0$ in (a)
(c) By $A_5$, this is the case $z = -x$ in (a)
(d) $-(-x) = (x+(-x))-(-x) = x + (-x - (-x)) = x + 0 = x$

1.15 Proposition From axioms for multiplication
(a) $(x \ne 0)\wedge (xy = xz) \implies (y = z)$
(b) $(x \ne 0) \wedge (x y = x) \implies (y = 1)$
(c) $(-x)y = - (xy) = x(-y)$
(d) $(-x)(-y) = xy$
Proof:
(a) $0x + 0x \overset{Distri.}= (0+0)x \overset{A_4}{=} 0 x \overset{1.14(b)}{\implies} 0x = 0$
(c) $(x\ne 0)\wedge (xy = 1)\implies (y=1/x)$
(d) $(x\ne 0)\implies 1/(1/x) = x$

The proof is parallel to that of Proposition 1.14.

1.16 Proposition $0$ and additive inverse with multiplication
(a) $0x = 0$
(b) $(x\ne 0)\wedge (y\ne 0)\implies (xy \ne 0)$
(c) $(-x)y =-(xy) = x(-y)$
(d) $(-x)(-y) = xy$
Proof:
(a) $\left(0x + 0x \overset{Distri}{=} (0+0) x \overset{A_5}{=} 0x \right ) \overset{1.14(b)}{\implies} (0x = 0)$
(b) If $x\ne 0 \ne y$ but $xy = 0$, then $1 = (1/x)x(1/y)y = xy(1/x)(1/y) = 0(1/x)(1/y)\overset{(a)}{=}0$
Let's prove (c') $(-x = (-1)x)$: $\bigg((-1)x + x = (-1)x + 1x = (-1+1) x = 0 \bigg)\overset{1.14c}{\implies} ((-1)x = -x)$
Now
(c) $(-x)y =((-1)x)y = x(-1)y = x(-y) = (-1)(xy) = -(xy)$, Also
(d) $(-x)(-y) = - (x(-y)) = -(-(xy)) \overset{1.14(d)}{=} xy$
 « Next Oldest | Next Newest »

Forum Jump: