陶哲轩实分析（上）8.1及习题-Analysis I 8.1

以我相对平庸的智商来说，Analysis I 的第8章是整本书中相对比较难的章节之一，由于（以个人的理解）这章实际是将集合论和实分析的起步阶段内容都介绍了，但是篇幅又不会展开到实分析教科书的程度，对于第一次接触的人毫无疑问是比较棘手的，我虽然之前接触过实分析，大致能理解countable和uncountable的区别，但是从没有非常仔细的思考过著名的选择公理Axiom of Choice，以及其各种等价结论，还有其应用。这一章提供了一个机会。

本章的习题也是 Analysis I 中数量比较大，难度也挺高的一部分，基本上如果没有hint，都很难做出来。这种熬脑的感觉在第1节还好，在第5节达到了顶峰。

Exercise 8.1.1

First, if there exists a proper subset

⊊

Y⊊X

$Y ⊊ X$ which has the same cardinality as

$X$ , then assume

$X$ is finite, then

$Y$ must be finite. Since

∩

(

)

X=Y∩(X\backslash Y)

$X=Y∩(X\Y)$ , we know that

$Y$ and

X\backslash Y

$X\Y$ are disjoint, further

X\backslash Y

$X\Y$ is not empty, so

(

)

≥

\#(X\backslash Y)≥1

$#(X\Y)≥1$ , and

(

)

(

)

(

)

(

)

(

)

≥

(

)

(

)

\#(X)=\#(Y)+\#(X\backslash Y)=\#(X)+\#(X\backslash Y)≥\#(X)+1>\#(X)

$#(X)=#(Y)+#(X\Y)=#(X)+#(X\Y)≥#(X)+1>#(X)$
which is a contradiction.
On the contrary, if

$X$ is infinite, then we can form a set

$S$ like this:
Choose

∀

∈

∀x∈X

$\forall x \in X$ , rename it

x_0

$x_{0}$ and let

{

}

S_0=\{x_0 \}

$S_{0} = {x_{0}}$ , then choose any element belongs to

X\backslash S_0

$X\S0$ , and rename it

x_1

$x_{1}$ , then let

∪

{

}

S_1=S_0∪\{x_1 \}

$S_{1} = S_{0} \cup {x_{1}}$ . This process shall never stop, since

$X$ is infinite and we have the Axiom of Choice. Doing this recursively, we can eventually get a set

$S$ (by letting

→

∞

n→∞

$n \to \infty$ ), obviously

⊆

S⊆X

$S \subseteq X$ .
We let

{

}

Y=X\backslash \{x_0\}

$Y=X\{x0}$ , then

$Y$ is a proper subset of

$X$ . For any

∈

x∈X

$x \in X$ , either

∈

x∈S

$x \in S$ or

∈

x∈X\backslash S

$x∈X\S$ , thus we define a function

$f$ from

$X$ to

$Y$ as follows:

(

)

{

∈

f(x)=\begin{cases}x_{n+1},&x=x_n∈S\\x,&x∈X\backslash S \end{cases}

$f(x)={xn+1,x,x=xn∈Sx∈X\S$
It’s easy to show that

$f$ is a bijection.

Exercise 8.1.2

For the sake of contradiction we assume Proposition 8.1.4 is false, then there’s a nonempty set

⊆

X⊆\mathbf N

$X \subseteq N$ , such that for any

∈

n∈X

$n \in X$ , there exists one

∈

m∈X

$m \in X$ s.t.

m<n

$m < n$ .
Since

≠

∅

X≠∅

$X \neq = \emptyset$ , we choose

∈

x∈X

$x \in X$ , and let

x=a_0

$x = a_{0}$ , by our assumption, we can find an element

∈

a_1∈X,a_1<x=a_0

$a_{1} \in X, a_{1} < x = a_{0}$ , once we find

a_n

$a_{n}$ , we can by assumption find an element

∈

a_{n+1}∈X

$a_{n + 1} \in X$ such that

a_{n+1}<a_n

$a_{n + 1} < a_{n}$ , thus we build a infinite descent sequence

(

)

