Rudin 【Principle of Mathematical Analysis】Notes & Solutions - Printable Version

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Rudin 【Principle of Mathematical Analysis】Notes & Solutions - elim - 10-04-2010 05:41 PM

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. $B^A$ $B^A = \{f\mid f:A\to B\}$ (A set of function/mappings).$\{a_n\} \in E^{\mathbb{J}}$ is called a sequence in $E$ $\text{In}(\mathscr{F})$ $\text{In}(\mathscr{F}) = \{f\in\mathscr{F}:\,f(A)\cap f(B)\ne\varnothing\implies A\cap B\ne\varnothing\}$(Injections in $\mathscr{F}$) $\mathscr{M}(E,G)$ $\mathscr{M}(E,G) = \{f\in G^E: f(x) < f(y)\implies x < y\}$ $f\in \mathscr{M}(E,G)$ 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 $\{n_j\}\in \text{In}(\mathscr{M}(\mathbb{J},\mathbb{J}))$. $\mathscr{C}^n (E, F)$ $\mathscr{C}^n (E, F)=\{f\in F^B:\,f^{(m)}\in\mathscr{C}(E,F)\;(m < n+1))\}$($\mathscr{C}(E,F)=\{f\in F^E:\,f$ is continuous on $E\},\;n\le\infty$) If $F$ is fixed in the context, write $\mathscr{C}^n(E)$ for short. If both $E,\; F$ are fixed in the context, $\mathscr{C}^n$ can simply stand for it. We use $f\in \mathscr{C}^n (\{x_1, x_2, \ldots, x_m\})$ to mean that "$f^{(m)}$ 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 without center) $\mathfrak{P}[a,b]$ $\mathfrak{P}[a,b] = \{P\subset [a,b]: a,\, b\in P,\; |P| < \infty\}\,$(All $[a, b]$-partitions) $\mathscr{R}([a,b],\alpha)$ $\small\mathscr{R}([a,b],\alpha)=\{f \mid f \,$is Riemann-Stieijes$\small\,\alpha$-integrable over$\small\,[a,b]\}$ If $[a, b]$ is fixed in the context, write $\mathscr{R}(\alpha)$ or $\mathscr{R}$ for short (when $\alpha(x) = x$).

Rudin [Principle of Mathematical Analysis] Notes - elim - 10-04-2010 05:41 PM

1.1 Example We now show that the equation
(1) $p^2 = 2$ is not satisfied by any rational $p$.
$\quad$If there were such a $p$, we could write $p = m/n$ where $m,$
$\quad n(\in\mathbb{N}^+)$ are not both even. Assuming this is done, then
$\quad$(1) implies
(2) $m^2 = 2n^2$. This shows that $m^2$ is even. Hence $m$ is even
$\quad$(if $m$ were odd, $m^2$ would be odd), and so $m^2$ is divisible
$\quad$by $4$. It follows that the right side of (2) is divisible by $4$, so
$\quad$that $n^2$ is even, which implies that n is even.
$\quad$The assumption that (1) holds thus leads to the conclusion
$\quad$that both $m$ and $n$ are even, contrary to our choice of $m$
$\quad$and $n$. Hence (1) is impossible for rational p.

We now examine the 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 some
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) $q = p-\small\dfrac{p^2-2}{p+2} = \dfrac{2p+2}{p+2}$ Then
(4) $q^2- 2 =2\scriptsize\dfrac{p^2-2}{(p+2)^2}$ .

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

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

The idea of $q = p - \frac{p^2-2}{p+2} = \frac{2p+2}{p+2}$ in Example1.1 - elim - 12-15-2010 02:32 PM

For each rational $p> 0\underset{\tiny\,}{,}$ we'd like to define a rational
$q > 0$ such that $0< \frac{q-\sqrt{2}}{p-\sqrt{2}}<1.\;$This implies that
$(p^2>2 \underset{\tiny\,}{\iff} q^2>2)\wedge(|q-\sqrt{2}|<|p-\sqrt{2}|)$
Let $q = p -R(p)$, then $\frac{q-\sqrt{2}}{p-\sqrt{2}}=1-\frac{R(p)}{p-\sqrt{2}},\,$so we've
$0<\frac{R(p)}{p-\sqrt{2}}<1.\quad R(p)$ must be rational with the
same sign as $p^2-2$, thus $R(p)=\frac{p^2-2}{Q(p)}\;$with some
rational $Q(p).\,$Now $\frac{R(p)}{p-\sqrt{2}}=\frac{p+\sqrt{2}}{Q(p)}\,$so $0<\frac{p+\sqrt{2}}{Q(p)}<1.\;$
Clearly the simplest $Q(p)$ meets the condition is $p+2$.
and we then have $q=p-\frac{p^2-2}{p+2}=\frac{2p+2}{p+2}>0\;\; (p>0).$

