• 1 Votes - 5 Average
• 1
• 2
• 3
• 4
• 5
 Rudin 【Principle of Mathematical Analysis】Notes & Solutions
05-17-2012, 05:45 AM
Post: #31
 elim Moderator Posts: 577 Joined: Feb 2010 Reputation: 0
Chap.II Basic Topology --Rudin [Principle of Mathematical Analysis] Notes
FINITE, COUNTABLE, AND UNCOUNTABLE SETS

2.1 Definition Given sets $A$ and $B$, if for each $x\in A$ there is associated somehow an $f(x)\in B$. Then $f$ is said to be a function from $A$ to $B$ (or a mapping of $A$ into $B$), write as $f: A\to B$. $A$ is called the domain of $f$ (we also say $f$ is defined on $A$), and elements $f(x)$ are called the values of $f$. The set $\;\{f(x)\mid x\in A\}$ is called the range of $f$.

2.2 Definition Let $f:A\to B$. If $E\subset A$ then denote $f(E) = \{f(x)\mid x\in E\}$ and call $f(E)$ the image of $E$ under $f$. Thus $f(A)$ is the range of $f$. It's clear that $f(A)\subset B$. If $f(A) = B$, we say that $f$ maps $A$ onto $B$
$\quad\quad$ If $E\subset B$, call $f^{-1}(E) = \{x\in A: f(x)\in E\}$ the inverse image of $E$ under $f$. For $y\in B$, define $f^{-1}(y)=\{x\in A: f(x)=y\}$. If for each $y\in B$, $f^{-1}(y)$ consists of at most one element of $A$, then $f$ is said to be 1-1 (one-to-one). This may also be expressed as: $f:A\to B$ is 1-1 provided that $\forall s,t\in A:\; (s\ne t)\implies (f(s)\ne f(t)).\quad\quad$ ($s\ne t$ means that $s$ and $t$ are distinct elements; otherwise we write $s = t$)

2.3 Definition If there exists a 1-1 mapping $f:A\to B$, we say $A$ and $B$ can be put in 1-1 correspondence, or that $A$ and $B$ have the same cardinal number, or briefly, that $A$ and $B$ are equivalent (write $A\sim B$). The relation has these properties:
$\quad\quad$ It's reflexive: $A\sim A$
$\quad\quad$ It's symmetric: $(A\sim B) \implies (B\sim A)$
$\quad\quad$ It's transitive: $(A\sim B)\wedge (B\sim C)\implies (A\sim C)$
Any relation with these three properties is called an equivalence relation.

2.4 Definition For $n\in J =\mathbb{N}^+$, let $J_n = \{1,\cdots,n\}$. For any set $A$, say:
$\quad\quad$ (a) $A$ is finite if $A\sim J_n$ for some $n\in J$ ($\varnothing$ is also considered to be finite)
$\quad\quad$ (b) $A$ is infinite if $A$ is not finite.
$\quad\quad$ (c) $A$ is countable if $A\sim J$.
$\quad\quad$ (d) $A$ is uncountable if $A$ is neither finite or countable.
$\quad\quad$ (e) $A$ is at most countable if $A$ is finite or countable.

$\quad\quad$ Countable sets are sometimes called enumerable, or denumerable.
$\quad\quad$ For two finite sets $A$ and $B$, we evidently have $A\sim B$ off $A$ and $B$ contain the same number of elements. For infinite sets, however, the idea of "having the same number of elements" becomes quite vague, whereas the notion of 1-1 correspondence retains its clarity.
05-17-2012, 05:45 AM
Post: #32
 elim Moderator Posts: 577 Joined: Feb 2010 Reputation: 0
Chapter 2 Sets and Counting (ii) --Rudin [Principle of Mathematical Analysis] Notes
2.5 Examples $\mathbb{Z}$ is countable. As $f(n)=\begin{cases} n/2 & (n \text{ even})\\ (1-n)/2 & (n \text{ odd}) \end{cases}$ defined a 1-1 correspondence $f: J\to \mathbb{Z} =f(J)$

2.6 Remark A finite set cannot be equivalent to one of its proper subsets. That this is, however, possible for infinite sets, is shown by Example 2.5 where $J\subsetneq \mathbb{Z}$
$\quad\quad$ Actually we can replace Definition 2.4(b) by the statement: $A$ is infinite if $A$ is equivalent to some of its proper subsets.
05-17-2012, 05:45 AM
Post: #33
 elim Moderator Posts: 577 Joined: Feb 2010 Reputation: 0
2.7-2.8 Sets and Counting (iii) --Rudin [Principle of Mathematical Analysis] Notes
2.7 Definition By a sequence, we mean a function $f$ defined on $J=\mathbb{N}^+$. If $f(n)=x_n,\; (\forall n\in J)$, it is customary to denote $f$ by $\{x_n\}$, or sometimes by $x_1,x_2,\cdots$. The values of $f$, that is, elements $x_n$, are called the terms of the sequence. If $f(J)\subset A$, then $\{x_n\}$ is said to be a sequence in $A$, or a sequence of elements of $A$
$\quad\quad$ Note that the terms $x_1,x_2,\cdots$ of a sequence need not be distinct.
