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陶哲轩实分析6.3及习题-Analysis I 6.3

2019年4月23日2019年12月13日christangdt留下评论

这一节更像个开胃菜，为后面比较难理解的limsup和liminf做准备。

Exercise 6.3.1. Verify the claim in Example 6.3.4.
Solution: To verify $\sup (a_n)_{n=1}^{\infty}=1$ , we first notice that $a_n\leq 1,\forall n$ , also for any number $k<1$ , since $a_1=1\in (a_n)_{n=1}^{\infty}$ , we have $k<a_1$ , thus $k$ is not an upper bound of ${n=1}^{\infty}$ .
To verify $\inf (a_n){n=1}^{\infty}=0$ , we first see that $a_n>0,\forall n$ , also for any number $k>0$ , by the Archimedean property we have a positive integer $N$ such that $Nk>1$ or $1/N<k$ , this means $a_N<k$ , so $k$ is not an lower bound of $(a_n)_{n=1}^{\infty}$ .

$\blacksquare$

Exercise 6.3.2. Prove Proposition 6.3.6.
Solution: We let $E=\{a_n:n\geq m\}$ , then $x=\sup E$ by Definition 6.3.1, and by Theorem 6.2.11 (a), for all $n\geq m$ we have $a_n\in E$ , thus $a_n\leq x$ .
If $M\in \mathbf R^*$ is an upper bound for $a_n$ , then $M$ is an upper bound for $E$ , by Theorem 6.2.11 (b) we have $x\leq M$ .
If $y\in \mathbf R^*,y<x$ , then $y\neq +{\infty}$ , assume for all $n\geq m$ we have $a_n\leq y$ , then $y$ is an upper bound of $E$ , also $+{\infty}\notin E$ , so $x=\sup E=\sup E\backslash\{-{\infty}\}$ , as $E\backslash \{-{\infty}\}\in \mathbf R$ , $x$ shall be no larger than $y$ by definition, this is an contradiction.

$\blacksquare$

Exercise 6.3.3. Prove Proposition 6.3.8.
Solution: Let $x=\sup (a_n)_{n=1}^{\infty}$ , then $x\leq M$ by Proposition 6.3.6. Let $\forall \varepsilon >0$ , we have $x-\varepsilon <x$ , so

$\exists N\geq m,\quad x-\varepsilon <a_N\leq x$

Thus we have

$x-\varepsilon <a_N\leq a_n\leq x,\quad \forall n\geq N$

$|a_n-x|<\varepsilon ,\quad \forall n\geq N$

$\blacksquare$

Exercise 6.3.4. Explain why Proposition 6.3.10 fails when $x>1$ . In fact, show that the sequence $(x^n)_{n=1}^{\infty}$ diverges when $x>1$ .
Solution: Suppose $x>1$ , assume $(x^n)_{n=1}^{\infty}$ converges to $L$ , then we have $0<1/x<1$ and

$\lim\limits_{n\to {\infty}}(1/x)^n=0$

Use the identity $(1/x)^n x^n=1$ , we let $n\to {\infty}$ and get $0\cdot L=0=1$ , which is a contradiction.

$\blacksquare$

陶哲轩实分析6.2及习题-Analysis I 6.2

2019年4月23日2019年12月13日christangdt2条评论

这一节严格的引入了无穷的概念，并且和上下确界关联起来。

Exercise 6.2.1. Prove Proposition 6.2.5.
Solution:
( a ) if $x$ is a real number then $x\leq x$ shall be true. If $x=+\infty$ , then any $y\in \mathbf R^*, y\leq x$ , and $x\leq x$ . If $x=-\infty$ , then any $y\in \mathbf R^*, y\geq x$ , and $x\geq x$ .

( b ) Trichotomy is true when $x,y$ are real numbers. Now consider the case when $x=+\infty$ or $x=-\infty$ :

Case I: , then
- i. $y=-\infty$ , then $y\leq x$ and $y\neq x$ , thus $y<x$
- ii. $y$ is a real number, then $y\leq x$ and $y\neq x$ , thus $y<x$
- iii. $y=+\infty$ , then $y=x$
Case II: , then
- i. $y=-\infty$ , then $y=x$
- ii. $y$ is a real number, then $x\leq y$ and $y\neq x$ , thus $x<y$
- iii. $y=+\infty$ , then $x\leq y$ and $y\neq x$ , thus $x<y$
We left with two cases when and is infinity:
- i. $y=-\infty$ , then $y\leq x$ and $y\neq x$ , thus $y<x$
- ii. $y=+\infty$ , then $x\leq y$ and $y\neq x$ , thus $x<y$

( c ) We inspect cases where at least one of $x,y,z$ is infinity, starting from the left

i. $x=-\infty$ , then $x\leq z$ is true whatever $z$ ’s value is.
ii. $x$ is real number, and $y$ is real and $z=+\infty$ , then $x\leq z$ by definition. $y=+\infty$ , then as $y\leq z$ we must have $z=+\infty$ , thus $x\leq z$
iii. $x=+\infty$ , then $y=z=+\infty$ , thus $x\leq z$ by definition.

