陶哲轩实分析7.2及习题-Analysis I 7.2

christangdt留下评论

将有限级数拓展到了无限，引入了绝对收敛、条件收敛、交错级数等等。

Exercise 7.2.1. Is the series \sum_{n=1}^{\infty}(-1)^n convergent or divergent? Justify your answer.
Solution: The sequence is divergent. We define S_N=\sum_{n=1}^N(-1)^n , then S_N=-1 if N is even, S_N=0 if N is odd, thus the upper limit and lower limit of (S_N) does not equal.

\blacksquare

Exercise 7.2.2. Prove Proposition 7.2.5.
Solution: Write S_N=\sum_{n=m}^Na_n , then \sum_{n=m}^{\infty}a_n converges iff (S_N)_{N=m}^{\infty} converges, which is true iff (S_N)_{N=1}^{\infty} is a Cauchy sequence, i.e. for \forall \varepsilon >0, \exists N\geq m such that

|S_q-S_{p-1}|=\left|\sum\limits_{n=p}^qa_n \right|\leq \varepsilon ,\quad \forall p,q\geq N

\blacksquare

Exercise 7.2.3. Use Proposition 7.2.5 to prove Corollary 7.2.6.
Solution: Assume \lim_{n\to {\infty}}a_n\neq 0, then for each \varepsilon >0, and every integer n, we can have a M\geq n such that |a_M |\geq \varepsilon . Now if \sum_{n=m}^{\infty}a_n converges, given an \varepsilon >0, we can find an N such that

\left|\sum\limits_{n=p}^qa_n \right|\leq \dfrac{\varepsilon}{2},\quad \forall p,q\geq N

For this N, we choose a M\geq N s.t. a_M\geq \varepsilon , then if p=q=M\geq N, we shall have

\left|\sum\limits_{n=p}^q a_n \right|=|a_M |\leq \dfrac{\varepsilon}{2}

Which leads to a contradiction.

\blacksquare

Exercise 7.2.4. Prove Proposition 7.2.9.
Solution: If \sum_{n=m}^{\infty}a_n is absolutely convergent, then \sum_{n=m}^{\infty}|a_n| is convergent, thus by Proposition 7.2.5, for every \varepsilon >0, \exists N\geq m such that

\left|\sum\limits_{n=p}^q|a_n|\right|\leq \varepsilon ,\quad \forall p,q\geq N

Recursively use the triangle inequality we have

\left|\sum\limits_{n=p}^qa_n \right|\leq \sum\limits_{n=p}^q|a_n|=\left|\sum\limits_{n=p}^q|a_n|\right|,\quad \forall p,q\geq N

Thus use Proposition 7.2.5 again, we know that \sum_{n=m}^{\infty}a_n is convergent. Now define

S_N=|\sum\limits_{n=m}^Na_n|,\quad T_N=\sum\limits_{n=m}^N|a_n|

We can easily show from Proposition 7.1.4(e) that S_N\leq T_N,\forall N\geq m, so we know that

|\sum\limits_{n=m}^{\infty}a_n|=\lim\limits_{N\to {\infty}}S_N\leq \lim\limits_{N\to {\infty}}T_N=\sum\limits_{n=m}^{\infty}|a_n|

\blacksquare

Exercise 7.2.5. Prove Proposition 7.2.14.
Solution:
( a ) We let S_N=\sum_{n=m}^Na_n, T_N=\sum_{n=m}^N b_n , then \lim_{N\to {\infty}} S_N=x,\lim_{N\to {\infty}} T_N=y, also we have

\sum\limits_{n=m}^N(a_n+b_n ) =\sum\limits_{n=m}^Na_n +\sum\limits_{n=m}^Nb_n =S_N+T_N

Thus

\sum\limits_{n=m}^{\infty}(a_n+b_n) =\lim\limits_{N\to {\infty}} \sum\limits_{n=m}^N (a_n+b_n ) =\lim\limits_{N\to {\infty}} (S_N+T_N )=x+y