∞

(a_n )_{n=0}^∞

$(a_{n})_{n = 0}^{\infty}$ of natural numbers, this contradicts the principle of infinite descent.
If we replace the natural numbers by the integers, then the well-ordering principle doesn’t work. For example the set

{

−

…

}

X=\{-1,-2,-3,…\}

$X = {- 1, - 2, - 3, \dots}$ doesn’t have a minimum element.
If we replace the natural numbers by the positive rationals, then the well-ordering principle doesn’t work. For example the set

{

∈

}

X=\{q∈\mathbf Q:q>0\}

$X = {q \in Q : q > 0}$ doesn’t have a minimum element.

Exercise 8.1.3

I will prove this Proposition in a complete manner.
Recursively define a sequence of natural numbers

…

a_0,a_1,a_2,…

$a_{0}, a_{1}, a_{2}, \dots$ using the formula below:

≔

min

⁡

{

∈

∀

≠

}

a_n≔\min \{x∈X:∀m<n,x≠a_m \}

$a_{n} : = min {x \in X : \forall m < n, x \neq = a_{m}}$
Since

$X$ is infinite, the set

{

∈

∀

≠

}

\{x∈X:∀m<n,x≠a_m \}

${x \in X : \forall m < n, x \neq = a_{m}}$ should also be infinite, assume not, then for some n this set is finite, and

{

…

−

}

\{a_0,a_1,…,a_{n-1} \}

${a_{0}, a_{1}, \dots, a_{n - 1}}$ is finite, by

{

∈

∀

≠

}

∪

{

…

−

}

X=\{x∈X:∀m<n,x≠a_m \}∪\{a_0,a_1,…,a_{n-1} \}

$X = {x \in X : \forall m < n, x \neq = a_{m}} \cup {a_{0}, a_{1}, \dots, a_{n - 1}}$
We deduce

$X$ is finite, which is a contradiction.
Since

{

∈

∀

≠

}

\{x∈X:∀m<n,x≠a_m \}

${x \in X : \forall m < n, x \neq = a_{m}}$ is infinite, it’s not empty for every

$n$ , so we can successfully define

min

⁡

{

∈

∀

≠

}

\min⁡\{x∈X:∀m<n,x≠a_m \}

$min {x \in X : \forall m < n, x \neq = a_{m}}$ by the well-ordering principle.
For

∀

∀n

$\forall n$ , we know that

≔

min

⁡

{

∈

∀

≠

}

a_{n+1}≔\min⁡\{x∈X:∀m<n+1,x≠a_m \}

$a_{n + 1} : = min {x \in X : \forall m < n + 1, x \neq = a_{m}}$
By the definition of

a_{n+1}

$a_{n + 1}$ we can see

≠

a_{n+1}≠a_n

$a_{n + 1} \neq = a_{n}$ and

∈

{

∈

∀

≠

}

a_{n+1}∈\{x∈X:∀m<n,x≠a_m \}

$a_{n + 1} \in {x \in X : \forall m < n, x \neq = a_{m}}$ , since

≔

min

⁡

{

∈

∀

≠

}

a_n≔\min⁡\{x∈X:∀m<n,x≠a_m \}

$a_{n} : = min {x \in X : \forall m < n, x \neq = a_{m}}$ , we can deduce

∀

a_n<a_{n+1},∀n

$a_{n} < a_{n + 1}, \forall n$ . Thus

⋯

a_0<a_1<a_2<⋯

$a_{0} < a_{1} < a_{2} < \dots$
Which means for all

≠

n≠m,a_n≠a_m

$n \neq = m, a_{n} \neq = a_{m}$ , otherwise we would get

a_n<a_n

$a_{n} < a_{n}$ or

a_m<a_m

$a_{m} < a_{m}$ . Also we know by the definition of

min

⁡

(

)