( d ) By definition 6.2.3, we have one of the three statements true.

i. $x,y$ are real and $x\leq y$ , then we shall have $-y\leq -x$ easily.
ii. $x=-\infty$ , then $-x=+\infty$ , thus $-y\leq -x$ by Definition 6.2.3 ( b )
iii. $y=+\infty$ , then $-y=-\infty$ , thus $-y\leq -x$ by Definition 6.2.3 ( c )

$\blacksquare$

Exercise 6.2.2. Prove Theorem 6.2.11.
Solution:
( a ) We shall split the case:

i. $+\infty \notin E$ and $-\infty \notin E$ , then the conclusion follows from Definition 5.5.10
ii. $+\infty \in E$ and $-\infty \notin E$ , then $\sup(E)=+\infty$ and we certainly have $x\leq \sup (E)$ , also since $-\infty \in -E$ , but $+\infty \notin -E$ , we shall have $\inf E=-\sup (-E\backslash \{-\infty \})$ , so $\begin{aligned}(\forall x\in E)&\implies (-x\in -E)\implies (-x\in -E\backslash \{-\infty \})\\&\implies (-x\leq \sup (-E\backslash \{-\infty \}) )\\&\implies x\geq -\sup (-E\backslash \{-\infty \})=\inf E\end{aligned}$
iii. $+\infty \notin E$ and $-\infty \in E$ , then $(\forall x\in E)\implies (x=-\infty )\wedge (x\in E\backslash \{-\infty \})$ , so that
$(x=-\infty )\wedge (\sup E=\sup (E\backslash \{-\infty \})\in R^* )\implies x\leq\sup E\\ (x\in E\backslash \{-\infty \})\implies x\leq \sup (E\backslash \{-\infty \})=\sup E$
Since $+\infty \in -E$ , we have $\sup (-E)=+\infty$ and $\inf E=-\infty$ , so $x\geq \inf E$ .
iv. $+\infty \in E$ and $-\infty \in E$ , then we have $\sup E=+\infty$ and $\inf E=-\infty$ , use Definition 6.2.3 we can get $\inf E\leq x\leq \sup E$ for all $x\in E$ .

( b ) we shall split the case:

i. $E$ is contained in $R$ , then the definition of $\sup E$ guarantees $\sup E\leq M$ .
ii. $+\infty \in E$ , so if $M$ is an upper bound of $E$ we shall have $+\infty \leq M$ , thus
$(M=+\infty )\implies (\sup E=+\infty \leq +\infty =M)$
iii. $+\infty \notin E$ and $-\infty \in E$ , then $\sup E=\sup (E\backslash \{-\infty \})$ , and we have $M$ is an upper bound for $E\backslash \{-\infty \}$ , since $E\backslash \{-\infty \}\in \mathbf R$ , we know that $M\geq \sup (E\backslash \{-\infty \})$ .

( c ) we shall split the case:

i. $E$ is contained in $R$ , then the definition of $\inf E$ guarantees $\inf E\geq M$ .
ii. $-\infty \in E,+\infty \in -E$ , so if $M$ is an lower bound of $E$ we shall have $M\leq -\infty$ , thus
$(M=-\infty )\implies (\inf E=-\infty \geq -\infty =M)$
iii. $-\infty \notin E$ and $+\infty \in E$ , then $\sup?(-E)=\sup (-E\backslash \{-\infty \})$ , and $M$ is an lower bound for $E$ means $-M$ is an upper bound for $-E\backslash \{-\infty \}$ , since $-E\backslash \{-\infty \}\in \mathbf R$ we know that $-M\geq \sup (-E\backslash \{-\infty \})=\sup (-E)$ , thus $M\leq -\sup (-E)=\inf E$ .

