陶哲轩实分析（上）8.2及习题-Analysis I 8.2

这一节的逻辑构成是这样的：先定义了在countable集合上的级数，并且在绝对收敛下有Fubini定理成立，而后在uncountable集合上也可以定义级数，只要满足任何有限集合的sup存在。如果f是一个定义在uncountable集合上的绝对收敛级数，那么f非退化的点是至多可数的（证明需要用选择公理）。绝对收敛级数有很好理解的series laws，条件收敛级数则有Riemann的结论：可以收敛到任何实数L。

这一节的习题主要是完善正文中的定理证明，然而也不易。

Exercise 8.2.1

Exercise 3.6.3 says if there’s a function whose domain is on a finite subset of N\mathbf NN, then the range of this function is also finite. I’ll use this result later.
Since XXX is at most countable, XXX may be finite or countable. If XXX is finite, then the statement is obviously true. Now we consider the case when XXX is countable.
First suppose the series ∑x∈Xf(x)∑_{x∈X}f(x)∑x∈Xf(x) is absolutely convergent, then there’s a bijection g:N→Xg:\mathbf N→Xg:N→X such that ∑n=0∞f(g(n))∑_{n=0}^∞f(g(n))∑n=0∞f(g(n)) is absolutely convergent. Given any element EEE in the set
S={∑x∈A∣f(x)∣∶A⊆X,Afinite}S=\left\{∑_{x∈A}|f(x)|∶A⊆X,A\text{ finite}\right\}S={x∈A∑∣f(x)∣∶A⊆X,A finite}
We know there’s a finite set AAA such that E=∑x∈A∣f(x)∣E=∑_{x∈A}|f(x)|E=∑x∈A∣f(x)∣ . So we can have a finite subset S′S'S′ of NNN such that g(S′)=Ag(S')=Ag(S′)=A, from Exercise 3.6.3 we can have S′S'S′ is bounded above by some natural number kkk, so
E=∑x∈A∣f(x)∣=∑n∈S′∣f(g(n))∣≤∑n=0k∣f(g(n))∣≤∑n=0∞∣f(g(n))∣E=∑_{x∈A}|f(x)| =∑_{n∈S'}|f(g(n))| ≤∑_{n=0}^k|f(g(n))| ≤∑_{n=0}^∞|f(g(n))| E=x∈A∑∣f(x)∣=n∈S′∑∣f(g(n))∣≤n=0∑k∣f(g(n))∣≤n=0∑∞∣f(g(n))∣
Since this is true for all EEE, we know that ∑n=0∞∣f(g(n))∣∑_{n=0}^∞|f(g(n))|∑n=0∞∣f(g(n))∣ is an upper bound of SSS, so
sup⁡⁡S≤∑n=0∞∣f(g(n))∣<+∞\sup⁡S≤∑_{n=0}^∞|f(g(n))| <+∞sup⁡S≤n=0∑∞∣f(g(n))∣<+∞
On the contrary, if M=sup⁡{∑x∈A∣f(x)∣∶A⊆X,Afinite}<+∞M=\sup\{∑_{x∈A}|f(x)|∶A⊆X,A\text{ finite}\}<+∞M=sup{∑x∈A∣f(x)∣∶A⊆X,A finite}<+∞, we assume the series ∑x∈Xf(x)∑_{x∈X}f(x)∑x∈Xf(x) is not absolutely convergent, then given a bijection g:N→Xg:\mathbf N→Xg:N→X, we have SN=∑n=0N∣f(g(n))∣S_N=∑_{n=0}^N|f(g(n))|SN=∑n=0N∣f(g(n))∣ is not bounded above, so there exists a k∈Nk∈\mathbf Nk∈N such that ∑n=0k∣f(g(n))∣>M∑_{n=0}^k|f(g(n))| >M∑n=0k∣f(g(n))∣>M, we let A=g({n:0≤n≤k})A=g(\{n:0≤n≤k\})A=g({n:0≤n≤k}), then AAA is finite, and
∑n=0k∣f(g(n))∣=∑x∈A∣f(x)∣>M∑_{n=0}^k|f(g(n))| =∑_{x∈A}|f(x)| >Mn=0∑k∣f(g(n))∣=x∈A∑∣f(x)∣>M
This leads to a contradiction.

Exercise 8.2.2

Since ∑x∈Xf(x)∑_{x∈X}f(x)∑x∈Xf(x) is absolutely convergent, the quantity MMM is well defined. For ∀n∈N∀n∈\mathbf N∀n∈N, consider the set {x∈X:∣f(x)∣>1/n}\{x∈X:|f(x)|>1/n\}{x∈X:∣f(x)∣>1/n}, this set is finite with cardinality at most MnMnMn, thus use Exercise 8.1.9, we know that
{x∈X:f(x)≠0}=⋃n=1∞{x∈X:∣f(x)∣>1/n}\{x∈X:f(x)≠0\}=⋃_{n=1}^∞\{x∈X:|f(x)|>1/n\}{x∈X:f(x)=0}=n=1⋃∞{x∈X:∣f(x)∣>1/n}
is at most countable.

