推 stberry:不好意思 我不太懂為什麼把二拿出來就會讓表法唯一@@ 12/17 20:58
推 stberry:或者我原來的那個函數有問題嗎？ 12/17 21:00
→ stberry:(我的想法看起來是有某個Σαn/3^n不屬於cantor set@@) 12/17 21:01
→ stberry:麻煩您了@@ 12/17 21:01
→ stberry:(Onto 應該只是大於 而沒有等於？) 12/17 21:02
※ 引述《stberry (鮮紅)》之銘言：
: "The function from C(原文已經認為Σαn/3^n就是cantor set,我先假裝不知道)
: to [0,1] is defined by taking the numeral that does
: consist entirely of 0s and 2s, replacing all the 2s by 1s, and interpreting
: the sequence as a binary representation of a real number. In a formula,
: f(Σαn/3^n)=Σ(αk/2)/2^n
: For any number y in [0,1], its binary representation can be translated into a
: ternary representation of a number x in C by replacing all the 1s by 2s. With
: this, f(x) = y so that y is in the range of f. For instance if y = 3/5 =
: 0.100110011001...2, we write x = 0.200220022002...3 = 7/10. Consequently f is
: surjective; however, f is not injective — interestingly enough, the values
: for which f(x) coincides are those at opposing ends of one of the middle
: thirds removed. For instance, 7/9 = 0.2022222...3 and 8/9 = 0.2200000...3 so
: f(7/9) = 0.101111...2 = 0.112 = f(8/9)."
: 我的想法是 既然這邊說f:Σαn/3^n→Σ(αk/2)/2^n是onto
: 那麼就代表Σαn/3^n的元素個數大於Σ(αk/2)/2^n(即[0,1]的個數)
: 但cantor set又屬於[0,1] 所以元素個數必小於等於[0,1]
: 那這樣豈不代表cantor set的個數小於Σαn/3^n嗎？為什麼兩者會相等？
: 麻煩了@@
你得到的應該是相等：因 #C ≧ #[0,1], 且 #[0,1] ≧#C.
所以，#C ＝ #[0,1] ＝ χ_1.
--(以下是我個人習慣)
我想應該是這樣：命 C 表示標準的 Cantor set in [0,1]. 於是利用三進位表示法，
∞ a_k
x in C ＜＝＞ x ＝ 2 Σ -----, where a_k ＝ 0, 1. (要求表法唯一).
k＝0 3^k
我們定義:
φ: C → [0,1]
∞ a_k
x |→ Σ -----
k＝0 2^k
則 φ(C)＝[0,1].
[注意：在我們要求底下（跟你上述不同)，我們會有 1-1 and onto].
於是， #C (＝ 集合 C 的個數) ＝ #[0,1] ＝ χ_1.
NOTE. 我有寫過關於 Cantor set 的 BBS 文，如果需要，我在找出來～
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