→ kyoiku : 謝啦，後來有用圖形畫出反例不過寫不出來 @@ 01/14 16:12
The proposition is trivially true if e >= 2.
Assume that 0<e<2 is fixed,
choose the sequence {r_n} of rational numbers in [0,1] such that
r_1 = 0,
r_2 < e/4,
r_3 < e/8, ... ,
r_k < e/(2^k) for some k depending on e to be fixed later.
Note that I_2, I_3, ..., I_k are subsets of I_1, and all these sets are
bounded above by e/2 < 1.
The length of I_n(e) is equal to 1/2^(n-1), so the infinite sum of the
lengths of I_(k+1), I_(k+2), ... is 1/2^(k-1).
Now choose k such that 1/2^(k-1) < 1 - e/2.
Suppose [0,1] is covered by {I_n(e)}, by compactness, we can take a finite
subcover {I_1, I_2, ..., I_m}.
If m <= k, then clearly 1 is not in the union.
If m > k, then the sum of the lengths of I_(k+1), ..., I_m is less than
1 - e/2, impossible to cover [e/2,1].
So the proposition may not hold if e<2.
※ 引述《kyoiku (生死間有大恐怖)》之銘言：
: [0,1] 上的有理點可列，令其為 {r_1,r_2,r_3,...}。
: 給定 e>0，令
: I_n(e) = (r_n - e/2^n , r_n + e/2^n)
: B(e) = the union of all open interval I_n(e) for n = 1,2,3,...
: 則 B(e) 包含 [0,1] for every e>0
: 上面的推論是否正確，是的話請證明之，否則給反例。
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