Exercise 8.2.3

Since both ∑x∈Xf(x)∑_{x∈X}f(x)∑x∈Xf(x) and ∑x∈Xg(x)∑_{x∈X}g(x)∑x∈Xg(x) are absolutely convergent, the sets S={x∈X:f(x)≠0}S=\{x∈X:f(x)≠0\}S={x∈X:f(x)=0} and B={x∈X:g(x)≠0}B=\{x∈X:g(x)≠0\}B={x∈X:g(x)=0} are at most countable. We denote ∑x∈Xf(x)=L∑_{x∈X}f(x)=L∑x∈Xf(x)=L and ∑x∈Xg(x)=M∑_{x∈X}g(x)=M∑x∈Xg(x)=M.
( a ) Let A⊆XA⊆XA⊆X be a finite set, then by proposition 7.1.11 we have
∑x∈A∣f(x)+g(x)∣≤∑x∈A∣f(x)∣+∑x∈A∣g(x)∣≤sup⁡⁡{∑x∈A∣f(x)∣∶A⊆X,Afinite}+sup⁡{∑x∈A∣g(x)∣∶A⊆X,Afinite}<∞\begin{aligned}∑_{x∈A} |f(x)+g(x)| &≤∑_{x∈A}|f(x)| +∑_{x∈A}|g(x)| \\&≤\sup⁡\left\{∑_{x∈A}|f(x)|∶A⊆X,A\text{ finite}\right\}+\sup \left\{∑_{x∈A}|g(x)|∶A⊆X,A\text{ finite}\right\}\\&<∞\end{aligned}x∈A∑∣f(x)+g(x)∣≤x∈A∑∣f(x)∣+x∈A∑∣g(x)∣≤sup⁡{x∈A∑∣f(x)∣∶A⊆X,A finite}+sup{x∈A∑∣g(x)∣∶A⊆X,A finite}<∞
which means
sup⁡⁡{∑x∈A∣f(x)+g(x)∣∶A⊆X,Afinite}≤sup⁡⁡{∑x∈A∣f(x)∣∶A⊆X,Afinite}+sup⁡{∑x∈A∣g(x)∣∶A⊆X,Afinite}\sup⁡\left\{∑_{x∈A}|f(x)+g(x)|∶A⊆X,A\text{ finite}\right\}\\\leq\sup⁡\left\{∑_{x∈A}|f(x)|∶A⊆X,A\text{ finite}\right\}+\sup \left\{∑_{x∈A}|g(x)|∶A⊆X,A\text{ finite}\right\}sup⁡{x∈A∑∣f(x)+g(x)∣∶A⊆X,A finite}≤sup⁡{x∈A∑∣f(x)∣∶A⊆X,A finite}+sup{x∈A∑∣g(x)∣∶A⊆X,A finite}
Use Definition 8.2.4 we know that ∑x∈X∣f(x)+g(x)∣∑_{x∈X}|f(x)+g(x)|∑x∈X∣f(x)+g(x)∣ is absolutely convergent. Thus
∑x∈X∣f(x)+g(x)∣=∑x∈X,f(x)+g(x)≠0(f(x)+g(x))=∑x∈S∪B(f(x)+g(x))∑_{x∈X}|f(x)+g(x)| =∑_{x∈X,f(x)+g(x)≠0} (f(x)+g(x)) =∑_{x∈S∪B} (f(x)+g(x))x∈X∑∣f(x)+g(x)∣=x∈X,f(x)+g(x)=0∑(f(x)+g(x))=x∈S∪B∑(f(x)+g(x))
S∪BS∪BS∪B is at most countable, if S∪BS∪BS∪B is finite we can use Proposition 7.1.11(f) to get
∑x∈S∪B(f(x)+g(x))=∑x∈S∪Bf(x)+∑x∈S∪Bg(x)=∑x∈Sf(x)+∑x∈Bg(x)∑_{x∈S∪B} (f(x)+g(x)) =∑_{x∈S∪B} f(x)+∑_{x∈S∪B} g(x) =∑_{x∈S} f(x) +∑_{x∈B} g(x)x∈S∪B∑(f(x)+g(x))=x∈S∪B∑f(x)+x∈S∪B∑g(x)=x∈S∑f(x)+x∈B∑g(x)
If S∪BS∪BS∪B is countable, we can have a bijection h:N→S∪Bh:\mathbf N→S∪Bh:N→S∪B, s.t.
∑x∈S∪B(f(x)+g(x))=∑n=0+∞(f(h(n))+g(h(n)))∑_{x∈S∪B} (f(x)+g(x)) =∑_{n=0}^{+∞}\bigg(f\Big(h(n)\Big)+g\Big(h(n)\Big)\bigg) x∈S∪B∑(f(x)+g(x))=n=0∑+∞(f(h(n))+g(h(n)))
