推 pobm :推用心 10/21 16:37
推 RaichueRen :感謝你的詳細解說! 10/21 22:25
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推 Lindemann :推好文 10/22 18:05
※ 引述《RaichueRen (雷丘仁)》之銘言：
: 要怎麼尋找一個[0,1]=[0,1]x[0,1]x[0,1]...(可數無窮個)
: 的1-1 onto 對應呢?(x是狄卡爾積)
: _
: 怎麼弄都會卡在實數展開成十位數的表達唯一性(0.9=1)
: 雖然似乎可以用Cantor–Bernstein–Schroeder定理避開
: 但是這樣似乎只是宣稱1-1 onto map存在
: 而沒有構造出來
: 謝謝!
提示（可能有錯誤）：
Step 1 （二進位表示＋可數大風吹）.
Ｎ
Find a bijection f : [0, 1] → {0, 1} as follows:
For x in [0, 1), say
(1) (2) (3) (4) (5)
0.x x x x x ...
is the unique binary expansion of x that does not end in
repeating 1s, and define
(1) (2)
b(x) = (x , x , ...).
Ｎ
Then b : [0, 1) → {0, 1} is one-to-one, and
Ｎ
S = {0, 1} ＼ b([0, 1)),
the set of sequences on {0, 1} ending in repeating 1s,
is countable, so enumerate S as (s ).
n
Define
e = (0, 0, ..., 0, 1 , 0, 0, ...)
i
where the nonzero element occurs in the ith position.
Define for x in [0, 1]
-i
╭ b(x) if x ≠ 2 for all i ≧ 0,
│ -i
f(x) = ┤ e if x ＝ 2 for some even i ≧ 0,
│ 1+i/2 -i
╰ s if x ＝ 2 for some odd i ＞ 0.
(i+1)/2
Step 2 （百川匯流）.
Ｎ×Ｎ Ｎ
Find a bijection g : {0, 1} → {0, 1} as follows:
Ｎ×Ｎ
For (x : n, m in Ｎ) in {0, 1} , define
n,m
g((x : n, m in Ｎ))
n,m
= (x , x , x , x , x , x , ...)
1,1 1,2 2,1 1,3 2,2 3,1
Step 3 （三招齊發）.
Ｎ Ｎ
Define F : [0, 1] → {0, 1} as
F((x : n in Ｎ)) = (f(x ) : n in Ｎ).
n n
Then
-1
h = f 。 g 。 F
Ｎ
is a bijection from [0, 1] to [0, 1].
Ｎ
╭─ [0, 1]
│
│ │
│ │ F
│ ↓
│ ╭ Ｎ╮Ｎ Ｎ×Ｎ
│ │{0, 1} │ ＝ {0, 1}
│ ╰ ╯
│ │
h │ │ g
│ ↓
│ Ｎ
│ {0, 1}
│
│ │ -1
│ │ f
│ ↓
│
╰→ [0, 1]
--
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◆ From: 75.62.128.123
※ 編輯: cgkm 來自: 75.62.128.123 (10/21 12:13)
※ 編輯: cgkm 來自: 75.62.128.123 (10/21 12:13)
※ 編輯: cgkm 來自: 75.62.128.123 (10/21 12:29)