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### 3 Answers

- LLv 71 decade agoFavorite 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.

- CopestoneLv 41 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。

小小批一下：

(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。