∈

∀

\min⁡(X), a_n∈X,∀n

$min (X), a_{n} \in X, \forall n$ .
Now we define

→

f:\mathbf N→X

$f : N \to X$ by

(

)

≔

f(n)≔a_n

$f (n) : = a_{n}$ , we already show

$f$ is 1-to-1. To prove

$f$ is surjective, choose

∈

x∈X

$x \in X$ , assume no

$n$ can let

(

)

f(n)=x

$f (n) = x$ , then this means

≠

∀

a_n≠x,∀n

$a_{n} \neq = x, \forall n$ , so for every

∈

{

∈

∀

≠

}

n, x∈\{x∈X:∀m<n,x≠a_m \}

$n, x \in {x \in X : \forall m < n, x \neq = a_{m}}$ , thus

≥

∀

x≥a_n,∀n

$x \geq a_{n}, \forall n$ . As

(

)

∞

(a_n )_{n=0}^∞

$(a_{n})_{n = 0}^{\infty}$ is an increasing sequence in natural numbers, we can prove by induction that

≥

a_n≥n

$a_{n} \geq n$ , thus

≥

x≥n

$x \geq n$ for every

$n$ , we can then deduce

≥

x≥x+1

$x \geq x + 1$ and get a contradiction. So

$f$ is surjective, thus bijective.
Last, we assume there’s another increasing bijective

→

≠

g:\mathbf N→X,g≠f

$g : N \to X, g \neq = f$ , then

{

∈

(

)

≠

(

)

}

\{n∈\mathbf N:g(n)≠f(n)\}

${n \in N : g (n) \neq = f (n)}$ is not empty, which means we can define

≔

min

⁡

{

∈

(

)

≠

(

)

}

m≔\min⁡\{n∈\mathbf N:g(n)≠f(n)\}

$m : = min {n \in N : g (n) \neq = f (n)}$
So

(

)

≠

(

)

g(m)≠f(m)=a_m

$g (m) \neq = f (m) = a_{m}$ , and for all

(

)

(

)

n<m,g(n)=f(n)=a_n

$n < m, g (n) = f (n) = a_{n}$ , in that case,

(

)

g(m)

$g (m)$ must equal some

∈

x∈X

$x \in X$ , since

$g$ is increasing, we have

(

)

∀

x>g(n),∀n<m

$x > g (n), \forall n < m$ , in particular:

∈

{

∈

∀

≠

}

⇒

≥

min

⁡

{

∈

∀

≠

}

x∈\{x∈X:∀n<m,x≠a_n \}⇒x≥\min⁡\{x∈X:∀n<m,x≠a_n \}=a_m

$x \in {x \in X : \forall n < m, x \neq = a_{n}} \Rightarrow x \geq min {x \in X : \forall n < m, x \neq = a_{n}} = a_{m}$
Together with

(

)

≠

g(m)≠a_m

$g (m) \neq = a_{m}$ we know that

(

)

∈

g(m)>a_m∈X

$g (m) > a_{m} \in X$ , since

$g$ is strictly increasing, this means

∉

(

)

a_m∉g(\mathbf N)

$a_{m} \in / g (N)$ , or

$g$ is not surjective, a contradiction.

Exercise 8.1.4

We define

$A$ as what Hint suggests. Consider

∣

→

(

)

f|_A:A→f(\mathbf N)

$f ∣_{A} : A \to f (N)$ , we prove

∣

f|_A

$f ∣_{A}$ is a bijection.
Let

∈

≠

m,n∈A,m≠n

$m, n \in A, m \neq = n$ , then we shall have

m<n

$m < n$ or

n<m

$n < m$ , in either case, we can deduce from the definition of

$A$ that

∣

(

)

≠

∣

(

)

f|_A (m)≠f|_A (n)

$f ∣_{A} (m) \neq = f ∣_{A} (n)$ , so

∣

f|_A

$f ∣_{A}$ is injective.
For

∀

∈

(

)

∀y∈f(\mathbf N)

$\forall y \in f (N)$ , we can have

∈

k∈\mathbf N

$k \in N$ , s.t.