Let SN=∑n=0N(f(h(n))+g(h(n)))=∑n=0Nf(h(n))+∑n=0Ng(h(n))S_N=∑_{n=0}^N\bigg(f\Big(h(n)\Big)+g\Big(h(n)\Big)\bigg) =∑_{n=0}^Nf(h(n)) +∑_{n=0}^Ng(h(n))SN=∑n=0N(f(h(n))+g(h(n)))=∑n=0Nf(h(n))+∑n=0Ng(h(n)) , we know that when N→∞N→∞N→∞, ∑n=0Nf(h(n))→L∑_{n=0}^Nf(h(n)) →L∑n=0Nf(h(n))→L since A⊆A∪BA⊆A∪BA⊆A∪B, by the same logic, ∑n=0Ng(h(n))→M∑_{n=0}^Ng(h(n)) →M∑n=0Ng(h(n))→M, thus the conclusion is valid.
( b ) If c=0c=0c=0 the conclusion is obviously true. Now suppose c≠0c≠0c=0, Let A⊆XA⊆XA⊆X be a finite set, then by proposition 7.1.11 we have
∑x∈A∣cf(x)∣=∣c∣∑x∈A∣f(x)∣≤∣c∣sup⁡⁡{∑x∈A∣f(x)∣∶A⊆X,Afinite}<∞∑_{x∈A}|cf(x)| =|c| ∑_{x∈A}|f(x)| ≤|c| \sup⁡\left\{∑_{x∈A}|f(x)|∶A⊆X,A\text{ finite}\right\}<∞x∈A∑∣cf(x)∣=∣c∣x∈A∑∣f(x)∣≤∣c∣sup⁡{x∈A∑∣f(x)∣∶A⊆X,A finite}<∞
Thus
sup⁡{∑x∈A∣cf(x)∣∶A⊆X,Afinite}≤∣c∣sup⁡⁡{∑x∈A∣f(x)∣∶A⊆X,Afinite}<∞\sup\left\{∑_{x∈A}|cf(x)|∶A⊆X,A\text{ finite}\right\}≤|c| \sup⁡\left\{∑_{x∈A}|f(x)|∶A⊆X,A\text{ finite}\right\}<∞sup{x∈A∑∣cf(x)∣∶A⊆X,A finite}≤∣c∣sup⁡{x∈A∑∣f(x)∣∶A⊆X,A finite}<∞
Use Definition 8.2.4 we know that ∑x∈Xcf(x)∑_{x∈X} cf(x)∑x∈Xcf(x) is absolutely convergent. Since we also have (f(x)≠0)⇔(cf(x)≠0)(f(x)≠0)⇔(cf(x)≠0)(f(x)=0)⇔(cf(x)=0), we know that
∑x∈Xcf(x)=∑x∈X,cf(x)≠0cf(x)=∑x∈Scf(x)∑_{x∈X} cf(x) = ∑_{x∈X,cf(x)≠0} cf(x) =∑_{x∈S} cf(x)x∈X∑cf(x)=x∈X,cf(x)=0∑cf(x)=x∈S∑cf(x)
If SSS is finite, use Proposition 7.1.11(g) we have ∑x∈Scf(x)=c∑x∈Sf(x)=c∑x∈Xf(x)∑_{x∈S} cf(x) =c∑_{x∈S} f(x) =c∑_{x∈X} f(x)∑x∈Scf(x)=c∑x∈Sf(x)=c∑x∈Xf(x) . If SSS is countable, we can have a bijection h:N→Sh:\mathbf N→Sh:N→S, s.t.
∑x∈Scf(x)=∑n=0+∞cf(h(n))=c∑n=0+∞f(h(n))∑_{x∈S} cf(x) =∑_{n=0}^{+∞}cf(h(n)) =c∑_{n=0}^{+∞}f(h(n)) x∈S∑cf(x)=n=0∑+∞cf(h(n))=cn=0∑+∞f(h(n))
The last equality comes from Proposition 7.2.14(b)
( c ) In the first case, let A⊆X1A⊆X_1A⊆X1 be any finite set, then A⊆XA⊆XA⊆X, thus
∑x∈A∣f(x)∣≤sup⁡{∑x∈A∣f(x)∣∶A⊆X,Afinite}∑_{x∈A}|f(x)| ≤\sup\left\{∑_{x∈A}|f(x)|∶A⊆X,A\text{ finite}\right\}x∈A∑∣f(x)∣≤sup{x∈A∑∣f(x)∣∶A⊆X,A finite}
Which gives
sup⁡{∑x∈A∣f(x)∣∶A⊆X1,Afinite}≤sup⁡{∑x∈A∣f(x)∣∶A⊆X,Afinite}\sup \left\{∑_{x∈A}|f(x)|∶A⊆X_1,A\text{ finite} \right \} ≤\sup\left\{∑_{x∈A}|f(x)|∶A⊆X,A\text{ finite}\right\}sup{x∈A∑∣f(x)∣∶A⊆X1,A finite}≤sup{x∈A∑∣f(x)∣∶A⊆X,A finite}
So ∑x∈X1f(x)∑_{x∈X_1}f(x)∑x∈X1f(x) is absolutely convergent.
