先謝謝你^^
不過我還是有個疑問 也就是證明Σαn/3^n 屬於cantor set的問題
我有點不懂為什麼這麼"clear"
又例如我在wiki看到的一段話:
"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@@
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