Cici Lin 發問時間： 科學數學 · 1 0 年前

# 請問厲害的大大”高微”問題

Let ｛Xn｝ be sequence in R.

(a) Xn is bounded below, liminf｛Xn｝= a iff (i) for allε>0, exist N s.t. a-ε<Xn, n≧N. (ii) for all ε>0, for all M, exist n≧M s.t. Xn<a+ε.

(b) Xn is bounded above, limsup｛Xn｝= b iff (i) for all ε>0, exist N s.t. Xn<b+ε, n≧N. (ii) for all ε>0, for all M, exist n≧M s.t. b-ε<Xn.

### 2 個解答

• prime
Lv 4
1 0 年前
最佳解答

Let An={Xn,Xn+1,....}.

(a)

For a given ε>0, there exist N1 s.t. a-ε<Xn for all n≧N1. ( by (1))

We have a-ε is a lower bound of A_k for all k≧N1.

It implies a-ε ≦ inf A_k for all k≧N1. ----- (*)

For each k≧N1, there exists an Nk≧ j s.t. X_(Nk)<a+ε.(by (2))

Then inf Ak ≦ X_(Nk) <a+ε for all k≧N1. ----(**)

Combiming (*) and (**), a-ε≦ inf A_k < a+ε for all k≧N1.

Thus, | inf A_k - a | ≦ ε for all k≧N1 .ie. liminf Xn = a.

(b)

For a given ε>0, there exist N1 s.t. b+ε>Xn for all n≧N1. ( by (1))

We have b+ε is a upper bound of A_k for all k≧N1.

It implies b+ε≧ inf A_k for all k≧N1. ----- (*)

For each k≧N1, there exists an Nk≧ j s.t. X_(Nk)>b-ε.(by (2))

Then b-ε< X_(Nk)≦ sup Ak for all k≧N1. ----(**)

Combiming (*) and (**), b-ε< sup A_k ≦ b+ε for all k≧N1.

Thus, | sup A_k - b | ≦ ε for all k≧N1 .ie. limsup Xn = b.

2006-11-26 01:35:22 補充：

(b)It implies b+ε≧ sup A_k for all k≧N1. ----- (*)

2006-11-26 20:18:12 補充：

lim sup的定義不是令 Rn = sup{X_n, X_(n+1), X_(n+2)............}limsup Xn = lim Rn我這邊是假設 Let An={Xn,Xn+1,....}.則 Rn = sup An直接去處理集合An會比較好證明....

2006-11-26 20:20:43 補充：

對不起打錯了, 那應該是k.

2006-11-26 20:23:09 補充：

For each k≧N1, there exists an Nk≧ k s.t. X_(Nk)<a+ε.(by (2))Then inf Ak ≦ X_(Nk) <a+ε for all k≧N1. ----(**)Combiming (*) and (**), a-ε≦ inf A_k < a+ε for all k≧N1.Thus, | inf A_k - a | ≦ ε for all k≧N1 .ie. liminf Xn = a.

2006-11-27 01:53:10 補充：

不是同一個 基本上這兩個證明幾乎完全一樣!你只要把對應數字的不等號反過來,上界改下界,a改成b.證明就轉過去了.其實如果{Xn}滿足(a), 則{-Xn}滿足(b).

• 1 0 年前

請問A_k是sequence Xn的等k項嗎

2006-11-26 20:09:01 補充：

請問~~~你解答裡的""j""是什麼？

2006-11-27 00:45:39 補充：

請問(a)中的N1與(b)中的N1是一樣的嗎