By the same logic ∑x∈X2f(x)∑_{x∈X_2}f(x)∑x∈X2f(x) is absolutely convergent.
Conversely, if ∑x∈X1h(x)∑_{x∈X_1} h(x)∑x∈X1h(x) and ∑x∈X2h(x)∑_{x∈X_2} h(x)∑x∈X2h(x) are absolutely convergent. Then for any finite set A⊆XA⊆XA⊆X, we can have A=(A∩X1)∪(A∩X2)A=(A∩X_1 )∪(A∩X_2)A=(A∩X1)∪(A∩X2), as X1X_1X1 and X2X_2X2 are disjoint, we have
∑x∈A∣f(x)∣=∑x∈A∩X1∣f(x)∣+∑x∈A∩X2∣f(x)∣≤sup⁡{∑x∈A∣f(x)∣∶A⊆X1,Afinite}+sup⁡{∑x∈A∣f(x)∣∶A⊆X2,Afinite}\begin{aligned} ∑_{x∈A}|f(x)| &=∑_{x∈A∩X_1}|f(x)| +∑_{x∈A∩X_2}|f(x)| \\&≤\sup\left\{∑_{x∈A}|f(x)|∶A⊆X_1,A\text{ finite}\right\}+\sup\left\{∑_{x∈A}|f(x)|∶A⊆X_2,A\text{ finite}\right\}\end{aligned}x∈A∑∣f(x)∣=x∈A∩X1∑∣f(x)∣+x∈A∩X2∑∣f(x)∣≤sup{x∈A∑∣f(x)∣∶A⊆X1,A finite}+sup{x∈A∑∣f(x)∣∶A⊆X2,A finite}
Then we have
sup⁡{∑x∈A∣f(x)∣∶A⊆(X1∪X2),Afinite}≤sup⁡{∑x∈A∣f(x)∣∶A⊆X1,Afinite}+sup⁡{∑x∈A∣f(x)∣∶A⊆X2,Afinite}\sup\left\{∑_{x∈A}|f(x)|∶A⊆(X_1∪X_2),A\text{ finite}\right\} \\≤\sup\left\{∑_{x∈A}|f(x)|∶A⊆X_1,A\text{ finite}\right\}+\sup\left\{∑_{x∈A}|f(x)|∶A⊆X_2,A\text{ finite}\right\}sup{x∈A∑∣f(x)∣∶A⊆(X1∪X2),A finite}≤sup{x∈A∑∣f(x)∣∶A⊆X1,A finite}+sup{x∈A∑∣f(x)∣∶A⊆X2,A finite}
To prove
∑x∈X1∪X2f(x)=∑x∈X1f(x)+∑x∈X2f(x)∑_{x∈X_1∪X_2}f(x) =∑_{x∈X_1} f(x) +∑_{x∈X_2} f(x)x∈X1∪X2∑f(x)=x∈X1∑f(x)+x∈X2∑f(x)
We only need to consider points where fff doesn’t vanish on, thus redefine
X={x∈X:f(x)≠0},X1={x∈X1:f(x)≠0},X2={x∈X2:f(x)≠0}X=\{x∈X:f(x)≠0\},\\X_1=\{x∈X_1:f(x)≠0\},\\X_2=\{x∈X_2:f(x)≠0\}X={x∈X:f(x)=0},X1={x∈X1:f(x)=0},X2={x∈X2:f(x)=0}
we first see the case is clear when X=X1∪X2X=X_1∪X_2X=X1∪X2 is finite. Now if one is countable and the other is finite, we may assume X2X_2X2 is finite, then there’s bijections g1:N→X1g_1:\mathbf N→X_1g1:N→X1 and g2:{i∈N:0≤i≤m}→X2g_2:\{i∈\mathbf N:0≤i≤m\}→X_2g2:{i∈N:0≤i≤m}→X2, such that
∑x∈X1f(x)=∑n=0∞f(g1(n)),∑x∈X2f(x)=∑n=0mf(g2(n))∑_{x∈X_1} f(x) =∑_{n=0}^∞f\Big(g_1 (n)\Big) ,\quad ∑_{x∈X_2} f(x) =∑_{n=0}^mf\Big(g_2 (n)\Big)x∈X1∑f(x)=n=0∑∞f(g1(n)),x∈X2∑f(x)=n=0∑mf(g2(n))
We define a bijection g:N→X1∪X2g:\mathbf N→X_1∪X_2g:N→X1∪X2 by
g(n)={g2(n),n≤mg1(n−m−1),n>mg(n)=\begin{cases}g_2 (n),&n≤m\\g_1 (n-m-1),&n>m\end{cases}g(n)={g2(n),g1(n−m−1),n≤mn>m

