城南讲马堂

子列的概念，最后得出经典的Bolzano-Weierstrass定理。

Exercise 6.6.1. Prove Lmma 6.6.4.
Solution: To show that $(a_n)_{n=0}^{\infty}$ is a subsequence of $(a_n)_{n=0}^{\infty}$ , we have $f(n)=n$ a strictly increasing function.
If $(b_n)_{n=0}^{\infty}$ is a subsequence of $(a_n)_{n=0}^{\infty}$ , $(c_n)_{n=0}^{\infty}$ is a subsequence of $(b_n)_{n=0}^{\infty}$ , then there’re two strictly increasing functions $f,g$ such that

$b_n=a_{f(n)} ,\quad c_n=b_{g(n)}$

Let $h=g\circ f$ , then $h$ is strictly increasing since

$f(n+1)>f(n) \implies g(f(n+1))>g(f(n)) \implies h(n+1)>h(n)$

And we have $c_n=a_{h(n)}$ .

$\blacksquare$

Exercise 6.6.2. Can you find two sequences $(a_n)_{n=0}^{\infty}$ and $(b_n)_{n=0}^{\infty}$ which are not the same sequence, but such that each is a subsequence of the other?
Solution: We let

$(a_n)_{n=0}^{\infty}=\{0,1,0,1,\cdots \},\quad (b_n)_{n=0}^{\infty}=\{1,0,1,0,\cdots \}$

$\blacksquare$

Exercise 6.6.3. Let $(a_n)_{n=0}^{\infty}$ be a sequence which is not bounded. Show that there exists a subsequence $(b_n)_{n=0}^{\infty}$ of $(a_n)_{n=0}^{\infty}$ such that $\lim_{n\to\infty}1/b_n$ exists and is equal to zero.
Solution: We let $n_0=1$ , and for every $j\geq 1$ , we recursively let

$n_j=\min \{n\in \mathbf N:|a_n |\geq j;n>n_{j-1} \}$

This set can’t be empty, since if we assume for some $J$ , the set $\{n\in \mathbf N:|a_n |\geq J;n>n_{J-1} \}$ is empty, then for every $n>n_{J-1}$ we have $|a_n |<J$ , this means $(a_n)_{n=0}^{\infty}$ is bounded by the number $\max \{|a_1 |,\cdots ,|a_{n_{J-1} }|,J\}$ , which is a contradiction to the fact that $(a_n)_{n=0}^{\infty}$ is unbounded.
Now let $b_j:=a_{n_j}$ , we see that $|b_j|\geq j,\forall j\in N$ , thus $\lim_{n\to {\infty}} 1/b_n=0$ .

$\blacksquare$

Exercise 6.6.4. Prove Proposition 6.6.5.
Solution: (b) means (a) is easy, since $(a_n)_{n=0}^{\infty}$ is a subsequence of itself.
To prove (a) means (b), we let $(a_{n_j})_{j=1}^{\infty}$ be a subsequence of $(a_n)_{n=0}^{\infty}$ , then as $a_n\to L$ , for $\forall \varepsilon >0$ there’s $N\geq 0$ s.t. $|a_n-L|<\varepsilon ,n\geq N$ . Since $n_N\geq N$ , we have

$|a_{n_j}-L|<\varepsilon ,\quad j\geq N$

This means $a_{n_j}\to L$ .

$\blacksquare$

Exercise 6.6.5. Prove Proposition 6.6.6.
Solution: To show (a) implies (b), we define $n_0:=0$ and for $j\geq 1$ :

$n_j := \min \{n>n_{j-1} :|a_n-L|\leq 1/j \}$

Since $L$ is a limit point of $(a_n)_{n=0}^{\infty}$ , for every $j$ we can find a $n$ such that $|a_n-L|\leq 1/j$ , so this set is never empty. Now the sequence $a_{n_j}$ converges to $L$ .
To show that (b) implies (a), we notice that $\forall \varepsilon >0$ , there’s $j\geq 0$ s.t. $|a_{n_j}-L|<\varepsilon$ , but $a_{n_j}$ is an item of $(a_n )_{n=0}^{\infty}$ , so the conclusion follows.

