推 TampaBayRays:感謝你 11/25 08:50
※ 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.
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(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
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