$\blacksquare$

陶哲轩实分析6.1及习题-Analysis I 6.1

2019年4月23日2019年12月13日christangdt7条评论

这一节开始讲实际的极限，建立metric概念（其实是实直线上的标准度量）和极限算律，从这一章起，内容与普通实分析差异不大，但由于有第2-5章的基础，从这章开始的进度是明显加快的，其后只用6章就把单变量微积分的严格构建以及实数型级数介绍完毕，还附带说了一些集合论的东西（第8章），由此可以看出基础打得好的作用，如果前面做题扎实，从第6章开始读完Analysis I的感觉应该是基本上每个概念都很清晰。本章起习题有所增加，且相对常规化，但仍都值得一做。

Exercise 6.1.1. Let $(a_n)_{n=0}^{\infty}$ be a sequence of real numbers, such that $a_{n+1}>a_n$ for each natural number $n$ . Prove that whenever $n$ and $m$ are natural numbers such that $m>n$ , then we have $a_m>a_n$ . (We refer to these sequences as increasing sequences.)
Solution: Given $n$ fixed and we induct on $m>n$ :
For $m=n+1$ , we have $a_{n+1}>a_n$ by condition given.
Suppose $m>n$ and $a_m>a_n$ , then for $m+1$ , we have $a_{m+1}>a_m>a_n$ , thus the induction hypothesis is valid.

$\blacksquare$

Exercise 6.1.2. Let $(a_n)_{n=m}^{\infty}$ be a sequence of real numbers, and let $L$ be a real number. Show that $(a_n)_{n=m}^{\infty}$ converges to $L$ if and only if, given any $\varepsilon >0$ , one can find an $N\geq m$ such that $|a_n-L|\leq\varepsilon$ for all $n\geq N$ .
Solution:

$\begin{aligned} (a_n)_{n=m}^{\infty}\text{ converges to }L&\iff (a_n )_{n=m}^{\infty}\text{ is eventually } \varepsilon \text{-close to } L \text{ for }\forall \varepsilon >0\\&\iff \exists N\geq m,\text{ s.t. }(a_n)_{n=N}^{\infty}\text{ is } \varepsilon \text{-close to }L\\&\iff \exists N\geq m, \text{ s.t. } a_n \text{ is } \varepsilon \text{-close to } L,\forall n\geq N\\&\iff \exists N\geq m,\text{ s.t. }|a_n-L|\leq \varepsilon ,\forall n\geq N \end{aligned}$

$\blacksquare$

Exercise 6.1.3. Let $(a_n)_{n=m}^{\infty}$ be a sequence of real numbers, let $c$ be a real number, and let $m'\geq m$ be an integer. Show that $(a_n)_{n=m}^{\infty}$ converges to $c$ if and only if $(a_n)_{n=m'}^{\infty}$ converges to $c$ .
Solution: Use Exercise 6.1.2 we have

$\begin{aligned}(a_n)_{n=m}^{\infty}\text{ converges to }c&\implies \exists N\geq m,\text{ s.t. }|a_n-c|\leq \varepsilon ,\forall n\geq N \\&\implies \exists N'=\max (N,m' )\geq m,\text{ s.t. }|a_n-c|\leq \varepsilon ,\forall n\geq N' \\&\implies \exists N'\geq m',\text{ s.t. }|a_n-c|\leq \varepsilon ,\forall n\geq N \end{aligned}$

The other direction is easy since any $N\geq m'$ we’ll have $N\geq m$ .

$\blacksquare$

Exercise 6.1.4. Let $(a_n)_{n=m}^{\infty}$ be a sequence of real numbers, let $c$ be a real number, and let $k\geq 0$ be a non-negative integer. Show that $(a_n)_{n=m}^{\infty}$ converges to $c$ if and only if $(a_{n+k})_{n=m}^{\infty}$ converges to $c$ . Solution: $(a_n)_{n=m}^{\infty}$ converges to $c$ means

$\exists N\geq m,\text{ s.t. } |a_n-c|\leq \varepsilon ,\forall n\geq N$

Thus if $n\geq N$ , then $n+k\geq N$ , thus $|a_{n+k}-c|\leq \varepsilon$ , so $(a_{n+k})_{n=m}^{\infty}$ converges to $c$ . Conversely, if $(a_{n+k} )_{n=m}^{\infty}$ converges to $c$ , then

$\exists N\geq m,\text{s.t. }|a_{n+k}-c|\leq \varepsilon ,\forall n\geq N$

Let $N'=N+k$ , then if $n>N'$ we have $|a_n-c|\leq \varepsilon$ , so $(a_n)_{n=m}^{\infty}$ converges to $c$ .