$\blacksquare$

limsup和liminf是（至少我认为）比较难以理解的概念，不过这个概念和极限点limit points的联系是很直接的。这一节用先定义一个新的序列，再取序列上下确界的方法定义limsup和liminf，能让这个东西显得直观一些。6.4.12-6.4.15是比较重要的结论，后面的章节也会经常用到。这一节最后提到并证明了实数集的完备性。

Exercise 6.4.1. Prove Proposition 6.4.5.
Solution: For $\forall \varepsilon >0$ , there’s $N_0\geq m$ s.t. $|a_n-c|<\varepsilon ,n\geq N_0$ . So that for any $N\geq m$ , we just need to choose $n\geq \max{N,N_0 }$ , then $|a_n-c|<\varepsilon$ . This shows $c$ to be a limit point of $(a_n)_{n=m}^{\infty}$ . To see that $c$ is the only limit point of $(a_n)_{n=m}^{\infty}$ , we let $c'\neq c$ , then as $a_n\to c$ , for $\varepsilon =|c-c'|/2>0$ , there’s $N_1\geq m$ , such that

$|a_n-c|<\varepsilon ,\quad n\geq N_1$

This means

$|a_n-c'|\geq |c-c'|-|a_n-c|\geq \varepsilon ,\quad \forall n\geq N_1$

So that $c'$ is not $\varepsilon$ -adherent to $(a_n)_{n=N_1}^{\infty}$ , which means $c'$ is not a limit point.

$\blacksquare$

Exercise 6.4.2. State and prove analogue of Exercises 6.1.3 and 6.1.4 for limit points, limit superior, and limit inferior.
Solution:

Statement I:
Let $(a_n)_{n=m}^{\infty}$ be a sequence of real numbers, let $c$ be a real number, and $m'\geq m$ be an integer. Show that $c$ is a limit point of $(a_n)_{n=m}^{\infty}$ if and only if $c$ is a limit point of $(a_n)_{n=m'}^{\infty}$ , and $\limsup_{n\to {\infty} }a_n=c$ for $(a_n)_{n=m}^{\infty}$ if and only if $\limsup_{n\to {\infty} } a_n=c$ for $(a_n)_{n=m'}^{\infty}$ , and $\liminf_{n\to {\infty} }a_n=c$ for $(a_n)_{n=m}^{\infty}$ if and only if $\liminf_{n\to {\infty} }a_n=c$ for $(a_n)_{n=m'}^{\infty}$ .

Proof:
For the limit point case:
$c$ is a limit point of $(a_n)_{n=m}^{\infty}$ means $\forall \varepsilon >0,\forall N\geq m,\exists n\geq N,|a_n-c|<\varepsilon$ , this includes the case that $\forall \varepsilon >0,\forall N\geq m',\exists n\geq N,|a_n-c|<\varepsilon$ , which means $c$ is a limit point of $(a_n)_{n=m'}^{\infty}$ .
On the other hand, $n\geq N\geq m'$ means $n\geq N\geq m$ , and once we find an $n$ when $N=m'$ , we can let this $a_n$ be the element we need when $m\leq N\leq m'$ .

For the limsup and liminf case:
By definition, $\limsup_{n\to {\infty} }a_n=\inf (a_N^+)_{N=m}^{\infty}$ , where $a_N^+=\sup (a_n)_{n=N}^{\infty}$ . As $m'\geq m$ , we have

$a_n^+\geq a_{m'}^+,\quad \forall m\leq n\leq m'$

So that $\inf (a_N^+)_{N=m}^{\infty} \geq \inf (a_N^+)_{N=m'}^{\infty}$ . Now assume $\inf (a_N^+)_{N=m}^{\infty} >\inf (a_N^+ )_{N=m'}^{\infty}$ , then $\exists y\in \mathbf R$ such that

$\inf (a_N^+)_{N=m}^{\infty} >y>\inf(a_N^+)_{N=m'}^{\infty}$

So $y$ is not a lower bound of $(a_N^+)_{N=m'}^{\infty}$ , which means there’s $p{\ } s.t.{\ } y>a_p^+$ , but obviously $a_p^+\in (a_N^+)_{N=m}^{\infty}$ , so $\inf(a_N^+)_{N=m}^{\infty} >a_p^+$ is a contradiction. Thus we must have

$\inf(a_N^+)_{N=m}^{\infty} =\inf(a_N^+)_{N=m'}^{\infty}$

The $\liminf$ case can be similarly proved.