Here is a better try:
We are trying to find a rational function $q = R(p)$ such
that $R^2(x) -2 = \frac{k(x^2 -2)}{g^2(x)}$ with rational $g(x)$ satisfies
$g^2 > k > 0\;(x > 0).$ For simplicity let $g(x) \underset{\tiny\,}{=}x + a$
and so $R^2(x) = \frac{k(x^2-2)+2(x+a)^2}{(x+a)^2} = \frac{(k+2)x^2+4ax +2(a^2-k)}{(x+a)^2}$.
Since $k = a = 2$ makes the numerator a perfect square
$4(x+1)^2$. We've found the simplest $R(x) = \small\dfrac{2x+2}{x+2}.$

In general, if $m\in\mathbb{N}^+$ is not a perfect square, then
$({\small R}(x))^2 -m = \frac{k(x^2-m)}{(x+b)^2}$ has solution $k = b^2 - m > 0,\;$
$b=\min\{j\in\mathbb{N}:\;j^2 > m\}$ thus ${\small R}(x) = \small\dfrac{bx +m}{x + b}$ i.e.
the corresponding recursion $c_{n+1} = \frac{bc_n+m}{c_n + b}\to \sqrt{m}.$

Let $c_n \underset{\tiny\,}{=} \small\beta_n+\sqrt{2}$, by the recursion formula,
$\,\beta_{n+1}^{-1}+\frac{1}{2\sqrt{2}}=(3+2\sqrt{2})(\beta_n^{-1}+\frac{1}{2\sqrt{2}}).$
$\,\beta_n^{-1}+\frac{1}{2\sqrt{2}}={\small(3+2\sqrt{2})}^{n-1}(\beta_1^{-1}+\frac{1}{2\sqrt{2}}).$
Let $\{c_n^{\pm}\}\;\,$be the resulting sequences with initial value
$\beta_1= 1-\sqrt{2},\;2-\sqrt{2}$ respectively. This yields
$\beta_{n+1}^{-1}=\frac{-1\pm(3+2\sqrt{2})^n}{2\sqrt{2}}\;$and$\;{\large c}_n^{\pm}=\sqrt{2}\pm\frac{2\sqrt{2}}{(3+2\sqrt{2})^n\mp 1}.\;\;$
Thus $\;c_1^{-}< c_2^{-}< \cdots< c_n^{-}< \sqrt{2}< c_n^{+}< \cdots< c_2^{+}< c_1^{+}$
or $\;\;1< \frac{4}{3}< \frac{7}{5}< \frac{24}{17}< \frac{41}{29}{\small <\cdots \underset{\tiny\,}{\;}}$
$\qquad \qquad\cdots< c_{n+1}^{-}< c_n^{-}< \sqrt{2}< c_n^{+}< c_{n+1}^{+}< \cdots\underset{\tiny\,}{\,}$
$\qquad\qquad\qquad\qquad\qquad {\small \cdots<} \frac{58}{41}< \frac{17}{12}< \frac{10}{7}< \frac{3}{2}< 2.$

Insert: Rudin [Principle of Mathematical Analysis] Notes - danmath - 12-15-2010 02:32 PM

(10-04-2010 05:41 PM)elim Wrote:  1.1 Example....

There is no problem to understand the text, but how the
expression $q = p -\small\dfrac{p^2-2}{p+2}=\dfrac{2p+2}{p+2}$ 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< q for every p on A
and q, if q > p then q= p + r for some r>0.
if $q^2$ < 2 then $p^2 + 2pr + r^2 \underset{\,}{<} 2,$ then $r< \small\dfrac{2-p^2}{2p + r}$
we know that r < 2 so we take $r = \small\dfrac{2-p^2}{2p+2}$ and then
the previously inequality is true for every p so
$q = p +\small \dfrac{2-p^2}{2p+2}$ is on A and is bigger than p.

1.2 Remark on Rationals --[Principle of Math Analysis] - elim - 12-18-2010 11:41 PM

1.2 Remark 1.1 Example Shows that $\mathbb{Q}$ has certain gaps
$\quad$in sprite off the fact that it's dense. The real number
$\quad$system fills these gaps.

1.3~1.4 Basic Set and Notations --Rudin [P.M.A] Notes - elim - 12-18-2010 11:41 PM

1.3 Definitions
$\quad$We write $x \in A$ when $x$ is a member(an element) of $A$
$\qquad x\not\in A$ means that $x$ is not a member of $A$.
$\quad$The set $\varnothing$ contains no element is called Empty set.
$\quad$A set $A$ is called nonempty if $A\ne\varnothing$.
$\quad A$ is a sebset of $B$(Write $\small A\subset B$ or $\small B\supset A$) if
$\qquad\qquad\forall x\in A \; (x\in B)$;
$\quad$Write $A=B$ if $(A\subset B)\wedge (B\subset A)$. Otherwise $A \ne B$
$\quad A$ is a proper sebset of $B$(Write $\small A \subsetneq B$) if
$\qquad\qquad(A \subset B) \wedge (A \ne B).$