(

)

y=f(k)

$y = f (k)$ . Now for all

≤

0≤n≤k

$0 \leq n \leq k$ , we can examine

(

)

f(n)

$f (n)$ recursively starting from

$0$ and choose the first

$m$ such that

(

)

f(m)=y

$f (m) = y$ . If

m=k

$m = k$ , this means

∈

k∈A

$k \in A$ . If

m<k

$m < k$ , then

∈

m∈A

$m \in A$ and

(

)

f(m)=y

$f (m) = y$ . In either case we have

∈

∣

(

)

y∈f|_A (A)

$y \in f ∣_{A} (A)$ , so

∣

f|_A

$f ∣_{A}$ is surjective.
Since

$A$ is a subset of

\mathbf N

$N$ and is at most countable,

(

)

f(\mathbf N)

$f (N)$ is at most countable.

Exercise 8.1.5

$X$ is countable, thus we can find a bijection

→

g:\mathbf N→X

$g : N \to X$ , then

∘

h=f∘g

$h = f \circ g$ is a function from

\mathbf N

$N$ to

$Y$ , thus use Proposition 8.1.8,

(

)

(

)

(

)

f(X)=f(g(\mathbf N))=h(\mathbf N)

$f (X) = f (g (N)) = h (N)$ is at most countable.

Exercise 8.1.6

If there exists an injective map

→

f:A→\mathbf N

$f : A \to N$ , then

$f$ is a bijection from

$A$ to

(

)

f(A)

$f (A)$ , since

(

)

f(A)

$f (A)$ is a subset of

\mathbf N

$N$ and is at most countable,

$A$ is at most countable as

−

f^{-1}

$f^{- 1}$ is a function from

(

)

f(A)

$f (A)$ to

$A$ and

−

(

)

A=f^{-1} (f(A))

$A = f^{- 1} (f (A))$ and use Corollary 8.1.9.
If

$A$ is at most countable, then

$A$ is finite or

$A$ is countable. Then
For the case

$A$ is finite, we can see

$A$ has cardinality

$n$ for some natural number

$n$ . We select a bijection

$g$ from

$A$ to

{

…

}

\{1,2,…,n\}

${1, 2, \dots, n}$ , and define

→

f:A→\mathbf N

$f : A \to N$ by

(

)

(

)

∀

∈

f(a)=g(a),∀a∈A

$f (a) = g (a), \forall a \in A$ , then

$f$ is injective since

$g$ is bijective.
For the case

$A$ is countable, we can find a bijection

→

f: A→\mathbf N

$f : A \to N$ , this function is injective.

Exercise 8.1.7

We use all the notations from Hint, then

→

∪

h:\mathbf N→X∪Y

$h : N \to X \cup Y$ is defined.
We need to show

(

)

∪

h(\mathbf N)=X∪Y

$h (N) = X \cup Y$ , notice first that

∀

∈

(

)

∈

∀n∈\mathbf N,h(n)∈X

$\forall n \in N, h (n) \in X$ if

$n$ is even, and

(

)

∈

h(n)∈Y

$h (n) \in Y$ if

$n$ is odd, thus

(

)

∈

∪

∀

∈

h(n)∈X∪Y,∀n∈\mathbf N

$h (n) \in X \cup Y, \forall n \in N$ , or

(

)

⊆

∪

h(\mathbf N)⊆X∪Y

$h (N) \subseteq X \cup Y$ .
Next, we have

(

∈

∪

)

⇒

(

∈

)

∨

(

∈

)

⇒

(

)

∨

(

)

⇒

(

)

∨

(

)

⇒

(

∈

(

)

\begin{aligned}(x∈X∪Y)&⇒(x∈X)∨(x∈Y)⇒(x=f(m))∨(x=g(n))\\&⇒(x=h(2m))∨(x=h(2n+1))⇒(x∈h(\mathbf N))\end{aligned}

