推 yueayase :所以重點是我們選擇用來逼近的數,是有理數吧 01/16 04:04
→ Madroach :對 01/16 14:47
推 isaac7910 :萬能的李神現身為我們這些平民解惑了 01/17 22:43
→ Madroach := =你怎麼會來這 01/18 01:02
推 isaac7910 :對不起我要來之前應該要先跟你報備的 是我的不好 01/20 23:35
※ 引述《yueayase (scrya)》之銘言：
: 看到這個敘述,我想到我看到得一題極限證明題:
: Let f be continous on [a,b] and let f(x) = 0 when x is rational.
: Prove that f(x) = 0 for every x in [a,b].
Claim: f(x) = 0 for all x in [a,b]
￣￣￣
Suppose not, there exists a number p in [a,b]
such that f(p) = m ≠ 0
Since f is continuous on [a,b], f is also continuous at p
By the definition of continuity: for all ε>0, there exists
δ>0 such that whenever |x-p|<δ implies |f(x)-f(p)|<ε.
Now choose a number ε', 0 <ε'< m, then for this ε', there
exists a number, say δ', such that whenever |x-p|<δ' implies
|f(x)-f(p)|<ε'< m. If the number x that we choose is a
rational number close to p in [a,b], then we have |x-p|<δ'
implies | 0-f(p)|= m <ε'< m, which is a contradiction.
Hence f(x) must equal to 0 for all x in [a,b].
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◆ From: 111.248.14.130
※ 編輯: Madroach 來自: 111.248.14.130 (01/16 03:17)