Re: [微積] 極限 & 連續

看板Math作者Madroach (小李)時間13年前 (2011/01/16 03:02)推噓3(3推 0噓 2→)

留言5則, 3人參與討論串4/4 (看更多)

※ 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]. -- ※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 111.248.14.130 ※ 編輯: Madroach 來自: 111.248.14.130 (01/16 03:17)

推

yueayase

01/16 04:04, , 1^F

01/16 04:04, 1^F

→

Madroach

01/16 14:47, , 2^F

01/16 14:47, 2^F

推

isaac7910

01/17 22:43, , 3^F

01/17 22:43, 3^F