( b ) Use the same notation of (a), we have

\sum\limits_{n=m}^Nca_n =c\sum\limits_{n=m}^Na_n \\ \implies \sum\limits_{n=m}^{\infty}ca_n=\lim\limits_{N\to {\infty}} \sum\limits_{n=m}^Nca_n=c\left(\lim\limits_{N\to {\infty}} \sum\limits_{n=m}^Na_n \right)=cx

( c ) Use Proposition 7.2.5 we know that for \forall \varepsilon >0:

\begin{aligned}\sum\limits_{n=m}^{\infty}a_n \text{ is convergent} &\iff \exists N\geq m,\text{ s.t. }\quad|\sum\limits_{n=p}^qa_n |\leq \varepsilon , \forall p,q\geq N\\&\iff \exists \max \{N,m+k\}\geq m+k,\text{ s.t. }\quad|\sum\limits_{n=p}^qa_n|\leq \varepsilon , \forall p,q\geq N\\&\iff \sum\limits_{n=m+k}^{\infty}a_n\text{ is convergent}\end{aligned}

And if both are convergent, we let S_N=\sum_{n=m}^Na_n , T_N=\sum_{n=m+k}^Na_n , we can have

S_N=\sum\limits_{n=m}^{m+k-1}a_n +T_N,\quad \forall N\geq m+k

Let N\to {\infty} we get the desired result.

( d ) From Lemma 7.1.4 (b) we have

S_N=\sum\limits_{n=m}^Na_n =\sum\limits_{n=m+k}^{N+k}a_(n-k),\quad \forall k\in Z

Thus when N\to {\infty}, we know N+k\to {\infty}, thus

x=\sum\limits_{n=m}^{\infty}a_n=\lim\limits_{N\to {\infty}} \sum\limits_{n=m}^Na_n=\lim\limits_{N+k\to {\infty}} \sum\limits_{n=m+k}^{N+k}a_{n-k}=\sum\limits_{n=m+k}^{\infty}a_{n-k}

\blacksquare

Exercise 7.2.6. Prove Lemma 7.2.15.
Solution:
We first calculate S_N=\sum_{n=0}^N(a_n-a_{n+1}) , it’s easy to see that

S_N=\sum\limits_{n=0}^N(a_n-a_{n+1})=a_0-a_{N+1}

So we have

\sum\limits_{n=0}^{\infty}(a_n-a_{n+1})=\lim\limits_{N\to {\infty}}S_N=\lim\limits_{N\to {\infty}} (a_0-a_{N+1})=a_0-\lim\limits_{N\to {\infty}} a_{N+1}=a_0

If a_n\to L\neq 0, then

\sum\limits_{n=0}^{\infty}(a_n-a_{n+1})=a_0-L

\blacksquare

陶哲轩实分析7.1及习题-Analysis I 7.1

christangdt5条评论

先建立了有限级数的各种性质（本质是部分和的性质），这样无限级数可以当作部分和的极限处理，就可以利用原来关于数列极限的各项结论。同时还建立了对有限集合上求和的概念，后续可以拓展到可数集合上。最后是一个很有用的结论即Fubini定理，有限级数的求和可以换序。

Exercise 7.1.1. Prove Lemma 7.1.4.
Solution:
( a ) For any m,n we use induction on p:
For p=m+1, we know that in this case n can only be m

\sum\limits_{i=m}^ma_i +\sum\limits _{i=m+1}^{m+1}a_i =a_m+a_{m+1}=\sum \limits _{i=m}^{m+1}a_i

Now suppose for p the statement holds, then for p+1: for all m\leq n<p we have

\sum \limits _{i=m}^na_i +\sum \limits _{i=n+1}^{p+1}a_i =\sum \limits _{i=m}^na_i+\sum \limits _{i=n+1}^pa_i+a_{p+1}=\sum \limits _{i=m}^pa_i+a_{p+1}=\sum \limits _{i=m}^{p+1}a_i

And for n=p:

\sum \limits _{i=m}^pa_i +\sum \limits _{i=p+1}^{p+1}a_i =\sum \limits _{i=m}^pa_i+a_{p+1}=\sum \limits _{i=m}^{p+1}a_i

Thus the conclusion holds.