$\quad\quad$ Since every countable set is the range of a 1-1 function defined on $J$, we may regard every countable set as the range of a sequence of distinct terms. Speaking more loosely, we may say that the elements of any countable set can be "arranged in a sequence".
$\quad\quad$ Sometimes it is convenient to replace $J=\mathbb{N}^+$ in this definition by $\mathbb{N}$ (all nonnegative integers), i.e., to start with $0$ rather than $1$.

2.8 Theorem Every infinite subset of a countable set $A$ is countable.
Proof Suppose $E\subset A$ is infinite. Sequence of distinct elements $\{x_n\}$ is an elements arrangement of $A$. Construct a sequence $\{n_k\}$ as follows:
$\quad\quad$ Let $n_1$ be the smallest positive integer such that $x_{n_1}\in E$. Having chosen $n_1,\cdots, n_{k-1}\;(k=2,3,4,\cdots)$, let $n_k$ be the smallest integer greater than $n_{k-1}$ such that $x_{n_k}\in E$.
$\quad\quad$ Putting $f(x)=x_{n_k}\; (k=1,2,3,\cdots)$, we obtain a 1-1 correspondence between $E$ and $J\quad\quad\square$.
$\quad\quad$ The theorem shows that, roughly speaking, countable sets represent the "smallest" infinity: No uncountable set can be a subset of a countable set.
05-17-2012, 05:45 AM
Post: #34
 elim Moderator Posts: 577 Joined: Feb 2010 Reputation: 0
2.9 Sets and Counting (iv) --Rudin [Principle of Mathematical Analysis] Notes
2.9 Definition Let $A$ and $\Omega$ be sets, suppose that with each element $\alpha\in A$ there is associated a subset of $\Omega$ which we denote by $E_{\alpha}$
$\quad\quad$ The set whose elements are sets $E_{\alpha}$ will be denoted by $\{E_{\alpha}\}$. We sometimes call a set of sets a collection or a family of sets.
$\quad\quad$ The union $S$ of the sets $E_{\alpha}$ is defined as $S = \bigcup_{\alpha\in A} E_{\alpha} = \{x\mid \exists \alpha\in A\;(x\in E_{\alpha})\}. \tag{1}$
$\quad\quad$ If A consists of the integers $1,2,\cdots,n$, we write $S = \bigcup_{\alpha\in A} E_{\alpha} = \bigcup_{m=1}^n E_m \tag{2}$ $\quad\quad$ or $S = E_1\cup E_2\cup \cdots\cup E_n. \tag{3}$ $\quad\quad$ If $A = \mathbb{N}^+$ is the set of all positive integers, the usual notation is $S = \bigcup_{m=1}^{\infty} E_m. \tag{4}$ $\quad\quad$ The symbol $\infty$ in (4) merely indicates the union of a countable collection of sets is taken, and should not be confused with the symbols $+\infty,\; -\infty$ introduced in Definition 1.2.3.

$\quad\quad$ The intersection of the sets $E_{\alpha}$ is defined to be the set $P\underset{\,}{\,}$ as
$P = \bigcap_{\alpha\in A} E_{\alpha} \tag{5}$
$\quad\quad$ or
$P = \bigcap_{m=1}^{\infty} E_m = E_1\cap E_2\cap\cdots \cap E_n \cap\cdots \tag{6}$
$\quad\quad$ or
$P = \bigcap_{m=1}^{\infty} E_m \tag{7}$

$\quad\quad$ as for unions. If $A\cap B$ is not empty, we say that $A$ and $B$ intersect; otherwise they are disjoint.
05-18-2012, 07:29 AM
Post: #35
 elim Moderator Posts: 577 Joined: Feb 2010 Reputation: 0
METRIC SPACES: 2.15~2.30 --Rudin [Principle of Mathematical Analysis] Notes
2.15 Definition A set $X$, whose elements we shall call points, is said to be a metric space if with any two points $p$ and $q$ of $X$ there is associated a real number $d(p,q)$, called the distance from $p$ to $q$, such that
$\quad\quad$ (a) $d(p,q)>0$ if $p\ne q$; $d(p,p) = 0$;
$\quad\quad$ (b) $d(p,q) = d(q,p)$;
$\quad\quad$ (c) $d(p,q)\le d(p,r) + d(r,q),\;(\forall r\in X)$
$\quad\quad$ Any function with these three properties is called a distance function, or a metric.