Then ∑x∈X1∪X2f(x)=∑n=0∞f(g(n))∑_{x∈X_1∪X_2} f(x) =∑_{n=0}^∞f\Big(g(n)\Big)∑x∈X1∪X2f(x)=∑n=0∞f(g(n)) , further we have for any N∈NN∈\mathbf NN∈N:
∑n=0N+mf(g(n))=∑n=0mf(g2(n))+∑n=0Nf(g1(n))∑_{n=0}^{N+m}f\Big(g(n)\Big) =∑_{n=0}^mf\Big(g_2 (n)\Big) +∑_{n=0}^Nf\Big(g_1 (n)\Big) n=0∑N+mf(g(n))=n=0∑mf(g2(n))+n=0∑Nf(g1(n))
Let N→+∞N→+∞N→+∞, we get the conclusion.

If both sets are countable, there’s bijections g1:N→X1g_1:\mathbf N→X_1g1:N→X1 and g2:N→X2g_2:\mathbf N→X_2g2:N→X2 such that
∑x∈X1f(x)=∑n=0∞f(g1(n)),∑x∈X2f(x)=∑n=0∞f(g2(n))∑_{x∈X_1} f(x) =∑_{n=0}^∞f(g_1 (n)) ,\quad ∑_{x∈X_2} f(x) =∑_{n=0}^∞f(g_2 (n)) x∈X1∑f(x)=n=0∑∞f(g1(n)),x∈X2∑f(x)=n=0∑∞f(g2(n))
We define a bijection g:N→X1∪X2g:\mathbf N→X_1∪X_2g:N→X1∪X2 by
g(n)={g1(k),n=2k,k∈Ng2(k),n=2k+1,k∈Ng(n)=\begin{cases}g_1 (k),& n=2k,k∈N \\g_2 (k),&n=2k+1,k∈N\end{cases}g(n)={g1(k),g2(k),n=2k,k∈Nn=2k+1,k∈N
Then ∑x∈X1∪X2f(x)=∑n=0∞f(g(n))∑_{x∈X_1∪X_2} f(x) =∑_{n=0}^∞f(g(n))∑x∈X1∪X2f(x)=∑n=0∞f(g(n)) , further we have for any N∈NN∈\mathbf NN∈N:
∑n=02N+1f(g(n))=∑n=0Nf(g2(n))+∑n=0Nf(g1(n))∑_{n=0}^{2N+1}f(g(n)) =∑_{n=0}^Nf(g_2 (n)) +∑_{n=0}^Nf(g_1 (n)) n=0∑2N+1f(g(n))=n=0∑Nf(g2(n))+n=0∑Nf(g1(n))
Let N→+∞N→+∞N→+∞, we get the conclusion.
The statement of h(x)h(x)h(x) can then be derived by substituting fff to hhh.
( d ) First we prove ∑y∈Yf(ϕ(y))∑_{y∈Y}f(ϕ(y))∑y∈Yf(ϕ(y)) is absolutely convergent. Choose any finite A⊆YA⊆YA⊆Y, as ϕϕϕ is a bijection, ϕ(A)⊆Xϕ(A)⊆Xϕ(A)⊆X is finite. Let

L=sup⁡⁡{∑x∈A∣f(x)∣∶A⊆X,Afinite}<∞L=\sup⁡\left\{∑_{x∈A}|f(x)|∶A⊆X,A\text{ finite}\right\}<∞L=sup⁡{x∈A∑∣f(x)∣∶A⊆X,A finite}<∞

Then ∑y∈A∣f(ϕ(y))∣=∑ϕ(y)∈ϕ(A)∣f(ϕ(y))∣≤L∑_{y∈A}|f(ϕ(y))| =∑_{ϕ(y)∈ϕ(A)}|f(ϕ(y))| ≤L∑y∈A∣f(ϕ(y))∣=∑ϕ(y)∈ϕ(A)∣f(ϕ(y))∣≤L, so
sup⁡⁡{∑y∈A∣f(ϕ(y))∣∶A⊆Y,Afinite}≤L<∞\sup⁡\left\{∑_{y∈A}|f(ϕ(y))|∶A⊆Y,A\text{ finite}\right\}≤L<∞sup⁡⎩⎨⎧y∈A∑∣f(ϕ(y))∣∶A⊆Y,A finite⎭⎬⎫≤L<∞

Next, to prove ∑y∈Yf(ϕ(y))=∑x∈Xf(x)∑_{y∈Y}f(ϕ(y)) =∑_{x∈X} f(x)∑y∈Yf(ϕ(y))=∑x∈Xf(x) , we see that for any y∈Yy∈Yy∈Y such that f(ϕ(y))≠0f(ϕ(y))≠0f(ϕ(y))=0, we can choose one x=ϕ(y)∈Xx=ϕ(y)∈Xx=ϕ(y)∈X such that f(x)≠0f(x)≠0f(x)=0, vice versa.(In this step we use the Axiom of Choice if YYY and XXX are uncountable) Thus the sets {y∈Y:f(ϕ(y))≠0}\{y∈Y: f(ϕ(y))≠0\}{y∈Y:f(ϕ(y))=0} and {x∈X:f(x)≠0}\{x∈X:f(x)≠0\}{x∈X:f(x)=0} have the same cardinality, thus are both finite or countable. In the finite case the statement is obviously true by Proposition 7.1.11( c ). If both sets are countable, we let a bijection g:N→{y∈Y:f(ϕ(y))≠0}g:\mathbf N→\{y∈Y: f(ϕ(y))≠0\}g:N→{y∈Y:f(ϕ(y))=0}, then ϕ∘g:N→{x∈X:f(x)≠0}ϕ∘g:\mathbf N→\{x∈X:f(x)≠0\}ϕ∘g:N→{x∈X:f(x)=0} is also a bijection, and we have