( b ) Given m,k we use induction on n:
For n=m, we have

\sum \limits _{i=m}^ma_i =a_m=a_{m+k-k}=\sum \limits _{j=m+k}^{m+k}a_{j-k}

Now suppose for n the statement holds, then for n+1, we have

\sum \limits _{i=m}^{n+1}a_i=\sum \limits _{i=m}^na_i+a_{n+1}=\sum \limits _{j=m+k}^{n+k}a_{j-k} +a_{n+k+1-k}=\sum \limits _{j=m+k}^{n+1+k}a_{j-k}

Thus the conclusion holds.

( c ) Given m we use induction on n:
For n=m, we have

\sum \limits _{i=m}^m(a_i+b_i ) =a_m+b_m=\sum \limits _{i=m}^ma_i+\sum \limits _{i=m}^mb_i

Now suppose for n the statement holds, then for n+1, we have

\begin{aligned}\sum \limits _{i=m}^{n+1}(a_i+b_i ) &=\sum \limits _{i=m}^n(a_i+b_i ) +(a_{n+1}+b_{n+1} )\\&=\sum \limits _{i=m}^na_i +\sum \limits _{i=m}^nb_i+a_{n+1}+b_{n+1}\\&=\sum \limits _{i=m}^na_i +a_{n+1}+\sum \limits _{i=m}^nb_i +b_{n+1}\\&=\sum \limits _{i=m}^{n+1}a_i +\sum \limits _{i=m}^{n+1}b_i \end{aligned}

Thus the conclusion holds.

( d ) Given m we use induction on n:
For n=m, we have

\sum \limits _{i=m}^m(ca_i)=ca_m=c\left(\sum \limits _{i=m}^ma_i \right)

Now suppose for n the statement holds, then for n+1, we have

\begin{aligned}\sum \limits _{i=m}^{n+1}(ca_i)&=\sum \limits _{i=m}^n(ca_i)+ca_{n+1}=c\left(\sum \limits _(i=m)^na_i \right)+ca_(n+1)\\&=c\left(\sum \limits _{i=m}^na_i +a_{n+1} \right)=c\left(\sum \limits _{i=m}^{n+1}a_i \right)\end{aligned}

Thus the conclusion holds.

( e ) Given m we use induction on n:
For n=m, we have

|\sum \limits _{i=m}^ma_i|=|a_m|=\sum \limits _{i=m}^m|a_i|\leq \sum \limits _{i=m}^m|a_i|

Now suppose for n the statement holds, then for n+1, we have

\begin{aligned}|\sum \limits _{i=m}^{n+1}a_i|=|\sum \limits _{i=m}^na_i +a_{n+1} |&\leq |\sum \limits _{i=m}^na_i |+|a_{n+1}|\\&\leq \sum \limits _{i=m}^n|a_i |+|a_{n+1} |=\sum \limits _{i=m}^{n+1}|a_i |\end{aligned}

Thus the conclusion holds.

( f ) Given m we use induction on n:
For n=m, we have

\sum \limits _{i=m}^ma_i =a_m\leq b_m=\sum \limits _{i=m}^mb_i

Now suppose for n the statement holds, then for n+1, we have

\sum \limits _{i=m}^{n+1}a_i =\sum \limits _{i=m}^na_i +a_{n+1}\leq \sum \limits _{i=m}^nb_i +b_{n+1}=\sum \limits _{i=m}^{n+1}b_i

Thus the conclusion holds.

