城南讲马堂

陶哲轩实分析5.3及习题-Analysis I 5.3

2019年4月20日2019年12月13日christangdt5条评论

实数是有理数的Cauchy序列形式极限所构成的，这一章通过不断运用Cauchy序列，构建了实数的加、乘和除等定义。习题解答如下

Exercise 5.3.1. Prove Proposition 5.3.3
Solution: We obviously have $(a_n)$ and $(a_n)$ are equivalent Cauchy sequences.
If $x=y$ , i.e. $(a_n)$ and $(b_n)$ are equivalent Cauchy sequences. Then $\forall \varepsilon >0$ , there’s $N$ , s.t.

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

This also means $(b_n)$ and $(a_n)$ are equivalent Cauchy sequences, thus $y=x$ . If $(x=y)\wedge (y=z)$ , then $\forall \varepsilon >0$ , we have
$(a_n)$ and $(b_n)$ are equivalent Cauchy sequences, so $\exists N_1,s.t.|a_n-b_n|<\varepsilon /2,\forall n\geq N_1$
$(b_n)$ and $(c_n)$ are equivalent Cauchy sequences, so $\exists N_2,s.t.|b_n-c_n|<\varepsilon /2,\forall n\geq N_2$
Thus for $N=\max(N_1,N_2)$ , we have, if $n\geq N$

$|a_n-c_n|\leq |a_n-b_n|+|b_n-c_n|<\varepsilon$

This means $(a_n)$ and $(c_n)$ are equivalent Cauchy sequences, thus $x=z$ .

$\blacksquare$

Exercise 5.3.2. Prove Proposition 5.3.10
Solution: To show $xy$ is a real number we have to show $(a_n b_n)$ is a Cauchy sequence. Let $\forall \varepsilon >0$ . We can assume $\varepsilon <3$ .
By Lemma 5.1.15, there’s $M>1$ such that

$|a_n |\leq M,\quad |b_n |\leq M,\quad \forall n$

Also since $(a_n)$ and $(b_n)$ are Cauchy sequences, we can see that $\exists N>0$ , such that for $m,n>N$ , $a_m$ and $a_n$ are $\varepsilon /3M$ -close, $b_m$ and $b_n$ are $\varepsilon /3M$ -close, so by Proposition 4.3.7(h), we can see that for $m,n\geq N$

$|a_m b_m-a_n b_n|\leq \left(\dfrac{\varepsilon}{3M}|b_m|+\dfrac{\varepsilon}{3M}|a_m|+\dfrac{\varepsilon}{3M}\dfrac{\varepsilon}{3M}\right)\leq \left(\dfrac{\varepsilon}{3}+\dfrac{\varepsilon}{3}+\dfrac{\varepsilon^2}{9}\right)<\varepsilon$

This means $(a_n b_n)$ is a Cauchy sequence.
If $x=x'$ , then $(a_n)$ and $(a_n')$ are equivalent Cauchy sequences, so $\exists N'>0, s.t.$

$|a_n-a_n'|<\dfrac{\varepsilon }{M},\quad n\geq N'$

Thus when $n\geq N$ :

$|a_n b_n-a_n' b_n|=|a_n-a_n'||b_n|<\dfrac{\varepsilon}{M}M=\varepsilon$

Which means $(a_n b_n)$ and $(a_n' b_n)$ are equivalent Cauchy sequences, or $xy=x'y$ .

$\blacksquare$

Exercise 5.3.3. Let $a,b$ be rational numbers. Show that $a=b$ if and only if $\text{LIM}_{n\to\infty}a=\text{LIM}_{n\to\infty}b$ .
Solution: If $a=b$ , then the two sequence are identical, and the conclusion holds.
If the Cauchy sequences $a,a,a,a,\dots$ and $b,b,b,b\dots$ are equivalent, and suppose $a\neq b$ , then let $\varepsilon =|a-b|/2$ , we will see the two sequences are not eventually $\varepsilon$ -close, a contradiction to the equivalence of the two sequences.

$\blacksquare$

Exercise 5.3.4. Let $(a_n)_{n=0}^{\infty}$ be a sequence of rational numbers which is bounded. Let $(b_n)_{n=0}^{\infty}$ be another sequence of rational numbers which is equivalent to $(a_n)_{n=0}^{\infty}$ . Show that $(b_n)_{n=0}^{\infty}$ is also bounded.
Solution: Let $\varepsilon >0$ , since $(b_n)_{n=0}^{\infty}$ is equivalent to $(a_n)_{n=0}^{\infty}$ , the two sequences are eventually $\varepsilon$ -close, thus by Exercise 5.2.2, $(b_n)_{n=0}^{\infty}$ is bounded.