∑y∈Yf(ϕ(y))=∑n=0∞f(ϕ(g(n)))∑x∈Xf(x)=∑n=0∞f(ϕ∘g(n))=∑n=0∞f(ϕ(g(n)))∑_{y∈Y}f(ϕ(y)) =∑_{n=0}^∞f(ϕ(g(n))) \\ ∑_{x∈X} f(x) =∑_{n=0}^∞f(ϕ∘g(n)) =∑_{n=0}^∞f(ϕ(g(n))) y∈Y∑f(ϕ(y))=n=0∑∞f(ϕ(g(n)))x∈X∑f(x)=n=0∑∞f(ϕ∘g(n))=n=0∑∞f(ϕ(g(n)))
Then we have ∑y∈Yf(ϕ(y))=∑x∈Xf(x)∑_{y∈Y}f(ϕ(y)) =∑_{x∈X} f(x)∑y∈Yf(ϕ(y))=∑x∈Xf(x) .

Exercise 8.2.4

Let ∑n=0∞an=L∑_{n=0}^∞a_n =L∑n=0∞an=L, we also know that ∑n=0∞∣an∣∑_{n=0}^∞|a_n |∑n=0∞∣an∣ doesn’t exist. Now assume the conclusion is wrong, then one of the three assertions below must be true:

i. ∑n∈A+an∑_{n∈A^+}a_n∑n∈A+an converges, but ∑n∈A−an∑_{n∈A^-}a_n∑n∈A−an is not conditionally convergent.
ii. ∑n∈A−an∑_{n∈A^-}a_n∑n∈A−an converges, but ∑n∈A+an∑_{n∈A^+}a_n∑n∈A+an is not conditionally convergent.
iii. both ∑n∈A+an∑_{n∈A^+}a_n∑n∈A+an and ∑n∈A−an∑_{n∈A^-}a_n∑n∈A−an converges.

In case i and ii, we know that ∑n=0∞an=∑n∈A+an+∑n∈A−an∑_{n=0}^∞a_n =∑_{n∈A^+}a_n +∑_{n∈A^-}a_n∑n=0∞an=∑n∈A+an+∑n∈A−an , so we can have ∑n∈A−an=∑n=0∞an−∑n∈A+an∑_{n∈A^-}a_n =∑_{n=0}^∞a_n -∑_{n∈A^+}a_n∑n∈A−an=∑n=0∞an−∑n∈A+an in case i and ∑n∈A+an=∑n=0∞an−∑n∈A−an∑_{n∈A^+}a_n =∑_{n=0}^∞a_n -∑_{n∈A^-}a_n∑n∈A+an=∑n=0∞an−∑n∈A−an in case ii, both are contradiction since the left side is not convergent but the right side is convergent.
In case iii, we let ∑n∈A+an=L1∑_{n∈A^+}a_n =L_1∑n∈A+an=L1 and ∑n∈A−an=L2∑_{n∈A^-}a_n =L_2∑n∈A−an=L2, thus we have:
∑n=0∞∣an∣=∑n∈A+∣an∣+∑n∈A−∣an∣=∑n∈A+an−∑n∈A−an=L1−L2∑_{n=0}^∞|a_n |=∑_{n∈A^+}|a_n |+∑_{n∈A^-}|a_n |=∑_{n∈A^+}a_n -∑_{n∈A^-}a_n =L_1-L_2 n=0∑∞∣an∣=n∈A+∑∣an∣+n∈A−∑∣an∣=n∈A+∑an−n∈A−∑an=L1−L2
Which is a contradiction to the fact that ∑n=0∞∣an∣∑_{n=0}^∞|a_n |∑n=0∞∣an∣ doesn’t exist. Thus the original conclusion must be true.