2.16 Examples The most important examples of metric spaces, from our standpoint, are the euclidean spaces $\mathbb{R}^+$, especially $\mathbb{R}^1$ (the real line) and $\mathbb{R}^2$ (the complex plane); the distance in $\mathbb{R}^k$ is defined by$d(\mathbf{x,y}) = |\mathbf{x-y}|\quad (\mathbf{x,y}\in\mathbb{R}^k) \tag{19}$ By Theorem 1.37, the conditions of Definition 2.15 are satisfied by (19).
$\quad\quad$ Every subset $Y$ of a metric space $X$ is a metric space in its own right, with the same distance function.
$\quad\quad$ Thus every subset of a euclidean space is a metric space. Other examples are the spaces $\mathscr{C}(K)$ and $\mathscr{L}^2(\mu)$, which are discussed in Chaps. 7 and 11, respectively.

2.17 Definition By the segment $(a,b)$ we mean the set of all real numbers $x$ such that $a < x < b$
$\quad\quad$ By the interval $[a,b]$ we mean the set of all real numbers $x$ such that $a\le x\le b$
$\quad\quad$ Occasionally we use "half-open intervals" $[a,b) = \{x\mid a\le x < b\},\;(a,b] = \{x\mid a< x\le b\}$
$\quad\quad$ If $a_i < b_i,\; i=\overline{1,k}$, the set $\{\mathbf{x}=(x_1,\cdots,x_k)\in\mathbb{R}^k: \; a_i\le x\le b_i\;(1\le x\le k)\}$ is called a $k-cell$. Thus a 1-cell is an interval, a 2-cell is a rectangle, etc.
$\quad\quad$ If $\mathbf{x}\in\mathbb{R}^k,\; r>0$, the open (or closed) ball $B$ with center at $\mathbf{x}$ and radius $r$ is defined to be the set of all $\mathbf{y}\in\mathbb{R}^k$ such that $|\mathbf{y-x}| < r$ (or $|\mathbf{y-x}|\le r$).
$\quad\quad$ We call a set $E\subset \mathbb{R}^k$ convex if $\lambda\mathbf{x}+(1-\lambda)\mathbf{y}\in E\quad(\forall \mathbf{x,y}\in E,\forall \lambda \in (0,1))$ $\quad\quad$ Since $|(\lambda\mathbf{y}+(1-\lambda)\mathbf{z}) -\mathbf{x}| \le \lambda|\mathbf{y-x}|+(1-\lambda)|\mathbf{z-x}|\le \max\{|\mathbf{y-x}|,|\mathbf{z-x}|\}$, balls are convex. It's easy to see that $k$-cells are also convex.

2.18 Definition Let $X$ be a metrix space. All points and sets mentioned below are understood to be elements and subsets of $X$.
$\quad\quad$ (a) A neighborhood of $p$ is a set $N_r(p) = \{q\in X: d(p,q)$\quad\quad$(b) A point$p$is a limit point of$E\subset X$if$\forall r>0,\; (N_r(p)\setminus \{p\})\cap E \ne\varnothing\quad\quad$(c) If$p\in E$and$p$is not a limit point of$E$, then$p$is an isolated point of$E$.$\quad\quad$(d)$E$is closed if every limit point of$E$is a point of$E$.$\quad\quad$(e) If$N_r(p)\subset E$for some$r > 0$, call$p$an interior point of$E$.$\quad\quad$(f)$E$is open if every point of$E$is an interior point of$E$.$\quad\quad$(g) The complement of$E$is the set$E^c = X\setminus E\quad\quad$(h)$E$is perfect if$E$is closed, and has no isolated points.$\quad\quad$(i)$E$is bounded if$E\subset N_M(p)\subset X$for some$M>0$and$p\in X$.$\quad\quad$(i)$E$is dense if$\forall p\in X,\; (p\in E)\vee (p$is a limit point of$E)$.$\quad\quad$Clearly neighborhoods in$\mathbb{R}$are segments, and they are interiors of circles in$\mathbb{R}^2$. 2.19 Theorem Neighborhood is open.$\begin{align}\textbf{Proof }\quad (q\in N_r(p)) & \implies (h = r-d(p,q) > 0)\\