$\blacksquare$

Exercise 5.3.5. Show that $\text{LIM}_{n\to\infty}1/n=0$ .
Solution: Let $\forall \varepsilon >0$ . Then there’s $N$ such that

$\left|\dfrac{1}{n}-0\right|=\dfrac{1}{n}<\varepsilon ,\quad \forall n\geq N$

thus the sequence $(1/n)_{n=1}^{\infty}$ is equivalent to the sequence $(0)_{n=1}^{\infty}$ , which is equal to $0$ .

$\blacksquare$

陶哲轩实分析5.2及习题-Analysis I 5.2

2019年4月16日2019年12月13日christangdt2条评论

等价Cauchy序列的概念，离实数又进一步。这一节也相对简单。

Exercise 5.2.1. Show that if $(a_n)_{n=1}^{\infty}$ and $(b_n)_{n=1}^{\infty}$ are equivalent sequences of rationals, then $(a_n)_{n=1}^{\infty}$ is a Cauchy sequence if and only if $(b_n)_{n=1}^{\infty}$ is a Cauchy sequence.
Solution: Let $(a_n)_{n=1}^{\infty}$ be a Cauchy sequence, then for $\forall \varepsilon>0, \exists N_1\geq 0$ , s.t.

$|a_m-a_n|\leq \varepsilon,\quad \forall m,n\geq N_1$

As $(a_n)_{n=1}^{\infty}$ and $(b_n)_{n=1}^{\infty}$ are equivalent, $\exists N_2\geq 0$ , s.t.

$|a_m-b_m |\leq \varepsilon,\quad \forall m\geq N_2$

Let $N=N_1+N_2$ , then $N\geq N_1,N\geq N_2$ and

$|b_m-b_n|\leq |b_m-a_m|+|a_m-a_n|+|a_n-b_n|\leq 3\varepsilon,\quad\forall m,n\geq N$

This means $(b_n)_{n=1}^{\infty}$ is a Cauchy sequence.
The other way can be similarly proved.

$\blacksquare$

Exercise 5.2.2. Show that if $(a_n)_{n=1}^{\infty}$ and $(b_n)_{n=1}^{\infty}$ are eventually $\varepsilon$ -close, then $(a_n)_{n=1}^{\infty}$ is bounded if and only if $(b_n)_{n=1}^{\infty}$ is bounded.
Solution: Let $(a_n)_{n=1}^{\infty}$ be bounded, thus $\exists M_1\geq 0$ , s.t.

$|a_n|\leq M_1,\quad \forall n$

As $(a_n)_{n=1}^{\infty}$ and $(b_n)_{n=1}^{\infty}$ are eventually $\varepsilon$ -close, $\exists N\geq 0$ , s.t.

$|a_m-b_m|\leq \varepsilon,\quad \forall m\geq N$

So that if $m\geq N$ , we have

$|b_m|\leq |a_m-b_m|+|a_m|\leq M_1+\varepsilon$

The sequence $(b_n)_{n=1}^{N-1}$ is finite, thus bounded, so $\exists M_2\geq 0$ , s.t.

$|b_n |\leq M_2,\quad \forall n<N$

Let $M=M_1+M_2+\varepsilon$ , then we have

$|b_n|\leq M,\quad \forall n\in \mathbf N$

This proves $(b_n)_{n=1}^{\infty}$ is bounded.
The other way can be similarly proved.

$\blacksquare$

陶哲轩实分析5.1及习题-Analysis I 5.1

2019年4月15日2019年12月13日christangdt留下评论

这一节讲Cauchy序列，更多的是为引入实数做准备，尽管Cauchy序列本身是一个非常重要的概念，个人认为在初步的实分析中是最重要的概念之一。习题只有一道，比较简单：

Exercise 5.1.1. Prove Lemma 5.1.15.
Solution: Let $(a_n)_{n=1}^{\infty}$ be a Cauchy sequence, then for $1$ there is a number $N$ such that

$|a_n-a_m|<1,\quad \forall m,n\geq N$

In particular we have

$|a_n-a_N|<1,\quad \forall n\geq N$

which means

$|a_n|=|a_n-a_N+a_N|\leq|a_n-a_N|+|a_N|<1+|a_N|,\quad \forall n \geq N$

Now the sequence $(a_n)_{n=1}^{N-1}$ is a finite sequence, thus bounded by Lemma 5.1.14, so there’s $M>0$ such that

$|a_n|\leq M,\quad n=1,\dots,N-1$

We then can have

$|a_n|\leq M+1+|a_N|,\quad \forall n\in \mathbf N$

$\blacksquare$

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