1.4 DefinitionWe use $\mathbb{Q}$ for the set of all rational numbers.

1.5 - 1.10 Ordered set: Definitions and Examples. Rudin [P.M.A] - elim - 12-27-2010 01:38 PM

ORDERED SETS
1.5 Definition An order on a set $S$ is a relation $<$ such that
$\quad\quad$ (i) $\forall x, y\in S$, one and only one of the following is true:
$\qquad\qquad\qquad x < y,\;\; x = y,\;\; y < x$
$\quad\quad$ (ii) $\forall x, y, z \in S, \quad (x < y) \wedge (y < z) \Rightarrow (x < z)$
$\quad$We read $x < y$ as "$x$ is less (smaller) than $y$" or "$y > x$
$\quad$($y$ is bigger (greater) than $x$)" or "$x$ precedes $y$"
$\quad$Write $x \le y$ for $(x < y) \vee (x = y)$
$\qquad\qquad\qquad$ i.e. $\small \lnot (y < x)\,$ i.e.$\,\small\lnot (x > y)$

1.6 Definition. An ordered set is a set with an order defined.
$\quad\quad \mathbb{Q}$ is an ordered set where $r < s$ means $s - r >0.$

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

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

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

1.10 Definition. An ordered set $S$ is said to have the
$\qquad$least-upper-bound property If
$\qquad\quad\forall E\subset S \; ((E \ne \varnothing) \wedge (E$ is bounded above$))$
$\qquad\qquad\implies(\exists \alpha \in S, \; \alpha = \sup E )$

1.11 Theorem [Symmetric Properties between $\inf$ and $\sup$] - elim - 12-28-2010 04:47 PM

1.11 Theorem Suppose $S$ has least-upper-bound property,
$\quad \varnothing \ne B\subset S,\quad \varnothing \ne L=\{x\in S \mid x \le y(\forall y \in B)\}$
$\quad$Then $\exists \;\alpha \in S\, (\inf B =\alpha = \sup L).$
Proof. Since $L(\ne\varnothing)$ is bounded above, $\alpha:=\sup L\in S.$
$\quad$If $\gamma< \alpha$ then $\exists\lambda\in L\,\forall x \in B\,(\gamma < \lambda \le x)$ and so
$\quad\gamma \not\in B$, therefore $\forall x \in B\,(\alpha \le x).$ i.e. $\alpha \in L$ or $\alpha$ is a
$\quad$lower bound of $B$.
$\quad$If $\alpha < \beta$, then $\beta \not\in L$ since $\alpha = \sup L$. Therefore $\alpha$ is
$\quad$the greatest lower bound of $B$, i.e. $\alpha = \inf B \quad\quad \square$

1.12 Fields, 1.13 Remarks Rudin [Principle of Math. Analysis] - elim - 01-10-2011 10:16 AM

1.12 Definition A field is a set $F$ with two operations
$\quad$addition and multiplication, satisfying axioms (A),(M)
$\quad$and (D) below:

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

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

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

1.13 Remarks
(a) In any field one usually write
$\qquad\displaystyle{x-y, \frac{x}{y}, x+y+z, xyz, x^2, x^3, 2x, 3x,\cdots}$
$\qquad$In place of
$\quad\quad \displaystyle{x+(-y), x\cdot \left(\frac{1}{y} \right),(x+y)+z,(xy)z, }$
$\qquad xx, xxx,x+x, x+x+x,\cdots$
(b) Rational numbers $\mathbb{Q}$ with its customary addition and
$\qquad$multiplication is a field. (c) We shall see that some
$\qquad$familiar properties of $\mathbb{Q}$ are actually the
$\quad\quad$consequences of field axioms thus hold in any field
$\qquad$(complex numbers, etc)

1.14 - 1.16 Propositions from Field Axioms. Rudin 【P.M.A】 - elim - 04-23-2012 11:04 AM

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) $y\overset{A_4}{=} 0+y \overset{A_5}{=} (-x + x) + y \overset{A_3}{=} -x + (x+y)$
$\qquad\overset{\small\text{asm}}{=} -x + (x+z) \overset{A_3}{=} (-x+x) + z \overset{A_5}= 0+z \overset{A_4}{=}z$
(b) This is the case $z = 0$ in (a)
(c) By $A_5$, this is the case $z = -x$ in (a)
(d) $-(-x) = (x+(-x))-(-x)$
$\qquad\qquad\;\;= x + (-x - (-x)) = x + 0 = x$

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
$\qquad\small 1 = (1/x)x(1/y)y = xy(1/x)(1/y) = 0(1/x)(1/y)\overset{(a)}{=}0$
Let's prove $(c')\quad (-x = (-1)x)$:
$\qquad\small((-1)x + x = (-1)x + 1x = (-1+1) x = 0)\overset{1.14c}{\implies} (c')$
Now
(c) $(-x)y =((-1)x)y = (-1)(xy) = -(xy)$, Also
(d) $(-x)(-y) = - (x(-y)) = -(-(xy)) \overset{1.14(d)}{=} xy$