& \implies ((d(q,s) < h)\implies (d(p,s)\le d(p,q)+d(q,s) & \implies (N_h(q)\subset N_r(p)) \end{align}$2.20 Theorem If$p$is a limit point of$E\subset X$then$N_r(p)\cap E$is infinite ($\forall r>0$). Proof If$A = (N\setminus\{p\})\cap E$is finite for some$N=N_{\lambda}(p)\;(\lambda>0)$, then$A\ne \varnothing$for$p$is a limit point.$\quad\quad$So$A = \{q_1,\cdots,q_n\}$for some$n\in\mathbb{N}^+$and$N_r(p)$is a neighborhood for$\displaystyle{r = \min_{1\le m\le n} d(p,q_m)} > 0.\quad\quad$Let$B = (N_r(p)\setminus \{p\}) \cap E$. Then$B \subset N\setminus\{p\}\cap E = A$hence$B = B\cap A = \varnothing$.$\quad\quad$This contradicts to the assumption that$p$is a limit point of$E$. Corollary A finite point set has no limit points. 2.21 Examples Consider some subsets of$\mathbb{R}^2\quad\quad$(a) The set of all complex$z$such that$|z| < 1$.$\quad\quad$(b) The set of all complex$z$such that$|z| \le 1$.$\quad\quad$(c) A nonempty finite set.$\quad\quad$(d) The set of all integers$\mathbb{Z}$.$\quad\quad$(e)$\{1/n \mid n\in\mathbb{N}^+\}$, whose limit point$0$does not belong to it.$\quad\quad$(f) The set of all complex numbers ($\mathbb{R}^2$)$\quad\quad$(g) The segment$(a,b)$.$\quad\quad$Note that (d),(e),(g) can be regarded also as subsets of$\mathbb{R}^1$.$\quad\quad$We have the following table of properties$\quad\quad$In (g), we left the second entry blank because$(a,b)$is open in$\mathbb{R}^1$but not open for space$\mathbb{R}^2$2.22 Theorem Let$\{E_{\alpha}\}$be a (finite or infinite) collection of sets$E_{\alpha}$. Then $\left(\bigcup_{\alpha} E_{\alpha} \right )^c = \bigcap_{\alpha} E_{\alpha}^c. \;\tag{20}$ Proof ($x\in$LHS)$\Leftrightarrow (\forall \alpha :\; x\notin E_{\alpha}) \Leftrightarrow (\forall \alpha: \; x\in E_{\alpha}^c)\Leftrightarrow (x\in RHS)$2.23 Theorem A set$E$is open if and only if its complement$E^c$is closed. Proof If$E^c$is closed and$x\in E$, then$x \not E^c$and$x$is not a limit point of$E^c$. Hence there is a neighborhood$H$of$x$such that$H\cap E^c = \varnothing$. That means$H\subset E$and so$E$is open.$\quad\quad$If$E$is open and$x$is a limit point of$E^c$, then any neighborhood of$x$contains a point of$E^c$thus$x$is not an interior point of open set$E$. So$x\in E^c$and thus$E^c$is closed. Corollary$E$is closed if and only if$E^c$is open. 2.24 Theorem$\quad\quad$(a)$G_{\alpha} \text{ is open }(\forall \alpha)\implies \bigcup_{\alpha} G_{\alpha}$is open.$\quad\quad$(b)$F_{\alpha} \text{ is closed }(\forall \alpha)\implies \bigcap_{\alpha} F_{\alpha}$is closed.$\quad\quad$(c) For any finite open sets collection$\{G_{\alpha}\}$,$\displaystyle{\bigcap_{\alpha} G_{\alpha}}$is open.$\quad\quad$(c) For any finite closed sets collection$\{F_{\alpha}\}$,$\displaystyle{\bigcup_{\alpha} F_{\alpha}}$is closed. Proof (a)$\displaystyle{\bigg(x\in G = \bigcup_{\alpha} G_{\alpha}\bigg)}\implies (\exists \alpha:\; x\in G_{\alpha})\implies (\exists r>0\; \exists \alpha:\; N_r(x)\subset G_{\alpha}\subset G)\quad\quad$(b) By Theorem 2.22, $\bigg(\bigcap_{\alpha} F_{\alpha}\bigg)^c = \bigcup_{\alpha} F_{\alpha}^c, \tag{21}$$\quad\quad\quad$and$F_{\alpha}^c$is open, by Theorem 2.23. Hence (a) implies (21) is open thus$\bigcap_{\alpha} F_{\alpha}$is closed.$\quad\quad\;\begin{align}(c )\; \bigg(x \in H=\bigcap_{i=1}^n G_i\bigg) & \implies (\exists r_i >0:\; N_{r_i}(x)\subset G_i\; (i=\overline{1,n}))\\
& \implies (\exists r=\min(r_1,\cdots,r_n) >0:\; N_r(x)\subset H)\end{align}\quad\quad\quad$so$\displaystyle{H=\bigcap_{i=1}^n G_i}$is open.$\quad\quad$(d) follows from (c) just like (b) follows from (a) since$\displaystyle{\bigg(\bigcup_{i=1}^n F_i \bigg )^c = \bigcap_{i=1}^n F_i^c}$2.25 Examples In (c), (d) of the theorem 2.24, finiteness of the collections is essential. Let$G_n = (-1/n, 1/n)\subset \mathbb{R}^1,\; n\in\mathbb{N}^+$, then$G_n$is open but$G = \bigcap_{n=1}^n G_n = \{0\}$is a single point set which is not open in$\mathbb{R}^1$.