Exercise 8.2.5

Why1: A+A^+A+ and A−A^-A− are infinite.
Assume not, then one of A+A^+A+ and A−A^-A− are finite, then either ∑n∈A+an∑_{n∈A^+}a_n∑n∈A+an or ∑n∈A−an∑_{n∈A^-}a_n∑n∈A−an is a finite series, thus must be convergent, a contradiction.
Why2: ∑m=0∞af+(m)∑_{m=0}^∞a_{f_+ (m)}∑m=0∞af+(m) and ∑m=0∞af−(m)∑_{m=0}^∞a_{f_- (m)}∑m=0∞af−(m) are not absolutely convergent.
Since af+(m)≥0,∀ma_{f_+ (m)}≥0,∀maf+(m)≥0,∀m and af−(m)<0,∀ma_{f_- (m)}<0,∀maf−(m)<0,∀m, if we assume ∑m=0∞af+(m)∑_{m=0}^∞a_{f_+ (m)}∑m=0∞af+(m) or ∑m=0∞af−(m)∑_{m=0}^∞a_{f_- (m)}∑m=0∞af−(m) is absolutely convergent, then we can know from Proposition 7.4.3 that ∑m=0∞af+(m)=∑n∈A+an∑_{m=0}^∞a_{f_+ (m)} =∑_{n∈A^+}a_n∑m=0∞af+(m)=∑n∈A+an or ∑m=0∞af−(m)=∑n∈A−an∑_{m=0}^∞a_{f_- (m)} =∑_{n∈A^-}a_n∑m=0∞af−(m)=∑n∈A−an , in particular either ∑n∈A+an∑_{n∈A^+}a_n∑n∈A+an or ∑n∈A−an∑_{n∈A^-}a_n∑n∈A−an is absolute convergent, which is a contradiction.
Why3: The map j→njj→n_jj→nj is injective.
Let k≠jk≠jk=j, then either k<jk<jk<j or j<kj<kj<k, if k<jk<jk<j, the definition of njn_jnj guarantees that nj≠nkn_j≠n_knj=nk. If k<jk<jk<j we can similarly prove nk≠njn_k≠n_jnk=nj, so this map is injective.
Why4: Both Case I and II occur an infinite number of times.
Assume Case I only occurs finite number of times, this means for some M∈NM∈\mathbf NM∈N, we’ll have
∑0≤i<jani≥L,∀j>M∑_{0≤i<j}a_{n_i } ≥L,\quad ∀j>M0≤i<j∑ani≥L,∀j>M
Thus we have ∑0≤i≤Mani+∑M<iani≥L∑_{0≤i≤M}a_{n_i} +∑_{M<i}a_{n_i} ≥L∑0≤i≤Mani+∑M<iani≥L, or ∑M<iani≥L−∑0≤i≤Mani∑_{M<i}a_{n_i } ≥L-∑_{0≤i≤M}a_{n_i}∑M<iani≥L−∑0≤i≤Mani , notice that as iii increases, ∑M<iani∑_{M<i}a_{n_i}∑M<iani decreases since all ani<0a_{n_i}<0ani<0, so ∑M<iani∑_{M<i}a_{n_i}∑M<iani converges, which means ∑n∈A−an∑_{n∈A^-}a_n∑n∈A−an converges, a contradiction.
When assume Case II occurs finite number of times, we can prove a contradiction similarly.

Why5: The map j→njj→n_jj→nj is surjective.
For ∀n∈N∀n∈\mathbf N∀n∈N, either n∈A+n∈A^+n∈A+ or n∈A−n∈A^-n∈A−. Assume no jjj can make nj=nn_j=nnj=n, this means either Case I or Case II stops at nnn, so can only occur finite number of times, which is a contradiction.
Why6: lim⁡j→∞anj=0\lim_{j→∞} a_{n_j}=0limj→∞anj=0
From Corollary 7.2.6 we have lim⁡n→∞an=0\lim_{n→∞} a_{n}=0limn→∞an=0, since both Case I and II occur an infinite number of times, when j→∞j→∞j→∞ we must have nj→∞n_j→∞nj→∞.

Why7: lim⁡j→∞∑0≤i<jani=L\lim_{j→∞} ∑_{0≤i<j}a_{n_i} =Llimj→∞∑0≤i<jani=L
From Why6 we know that for ∀ϵ>0,∃J∈N∀ϵ>0,∃J∈\mathbf N∀ϵ>0,∃J∈N,s.t.∣anj∣<ϵ,(j>J)|a_{n_j} |<ϵ,(j>J)∣anj∣<ϵ,(j>J). Start from nJn_JnJ, we search for a number K>JK>JK>J such that ∑0≤i<Kani<L∑_{0≤i<K}a_{n_i} <L∑0≤i<Kani<L and ∑0≤i<K+1ani≥L∑_{0≤i<K+1}a_{n_i} ≥L∑0≤i<K+1ani≥L, this KKK must exist, otherwise we’ll get a contradiction with Why4.
We denote Sk=∑0≤i<kaniS_k=∑_{0≤i<k}a_{n_i}Sk=∑0≤i<kani , notice that:
If Sk<LS_k<LSk<L, we shall add positive ania_{n_i}ani to SkS_kSk, the positive adding shall stop once Sk≥LS_k≥LSk≥L, so Sk≤L+∣ani∣S_k≤L+|a_{n_i} |Sk≤L+∣ani∣, if k>K>Jk>K>Jk>K>J, we can have Sk≤L+ϵS_k≤L+ϵSk≤L+ϵ.
If Sk≥LS_k≥LSk≥L, we shall add negative ania_{n_i}ani to SkS_kSk, the negative adding shall stop once Sk<LS_k<LSk<L, so Sk>L−∣ani∣S_k>L-|a_{n_i} |Sk>L−∣ani∣, if k>K>Jk>K>Jk>K>J, we can have Sk>L−ϵS_k>L-ϵSk>L−ϵ.
Combined we get ∣Sk−L∣<ϵ,k>K|S_k-L|<ϵ,k>K∣Sk−L∣<ϵ,k>K, this means lim⁡k→∞Sk=L\lim_{k→∞} S_k=Llimk→∞Sk=L.

