2-7. 計數問題

Theorem of Finite Set Relation

∣A∣ |A| = m m < ∞ \infty , ∣B∣ |B| = n n < ∞ \infty ,

  • f f : A→B A \rightarrow B 1-1 ⇒m≤n \Rightarrow m \le n
  • f f : A→B A \rightarrow B onto ⇒m≥n \Rightarrow m \ge n
  • f f : A→B A \rightarrow B 1-1 且 onto ⇒m \Rightarrow m = n n

Define Cardinality

A A , B B : sets, 若 ∃f \exists f : A→B A \rightarrow B 為 1-1 且 onto, 稱 A A 與 B B 具相同之基數(cardinality): ∣A∣ |A| = ∣B∣ |B| , 記作 A∼B A \sim B

∼ \sim 為一等價關係

Define Finite and Infinite Set

A A : set, A A = ∅ \emptyset 或 ∃n∈Z+ \exists n \in Z^+ s.t. A∼ A \sim { 1 1 , 2 2 , ..., n n }, 稱 A A 為有限集(finite set) ↔A≠∅ \leftrightarrow A \ne \emptyset 且 ∄n∈Z+ \nexists n \in Z^+ s.t. A∼ A \sim { 1 1 , 2 2 , ..., n n } 稱 A A 為無限集(infinite set)

Define Countable and Uncountable Set

A A : set, A A 為有限集或 A∼Z+ A \sim Z^+ , 則稱 A A 為可數集(countable set) ↔A \leftrightarrow A 為無限集且 A≁Z+ A \nsim Z^+ , 稱 A A 為不可數集(uncountable set)

Z+ Z^+ 為無限可數集(countably infinite set)

Theorem of Z is countable

Z Z 為可數集

proof:

定義 f f : Z→Z+ Z \rightarrow Z^+ by f f (x x ) ={1, if x=02x, if x>0−2x+1, if x<0 = \begin{cases} 1 &\text{, if } x = 0 \\ 2x &\text{, if } x > 0 \\ -2x + 1 &\text{, if } x < 0 \end{cases}

x x ... -2 -1 0 1 2 ...
f f (x x ) ... 5 3 1 2 4 ...

⇒f \Rightarrow f (x x ): 1-1 且 onto ∴Z∼Z+ \therefore Z \sim Z^+

Properties of Countable and Uncountable Set

  • 有限集 ⇒ \Rightarrow 可數集

    可數集 ⇏ \nRightarrow 有限集, ex.Z+ ^{ex.} Z^+

  • 不可數集 ⇒ \Rightarrow 無限集

    無限集 ⇏ \nRightarrow 不可數集, ex.Z+ ^{ex.} Z^+

  • A A , B B : sets, A⊆B A \subseteq B
    • B B is countable ⇒A \Rightarrow A is countable.
    • B B is uncountable ⇒A \Rightarrow A is uncountable.

Corollary of Countably Infinite Set

A A : infinite set, ∃f \exists f : A→Z+ A \rightarrow Z^+ 1-1 ⇒A \Rightarrow A is countable.

Theorem of Positive Z Product

Countable

Z+×Z+ Z^+ \times Z^+ is countable.

proof:

定義 f f : Z+×Z+→Z+ Z^+ \times Z^+ \rightarrow Z^+ by f f (a a , b b ) = 2a3b 2^a 3^b
∵ \because 質因數分解必唯一 ∴f \therefore f is 1-1
By Corollary of Countably Infinite Set, f f : Z+×Z+ Z^+ \times Z^+ is countable.

Same Cardinality with Positive Z

Z+×Z+∼Z+ Z^+ \times Z^+ \sim Z^+

proof:

定義 f f : Z+×Z+→Z+ Z^+ \times Z^+ \rightarrow Z^+ by f f (i i , j j ) = (∑k=1i+j−2k \displaystyle\sum_{k=1}^{i + j - 2} k ) + j j = (i+j−2)(i+j−1)2 \dfrac{(i + j - 2)(i + j - 1)}{2} + j j
⇒ \Rightarrow 2 維陣列 → \rightarrow 1 維陣列,依對角線做編號

(i i , j j ) (1, 1) (2, 1) (1, 2) (3, 1) (2, 2) (1, 3) ...
f f (i i , j j ) 1 2 3 4 5 6 ...

⇒f \Rightarrow f : 1-1 且 onto ∴Z+×Z+∼Z+ \therefore Z^+ \times Z^+ \sim Z^+

Corollary of Positive Z Product

  • A A , B B : countable ⇒A×B \Rightarrow A \times B : countable.
  • Ai A_i is countable, ∀i \forall i = 1 1 , 2 2 , ... ⇒⋃i=1∞Ai \Rightarrow \displaystyle\bigcup_{i=1}^\infty A_i is countable.
  • Q+ Q^+ = { qp \dfrac{q}{p} | p p , q∈Z+ q \in Z^+ } ⊆X \subseteq X = { qp \dfrac{q}{p} | p p , q∈Z+ q \in Z^+ , 不考慮約分 } ∼ \sim { (p p , q q ) | p p , q∈Z+ q \in Z^+ } = Z+×Z+∼Z+ Z^+ \times Z^+ \sim Z^+ is countable ⇒Q+ \Rightarrow Q^+ is countable.