Statement II:
Let $(a_n)_{n=m}^{\infty}$ be a sequence of real numbers, let $c$ be a real number, and $k\geq 0$ be an non-negative integer. Show that $c$ is a limit point of $(a_n)_{n=m}^{\infty}$ if and only if $c$ is a limit point of $(a_n)_{n=m+k}^{\infty}$ , and $\limsup_{n\to {\infty} }a_n=c$ for $(a_n)_{n=m}^{\infty}$ if and only if $\limsup_{n\to {\infty} }a_n=c$ for $(a_n)_{n=m+k}^{\infty}$ , and $\liminf_{n\to {\infty} } a_n=c$ for $(a_n)_{n=m}^{\infty}$ if and only if $\liminf_{n\to {\infty} }a_n=c$ for $(a_n)_{n=m+k}^{\infty}$ .

Proof:
For the limit point case:
$c$ is a limit point of $(a_n)_{n=m}^{\infty}$ means $\forall \varepsilon >0,\forall N\geq m,\exists n\geq N,|a_n-c|<\varepsilon$ , this includes the case that $\forall \varepsilon >0,\forall N\geq m+k,\exists n\geq N,|a_n-c|<\varepsilon$ , which means $c$ is a limit point of $(a_n)_{n=m+k}^{\infty}$ .
On the other hand, $n\geq N\geq m+k$ means $n\geq N\geq m$ , and once we find an $n$ when $N=m+k$ , we can let this $a_n$ be the element we need when $m\leq N\leq m+k$ .
For the limsup and liminf case:
By definition, $\limsup_{n\to {\infty} } a_n=\inf(a_N^+)_{N=m}^{\infty}$ , where $a_N^+=\sup(a_n)_{n=N}^{\infty}$ . As $m+k\geq m$ , we have

$a_n^+\geq a_{m+k}^+,\quad \forall m\leq n\leq m+k$

So that $\inf (a_N^+)_{N=m}^{\infty} \geq \inf (a_N^+)_{N=m+k}^{\infty}$ . Now assume $\inf(a_N^+)_{N=m}^{\infty} >\inf(a_N^+)_{N=m+k}^{\infty}$ , then $\exists y\in \mathbf R$ such that

$\inf(a_N^+)_{N=m}^{\infty} >y>\inf(a_N^+)_{N=m+k}^{\infty}$

So $y$ is not a lower bound of $(a_N^+)_{N=m+k}^{\infty}$ , which means there’s $p$ , s.t. $y>a_p^+$ , but obviously $a_p^+\in (a_N^+)_{N=m}^{\infty}$ , so $\inf(a_N^+)_{N=m}^{\infty} >a_p^+$ is a contradiction. Thus we must have

$\inf (a_N^+)_{N=m}^{\infty} =\inf(a_N^+ )_{N=m+k}^{\infty}$

The $\liminf$ case can be similarly proved.

$\blacksquare$

Exercise 6.4.3. Prove parts (c),(d),(e),(f) of Proposition 6.4.12.
Solution:
( c ) $\inf (a_n)_{n=m}^{\infty} =a_m^-\leq \sup (a_N^-)_{N=m}^{\infty} =L^-$ , $\sup (a_n)_{n=m}^{\infty} =a_m^+\geq \inf(a_N^+)_{N=m}^{\infty} =L^+$
To prove $L^-\leq L^+$ , we assume $L^->L^+$ , then $\inf (a_N^+)_{N=m}^{\infty} <\sup (a_N^-)_{N=m}^{\infty}$ , by the definition of infimum, we can find an $a_N^+<\sup(a_N^-)_{N=m}^{\infty}$ , again by the definition of supremum, we can find an $a_M^-$ s.t. $a_N^+<a_M^-<\sup (a_N^-)_{N=m}^{\infty}$ , now choose $p=N+M+1$ , then $p>N,p>M$ , so we get a contradiction from