$(x \in X \cup Y) \Rightarrow (x \in X) \lor (x \in Y) \Rightarrow (x = f (m)) \lor (x = g (n)) \Rightarrow (x = h (2 m)) \lor (x = h (2 n + 1)) \Rightarrow (x \in h (N))$
Thus

∪

⊆

(

)

X∪Y⊆h(\mathbf N)

$X \cup Y \subseteq h (N)$ . Then we can have

∪

(

)

X∪Y=h(\mathbf N)

$X \cup Y = h (N)$
By Corollary 8.1.9 we know that

∪

X∪Y

$X \cup Y$ is at most countable, we remains to show

∪

X∪Y

$X \cup Y$ is not finite. Assume so, then

∪

X∪Y

$X \cup Y$ has cardinality

$n$ for some natural number

$n$ . So

⊆

∪

⇒

(

)

≤

X⊆X∪Y ⇒ \#(X)≤n

$X \subseteq X \cup Y \Rightarrow # (X) \leq n$
But if we let

{

…

}

S=\{1,2,…,n+1\}

$S = {1, 2, \dots, n + 1}$ , then

→

(

)

f:S→f(S)

$f : S \to f (S)$ is a bijection, thus

(

)

f(S)

$f (S)$ has cardinality

n+1

$n + 1$ , however,

(

)

⊆

f(S)⊆X

$f (S) \subseteq X$ , so

(

)

≥

\#(X)≥n+1

$# (X) \geq n + 1$ , this is a contradiction.

Exercise 8.1.8

By hypothesis, we have a bijection

→

f:\mathbf N→X

$f : N \to X$ and a bijection

→

g:\mathbf N→Y

$g : N \to Y$ . We define a function

→

h:\mathbf N×\mathbf N→X×Y

$h : N \times N \to X \times Y$ by

(

)

(

)

(

)

h(m,n)=(f(m),g(n))

$h (m, n) = (f (m), g (n))$ , then we have

(

)

(

′

)

⇒

(

)

(

′

)

∧

(

)

(

′

)

⇒

(

′

)

∧

(

′

)

⇒

(

)

(

′

)

\begin{aligned}(h(m,n)=h(m’,n’))&⇒((f(m)=f(m’ ))∧(g(n)=g(n’ )))\\&⇒(m=m’ )∧(n=n’ )⇒(m,n)=(m’,n’ )\end{aligned}

$(h (m, n) = h (m^{'}, n^{'})) \Rightarrow ((f (m) = f (m^{'})) \land (g (n) = g (n^{'}))) \Rightarrow (m = m^{'}) \land (n = n^{'}) \Rightarrow (m, n) = (m^{'}, n^{'})$
Thus

$h$ is injective. And for

∀

(

)

∈

∀(x,y)∈X×Y

$\forall (x, y) \in X \times Y$ , we have

∈

x∈X

$x \in X$ and

∈

y∈Y

$y \in Y$ , so

∃

∈

∃m,n∈N

$\exists m, n \in N$ , s.t.

(

)

(

)

f(m)=x,g(n)=y

$f (m) = x, g (n) = y$ , thus

(

)

(

)

h(m,n)=(x,y)

$h (m, n) = (x, y)$ , which means

$h$ is surjective.
Now that

$h$ is a bijection, from Corollary 8.1.13 we know

X×Y

$X \times Y$ is countable.