    所有正有理數所成集合 Q+ Q^+ 為可數集

  • Q Q = Q+∪Q−∪ Q^+ \cup Q^- \cup { 0 0 } is countable.

    所有有理數所成集合 Q Q 為可數集

Theorem of Zero-one Interval

[0 0 , 1 1 ] 為不可數集

proof:

設 [0 0 , 1 1 ] is countable ⇒Z+∼ \Rightarrow Z^+ \sim [0 0 , 1 1 ] ⇒∃f \Rightarrow \exists f : Z+→ Z^+ \rightarrow [0 0 , 1 1 ] 1-1 且 onto
令 f f (i i ) = ri r_i = 0.ai1ai2... 0.{a_{i1}}{a_{i2}}... , ∀i \forall i = 1 1 , 2 2 , ...

r1 r_1 = 0.a11a12a13… 0.{\color{red}{a_{11}}}{a_{12}}{a_{13}} \ldots
r2 r_2 = 0.a21a22a23… 0.{a_{21}}{\color{red}{a_{22}}}{a_{23}} \ldots
r1 r_1 = 0.a31a32a33… 0.{a_{31}}{a_{32}}{\color{red}{a_{33}}} \ldots
⋮⋮⋮⋮⋱ \; \vdots \qquad \, \enspace \vdots \quad \vdots \quad \vdots \, \ddots

取 X X = 0.x1x2x3… 0.{x_1}{x_2}{x_3} \ldots , 其中 xi={5, if aii=44, if aii≠4 x_i = \begin{cases} 5 &\text{, if } {\color{red}{a_{ii}}} = 4 \\ 4 &\text{, if } {\color{red}{a_{ii}}} \ne 4 \end{cases} , ∀i \forall i = 1 1 , 2 2 , ..., 則 X∈ X \in [0 0 , 1 1 ]
但 X≠f X \ne f (i i ), ∀i \forall i = 1 1 , 2 2 , ... ⇒ \Rightarrow 與 f f 為 onto 矛盾 ∴ \therefore [0 0 , 1 1 ] is uncountable.

Theorem of R and Zero-one Interval

R R 為不可數集且 [0 0 , 1 1 ] ∼R \sim R

proof:

→ \rightarrow 用漸近線
定義 f f : [0 0 , 1 1 ] → \rightarrow [-π2 \dfrac{\pi}{2} , π2 \dfrac{\pi}{2} ] 為 f f (x x ) = πx \pi{x} - π2 \dfrac{\pi}{2}

[-π2 \dfrac{\pi}{2} , π2 \dfrac{\pi}{2} ]: y y = tanx \tan{x}

又定義 g g : [-π2 \dfrac{\pi}{2} , π2 \dfrac{\pi}{2} ] →R \rightarrow R 為 g g (x x ) = tanx \tan{x} , 則 f f , g g : 1-1 且 onto
取 h h : [0 0 , 1 1 ] →R \rightarrow R 為 h h (x x ) = (g g ∘ \tiny{\circ} f f )(x x ) = tan \tan (πx \pi{x} - π2 \dfrac{\pi}{2} ), 則 h h 亦為 1-1 且 onto
∴ \therefore [0 0 , 1 1 ] ∼R \sim R

Comparison of All Numbers

  • A A is uncountable, B B is countable ⇒A \Rightarrow A - B B is uncountable.

    ∵A \because A (uncountable) = (A A - B B ) ∪ \cup (A∩B A \cap B )(countable) ∴A \therefore A - B B is uncountable.

  • Q‾ \overline{Q} = R R (uncountable) - Q Q (countable) ⇒ \Rightarrow uncountable
  • C∼R2∼R⇒ C \sim R^2 \sim R \Rightarrow uncountable

    C∼R2 C \sim R^2 : a a + bi⇒ bi \Rightarrow (a a , b b )

→∣Z+∣ \rightarrow \Large{|Z^+|} = ∣Z∣ \Large{|Z|} = ∣Q∣ \Large{|Q|} < \Large{\color{red}{<}} ∣Q‾∣ \Large{|\overline{Q}|} = ∣R∣ \Large{|R|} = ∣C∣ \Large{|C|}

  • countable ← \leftarrow < \Large{\color{red}{<}} → \rightarrow uncountable
  • ∣P(Z+)∣ |P(Z^+)| = ∣R∣ |R|
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