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# 高等微積分│定理證明│很急很急│拜託幫幫忙│感謝感恩│２０點

1. A為閉集⇔A包含其所有聚點

2. cl(A)=A∪accu(A)

3. x∈cl(A) ⇔inf{d(y,x)│y∈A}=0

4. 設{Xn}為Cauchy序列，則 (1){Xn}為有界 (2)若存在子序列Xnk→X⇒Xn→X

≠ ＜ ＞ ≤ ≥ | | ! ~ ⇒

⇔ → ↔ ∧ ∨ ⊕ ∀ ∃

≡ ∈ ⊆ ⊂ ⊇ ⊃ ∪ ∩ \ ∞

π || || ∑ ∏ ∫ ∇ ∂ ⊥ ≈ √

∵ ∴ ε ω ρ σ α β γ δ

### 1 Answer

- Scharze spaceLv 71 decade agoFavorite Answer
1.(=>)Suppose A is closed,then M\A is open(M denoted the metric space),Now if x€M\A,there is ε＞０ such that B(x,ε)∩A=空集合

so x is not an accmulation point of A

so A contains all its accumulation points

(<=)Suppose A contains all its accmulation points,if x €M\A,x不屬於A and x is not accmulation point of A

so there is ε＞０,such that B(x,ε)∩A=空集合

＝＞B(x,ε) contained in M\A

=>M\A is open

=>A is closed

2Let B=A∪accu(A),Note that for any C is closed set containing A must be contains B

since if x€B,then x€A or x is an accmulation point of A,x€A=>x€C

x is an accmulation point of A,for ε＞０,the open set B(x,ε)∩A\{x} is not empty

and B(x,ε)∩C\{x} is also not empty,x is accmulation point of C,C is closed

=>x€C

Hence if B is closed,then it must the smallest closed set containing A

and b=cl(A)

so it suffices to show that B is closed

Consider y is an accmulation point of B,for ε＞０,B(y,ε) contains other point of B,say z,then z€A or z is an accmulation point of A

In the latter case,choose δ＝ε-d(y,z),then B(z,ε-d(y,z)) contains other point of A,say z',we have z'≠y,if not ,then z' can not be belong to B,a contracdits our hypothesis

so z'≠y,and d(y,z')<=d(y,z)+d(z,z')<ε

=>y is an accmulation point of A

=>y€B

2008-01-17 17:33:31 補充：

第三題的證明 你要先知道以下這個事實:

http://tw.knowledge.yahoo.com/question/question?qi...

第一題的證明

3(only if) Let x€cl(A),by definition,for any η＞０,choose y€A,such that d(x,y)<η

this is implies that inf{d(x,y)|y€A}=0

(if) Suppose that inf{d(x,y)|y€A}=0

given anyη＞０,choose y€A,such that d(x,y)<0+η＝η

=>x€cl(A)

Source(s): 剩下的明天再寫