\blacksquare

Exercise 7.1.2. Prove Proposition 7.1.11.
Solution:
( a ) We assign a function g:\{n:1\leq n\leq 0\}\to X, then g is a bijection, so

\sum\limits_{x\in X}f(x)=\sum \limits _{i=1}^0f(g(i)) =0

( b ) We assign a function g:\{n:1\leq n\leq 0\}\to X by g(1)=x_0, then g is a bijection, so

\sum \limits _{x\in X}f(x)=\sum \limits _{i=1}^1f(g(i)) =f(g(1))=f(x_0)

( c ) Suppose X has n elements, then Y must have n elements. We assign a bijective function h:\{i\in \mathbf N:1\leq i\leq n\}\to Y, then the function g\circ h: \{i\in \mathbf N:1\leq i\leq n\}\to X is a bijection, thus we have

\sum\limits_{x\in X}f(x) =\sum\limits_{i=1}^nf\left((g\circ h)(i)\right) =\sum\limits_{i=1}^nf(g(h(i)))

On the other hand, h is a bijection means given the function f\circ g, we have

\sum\limits_{y\in X}f(g(y)) =\sum\limits_{y\in X}(f\circ g)(y) =\sum\limits_{i=1}^n(f\circ g)(h(i)) =\sum\limits_{i=1}^nf(g(h(i)))

( d ) We use the bijection g: X\to \{i\in \mathbf Z:n\leq i\leq m\} to be g(i)=i,i\in X, then

\sum\limits_{i\in X}a_i =\sum\limits_{i=n}^ma_{g(i)} =\sum\limits_{i=n}^ma_i

( e ) Suppose X has m elements and Y has n elements, then we can find bijections:

g_1:\{i\in \mathbf N: 1\leq i\leq m\}\to X,\quad g_2:\{i\in \mathbf N:1\leq i\leq n\}\to Y

So we can have

\sum\limits_{x\in X}f(x) =\sum\limits_{i=1}^mf(g_1 (i)) ,\quad \sum\limits_{y\in Y}f(y) =\sum\limits_{i=1}^nf(g_2 (i))

Now we define h: \{i\in \mathbf N: 1\leq i\leq m+n\}\to X\cup Y as follows:

h(i)=\begin{cases}g_1 (i),\quad 1\leq i\leq m\\g_2 (i-m),\quad m+1\leq i\leq m+n \end{cases}

It’s easy to see that h is a bijection, so use Lemma 7.1.4 (a) and (b) we have

\begin{aligned}\sum\limits_{z\in X\cup Y}f(z) &=\sum \limits _{i=1}^{m+n}f(h(i)) =\sum \limits _{i=1}^mf(h(i)) +\sum \limits _{i=m+1}^{m+n}f(h(i))\\& =\sum \limits _{i=1}^mf(g_1 (i)) +\sum \limits _{i=m+1}^{m+n}f(g_2 (i-m)) \\&=\sum \limits _{x\in X}f(x) +\sum \limits _{i=1}^nf(g_2 (i)) =\sum \limits _{x\in X}f(x) +\sum \limits _{y\in Y}f(y) \end{aligned}

( f ) Suppose X has n elements, we assign a bijection h:\{i\in \mathbf N: 1\leq i\leq n\}\to X, so

\begin{aligned}\sum\limits_{x\in X}(f(x)+g(x))&=\sum \limits _{i=1}^n\big(f(h(i))+g(h(i))\big) \\&=\sum \limits _{i=1}^nf(h(i)) +\sum \limits _{i=1}^ng(h(i)) =\sum \limits _{x\in X}f(x) +\sum \limits _{x\in X}g(x) \end{aligned}

( g ) Suppose X has n elements, we assign a bijection g:\{i\in \mathbf N: 1\leq i\leq n\}\to X. So by Lemma 7.1.4(d)

\sum \limits _{x\in X}cf(x) =\sum \limits _{i=1}^ncf(g(i)) =c\sum \limits _{i=1}^nf(g(i)) =c\sum \limits _{x\in X}f(x)

( h ) Suppose X has n elements, we assign a bijection h:\{i\in \mathbf N: 1\leq i\leq n\}\to X, so

f(h(i))\leq g(h(i)),\quad \forall 1\leq i\leq n

Which means by Lemma 7.1.4(f)

\sum \limits _{x\in X}f(x) =\sum \limits _{i=1}^nf(h(i)) \leq \sum \limits _{i=1}^nf(g(i)) =\sum \limits _{x\in X}g(x)

