Re: [高微] 高微證明

看板Math作者 (QQ)時間14年前 (2011/11/25 03:00), 編輯推噓1(100)
留言1則, 1人參與, 最新討論串2/2 (看更多)
※ 引述《TampaBayRays (坦帕灣光芒)》之銘言: : 1.Prove that every infinite set of E has a limit point in E. : 2.Prove that every sequence in E contains a subsequence which converges in E. : 請各位高手幫幫我 : 謝謝 應該是要證1.和2.是equivalent吧 (1) => (2): Let a_n is a sequence in E Case1.:Let S={a_n€E│n€Natural number} with │S│< ∞ then there exists N>0 , s.t. a_n = constant€E , for all n≧N (如果把數列看成集合後 只有有限個元素 則在第N項後必定每項都一樣) so a_n converges to this constant€E , satisfying (2) Case2.:│S│= ∞ , from (1) ,we know S has a limit point in E , denoted by p so we can pick up a subsequence of a_n , denoted by a_n_j s.t. a_n_j converges to p€E , satisfying (2) (去驗證確實可挑出,方法:因為他是limit pt.所以你不論半徑取多小 圓圈裡面必有無限多個點,意思就是你能夠找到越來越大的下標 若否,則必存在一個圓圈使得裡面只有有限個點,與limit pt.矛盾) (2) => (1): Let S is an infinite set of E , i.e. │S│= ∞ so we can pick up a sequence from S , denoted by a_n s.t. a_n ≠ a_m , if n≠m from (2) , a_n has a convergent subsequence , denoted by a_n_j which converges to p€E Since a_n ≠ a_m , if n≠m , so a_n_j ≠ a_n_i , if j≠i so p€E is the limit pt. ------------------------------- (2)就是sequenctially compact的定義 還有這題好像不需要R^k?? 因為沒用到Bolzano-Weierstrass 感覺 E is a subset of a metric space 皆可 -- ※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 114.25.182.41

11/25 08:50, , 1F
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11/25 08:50, 1F
文章代碼(AID): #1EpfHKA1 (Math)
文章代碼(AID): #1EpfHKA1 (Math)