Exercise 8.1.9

$I$ is finite and

A_α

$A_{α}$ is finite for every

$α$ , then the statement is obvious.
Now suppose

$I$ is countable, then we have a bijection

→

g:\mathbf N→I

$g : N \to I$ , for each

∈

α∈I

$α \in I$ such that

≠

∅

A_α≠∅

$A_{α} \neq = \emptyset$ , we assign a function

→

f_α:\mathbf N→A_α

$f_{α} : N \to A_{α}$ as follows:
If

A_α

$A_{α}$ is finite, then it’s possible to write

{

…

}

∈

A_α=\{x_1,…,x_n \},n∈\mathbf N

$A_{α} = {x_{1}, \dots, x_{n}}, n \in N$ , then let

(

)

{

≤

f_α (i)=\begin{cases}x_i,&1≤i≤n\\x_1,&i>n\end{cases}

$f_{α} (i) = {x_{i}, x_{1}, 1 \leq i \leq n i > n$
If

A_α

$A_{α}$ is infinite or countable, then we can find a bijection

′

→

f’:\mathbf N→A_α

$f^{'} : N \to A_{α}$ , we let

′

f_α=f’

$f_{α} = f^{'}$ .
So as long as

A_α

$A_{α}$ is nonempty, we have assigned a function

→

f_α:\mathbf N→A_α

$f_{α} : N \to A_{α}$ . Next, if denote

{

∈

≠

∅

}

J=\{α∈I:A_α≠∅\}

$J = {α \in I : A_{α} \neq = \emptyset}$ we define

→

⋃

∈

h: N×N→⋃_{α∈J}A_α

$h : N \times N \to ⋃_{α \in J} A_{α}$ as follows:

(

)

(

)

(

)

h(m,n)=f_{g(m)} (n)

$h (m, n) = f_{g (m)} (n)$
Then use Corollary 8.1.13 and Corollary 8.1.9, we can see that

⋃

∈

⋃_{α∈J}A_α

$⋃_{α \in J} A_{α}$ is at most countable. The final statement results from

⋃

∈

⋃

∈

⋃_{α∈J}A_α =⋃_{α∈I}A_α

$⋃_{α \in J} A_{α} = ⋃_{α \in I} A_{α}$ .

Exercise 8.1.10

We define

{

∣

≤

∈

}

Q_n=\{a/b:|a|+b≤n,a∈\mathbf Z,b∈\mathbf N^+\}

$Q_{n} = {a / b : ∣ a ∣ + b \leq n, a \in Z, b \in N^{+}}$ for every

∈

n∈\mathbf N

$n \in N$ , then

∅

{

}

{

−

}

Q_0=∅,Q_1=\{0\}, Q_2=\{0,1,-1\}

$Q_{0} = \emptyset, Q_{1} = {0}, Q_{2} = {0, 1, - 1}$ , etc. Each

Q_n

$Q_{n}$ is finite, thus have cardinality

c_n

$c_{n}$ , we can see

c_0=0,c_1=1,c_2=3

$c_{0} = 0, c_{1} = 1, c_{2} = 3$ , etc. As

⊆

Q_n⊆Q_{n+1}

$Q_{n} \subseteq Q_{n + 1}$ , we define

(

⋃

−

)

∈

S_n=Q_n\backslash \left(⋃_{i=0}^{n-1}Q_i \right),\quad n∈\mathbf N^+

$Sn=Qn\(i=0⋃n−1Qi),n∈N+$
since

−

∈

\frac{1}{n-1}∈Q_n

$\frac{1}{n - 1} \in Q_{n}$ , but

−

∉

⋃

−

\frac{1}{n-1}∉⋃_{i=0}^{n-1}Q_i

$\frac{1}{n - 1} \in / ⋃_{i = 0}^{n - 1} Q_{i}$ , each

S_n

$S_{n}$ is nonempty for

n>1

$n > 1$ , and

{

}

S_1=Q_1=\{0\}

$S_{1} = Q_{1} = {0}$ , so each

S_n

$S_{n}$ is nonempty for

∈

n∈\mathbf N^+

$n \in N^{+}$ .
Now we define

−

q_n=c_n-c_{n-1}

$q_{n} = c_{n} - c_{n - 1}$ , then each

S_n

$S_{n}$ has cardinality

q_n

$q_{n}$ , so

(

)

∞

(q_n )_{n=1}^∞

$(q_{n})_{n = 1}^{\infty}$ is a positive increasing sequence.
We define a function

