Re: [微積] 一題高微證明

看板Math作者 (QQ)時間13年前 (2012/03/12 22:23), 編輯推噓1(1推 0噓 0→)
留言1則, 1人參與, 最新討論串3/5 (看更多)
Denote f([a,b]) = Imf Since [a,b] is compact, f is continuous by extreme value theorem, there exists c,d€[a,b] s.t. f(c)=<f(x)<=f(d) , for all x€[a,b] ---(*) so Imf⊆[f(c),f(d)] Since [a,b] is connected, f is continuous we have Imf is connected, too. Since from (*) we know f(c),f(d)€Imf so [f(c),f(d)] ⊆ Imf (if not, there exists two open sets seperate Imf) -- ※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 114.25.179.245
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03/12 22:38, , 1F
感謝Z大!!
03/12 22:38, 1F
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文章代碼(AID): #1FNWRFky (Math)