$\quad\quad$Thus the intersection of an infinite collection of open sets need not be open, ans the union of an infinite collection of closed sets need not be closed. 05-25-2012, 11:24 AM Post: #36  elim Moderator Posts: 577 Joined: Feb 2010 Reputation: 0 More for matiric space --- Rudin [Principle of Mathematical Analysis] Notes 2.26 Definition Let$X$be a metric space and$E\subset X$, write$E'$the set of all limit points of$E$in$X$,$\quad\quad$and call$\overline{E} = E\cup E'$the closure of$E$. 2.27 Theorem Let$X$be a metric space and$E\subset X$, then$\quad\quad$(a)$\overline{E}$is closed,$\quad\quad$(b)$E = \overline{E}$if and only if$E$is closed,$\quad\quad\; (c )\; \overline{E}\subset F$for every closed set$F\subset X$such that$E\subset F$.$\quad\quad$By (a) and (c),$\overline{E}$is the smallest closed subset of$X$containing$E$Proof (a)$p\in (\overline{E})^c \implies \exists r>0:\; N_r(p)\cap E = \varnothing \implies N_r(p)\subset (\overline{E})^c \implies \overline{E}$is closed.$\quad\quad$(b)$E = \overline{E} = E\cup E' \Leftrightarrow E'\subset E\quad\quad\; (c )$If$F$is closed and$E\subset F$, then$E' \subset F' \subset F$hence$\overline{E}\subset F$. 2.28 Theorem If$\varnothing \ne E\subset\mathbb{R} \ni y=\sup E$, then$y\in\overline{E}$hence$y\in E$if$E$is closed. Proof For any$h>0$,$(y-h,y] \cap E \ne \varnothing$hance$y \in E\cup E' = \overline{E}.\;\square\quad$(compare 1.9 examples) 2.29 Remark We see (Sec. 2.16) that if$E\subset Y\subset X$with$X$a metric space,$Y$is also a metric space and$N_r(p, Y) = N_r(p,X)\cap Y$is the neighborhood of$p$with respect to metric space$Y$; and$E$is open with respect to$Y$need not to be open in$X$as in Example 2.21(g). But there is a simple relation below. 2.30 Theorem Let$Y$be a sub-metric space of$X$, then$\quad\quad\; E\subset Y$is open in$Y$if and only if$E = Y\cap G$for some$G$open in$X$. Proof$E\in\mathscr{O}(Y)\Leftrightarrow \forall p\in E\; \exists r_p > 0:\; E= \bigcup_{p\in E} N_{r_p}(p,Y) = (\bigcup_{p\in E} N_{r_p}(p)) \cap Y$while$\quad\quad\;G\in\mathscr{O}(X) \Leftrightarrow \forall p\in G;\exists \gamma_p > 0:\; G = \bigcup_{p\in G} N_{\gamma_p}(p).\quad\square\quad\quad\;$Where$N_r(p,Y) = N_r(p)\cap Y$and$\mathscr{O}(M)$is the collection of all open sets for$M$. 05-26-2012, 06:24 PM Post: #37  elim Moderator Posts: 577 Joined: Feb 2010 Reputation: 0 RE: Rudin [Principle of Mathematical Analysis] Notes COMPACT SETS 2.31 Definition An open cover of a set$E$in a metric space$X$is a collection$\{G_{\alpha}\}$of open sets in$X$such that$E\subset \bigcup_{\alpha} G_{\alpha}$2.32 Definition A subset$K$of a metric space$X$is said to be compact if every open cover of$K$contains a finite subcover.$\quad\quad$The notion of compactness is of great importance in analysis, especially with continuity (Chap.4).$\quad\quad$Write$\mathscr{O}(S),\;\mathscr{F}(S)$the class of all open (closed) sets of$S$respectively.$\quad\quad$Finite set is compact. There is a large class of infinite compact sets in$\mathbb{R}^k$by Theorem 2.41.$\quad\quad$We see (in 2.29) that when$E\subset Y\subset X$, it may happen that$E\in\mathscr{O}(Y)\setminus \mathscr{O}(X)$. The property of being open thus depends on the space containing$E$. The same is true for the property of being closed.$\quad\quad$Compactness behaves better: 2.33 Theorem Let$K\subset Y\subset X$. Then ($K$is compact relative to$X$)$\Leftrightarrow$($K$is compact relative to$Y$) Proof Since$U\subset \mathscr{O}(Y)\Leftrightarrow \exists G\in\mathscr{O}(X):\; U = G\cap Y$, the property of having finite subcover for any open covering can be mutually deduced between$Y$and$X$is obvious.