Scharze space asked in 科學數學 · 1 decade ago





3 Answers

  • L
    Lv 7
    1 decade ago
    Favorite Answer

    Claim :

    Every sequence in T has a convergent subsequence with limit in T.

    pf of claim :

    Let {x_n} be a sequence in T which has no convergent subsequence, then for each x in T, there exist B(x,r_x) so that B(x,r_x) contains finitely many elements of {x_n} (otherwise, x will be a limit of some subsequence of {x_n}). The union of all B(x,r_x) convers T, since T is compact, there are finitely many B(x,r_x) so that their union convers T, this implies {x_n} is finite, contradiction. The claim is true.

    Since a Cauchy sequence which has a convergent subsequence must converge and the limit as same as the limt of its subsequence.

    From above, the result follows.

  • 1 decade ago


  • 1 decade ago

    這題不是難不難的問題,而是太基本了,應該是當成數學的常識,沒證明過的人可以自己去證明,證明相當直觀直接,也可以通過先建立 metric space 的 sequentially compactness 概念來理解。


    (1) completeness 不是拓璞性質,卻由於空間有度量,變成 compactness 的一個必然結論,這本身相當有趣。

    (2) 可以證明在 metric space, compact 等價於 complete 及 totally bounded

    推廣了歐氏空間內,compact 等價於 closed and bounded 的結果。

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