$\blacksquare$

Exercise 6.1.5. Prove Proposition 6.1.12.
Solution: For $\forall \varepsilon >0$ , let $L=\lim_{n\to {\infty}}a_n$ , then $\exists N\geq m,s.t.|a_n-L|\leq \varepsilon /2,\forall n\geq N$ , thus if $p,q\geq N$ ,

$|a_p-a_q |\leq |a_p-L|+|a_q-L|\leq \varepsilon$

This means $(a_n)_{n=m}^{\infty}$ is a Cauchy sequence.

$\blacksquare$

Exercise 6.1.6. Prove Proposition 6.1.15.
Solution: We assume $a_n$ is not eventually $\varepsilon$ -close to $L$ , then first there’s a $N'$ such that

$|a_n-a_m |<\varepsilon /2,\quad \forall n,m\geq N'$

Next, we can find a $N>N'$ such that

$|a_N-L|\geq \varepsilon$

Let $n\geq N$ , use both results we can have

$|a_n-L|\geq |a_N-L|-|a_n-a_N |\geq \varepsilon -\varepsilon /2=\varepsilon /2$

This means for all $n\geq N$ , either $a_n>L+\varepsilon /2$ or $a_n<L-\varepsilon /2$ , by Exercise 5.4.8, we shall have $L\geq L+\varepsilon /2$ or $L\leq L-\varepsilon /2$ , both leads to a contradiction.

$\blacksquare$

Exercise 6.1.7. Show that Definition 6.1.16 is consistent with Definition 5.1.12.
Solution: We have to show the following result is true:
Let $(a_n )_{n=m}^{\infty}$ be a sequence of rational numbers starting at some integer index $m$ , then $(a_n )_{n=m}^{\infty}$ is bounded in the sense of Definition 6.1.16 iff $(a_n )_{n=m}^{\infty}$ is bounded in the sense of Definition 5.1.12.

We first suppose $(a_n)_{n=m}^{\infty}$ is bounded in the sense of Definition 6.1.16, i.e. $\exists M>0$ , s.t.

$|a_n|\leq M,\quad \forall n\geq m$

As $M$ is a positive real, we can have a positive integer (thus rational) $N$ such that $M\leq N$ , so we can have

$|a_n|\leq N,\quad \forall n\geq m$

This shows that $(a_n)_{n=m}^{\infty}$ is bounded in the sense of Definition 5.1.12.
On the other hand, suppose $(a_n)_{n=m}^{\infty}$ is bounded in the sense of Definition 5.1.12. Then we can find a rational $M\geq 0$ such that

$|a_n|\leq M,\quad \forall n\geq m$

Since $M$ is also a real number, we get the result directly.

$\blacksquare$

Exercise 6.1.8. Prove Theorem 6.1.19.
Solution: We let $\forall \varepsilon >0$ : then have

$\left(x=\lim_{n\to {\infty}}a_n \right)\implies (\exists N>0,s.t.|a_n-x|<\varepsilon ,\forall n\geq N) \\ \left(y=\lim_{n\to {\infty})}b_n \right)\implies (\exists M>0,s.t.|b_n-y|<\varepsilon ,\forall n\geq M)$

Further, convergent sequences are bounded, so we have $M_1,M_2>0$ , and

$|a_n |<M_1,\quad |b_n |<M_2$

( a ) The result comes from

$|a_n+b_n-x-y|\leq |a_n-x|+|b_n-y|\leq 2\varepsilon ,\quad \forall n\geq \max(N,M)$

( b ) We have, for all $n\geq \max (N,M)$ :

$\begin{aligned}|a_n b_n-xy|=|a_n b_n-a_n y+a_n y-xy|&\leq M_1 |b_n-y|+y|a_n-x|\\&\leq (M_1+y)\varepsilon \end{aligned}$

( c ) Let $b_n=c,\forall n$ and use (b).

( d ) From (c) we can get

$\lim\limits_{n\to {\infty}}{-b_n}=-\lim\limits_{n\to {\infty}} b_n$

And then use (b) to get the result.