Exercise 8.2.6

We only have to prove lim inf⁡N→∞∑m=N∞af(m)=+∞\liminf_{N→∞}∑_{m=N}^∞a_f(m) =+∞N→∞liminf∑m=N∞af(m)=+∞. With the same definition of A+A^+A+ and A−A^-A− in Lemma 8.2.7, we find increasing bijections f+:N→A+f_+:N→A^+f+:N→A+ and f−:N→A−f_-:N→A^-f−:N→A−. We define a sequence n0,n1,n2,…n_0,n_1,n_2,…n0,n1,n2,… of natural numbers recursively as below:
Let n0=f+(0),n1=f−(0)n_0=f_+ (0),n_1=f_- (0)n0=f+(0),n1=f−(0). Suppose that j≥2j≥2j≥2 is a natural number, and nin_ini has already be defined for all i<ji<ji<j. Then we let
k=max⁡{n:f−(n)∈{ni:i<j}}k=\max \Big\{n:f_- (n)∈\{n_i:i<j\}\Big\}k=max{n:f−(n)∈{ni:i<j}}
For example, if j=2j=2j=2, then n0n_0n0 and n1n_1n1 has been defined, then k=1k=1k=1. Now
(I) If ∑0≤i<jani<k∑_{0≤i<j}a_{n_i} <k∑0≤i<jani<k, we set nj=min⁡{n∈A+:n≠nifor all i<j}n_j=\min\{n∈A^+:n≠n_i\text{ for all }i<j\}nj=min{n∈A+:n=ni for all i<j}
(II) If ∑0≤i<jani≥k∑_{0≤i<j}a_{n_i}≥k∑0≤i<jani≥k, we set nj=min⁡{n∈A−:n≠nifor all i<j}n_j=\min\{n∈A^-:n≠n_i\text{ for all }i<j\}nj=min{n∈A−:n=ni for all i<j}
Intuitively, we compare ∑0≤i<jani∑_{0≤i<j}a_{n_i}∑0≤i<jani with an increasing KKK, which increases by 111 once we add an negative number to the sum. From the proof of Exercise 8.2.5 we can see that

A+A^+A+ and A−A^-A− are infinite.
∑m=0∞af+(m)∑_{m=0}^∞a_{f_+ (m)}∑m=0∞af+(m) and ∑m=0∞af−(m)∑_{m=0}^∞a_{f_- (m)}∑m=0∞af−(m) are not absolutely convergent.
The map j→njj→n_jj→nj is injective.

And we can prove the following results:

Both Case I and II occur an infinite number of times.

Assume Case I occur only finite number of times, then from some JJJ and KKK we’ll get ∑0≤i<Jani≥K∑_{0≤i<J}a_{n_i} ≥K∑0≤i<Jani≥K and Case II happens infinitely, but as we add negative numbers to ∑0≤i<Jani∑_{0≤i<J}a_{n_i}∑0≤i<Jani , we’ll have ∑0≤i<jani≤∑0≤i<Jani∑_{0≤i<j}a_{n_i} ≤∑_{0≤i<J}a_{n_i}∑0≤i<jani≤∑0≤i<Jani , and K→∞K→∞K→∞, this will lead to a contradiction since ∑0≤i<Jani∑_{0≤i<J}a_{n_i}∑0≤i<Jani is finite.
Assume Case II occur only finite number of times, then from some JJJ and KKK we’ll get ∑0≤i<Jani<K∑_{0≤i<J}a_{n_i} <K∑0≤i<Jani<K, and ∑0≤i<jani<K,∀j>J∑_{0≤i<j}a_{n_i} <K,∀j>J∑0≤i<jani<K,∀j>J, this means ∑m=0∞af+(m)∑_{m=0}^∞a_{f_+ (m)}∑m=0∞af+(m) is convergent, a contradiction.

The map j→njj→n_jj→nj is surjective.
lim⁡j→∞anj=0\lim_{j→∞} a_{n_j}=0limj→∞anj=0.

The proofs are similar to that of Exercise 8.2.5.
Finally we shall prove lim⁡j→∞∑0≤i<jani>M\lim_{j→∞} ∑_{0≤i<j}a_{n_i} >Mlimj→∞∑0≤i<jani>M for any M>0M>0M>0, thus complete the proof. For any M>0M>0M>0, we can find a natural number K>MK>MK>M, and as lim⁡j→∞anj=0\lim_{j→∞} a_{n_j}=0limj→∞anj=0, there’s JJJ such that

∣anj∣<1,∀j≥J|a_{n_j} |<1,\quad ∀j≥J∣anj∣<1,∀j≥J
We’ll eventually find a j′>Jj'>Jj′>J such that ∑0≤i<j′ani≥K+1∑_{0≤i<j'}a_{n_i} ≥K+1∑0≤i<j′ani≥K+1, and adding negative ania_{n_i}ani to the sum would not let the new sum falls below KKK, once the sum is between KKK and K+1K+1K+1, we would continue to add positive ania_{n_i}ani and push the sum to K+2K+2K+2 and even larger, thus we have ∑0≤i<jani≥K>M,∀j>j′∑_{0≤i<j}a_{n_i} ≥K>M,∀j>j'∑0≤i<jani≥K>M,∀j>j′, and the conclusion follows.