※ 引述《yueayase (scrya)》之铭言： : 看到这个叙述,我想到我看到得一题极限证明题: : 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]. : 尝试去做证明: : Proof: : Assume there is a irrational number x' in [a,b] such that f(x') = a ≠ 0 : Then, we want to show that lim f(x) = 0 ? : x->x' : or want to show that f is not continuous? : 到底该怎麽取? Since f is continuous on [a,b] then for all x in (R\Q)∩[a,b] implies lim f(r) = f(x) r→x make a sequence{r_n} which r_n→x as n→∞ each r_n is in Q for n in N then lim f(r_n) = lim f(r_n) = f(x) n→∞ r_n→x but each f(r_n)=0, hence f(x)=0 for all x in (R\Q)∩[a,b] so f(x)=0 for all x in [a,b] --

※ 发信站: 批踢踢实业坊(ptt.cc) ◆ From: 111.248.14.130 ※ 编辑: Madroach 来自: 111.248.14.130 (01/16 03:19)