( i ) Suppose X has n elements, we assign a bijection g:\{i\in \mathbf N: 1\leq i\leq n\}\to X.
So by Lemma 7.1.4(e)

|\sum \limits _{x\in X}f(x) |=|\sum \limits _{i=1}^nf(g(i)) |\leq \sum \limits _{i=1}^n|f(g(i))| =\sum \limits _{x\in X}|f(x)|

\blacksquare

Exercise 7.1.3. Form a definition for the finite products \prod_{i=1}^na_i and \prod_{x\in X}f(x). Which of the above results for finite series have analogues for finite products?
Solution:
(Finite products). Let m,n be integers, and let (a_i)_{i=m}^n be a finite sequence of real numbers, assigning a real number a_i to each integer i between m and n inclusive. Then we define the finite product \prod_{i=m}^na_i by the recursive formula

\prod\limits_{i=m}^na_i :=1\text{ whenever }n<m \\ \prod \limits _{i=m}^{n+1}a_i :=(\prod \limits _{i=m}^na_i )\cdot a_{n+1}\text{ whenever }n\geq m-1

(Production over finite sets). Let X be a finite set with n elements (n\in \mathbf N), and let f:X\to R. Then we can select any bijection g:\{i\in \mathbf N: 1\leq i\leq n\}\to X, we then define

\prod \limits _{x\in X}f(x):=\prod \limits _{i=1}^nf(g(i))

We have Lemma 7.1.4’:
( a ) (\prod_{i=m}^na_i )(\prod_{i=n+1}^pa_i )=\prod_{i=m}^pa_i
( b ) \prod_{i=m}^na_i =\prod_{j=m+k}^{n+k}a_{j-k}
( c ) \prod_{i=m}^n(a_i b_i)=(\prod_{i=m}^na_i )(\prod_{i=m}^nb_i )
( d ) \prod_{i=m}^nca_i =c^{n-m+1} \prod_{i=m}^na_i
( e ) |\prod_{i=m}^na_i |\leq \prod_{i=m}^n|a_i |
Also Lemma 7.1.11 is valid through (a)-(g) and (i) with summation replaced by products.

\blacksquare

Exercise 7.1.4. Define the factorial function n! for natural numbers n by the recursive definition 0!:=1 and (n+1)!:=n!\times (n+1). If x and y are real numbers, prove the binomial formula

(x+y)^n=\sum\limits_{j=0}^n \dfrac{n!}{j!(n-j)!}x^jy^{n-j}

for all natural numbers n.
Solution: We induct on n:
When n=0, we have

(x+y)^0=1,\quad \sum\limits_{j=0}^0\dfrac{n!}{j!(n-j)!} x^j y^{n-j}=x^0 y^0=1

Now suppose the case is true for n, consider n+1:

\begin{aligned}(x+y)^{n+1}&=(x+y)^n (x+y)=\left[\sum_{j=0}^n\dfrac{n!}{j!(n-j)!} x^j y^{n-j}\right](x+y)\\&=\sum_{j=0}^n\dfrac{n!}{j!(n-j)!} x^{j+1} y^{n-j}+\sum_{j=0}^n\dfrac{n!}{j!(n-j)!} x^j y^{n+1-j}\\&=\sum_{j=0}^{n-1}\dfrac{n!}{j!(n-j)!} x^{j+1} y^{n-j}+x^{n+1}+y^{n+1}+\sum_{j=1}^n\dfrac{n!}{j!(n-j)!} x^j y^{n+1-j}\\&=x^{n+1}+y^{n+1}+\sum_{j=1}^n\dfrac{n!}{(j-1)!(n-j+1)!} x^j y^{n-j+1}+\sum_{j=1}^n\dfrac{n!}{j!(n-j)!} x^j y^{n+1-j}\end{aligned}

We have

\begin{aligned}\dfrac{n!}{(j-1)!(n-j+1)!}+\dfrac{n!}{j!(n-j)!}&=\dfrac{n!}{(j-1)!(n-j)!} \left(\dfrac{1}{n-j+1}+\dfrac{1}{j}\right)\\&=\dfrac{(n+1)!}{(j)!(n-j+1)!}\end{aligned}

So the expression equals to

x^{n+1}+y^{n+1}+\sum\limits_{j=1}^n\dfrac{(n+1)!}{(j)!(n-j+1)!} x^j y^{n+1-j}=\sum\limits_{j=0}^{n+1}\dfrac{(n+1)!}{(j)!(n-j+1)!}x^j y^{n+1-j}

Thus the conclusion holds.

