2-7. 計數問題
Theorem of Finite Set Relation
∣A∣ = m < ∞, ∣B∣ = n < ∞,
- f: A→B 1-1 ⇒m≤n
- f: A→B onto ⇒m≥n
- f: A→B 1-1 且 onto ⇒m = n
Define Cardinality
A, B: sets, 若 ∃f: A→B 為 1-1 且 onto, 稱 A 與 B 具相同之基數(cardinality): ∣A∣ = ∣B∣, 記作 A∼B
∼ 為一等價關係
Define Finite and Infinite Set
A: set, A = ∅ 或 ∃n∈Z+ s.t. A∼ { 1, 2, ..., n }, 稱 A 為有限集(finite set) ↔A≠∅ 且 ∄n∈Z+ s.t. A∼ { 1, 2, ..., n } 稱 A 為無限集(infinite set)
Define Countable and Uncountable Set
A: set, A 為有限集或 A∼Z+, 則稱 A 為可數集(countable set) ↔A 為無限集且 A≁Z+, 稱 A 為不可數集(uncountable set)
Z+ 為無限可數集(countably infinite set)
Theorem of Z is countable
Z 為可數集
proof:
定義 f: Z→Z+ by f(x) =⎩⎪⎨⎪⎧12x−2x+1, if x=0, if x>0, if x<0
x |
... |
-2 |
-1 |
0 |
1 |
2 |
... |
f(x) |
... |
5 |
3 |
1 |
2 |
4 |
... |
⇒f(x): 1-1 且 onto ∴Z∼Z+
Properties of Countable and Uncountable Set
- 有限集 ⇒ 可數集
可數集 ⇏ 有限集, ex.Z+
- 不可數集 ⇒ 無限集
無限集 ⇏ 不可數集, ex.Z+
- A, B: sets, A⊆B
- B is countable ⇒A is countable.
- B is uncountable ⇒A is uncountable.
Corollary of Countably Infinite Set
A: infinite set, ∃f: A→Z+ 1-1 ⇒A is countable.
Theorem of Positive Z Product
Countable
Z+×Z+ is countable.
proof:
定義 f: Z+×Z+→Z+ by f(a, b) = 2a3b
∵ 質因數分解必唯一 ∴f is 1-1
By Corollary of Countably Infinite Set, f: Z+×Z+ is countable.
Same Cardinality with Positive Z
Z+×Z+∼Z+
proof:
定義 f: Z+×Z+→Z+ by f(i, j) = (k=1∑i+j−2k) + j = 2(i+j−2)(i+j−1) + j
⇒ 2 維陣列 → 1 維陣列,依對角線做編號
(i, j) |
(1, 1) |
(2, 1) |
(1, 2) |
(3, 1) |
(2, 2) |
(1, 3) |
... |
f(i, j) |
1 |
2 |
3 |
4 |
5 |
6 |
... |
⇒f: 1-1 且 onto ∴Z+×Z+∼Z+
Corollary of Positive Z Product
- A, B: countable ⇒A×B: countable.
- Ai is countable, ∀i = 1, 2, ... ⇒i=1⋃∞Ai is countable.
- Q+ = { pq | p, q∈Z+ } ⊆X = { pq | p, q∈Z+, 不考慮約分 } ∼ { (p, q) | p, q∈Z+ } = Z+×Z+∼Z+ is countable ⇒Q+ is countable.
所有正有理數所成集合 Q+ 為可數集
- Q = Q+∪Q−∪ { 0 } is countable.
所有有理數所成集合 Q 為可數集
Theorem of Zero-one Interval
[0, 1] 為不可數集
proof:
設 [0, 1] is countable ⇒Z+∼ [0, 1] ⇒∃f: Z+→ [0, 1] 1-1 且 onto
令 f(i) = ri = 0.ai1ai2..., ∀i = 1, 2, ...
r1 = 0.a11a12a13…
r2 = 0.a21a22a23…
r1 = 0.a31a32a33…
⋮⋮⋮⋮⋱
取 X = 0.x1x2x3…, 其中 xi={54, if aii=4, if aii≠4, ∀i = 1, 2, ..., 則 X∈ [0, 1]
但 X≠f(i), ∀i = 1, 2, ... ⇒ 與 f 為 onto 矛盾 ∴ [0, 1] is uncountable.
Theorem of R and Zero-one Interval
R 為不可數集且 [0, 1] ∼R
proof:
→ 用漸近線
定義 f: [0, 1] → [-2π, 2π] 為 f(x) = πx - 2π
[-2π, 2π]: y = tanx
又定義 g: [-2π, 2π] →R 為 g(x) = tanx, 則 f, g: 1-1 且 onto
取 h: [0, 1] →R 為 h(x) = (g ∘ f)(x) = tan(πx - 2π), 則 h 亦為 1-1 且 onto
∴ [0, 1] ∼R
Comparison of All Numbers
- A is uncountable, B is countable ⇒A - B is uncountable.
∵A(uncountable) = (A - B) ∪ (A∩B)(countable) ∴A - B is uncountable.
- Q = R(uncountable) - Q(countable) ⇒ uncountable
- C∼R2∼R⇒ uncountable
C∼R2: a + bi⇒ (a, b)
→∣Z+∣ = ∣Z∣ = ∣Q∣ < ∣Q∣ = ∣R∣ = ∣C∣
- countable ← < → uncountable
- ∣P(Z+)∣ = ∣R∣