$\quad\quad$By virtue of this theorem, we can regard compact sets as metric spaces in their own right without paying any attention to any embedding space. 2.34 Theorem Compact subsets of metric spaces are closed. Proof Let$X$be a metric space and$K (\subset X)$be compact. We'll show that$K^c \in\mathscr{O}(X)\quad\quad \begin{align} (p\in K^c = X\setminus K) & \implies K\subset\bigcup_{q}N_{r(p,q)}(q),\;N_{r(p,q)}(p)\cap N_{r(pq)}(q)=\varnothing\; (r(p,q)=\frac{1}{2}d(p,q))\\
& \implies \exists \{q_1,\cdots,q_n\}\subset K:\; K\subset W =\bigcup_1^n N_{r(p,q_j)}(q_j)\quad(\text{a subcover})\\
&\implies (N(p)=\bigcap_1^n N_{r(p,q_j)}(p)\in\mathscr{O}(X),\; N(p)\cap W=\varnothing)\implies K^c\in\mathscr{O}(X) \end{align}$2.35 Theorem Closed subsets of compact sets are compact. Proof Suppose$F\subset K\subset X,\; F\in\mathscr{O}(X)$and$K$is compact. If$\{V_{\alpha}\}$is an open cover of$F$, then$\quad\quad\; \{U_{\beta}\} = \{V_{\alpha}\}\cup \{F^c\}$is an open cover of$K$and so$K$is covered by a finite sub cover$\quad\quad\; \{U_1,\cdots,U_n\}\subset \{U_{\beta}\}$, therefore$\{U_1,\cdots,U_n\}\setminus \{F^c\} (\subset \{V_{\alpha}\})$is a finite subcover of$F$. Corollary If$F$is closed and$K$is compact, then$F\cap K$is compact. Proof By Theorems 2.24(b) and 2.34,$F\cap K (\subset K)$is closed and so compact by Theorem 2.35. 2.36 Theorem If$\{K_{\alpha}\}$is a collection of compact subsets of a metric space$X$such that the intersection of every finite sub collection of$\{K_{\alpha}\}$is nonempty, the$\bigcap K_{\alpha}$is nonempty. Proof Let$G_{\alpha} = F_{\alpha}^c$. Assume no point of$K_1$belongs to every$K_{\alpha}$, then sets$G_{\alpha}$form an open cover$\quad\quad$of$K_1$thus$K_1\subset G_{\alpha_1}\cup\cdots\cup G_{\alpha_n}$for finite many indexes$\{\alpha_1,\cdots,\alpha_n\}$. Then we have a finite$\quad\quad$intersection$K_1\cap K_{\alpha_1}\cap\cdots\cap K_{\alpha_n}=\varnothing$, a contradiction! Corollary If$\{K_n\}$is a sequence of nonempty compact sets such that$K_n\supset K_{n+1}\;(n=1,2,3,\cdots)$, then$\bigcup_1^{\infty} K_n $is not empty. 2.37 Theorem If$E$is an infinite subset of a compact set$K$, then$E' \cap K\ne \varnothing$Proof If$E' \cap K = \varnothing$, then each$q\in K$would have a neighborhood$V_q$such that$V_q\cap E\subset \{q\}$. So no finite sub collection of$\{V_q\}$can cover$E$nor$K$since$E\subset K$. This contradicts the compactness of$K$. 2.38 Theorem If$\{I_n\}$is a sequence of intervals in$\mathbb{R}^1$with$I_n\supset I_{n+1}\; (\forall n)$, then$\bigcap_1^\infty \ne \varnothing$Proof Let$I_n = [a_n,\;b_n],\; E = \{a_n \mid n\in\mathbb{N}^+\}$, then$E\ne\varnothing$and$E$is bounded above (by each$b_m$)$\quad\quad$since$a_n\le a_{m+n}\le b_{m+n}\le b_m$. So$a_n \le x = \sup E\le b_n\; (\forall n).\quad\square$2.39 Theorem Let$k\in\mathbb{N}^+$and$\{I_n\}$be a sequence of$k$-cells with$I_n\supset I_{n+1}\; (\forall n)$, then$\bigcap_1^\infty I_n \ne \varnothing.$2.40 Theorem$k$-cell is compact. Proof For$k$-cell$I = \{(x_1,\cdots,x_n)\mid a_j\le x_j\le b_j\; (j=\overline{1,k})\}$, let$\delta = \delta(I)=\left\{\sum_1^k (b_j-a_j)^2\right\}^{1/2}\quad\quad$Suppose, to get a contradiction, there is no finite$I$-subcover for some open cover$\{G_{\alpha}\}$of$I$.