→

f:\mathbf N→\mathbf Q

$f : N \to Q$ inductively as follows:
Start from

$0$ , we let

(

)

f_1 (0)=0

$f_{1} (0) = 0$ , which is in fact a bijection from

{

∈

≤

}

\{i∈\mathbf N:c_0≤i<c_1\}

${i \in N : c_{0} \leq i < c_{1}}$ to

S_1

$S_{1}$ .
Then for any

S_n

$S_{n}$ , we can find a bijection

{

∈

≤

}

→

g_n:\{i∈\mathbf N:0≤i<q_n \}→S_n

$g_{n} : {i \in N : 0 \leq i < q_{n}} \to S_{n}$ , we then define

(

)

(

−

)

−

≤

f_n (i)=g_n (i-c_{n-1} ),\quad c_{n-1}≤i<c_n

$f_{n} (i) = g_{n} (i - c_{n - 1}), c_{n - 1} \leq i < c_{n}$
So we know

f_n

$f_{n}$ is a bijection from

{

∈

−

≤

}

\{i∈\mathbf N:c_{n-1}≤i<c_n\}

${i \in N : c_{n - 1} \leq i < c_{n}}$ to

S_n

$S_{n}$ .
Finally we can define

(

)

(

)

f(k)=f_n (k)

$f (k) = f_{n} (k)$ , in which

$n$ is the positive natural number which satisfies

−

≤

c_{n-1}≤k<c_n

$c_{n - 1} \leq k < c_{n}$ , since if

≠

m≠n

$m \neq = n$ , we have

{

∈

−

≤

}

∩

{

∈

−

≤

}

∅

\{i∈\mathbf N:c_{n-1}≤i<c_n \}∩\{i∈\mathbf N:c_{m-1}≤i<c_m \}=∅

${i \in N : c_{n - 1} \leq i < c_{n}} \cap {i \in N : c_{m - 1} \leq i < c_{m}} = \emptyset$
Since any for natural number

$n$ , we have

{

…

−

}

⊆

\{0,1,…,n-1\}⊆Q_n

${0, 1, \dots, n - 1} \subseteq Q_{n}$ , so

≥

c_n≥n

$c_{n} \geq n$ , which means

⋃

∞

{

∈

−

≤

}

⋃_{n=1}^∞\{i∈\mathbf N:c_{n-1}≤i<c_n\}=\mathbf N

$n = 1 ⋃ \infty {i \in N : c_{n - 1} \leq i < c_{n}} = N$
Thus

$f$ is truly a function defined on

\mathbf N

$N$ , and the injectivity of

$f$ is clear from the induction process, as we form

$f$ from a sequence of bijective functions on disjoint subsets of

\mathbf N

$N$ . We only remains to show

$f$ is surjective. Choose any

∈

q∈\mathbf Q

$q \in Q$ , then we can write

q=a/b

$q = a / b$ , in which

$b$ is a positive natural number, we can be sure that

∈

∣

q∈Q_{|a|+b}

$q \in Q_{∣ a ∣ + b}$ , so

∈

≤

∣

q∈S_m,\quad 0<m≤|a|+b

$q \in S_{m}, 0 < m \leq ∣ a ∣ + b$
Thus there’s a number

∈

{

∈

−

≤

}

k∈\{i∈\mathbf N:c_{m-1}≤i<c_m\}

$k \in {i \in N : c_{m - 1} \leq i < c_{m}}$ such that

(

)

(

)

f_m (k)=f(k)=q

$f_{m} (k) = f (k) = q$ . This means

$f$ is surjective.

原文链接：https://blog.csdn.net/christangdt/article/details/102583595

Exercise 8.1.1

Exercise 8.1.2

Exercise 8.1.3

Exercise 8.1.4

Exercise 8.1.5

Exercise 8.1.6

Exercise 8.1.7

Exercise 8.1.8

Exercise 8.1.9

Exercise 8.1.10

你可能也喜欢