實數系的性質（一）

Question

假設在實數系中,{xn}為一數列試證:數列{xn}收斂的充分必要條件是:該數列是柯西數列

如果用Bolzano-Weierstrass Theorem去證要如何寫？

Answer 1

Assume that x_n -> L as n -> oo, then
for any e > 0, there is N > 0 such that |x_n - L| < e/2 whenever n ≧ N
thus |x_n - x_m| ≦ |x_n - L| + |L - x_m| < e whenever n,m ≧ N
hence x_n is Cauchy.

If x_n is Cauchy, then |x_n| ≦ 1 + |x_N| for some N and for all n ≧ N
thus |x_i| ≦ Max{|x_1| , ... , |x_(n-1)| , 1 + |x_N|} for all i
hence x_n is bounded.

Now, we consider such m such that x_n ≦ x_m for n ≧ m
1) if there are infinitely many m, then {x_n} has a decreasing subsequence.
2) if there are finitely many m, then we can choose a increasing subsequence.
hence {x_n} has a monotone subsequence.

So {x_n} has a bounded and monotone subsequence, call x_n(k).
x_n(k) converges, let L be a limit value of x_n(k) as k -> oo
for any e>0, we can choos K such that

|x_n(k) - L| < e/2 and |x_n - x_n(K)| < e/2 whenever k ≧ K and n ≧ n(K)

thus |x_n - L| ≦ |x_n - x_n(K)| + |x_n(K) - L| < e whenever n ≧ n(K)
hence x_n -> L as n -> oo

其實就是實數完備性的證明在繞來繞去, 中間用 Bolzano-Weierstrass 定理去說明也可以.

2006-08-31 20:44:35 補充：
因為 {x_n} 是 Cauchy 且有界考慮 S = {x_1 , x_2 , ...}, 若 S 是有限集, 則 x_n 有一收斂子列. 若 S 是無限集則由 Bolzano-Weierstrass 定理可知 S 存在 limit point, 故必然存在一收斂子列, 然後就可以跳到下面的證明了 QQ 其實也可以拿中間那段來證 B-W 定理 ...

Answer 2

這個網頁裡有詳細的定義和證明：

http://www.gap-system.org/~john/analysis/Lectures/L10.html

Answer 3

直接用定義去證!

2006-08-21 10:12:34 補充：
如果直接用完備性性質,那就沒什麼好證了

Answer 4

是否可以由完備性(complete)來探討呢??

因為完備性的定義是 每個柯西數列都收斂

我想到的~~不知道對不對?!