$\quad\quad$Put$c_j = (a_j+b_j)/2,$then intervals$[a_j,c_j],[c_j,b_j]$determine$2^k\; k$-cells$Q_i$whose union is$I$,$\quad\quad$at least one of these$Q_i$, call it$I_1$cannot be covered by any finite subcollection of$\{G_{\alpha}\}$and$\quad\quad$we can subdivide$I_1$and continue the process, to obtain a sequence$\{I_n\}$such that:$\quad\quad$(a)$I\supset I_1\supset I_2\supset I_3\supset \cdots\quad\quad$(b)$I_n$is not covered by any finite sub collection of$\{G_{\alpha}\}\quad\quad\;(c )\; \mathbf{x,y}\in I_n \implies |\mathbf{x-y}|\le 2^{-n}\delta\quad\quad$By (a) and Theorem 3.29,$\exists \mathbf{x}^*\in\bigcap I_n$and$\mathbf{x}^*\in G_{\alpha}$for some$\alpha$. Thus$\exists r> 0:\;N_r(\mathbf{X}^*)\subset G_{\alpha}\quad\quad$Choose$n > \delta/r$then$2^{-n}\delta < r$and$I_n\subset N_r(\mathbf{X}^*)\subset G_{\alpha}$which contradicts (b)$\quad\square$.$\quad\quad$The equivalence of (a) and (b) in the next theorem is known as the Heine-Borel theorem. 2.41 Theorem In$\mathbb{R}^k$the following properties are equivalent:$\quad\quad$(a)$E$is closed and bounded.$\quad\quad$(b)$E$is compact.$\quad\quad\;(c )$Every infinite subset of$E$has a limit point in$E$. Proof If (a) holds, then$E\subset I$for some$k$-cell$I$and (b) follows from Theorems 2.40 and 2.35.$\quad\quad$Theorem 2.37 shows that (b) implies (c). We need to show$(c )\implies$(a).$\quad\quad$If$E$is not bounded, then$E$contains points$\mathbf{x}_n$with$|\mathbf{x}_n| > n\;(n=1,2,3,\cdots)$. The set$S$with$\quad\quad$those points has no limit point in$\mathbb{R}^k$hence has none in$E$. So (c) implies$E$is bounded.$\quad\quad$If$E$is not closed,$\exists \mathbf{x}_0\in E'\setminus E$. Choose$S=\{\mathbf{x}_n\}\subset E$such that$|\mathbf{x}_n-\mathbf{x}_0| < 1/n$, then$S\quad\quad$is infinite and$S$has unique limit point$\mathbf{x}_0$since$\quad\quad\;\mathbf{y}\ne \mathbf{x}_0\implies |\mathbf{x}_n-\mathbf{y}| \ge |\mathbf{x}_0-\mathbf{y}|-|\mathbf{x}_n-\mathbf{x}_0|\ge |\mathbf{x}_0-\mathbf{y}|-\frac{1}{n}\ge\frac{1}{2}|\mathbf{x}_0-\mathbf{y}|$holds for all$\quad\quad$but finitely many$n$and so$\mathbf{y}\notin S'$. Thus$S$has no limit point in$E$hence (c)$\implies E$is closed.$\quad\quad$We should remark that (b) and (c) are equivalent in any metric space (Exercise 26) but (a) doesn't in general, imply (b) and (c). Examples are furnished by Exercise 16 and by$\mathscr{L}^2$studied in Chap.11. 2.42 Theorem (Weierstrass) Every bounded infinite subset$E$of$\mathbb{R}^k$has a limit point in$\mathbb{R}^k$Proof$E\subset I$for some$k$-cell$I$that is compact by Theorem 2.40 hence$E'\cap I\ne \varnothing$by Theorem 2.37. 05-30-2012, 06:16 PM Post: #38  elim Moderator Posts: 577 Joined: Feb 2010 Reputation: 0 Perfect Sets --Rudin [Principle of Mathematical Analysis] Notes PERFECT SETS 2.43 Theorem Nonempty perfect set in$\mathbb{R}^k$is uncountable. Proof Since$P$has limit points, it's infinite. Suppose$P$is countable with points$\mathbf{x}_1,\mathbf{x}_2,\cdots\quad\quad$Denote$\overline{N}=\{\mathbf{x}: |\mathbf{x-p}| \le r\}$if$N=N_r(p)$. Let$V_1$be any neighborhood of$\mathbf{x}_1$, suppose$V_n\quad\quad$has been chosen so that$V_n\cap P\ne\varnothing$, since$P' = P$, there is a neighborhood$V_{n+1}$such that$\quad\quad\; (i)\; \overline{V}_{n+1}\subset V_n,\; (ii)\;\mathbf{x}_n\notin \overline{V}_{n+1},\; (iii)\; V_{n+1}\cap P\ne \varnothing.$By (iii), the inductive construction can$\quad\quad$continue and we get a sequence of neighborhoods$\{V_n\}$.