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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⇔∀{Xn}為A上的序列，且Xn→X∈M⇒X∈A

2. A為cpt⇒A為cld

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### 2 Answers

- ?Lv 71 decade agoFavorite Answer
請問一下 你的A是那一種空間？

2008-01-13 00:48:09 補充：

1.(pf)(=>)Suppose otherwise,that is ,there exist a sequence {x_n} in A

x_n->x,but x不屬於A,so x€M\A

Since x_n->x,for any ε＞０,choose N, such that for n>=N,we have

||x_n-x||<ε,x_n in A,so the open ball B(x,ε) does not contained in M\A

=>M\A is not open

=>A is not closed-----------><------------

so we done

(<=)Let x be an accumnlation point of A,we must show that x€A

Now for any positive integer n,choose x_n in A with x_n≠x

such that ||x_n-x||<1/n

We claim that x_n->x as n->∞:

Given any ε＞０, choose 1/N<ε, then for n>=N

=>||x_n-x||<1/n<=1/N<ε

so x_n->x as n->∞

By hypothesie,x€A

and so A is closed

2.Suppose A is compact ,first prove that A is bounded,for p€A.B(p,k) k=1,2,.... form an open covering of A,By compactness of A,there exist M with A is covering by ∪_(i=1~M)B(p,i)

Hence A is bounded

Next we will show that A is closed,Suppose A is not closed,that is there is an accumnlation point y of A but y is not belong to A

Let r_x=||x-y||,x€A,hence r_x>0,then {B(x,r_x),x€A} form an open covering of A,By compastness of A again,there is N such that ∪(i=1~N)B(xi,r_i) also covers A

Letting r=min{r_i:i=1,...,N} then for x€B(y,r),||x-y||<r<=r_i

=>||y-x_i||<=||y-x||+||x-x_i||

=>||x-x_i||>=||y-x_i||-||y-x||>2r_i-r_i=r_i

=>x is not belong to B(x_i,r_i) for all i=1,...,N

=>B(y,r)∩A is empty

this is contracditis y is an accmulation point of A

so A is closed

2008-01-13 23:30:29 補充：

有一個地方寫錯了：

r_x=||x-y||/2才

- ALSONLv 41 decade ago
題目上沒說，他題目就寫這樣!

The topology of euclidean space

和

Compact and Connected sets

範圍在以上這2章裡面