\blacksquare

Exercise 7.1.5. Let X be a finite set, let m be an integer, and for each x\in X let (a_n(x))_{n=m}^{\infty} be a convergent sequence of real numbers. Show that the sequence (\sum_{x\in X}a_n(x))_{n=m}^{\infty} is convergent, and

\lim\limits_{n\to\infty}\sum_{x\in X}a_n(x)=\sum_{x\in X}\lim\limits_{n\to\infty}a_n(x).

Solution: Suppose X has n elements, we induct on n.
If n=0, then X=\emptyset, since a_n (x) and \lim_{n\to \infty}a_n (x) are both functions, use Proposition 7.1.11(a) we know that the left side equals \lim_{n\to \infty} 0=0, and the right side is 0.
Now suppose the statement is true for n consider n+1. As n+1\geq 1, we can choose p\in X, and (a_n (p))_{n=m}^\infty is a convergent sequence of real numbers. Let Y=X-\{p\}, then Y has n elements, so we have

\sum_{x\in X}a_n (x)=\sum_{y\in Y}a_n (y)+a_n (p)

By the induction hypothesis \sum_{y\in Y}a_n (x) is convergent, thus \sum_{x\in X}a_n (x) is convergent, and we have

\begin{aligned}\lim\limits_{n\to \infty} \sum\limits_{x\in X}a_n (x)&=\lim \limits _{n\to \infty} \left(\sum\limits_{y\in Y}a_n (y)+a_n (p)\right)=\lim \limits _{n\to \infty} \sum \limits _{y\in Y}a_n (y)+\lim \limits _{n\to \infty}a_n (p)\\&=\sum \limits _{y\in Y}\lim \limits _{n\to \infty}a_n (y) +\lim \limits _{n\to \infty}a_n (p)\\&=\sum \limits _{y\in Y}\lim \limits _{n\to \infty}a_n (y) +\sum \limits _{x\in \{p\}}\lim \limits _{n\to \infty}a_n (x) =\sum \limits _{x\in X}\lim \limits _{n\to \infty}a_n(x) \end{aligned}

\blacksquare

陶哲轩实分析6.7及习题-Analysis I 6.7

christangdt留下评论

将指数运算拓展到了实数幂数，指数运算的相等关系和序关系保持不变。

Exercise 6.7.1. Prove the remaining components of Proposition 6.7.3.
Solution: To prepare, we first suppose there’s (q_n)_{n=1}^{\infty} and (r_n)_{n=1}^{\infty} in \mathbf Q, and q_n\to q,r_n\to r.

( a ) x^q is a positive real
As (q_n)_{n=1}^{\infty} converges to q, it is bounded by M\in \mathbf R, we can select a natural number N\geq M, so (q_n)_{n=1}^{\infty} is bounded by N, so we have

x^{-N}\leq x^{q_n}\leq x^N,\quad x\geq 1 \\ x^N\leq x^{q_n}\leq x^{-N},\quad 0<x<1

Thus we have x^q\geq x^{-N},x\geq 1 and x^q\geq x^N,0<x<1, in both cases x^q>0.

(b) (x^q)^r=x^qr
We first prove the case when q\in \mathbf R,r\in \mathbf Q, write r=n/m, where n\in \mathbf Z,m\in \mathbf N^+, then

\left((x^q)^{\frac{n}{m}} \right)^m=(x^q)^n=x^qn=\left(x^{\frac{qn}{m}} \right)^m

where we gain the first equality by Lemma 5.6.9(b), the third equality by Definition 5.6.1 and the first part of Proposition 6.7.3(b), to prove the second equality, notice when n>0 we have by Definition 5.6.1 and the first part of Proposition 6.7.3(b) that

(x^q)^n:= \underbrace{x^q \times \dots \times x^q}_{n \text{ times}}=x^{\overbrace{q+\cdots +q}^{n \text{ times}}} =x^{qn}

When n<0, we have n=-p,p\in N^+, then use Lemma 5.6.9(c) and Definition 6.7.2, we have

(x^q)^n=(x^q )^{-p}=\dfrac{1}{(x^q)^p} =\dfrac{1}{x^{qp}} =\dfrac{1}{\lim\limits_{k\to {\infty}} x^{p\cdot q_k}}
x^{qn}=\lim\limits_{k\to {\infty}}x^{nq_k}=\lim\limits_{k\to {\infty}}x^{q_k (-p)}=\lim\limits_{k\to {\infty}}\dfrac{1}{x^{p\cdot q_k }}=\dfrac{1}{\lim\limits_{k\to {\infty}}x^{q_k p}}

thus (x^q)^n=x^{qn}, and the case is obvious when n=0, we conclude the second equality is valid, and now we have ((x^q)^r )^m=(x^{qr})^m, by Proposition 5.6.3(Proposition 4.3.12(c)), (x^q )^r=x^qr.
Finally we extend the case to q\in \mathbf R,r\in \mathbf R, we have

(x^q)^r=\lim\limits_{n\to {\infty}}(x^q)^{r_n}=\lim\limits_{n\to {\infty}}x^{qr_n}

To finish the prove, notice that (q_n r_n)_{n=1}^{\infty} is equivalent to (qr_n)_{n=1}^{\infty}, thus x^{qr_n-q_n r_n}\to 1, thus

1=x^0=\lim\limits_{n\to {\infty}} x^{qr_n-q_n r_n }=\lim\limits_{n\to {\infty}} \dfrac{x^{qr_n}}{x^{q_n r_n}}=\dfrac{\lim\limits_{n\to {\infty}}x^{qr_n}}{\lim\limits_{n\to {\infty}} x^{q_n r_n}} =\dfrac{(x^q)^r}{x^{qr}} \implies (x^q)^r=x^qr

( c ) x^{-q}=1/x^q
Since -q_n\to -q we know that \lim_{n\to {\infty}}x^{-q_n}=x^{-q}, and \lim_{n\to {\infty}} x^{-q_n }=1/\lim_{n\to {\infty}} x^{q_n} =1/x^q.

(d) If q>0, then x>y if and only if x^q>y^q
If q>0, we can have (q_n)_{n=1}^{\infty} bounded away from 0.
First if x^q>y^q, and assume 0y, then x\geq y and x\neq y, from x\geq y we can know x^q\geq y^q. Assume x^q=y^q, then let r=1/q, r is real since q\neq 0, so by the results of (b) we have (x^q )^r=(y^q )^r, or

x=x^qr=y^qr=y

This contradicts the fact that x\neq y.

(e) If x>1, then x^q>x^r if and only if q>r. If x<1, then x^q>x^r if and only if q<r.
We have when x>1:

(x^q>x^r )\iff (x^r (x^{q-r}-1)>0)\iff (x^{q-r}-1>0)\iff (x^{q-r}>1)

Now if q>r,then q-r>0, thus by (d), x^{q-r}>1^{q-r}=1.
If x^{q-r}>1, then obviously q\neq r, and if we assume q-r<0, then from (d) again

x^{q-r}>1 \implies x^{r-q}<1=1^{r-q} \implies x<1

This is a contradiction, so we must have q>r by the trichotomy of real numbers.
The case x<1 can be proved since

(x^q>x^r )\iff (x^{-r}>x^{-q})\iff ((x^{-1})^r>(x^{-1})^q)\iff (r>q)

We use the fact that x^{-q-r}>0, x^{-1}>1 and the first part of (e).