( e ) Since $y\neq 0$ , we know that there’s $N_1$ such that

$|b_n-y|\leq |y|/2,\quad \forall n\geq N_1$

Which means when $n\geq N_1$

$(-\dfrac{|y|}{2}\leq b_n-y\leq \dfrac{|y|}{2})\implies \left(\dfrac{|y|}{2}\leq b_n\leq \dfrac{3|y|}{2}\right)\vee \left(-\dfrac{3|y|}{2}\leq b_n\leq -\dfrac{|y|}{2}\right)$

As $b_n\neq 0,\forall m\leq n\leq N_1$ , we show that $b_n$ is bounded away from $0$ , so there’s $c>0 s.t.$

$|b_n |\geq c>0$

Thus

$|\dfrac{1}{b_n} -\dfrac{1}{y}|=\dfrac{|y-b_n |}{|b_n y|} \leq \dfrac{\varepsilon }{c|y|} ,\quad \forall n\geq M$

( f ) We use the notation of $|b_n |\geq c>0$ in (e), then

$\begin{aligned}|a_n/b_n -x/y|=\dfrac{|a_n y-b_n x|}{|b_n y|} &\leq \dfrac{|a_n y-a_n b_n+a_n b_n-b_n x|}{c|y|} \\&\leq \frac{(M_1+M_2)}{c|y|} \varepsilon ,\quad \forall n\geq \max(N,M)\end{aligned}$

( g ) If $x=y$ , then $a_n$ and $b_n$ are equivalent Cauchy sequence, thus when $n\geq \max(N,M)$ :

$(|a_n-x|<\varepsilon )\wedge (|b_n-x|<\varepsilon )\implies |\max (a_n,b_n)-x|<\varepsilon$

Which means

$\lim\limits_{n\to {\infty}} \max (a_n,b_n) =x=\max (x,y)=\max (\lim\limits_{n\to {\infty}}a_n,\lim\limits_{n\to {\infty}}b_n)$

Now assume $x>y$ , then there’s $N_2>0 s.t.$

$|a_n-x|\leq (x-y)/3,\quad |b_n-y|\leq (x-y)/3,\quad \forall n\geq N_2$

Which will lead to

$a_n\geq x-\dfrac{x-y}{3}=\dfrac{2x}{3}+\dfrac{y}{3}>\dfrac{x}{3}+\dfrac{2y}{3}=\dfrac{x-y}{3}+y\geq b_n,\quad \forall n\geq N_2$

So that $\max(a_n,b_n)=a_n,\forall n\geq N_2$ , using Exercise 6.1.3 we know that

$\lim\limits_{n\to {\infty}}\max (a_n,b_n) =\lim\limits_{n\to {\infty}}a_n =x=\max (x,y)$

( h ) A similar proof like (g) can be made by the same logic.

$\blacksquare$

Exercise 6.1.9. Explain why Theorem 6.1.19(f) fails when the limit of the denominator is 0.
Solution: Let $a_n=1/n^2$ and $b_n=1/n$ , then by Proposition 6.1.11:

$\lim\limits_{n\to {\infty}} \dfrac{a_n}{b_n}=\lim\limits_{n\to {\infty}} \dfrac{1}{n}=0$

But $x=y=0$ , so the expression $\dfrac{\lim_{n\to {\infty}}a_n}{\lim_{n\to {\infty}}b_n}$ is meaningless.

$\blacksquare$

Exercise 6.1.10. Show that the concept of equivalent Cauchy sequence, as defined in Definition 5.2.6, does not change if $\varepsilon$ is required to be positive real instead of positive rational. More precisely, if $(a_n)_{n=0}^{\infty}$ and $(b_n)_{n=m}^{\infty}$ are sequences of reals, show that $(a_n)_{n=0}^{\infty}$ and $(b_n)_{n=m}^{\infty}$ are eventually $\varepsilon$ -close for every rational $\varepsilon >0$ if and only if they are eventually $\varepsilon$ -close for every real $\varepsilon >0$ .
Solution: Eventually $\varepsilon$ -close for every real $\varepsilon >0$ implies eventually $\varepsilon$ -close for every rational $\varepsilon >0$ .
To prove the converse, suppose $(a_n )_{n=0}^{\infty}$ and $(b_n )_{n=0}^{\infty}$ are eventually $\varepsilon$ -close for every rational $\varepsilon >0$ , then given any real $\delta >0$ , we can have a rational $\varepsilon$ such that $0<\varepsilon <\delta$ , and an integer $N$ s.t.

$|a_n-b_n |<\varepsilon <\delta ,\quad \forall n\geq N$

This means $(a_n )_{n=0}^{\infty}$ and $(b_n )_{n=0}^{\infty}$ are eventually $\delta$ -close.

$\blacksquare$

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