$a_p\leq a_N^+<a_M^-\leq a_p \implies a_p<a_p$

( d ) With the help of (c) we only need to show $L^-\leq c$ and $c\leq L^+$ . Assume $c<L^-<{\infty}$ , then let $\varepsilon=(L^- -c)/2>0$ , by the definition of $L^-$ we can have some $N\geq m$ such that $c+\varepsilon<a_N^-<L^-$ , which means

$\inf (a_n)_{n=N}^{\infty} >c+\varepsilon \implies a_n>c+\varepsilon,\forall n\geq N$

This contradicts $c$ to be a limit point of $(a_n)_{n=m}^{\infty}$ . Thus $L^-\leq c$ . If $L^-={\infty}$ we let $\varepsilon=1$ to complete the proof. Finally the case in which $c\leq L^+$ can be similarly proved.

( e ) $\forall \varepsilon >0$ and $\forall N\geq m$ , by ( a ) we can find $N'\geq m$ such that $a_{n'}<L^++\varepsilon$ for all ${n'}\geq N'$ , by ( b ) we can find $n\geq \max(N,N')$ such that $a_{n'}>L^+-\varepsilon$ . Then we have

$L^+-\varepsilon<a_n<L^++\varepsilon,\qquad n\geq N$

which means $L^+$ is a limit point of $(a_n)_{n=m}^{\infty}$ . The case of $L^-$ is proved similarly.

( f ) If $(a_n)_{n=m}^{\infty}$ converges to $c$ , then by (e), we have $L^+=L^-=c$ . Conversely if $L^+=L^-=c$ , then by (d) the limit point of $(a_n)_{n=m}^{\infty}$ can only be $c$ , thus $(a_n)_{n=m}^{\infty}$ converges to $c$ .

$\blacksquare$

Exercise 6.4.4. Prove Lemma 6.4.13.
Solution: We have, for all $n\geq m$ :

$(a_n\leq b_n)\implies (a_n\leq \sup (b_n)_{n=m}^{\infty} )\implies (\sup (a_n)_{n=m}^{\infty} \leq \sup (b_n)_{n=m}^{\infty})\\ (a_n\leq b_n)\implies (\inf (a_n)_{n=m}^{\infty} \leq b_n \implies (\inf (a_n)_{n=m}^{\infty} \leq \inf(b_n)_{n=m}^{\infty})$

The above inequality is independent of the starting number $m$ , thus we can have

$a_N^+\leq b_N^+,\quad a_N^-\leq b_N^-,\quad \forall N\geq m$

Again using the first two statements, we can get the remaining two statements.

$\blacksquare$

Exercise 6.4.5. Use Lemma 6.4.13 to prove Corollary 6.4.14.
Solution: From $a_n\leq b_n$ we can have $\liminf_{n\to {\infty} }a_n =L\leq \liminf_{n\to {\infty} }b_n$ .
From $b_n\leq c_n$ we can have $\limsup_{n\to {\infty}}b_n \leq \limsup_{n\to {\infty} }c_n =L$ .
Together we have $\liminf_{n\to {\infty} }b_n =\limsup_{n\to {\infty}} b_n =L$ , using Proposition 6.1.12 (f) we are done.

$\blacksquare$

Exercise 6.4.6. Give an example of two bounded sequences $(a_n)_{n=1}^{\infty}$ and $(b_n)_{n=1}^{\infty}$ such that $a_n<b_n$ for all $n\geq 1$ , but that $\sup (a_n)_{n=1}^{\infty}\nless \sup (b_n)_{n=1}^{\infty}$ . Explain why this does not contradict Lemma 6.4.13.
Solution: We can have $a_n=1-1/n$ and $b_n=1$ for all $n\geq 1$ , then obviously $a_n<b_n,\forall n$ , but

$\sup (a_n)_{n=1}^{\infty} =\sup(b_n)_{n=1}^{\infty} =1$

Lemma 6.4.13 requires $\leq$ not $<$ , so this is not a violation.

$\blacksquare$

Exercise 6.4.7. Prove Corollary 6.4.17. Is the corollary still true if we replace zero in the statement of this Corollary by some other number?
Solution: If $\lim_{n\to {\infty} } |a_n|=0$ , then $\lim_{n\to {\infty}}-|a_n |=0$ , by $-|a_n|\leq a_n\leq |a_n|$ and squeeze test, $\lim_{n\to {\infty} }a_n=0$ .
If $\lim_{n\to {\infty}} a_n=0$ , then $\forall \varepsilon >0$ , there’s $N\geq M$ such that

$|a_n-0|<\varepsilon ,\quad \forall n\geq N$

Which means $|a_n|=|(|a_n|-0)|<\varepsilon ,\forall n\geq N$ , thus $\lim_{n\to {\infty}} |a_n|=0$ .
If we replace $0$ by other number the Corollary can be wrong.

$\blacksquare$

Exercise 6.4.8. Let us say that a sequence $(a_n)_{n=M}^{\infty}$ of real numbers has $+\infty$ as a limit point iff it has no finite upper bound, and that it has $-\infty$ as a limit point iff it has no finite lower bound. With this definition, show that $\limsup_{n\to\infty}a_n$ is a limit point of $(a_n)_{n=M}^{\infty}$ , and futthermore that it is larger than all the other limit points of $(a_n)_{n=M}^{\infty}$ ; in other words, the limit superior is the largest limit point of a sequence. Similarly, show that the limit inferior is the smallest limit point of a sequence.
Solution: Given $(a_n)_{n=M}^{\infty}$ , we know it may or may not have finite upper bound. If it does have, then

$a_M^+=\sup (a_n)_{n=M}^{\infty} <+{\infty} \implies \limsup_{n\to {\infty} }a_n<+{\infty}$

Then by Proposition 6.4.12(e) we know $\limsup_{n\to {\infty} } a_n$ is a limit point of $(a_n)_{n=M}^{\infty}$ , and by Proposition 6.4.12(d) we know it’s larger than all other limit points of $(a_n)_{n=M}^{\infty}$ .
If it has no finite upper bound, then $+{\infty}$ is a limit point of $(a_n)_{n=M}^{\infty}$ by the condition given in the exercise, and if $c$ is a limit point of $(a_n)_{n=M}^{\infty}$ , we must have $c\leq +{\infty}$ .
The case of limit inferior can be proved similarly.

$\blacksquare$

Exercise 6.4.9. Using the definition in Exercise 6.4.8, construct a sequence $(a_n)_{n=1}^{\infty}$ which has exactly three limit points, at $-\infty,0$ and $+\infty$ .
Solution: The sequence $(a_n)_{n=1}^{\infty}$ which is

$1,1/2,-3,-1/4,5,1/6,-7,-1/8,\cdots$

$\blacksquare$

Exercise 6.4.10. Let $(a_n)_{n=N}^{\infty}$ be a sequence of real numbers, and let $(b_m)_{m=M}^{\infty}$ be another sequence of real numbers such that each $b_m$ is a limit point of $(a_n)_{n=N}^{\infty}$ . Let $c$ be a limit point of $(b_m)_{m=M}^{\infty}$ . Prove that $c$ is also a limit point of $(a_n)_{n=N}^{\infty}$ .
Solution: For $\forall \varepsilon >0$ , there’s $p\geq M$ such that $|b_p-c|<\varepsilon /2$ , for this $b_p$ , which is a limit point of $(a_n)_{n=N}^{\infty}$ , there’s $q\geq N$ such that $|a_q-b_p |<\varepsilon /2$ , so we have

$|a_q-c|\leq |a_q-b_p |+|b_p-c|<\varepsilon /2+\varepsilon /2=\varepsilon$

Which means $c$ is a limit point of $(a_n )_{n=N}^{\infty}$ .

$\blacksquare$

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城南讲马堂

一个讲点math的地方

陶哲轩实分析6.6及习题-Analysis I 6.6

陶哲轩实分析6.5及习题-Analysis I 6.5

陶哲轩实分析6.4及习题-Analysis I 6.4