$\quad\quad\quad$Put$K_n = \overline{V}_n\cap P$, then$K_n\ne\varnothing$is compact and$\mathbf{x}_n\notin K_n\subset P\; (\forall n)$thus$\bigcap_1^\infty K_n = \varnothing\quad\quad$but by (i)$K_n\supset K_{n+1}\; (\forall n)$that contradicts the Corollary of Theorem 2.36. Corollary Every interval$[a,\;b]\; (a< b)$is uncountable.$\mathbb{R}$is uncountable. 2.44 The Cantor set We shall see a perfect set in$\mathbb{R}^1$which contains no segment$\quad\quad$Let$E_0 = [0,\; 1],\; E_1 = [0,\frac{1}{3}]\cup [\frac{2}{3},1]$, and$E_2$the union of$[0,\frac{1}{9}],[\frac{2}{9},\frac{3}{9}],[\frac{6}{9},\frac{7}{9}],[\frac{8}{9},1]\quad\quad$in general we have$(a)\;E_1\supset E_2\supset E_3\supset\cdots;\;(b)E_n$is the union of$2^n$intervals(each of length$\quad\quad3^{-n}$) obtained by removing the middle thirds from each of$2^{n-1}$intervals that union up$E_{n-1}\quad\quad$The set $P=\bigcap_1^\infty E_n$$\quad\quad$is called the Cantor set.$P\ne \varnothing$is compact (Theorem 2.36)$\quad\quad$No segment of the form $\left(\frac{3k+1}{3^m},\frac{3k+2}{3^m} \right )\;\tag{24}$$\quad\quad$where$k$and$m$are positive integers, has a point in common with$P$. Since every segment$(\alpha,\beta)$contains a segment of the form (24), if $3^{-m} <\frac{\beta -\alpha}{6},$$\quad\quadP$contains no segment.$\quad\quadP$is perfect means that$P$contains no isolated points: Let$x\in P$and$S$be any segment containning$x$. Let$I_n$be that interval of$E_n$which contains$x$. Choice$n$large enough such that$I_n\subset S$. Let$x_n$be an endpoint of$I_n$such that$x_n\ne x$. Clearly$x_n\in P$thus$x$is a limit point of$P$.$\quad\quad$One of the most interesting properties of the Cantor set is that it is an example of an uncountable set of measure zero (see Chap.11). 06-07-2012, 07:23 AM Post: #39  elim Moderator Posts: 577 Joined: Feb 2010 Reputation: 0 RE: Rudin [Principle of Mathematical Analysis] Notes danmath's comment to my earlier post is moved to a better location (link below)... The idea of an expression in Example 1.1 Thanks danmath, Good points, Welcome to the forum! 06-21-2012, 08:45 AM Post: #40  elim Moderator Posts: 577 Joined: Feb 2010 Reputation: 0 Connected Sets 2.45~2.47: Rudin [Principle of Mathematical Analysis] Notes CONNECTED SETS 2.45 Definition If$(A,B\subset X) \wedge (A\cap \overline{B} = \overline{A}\cap B = \varnothing)$, then$A,\;B$are separated.$\quad\quad$A set$E$is said to be connected if$E$is not a union of two nonempty separated sets. 2.46 Remark Separated sets are disjoint but disjoint sets may not be separated:$(0,1),\;(1,2)$are separated but$[0,1],\; (1,2)$are not.$\quad\quad$In line$\mathbb{R}^1$, the structure of connected sets is very simple: 2.47 Theorem$E(\subset \mathbb{R}^1)$is connected if and only if$\forall x,y\in E\;\forall z\in\mathbb{R}^1\; ((x Proof If there are $x,y\in E$ and some $z\in (x,y)$ such that $z\notin E$, then $E = A_z\cup B_z$ with $A_z = E\cap (-\infty,z),\quad B_z = E\cap (z,\infty)$ $\quad\quad$ Since $x\in A_z,\; y\in B_z,\; A_z\ne\varnothing\ne B_z$ and $(-\infty,z),\; (z,\infty)$ are separated, $E$ is disconnected.
$\quad\quad$ Conversely, if $E$ is not connected, $E = A\cup B$ for some nonempty separated sets $A,\;B$.
$\quad\quad$ Pick $x\in A,\; y\in B$, assume (without loss of generality) that $x$\quad\quad$By Theorem 2.28,$z\in\overline{A}$hence$z\notin B$and so$x\le z < y$. We have either$\quad\quadz\notin A\implies (x < z < y)\wedge (z\notin E)$; or$z\in A\setminus \overline{B}\Rightarrow (\exists z_1\;(z $\quad\quad\quad\quad\quad \implies (x < z_1 < y) \wedge (z_1\notin E). \quad\square$
 « Next Oldest | Next Newest »

Forum Jump: