城南讲马堂

陶哲轩实分析5.6及习题-Analysis I 5.6

2019年4月22日2019年12月13日christangdt3条评论

用上确界定义正实数的1/n次幂，进而定义正实数的有理数次幂，这样的处理之前没有见过，确实很严格，下一章还会拓展到实数的实数次幂。

Exercise 5.6.1. Prove Lemma 5.6.6.
Solution:
( a ) if $y=x^{1/n}$ , then we know that

$y=\sup \{z\in \mathbf R:z\geq 0,z^n\leq x\}$

Also since $y^n\in \mathbf R$ , we must have exactly one of $y^nx$ to be true. Further we can compare $y$ with 1, we have either $y\leq 1$ or $y>1$ .
Now assume $y^n<x$ , then if $0<\varepsilon <1$ , we can have

$(y+\varepsilon )^n=y^n+ny^{n-1} \varepsilon +\dots +ny\varepsilon ^{n-1}+\varepsilon ^n<y^n+\varepsilon (ny^{n-1}+\dots +ny+1)$

If $y\leq 1$ , we can get

$(y+\varepsilon )^n<y^n+\varepsilon \left(n+\binom{n}{2}+\dots +n+1\right)$

If $y>1$ , we know that $y^k<y^n<x$ , thus

$(y+\varepsilon )^n<y^n+\varepsilon \left(n+\binom{n}{2}+\dots +n+1\right)x$

In either case, we can set $\varepsilon$ small enough to let

$\varepsilon \left(n+\binom{n}{2}+\dots +n+1\right)<x-y^n$

$\varepsilon \left(n+\binom{n}{2}+\dots +n+1\right)x<x-y^n$

Thus to get $(y+\varepsilon )^n<x$ , which means $y+\varepsilon \in \sup \{z\in \mathbf R:z\geq 0,z^n\leq x\}$ , but $y+\varepsilon >y$ , a contradiction to the definition of $y$ .
Now assume $y^n>x$ , then if $0<\delta <1$ , we can have

$(y-\delta )^n=y^n-ny^{n-1} \delta +\dots +ny(-\delta )^{n-1}+(-\delta )^n$

Drop all positive items we can have

$(y-\delta )^n>y^n-\delta \left(ny^{n-1}+\binom{n}{3}y^{n-3}+\dots \right)$

We expand each $y^k$ to be $1$ if $y\leq 1$ , and $x$ if $y>1$ , then choose $\delta$ small enough, we can have

$\delta \left(ny^{n-1}+\binom{n}{3}y^{n-3}+\dots \right)<y^n-x$

Then

$(y-\delta )^n>y^n-(y^n-x)=x$

Thus $y-\delta <y$ is an upper bound of the set $\{z\in \mathbf R:z\geq 0,z^n\leq x\}$ , again a contradiction to the definition of $y$ .

( b ) if $y^n=x$ , then $y\in \{z\in \mathbf R:z\geq 0,z^n\leq x\}$ , we need to prove $y$ is the supreme of this set.
Suppose $z\in \{z\in \mathbf R:z\geq 0,z^n\leq x\}$ , we assert $z\leq y$ , assume not, then $z>y$ , thus by Proposition 5.6.3:

$z^n>y^n=x\geq 0$

which means $z\notin \{z\in \mathbf R:z\geq 0,z^n\leq x\}$ , a contradiction. Thus $y$ is an upper bound of the set.
Now if $y'$ is any upper bound of the set, then $\forall z\in \{z\in \mathbf R:z\geq 0,z^n\leq x\},z\leq y'$ , in particular $y\leq y'$ .

( c ) as $x^{1/n}=\sup \{z\in \mathbf R:z\geq 0,z^n\leq x\}$ , we know that

$0\in \sup\{z\in \mathbf R:z\geq 0,z^n\leq x\} \implies x^{1/n}\geq 0$

So to prove the conclusion we need to show $x^{1/n}\neq 0$ , assume so, then by (a), we have

$(x^{1/n} )^n=0^n=0=x$

This is a contradiction since $x$ is a positive real number.

( d ) that $x^{1/n}>y^{1/n} \implies x>y$ is clear from Proposition 5.6.3 and (a). Now suppose $x>y$ , assume $x^{1/n}\leq y^{1/n}$ , again from Proposition 5.6.3 and (a) we can get $x\leq y$ , a contradiction.

( e ) use (d) we can get that $\forall n\geq 1$

$x>1 \implies x^{1/n})>1^{1/n}=1,\quad x<1 \implies x^{1/n}<1^{1/n}=1$

Now if $x>1$ , let $z_1=x^{1/k},z_2=x^{1/(k+1)}$ , then by (a), $z_1^k=z_2^{k+1}=x$ , since $z_2>1$ , we have

$(z_2^{k+1}=z_2^k\cdot z_2>z_2^k )\implies (z_1^k>z_2^k )\implies z_1>z_2$

Where the last arrow comes from (d) and (b), this means the function is decreasing.
Now if $x<1$ , by the same argument, we have $z_2<1$ , and

$(z_2^{k+1}=z_2^k\cdot z_2<z_2^k )\implies (z_1^k<z_2^k )\implies z_1<z_2$

Thus the function is increasing.
If $x=1$ , then for any $k$ , use the equality $(x^{1/k} )^k=x$ we must have $x^{1/k}=1$ , otherwise we’ll get a contradiction with (d).

( f ) let $z=(xy)^{1/n}, p=x^{1/n}, q=y^{1/n}$ , then by (a) $z^n=xy, p^n=x, q^n=y$ , and by (c), $z,p,q$ are positive. Thus

$z^n=p^n q^n=(pq)^n \implies z=pq$

( g ) let $p=(x^{1/n} )^{1/m}, q=x^{1/mn}$ , then by (a), $q^mn=x, p^m=x^{1/n}$ , by (c), $p,q$ are positive. Thus by (a) again,

$(p^m )^n=p^mn=x=q^mn \implies p=q$

$\blacksquare$

Exercise 5.6.2. Prove Lemma 5.6.9.
Solution: Uniformly we set $q=a/b,r=c/d$ , where $a,c\in \mathbf Z$ and $b,d\in \mathbf N^+$ .

( a ) by Lemma 5.6.6(a) we have $x^{1/b}$ a positive real. Thus $x^q$ a positive real no matter what value $a$ is.

( b ) by definition: $x^(q+r)=(x^{1/bd} )^{ad+bc}$ , and use Lemma 5.6.8 we have:

$x^q=(x^{1/b} )^a=(x^1{bd} )^{ad},\quad x^r=(x^{1/d} )^c=(x^{1/bd} )^{bc}$

Thus

$x^q x^r=(x^{1/bd} )^{ad} (x^{1/bd} )^{bc}=(x^{1/bd} )^{ad+bc}=x^{q+r}$

Also by definition: $x^{qr}=(x^{1/bd} )^{ac}$ , and $(x^q )^r=((x^q )^{1/d} )^c$ , let $y=x^{1/bd}$ , then $x^qr=y^ac$ , also $x^{1/b}=x^{d/bd}=y^d$ , so $x^q=y^{ad}$ . Thus $(x^q )^r=((y^{ad} )^{1/d} )^c$ . We can know from Lemma 5.6.6 (a) that

$\begin{aligned}((y^ad )^{\frac{1}{d}} )^d=y^{ad}=(y^a )^d &\implies (y^ad )^{\frac{1}{d}}=y^a \\&\implies \left(y^{ad} )^{\frac{1}{d}} \right)^c=(y^a )^c=y^{ac}\\&\implies x^{qr}=(x^q )^r\end{aligned}$

( c ) $x^{-q}=\left(x^{1/b} \right)^q=\dfrac{1}{\left(x^{1/b} \right)^a} =1/x^q$

( d ) $q>0 \implies a>0$ , thus by Lemma 5.6.6(c),(d) we have

$\begin{aligned}(x>y>0)&\iff (x^{1/b}>y^{1/b}>0)\\&\iff (x^{1/b} )^a>(y^{1/b} )^a>0\\&\iff x^q>y^q>0\end{aligned}$

( e ) If $x>1$ , then $y:=x^{1/bd}>1$ , so

$\begin{aligned}(x^q=y^{ad}>x^r=y^{bc})&\iff (ad>bc)\iff (\dfrac{ad-bc}{bd}>0)\\&\iff (q-r>0)\iff q>r\end{aligned}$

If $x<1$ , then $y:=x^{1/bd}<1$ , and the conclusion can be similarly proved.

$\blacksquare$

Exercise 5.6.3. If $x$ is a real number, show that $|x|=(x^2)^{1/2}$ .
Solution: If $x=0$ , then $|x|=0$ and $(x^2 )^{1/2}=0^{1/2}=0$ .
If $x>0,|x|=x$ . Let $y=(x^2 )^{1/2}$ , then $y>0$ and we have $y^2=x^2$ , which means $y=x$ .
If $x<0,x|=-x$ . Let $y=(x^2 )^{1/2},z=-x$ , then $y^2=x^2=(-x)^2=z^2$ , as both $y,z>0$ , from cancellation law we get $y=z=|x|$ .

$\blacksquare$

陶哲轩实分析5.5及习题-Analysis I 5.5

2019年4月22日2019年12月13日christangdt3条评论

上确界没有太多好说的，非常重要之概念以至于必须学好，也有助于后面对limsup和liminf的理解。

Exercise 5.5.1. Let $E$ be a subset of the real numbers $\mathbf R$ , and suppose that $E$ has a least upper bound $M$ which is a real number, i.e., $M=\sup (E)$ . Let $-E$ be the set

$-E:=\{-x:x\in E\}$

Show that $-M$ is the greatest lower bound of $-E$ , i.e., $-M=\inf (-E)$ .
Solution: First show $-M$ is lower than any element of $-E$ :
For $\forall y\in -E,\exists x\in E,y=-x$ , as $M=\sup (E)$ , we have $x\leq M$ , thus $-x=y\geq -M$ .
Next show $-M$ is the greatest lower bound of $-E$ :
Assume $A$ is a lower bound of $-E$ and $A>-M$ , then $\forall x\in E$ , we have

$(-x\in -E)\implies -x\geq A\implies x\leq -A$

Which means $-A$ is an upper bound of $E$ , and as $A>-M, -A<M$ . This contradicts the condition that $M=\sup (E)$ . Thus $A\leq -M$ .

$\blacksquare$

Exercise 5.5.2. Let $E$ be a non-empty subset of $\mathbf R$ , let $n\geq 1$ be an integer, and let $L<K$ be integers. Suppose that $K/n$ is an upper bound for $E$ , but that $L/n$ is not an upper bound for $E$ . Without using Theorem 5.5.9, show that there exists an integer $L<m\leq K$ such that $m/n$ is an upper bound for $E$ , but that $(m-1)/n$ is not an upper bound for $E$ .
Solution: First notice that when $m=K$ , the number $m/n$ is an upper bound of $E$ .
Assume that for all $L<m\leq K$ , $m/n$ and $(m-1)/n$ are upper bound of $E$ , or are not upper bound of $E$ . We use induction to show that in this case there’s a contradiction:
As $m/n$ is an upper bound of $E, (m-1)/n$ is an upper bound of $E$ .
If $k/n$ is an upper bound of $E$ , then by our assumption, $(k-1)/n$ is an upper bound of $E$ . Eventually we will get $L/n$ an upper bound of $E$ , this is false.

$\blacksquare$

Exercise 5.5.3. Let $E$ be a non-empty subset of $\mathbf R$ , let $n\geq 1$ be an integer, and let $m,m'$ be integers with the properties that $m/n$ and $m'/n$ are upper bounds for $E$ , but $(m-1)/n$ and $(m'-1)/n$ are not upper bounds for $E$ . Show that $m=m'$ .
Solution: Suppose $m\neq m'$ , assume $m<m'$ , then $m\leq m'-1$ , as $(m'-1)/n$ is not an upper bound of $E$ , there’s $x\in E,x>(m'-1)/n$ , thus $x>m/n$ , a contradiction to the condition that $m/n$ is an upper bound. The case $m>m'$ can be similarly proved.

$\blacksquare$

Exercise 5.5.4. Let $q_1,q_2,q_3,\dots$ be a sequence of rational numbers with the property that $|q_n-q_{n'}|\leq\frac{1}{M}$ whenever $M\geq 1$ is an integer and $n,n'\geq M$ . Show that $q_1,q_2,q_3,\dots$ is a Cauchy sequence. Furthermore, if $S:=\text{LIM}_{n\to\infty}q_n$ , show that $|q_M-S|\leq \frac{1}{M}$ for every $M\geq 1$ .
Solution: For $\forall \varepsilon >0$ , there’s a $M>0, s.t. M\varepsilon >1$ , so that

$|q_n-q_{n'}|\leq \dfrac{1}{M}<\varepsilon ,\forall n,\quad n'\geq M$

So $(q_n )_{n=1}^\infty$ is a Cauchy sequence.
For every $M\geq 1$ , we have

$|q_M-q_n |\leq \dfrac{1}{M},\quad \forall n\geq M$

Let $(a_n )_{n=1}^\infty$ be a sequence that $a_n=q_M,1\leq n<M,a_n=q_n,n\geq M$ , then we see $(a_n)_{n=1}^\infty$ is equivalent to $(q_n)_{n=1}^\infty$ , thus $S=\text{LIM}_{n\to \infty} a_n$ , and

$q_M-\dfrac{1}{M}\leq a_n\leq q_M+\dfrac{1}{M},\quad \forall n\geq 1$

Use Exercise 5.4.8, we have

$q_M-\dfrac{1}{M}\leq S\leq q_M+\dfrac{1}{M} \implies |q_M-S|\leq \dfrac{1}{M}$

$\blacksquare$

Exercise 5.5.5. Establish an analogue of Proposition 5.4.14, in which “rational” is replaced by “irrational.”
Solution:
Proposition: Given any two real numbers $x<y$ , we can find a irrational number $ir$ such that $x<r<y$ .
Proof: first we can find a rational number $q$ such that $x<q<y$ .
Let $q=a/b$ , in which $a\in \mathbf Z,b\in \mathbf N^+$ , as we know that $y-q>0$ , there’s a $M\in \mathbf Z^+$ such that

$M(y-q)>1 \implies y-q>1/M$

Since $b$ is a positive integer we have $b\geq 1$ , so $bM\geq M$ , we can see from definition that

$a/b=\dfrac{aM}{bM}=\dfrac{2aM}{2bM}=q \\ \implies ir:=\dfrac{2aM+\sqrt{2}}{2bM}=q+\dfrac{\sqrt{2}}{2bM}\leq q+\dfrac{\sqrt{2}}{2M}<q+\dfrac{1}{M}<q+y-q=y$

On the other hand, we have

$ir=\dfrac{2aM+\sqrt{2}}{2bM}=q+\dfrac{\sqrt{2}}{2bM}>q>x$

To see that $ir$ is irrational, assume $ir$ is rational, then $ir=m/n$ , which $m$ and $n$ are integers, then we have

$2aM+\sqrt{2}=\dfrac{2bmM}{n} \implies \sqrt{2}=\dfrac{2bmM}{n}-2aM\in \mathbf Q$

this is a contradiction.

$\blacksquare$

陶哲轩实分析5.4及习题-Analysis I 5.4

2019年4月22日2019年12月13日christangdt2条评论

这一节讲实数的序，构造得非常仔细，关于绝对值、倒数的处理可以说是非常小心了。本节还引入了Archimedean原理和有理数的间隔性原理。习题数量增加了

Exercise 5.4.1. Prove Proposition 5.4.4.
Solution: Let $x$ be a real number, then $x=\text{LIM}_{n\to \infty } a_n,a_n\in \mathbf Q$ , if for every $\varepsilon >0$ , we can find $N$ s.t.

$|a_n|<\varepsilon ,\quad n\geq N$

Then this means $x=0$ . If not, then $(a_n)_{n=1}^\infty$ can’t be eventually $\varepsilon$ -close to $(0)_{n=1}^\infty$ , thus there exists a $c>0$ such that no matter how large $N$ we select, there’s a $M\geq N$ s.t. $|a_M |\geq c$ .

As $(a_n)_{n=1}^\infty$ is a Cauchy sequence, we can have $N'$ s.t.

$|a_n-a_m|<\dfrac{c}{2},\quad n, m\geq N'$

Then choose a $M\geq N'$ and $|a_M|\geq c$ , we can see that

$|a_n|=|a_M+a_n-a_M |\geq |a_M|-|a_n-a_M|\geq c-\dfrac{c}{2}=\dfrac{c}{2}>0,\quad n\geq N'$

This means $(a_n)_{n=1}^\infty$ is eventually either positively bounded away from zero or negatively bounded away from zero, which means $x$ is either positive or negative (substitute the first $N'$ items of $(a_n)$ to be $c$ or $-c$ if necessary).
$x$ is negative in and only if $x=\text{LIM}_{n\to \infty } a_n$ and the Cauchy sequence $(a_n)_{n=1}^\infty$ is negatively bounded away from zero, this is true if and only if the Cauchy sequence $(-a_n)_{n=1}^\infty$ is positively bounded away from zero, this is true if and only if

$-x=-\text{LIM}_{n\to \infty} a_n=\text{LIM}_{n\to \infty } (-a_n)$

is positive.
If $x$ and $y$ are positive, then we have $x=\text{LIM}_{n\to \infty } a_n,y=\text{LIM}_{n\to \infty } b_n,a_n,b_n\in \mathbf Q$ and

$a_n\geq c_1>0,\quad b_n\geq c_2>0,\quad n\geq 1$

Then we have $a_n+b_n\geq c_1+c_2>0$ and $a_n b_n\geq c_1 c_2>0$ , which means $x+y$ and $xy$ are positive.

$\blacksquare$

Exercise 5.4.2. Prove the remaining claims in Proposition 5.4.7.
Solution:
( a ) Consider the real number $x-y$ , then by Proposition 5.4.4, exactly one of the three statements: $x-y=0,x-y\text{ is positive},x-y\text{ is negative}$ is true, and the conclusion follows.
( b ) $x<y \iff x-y<0 \iff -(x-y)=-x+y>0 \iff y>x$
( c ) We have

$(x<y)\wedge (y<z)\implies (y-x>0)\wedge (z-y>0)\\ \implies (z-x=z-y+y-x>0+y-x>0)\implies z>x$

( d ) We have

$(x<y)\implies (y-x>0)\\ \implies (y+z-z-x=(y+z-(x+z))>0)\implies y+z>x+z$

$\blacksquare$

Exercise 5.4.3. Show that for every real number $x$ there is exactly one integer $N$ such that $N\leq x<N+1$ . (This integer $N$ is called the integer part of $x$ , and is sometimes denoted $N=[x]$ .)
Solution: Given real number $x=\text{LIM}_{n\to \infty } a_n$ , we can find $M\in \mathbf N$ such that

$|a_n-a_M|<\dfrac{1}{2},\quad \forall n\geq M$

Specially by Proposition 4.4.1, for this $a_M$ we have an integer $K$ such that

$K\leq a_M<K+1$

Then we can deduce that

$K-\dfrac{1}{2}\leq x\leq K+1+\dfrac{1}{2}$

Comparing $x$ with $K$ and $K+1$ , we can see that by Proposition 5.4.7, exactly one of the cases below is true:
1) $x<K$ , in this case let $N=K-1$ .
2) $x=K$ , let $N=K$
3) $K<x<K+1$ , let $N=K$
4) $x=K+1$ , let $N=K+1$
5) $K+1<x$ , let $N=K+1$
We proved the existence of $N$ , for the uniqueness part, suppose $N'\leq x<N'+1,N'\neq N$ , then we must have $N<N'$ or $N'<N$ , however:

$N<N'\implies N+1\leq N'\implies x<N+1\leq N'\leq x\implies x<x \\N'<N\implies N'+1\leq N\implies x<N'+1\leq N\leq x\implies x<x$

So we get a contradiction in both cases.

$\blacksquare$

Exercise 5.4.4. Show that for any positive real number $x>0$ there exists a positive integer $N$ such that $x>1/N>0$ .
Solution: By Corollary 5.4.13, as $x$ and $1$ are both positive real numbers, we have a positive integer $N$ such that

$Nx>1\implies x>\dfrac{1}{N}>0$

$\blacksquare$

Exercise 5.4.5. Prove Proposition 5.4.14.
Solution: As $x0$ , so there exists a positive integer $N$ such that

$p=y-x>1/N>0$

Also consider the real number $Nx$ , there’s one integer $n$ such that

$n\leq Nx<n+1 \implies \dfrac{n}{N}\leq x<\dfrac{n+1}{N}$

then

$y-x>\dfrac{1}{N} \implies y>x+\dfrac{1}{N}\geq \dfrac{n}{N}+\dfrac{1}{N}=\dfrac{n+1}{N}$

Thus the rational number $q=(n+1)/N$ satisfies $x<q<y$ .

$\blacksquare$

Exercise 5.4.6. Let $x,y$ be real numbers and let $\varepsilon >0$ be a positive real. Show that $|x-y|<\varepsilon$ if and only if $y-\varepsilon <x<y+\varepsilon$ , and that $|x-y|\leq \varepsilon$ if and only if $y-\varepsilon \leq x\leq y+\varepsilon$ .
Solution: First we have

$y-\varepsilon <x<y+\varepsilon \implies (y-x<\varepsilon)\wedge (x-y<\varepsilon) \implies |x-y|<\varepsilon$

If $|x-y|<\varepsilon$ , then we split the case of $x=y,x>y,x<y$
When $x=y, x-y=0<\varepsilon$ and we have $y-\varepsilon <y<y+\varepsilon$
When $x>y$ , we have $|x-y|=x-y<\varepsilon \implies y-\varepsilon <y<x<y+\varepsilon$
When $x<y$ , we have $|x-y|=y-x<\varepsilon \implies y-\varepsilon <x<y<y+\varepsilon$
In conclusion, , the statement is true.
Next to prove $|x-y|\leq \varepsilon \iff y-\varepsilon \leq x\leq y+\varepsilon$ , we first see that

$|x-y|=\varepsilon \iff (y-\varepsilon =x)\vee (y+\varepsilon =x)$

Thus

$\begin{aligned}|x-y|\leq \varepsilon &\iff (|x-y|<\varepsilon )\vee (|x-y|=\varepsilon)\\&\iff (y-\varepsilon <x<y+\varepsilon )\vee (y-\varepsilon =x)\vee (y+\varepsilon =x)\\&\iff y-\varepsilon \leq x\leq y+\varepsilon \end{aligned}$

$\blacksquare$

Exercise 5.4.7. Let $x,y$ be real numbers. Show that $x\leq y+\varepsilon$ for all real numbers $\varepsilon >0$ if and only if $x\leq y$ . Show that $|x-y|\leq \varepsilon$ for all real numbers $\varepsilon >0$ if and only if $x=y$ .
Solution: First, $(x\leq y)\implies (x\leq y+\varepsilon ,\forall \varepsilon >0)$ , and now if $(x\leq y+\varepsilon ,\forall \varepsilon >0)$ , suppose $x>y$ , then select

$q=\dfrac{x-y}{2}>0$

We see that by the condition given

$(x\leq y+q)\implies (x\leq y+\dfrac{x-y}{2}=\dfrac{x+y}{2})\implies 2x\leq x+y\implies x\leq y$

This is a contradiction to the hypothesis.
Next, we see that $(|x-y|\leq \varepsilon )\implies (y-\varepsilon \leq x\leq y+\varepsilon )\implies (x\leq y+\varepsilon )\wedge (y\leq x+\varepsilon )$ , as this is true for all real numbers $\varepsilon >0$ , we see that $x\leq y$ and $y\leq x$ are both true, thus $x=y$ .
The other direction is easy to prove.

$\blacksquare$

Exercise 5.4.8. Let $(a_n)_{n=1}^{\infty}$ be a Cauchy sequence of rationals, and let $x$ be a real number. Show that if $a_n\leq x$ for all $n\geq 1$ , then $\text{LIM}_{n\to\infty}a_n\leq x$ . Similarly, show that if $a_n\geq x$ for all $n\geq 1$ , then $\text{LIM}_{n\to\infty}a_n\geq x$ .
Solution: For the sake of contradiction assume $\text{LIM}_{n\to \infty } a_n>x$ , then by Proposition 5.4.14 there’s a rational number $q$ such that

$x<q<\text{LIM}_{n\to \infty } a_n$

Now let $b_n=q,\forall n$ , then $(b_n)_{n=1}^\infty$ is a Cauchy sequence, and

$a_n\leq x<q=b_n,\quad \forall n\geq 1$

Use Corollary 5.4.10, $\text{LIM}_{n\to \infty} a_n\leq \text{LIM}_{n\to \infty} b_n=q<\text{LIM}_{n\to \infty } a_n$ , which is a contradiction.
The other case can be proved similarly.

$\blacksquare$

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