L
Lv 7
L asked in 科學數學 · 9 years ago

# [高微] 連通性相關證明

R^n 中的子集 A 為連通 if and only if

1) U∩A 和 V∩A 都非空集合

2) (U∩A)∪(V∩A) = A

3) (U∩A)∩(V∩A) = empty set

Rating
• Sam
Lv 6
9 years ago

試以最下方之連通集的定義證明

R^n 中的子集 A 為連通 if and only if

每一個 A 上的實數值連續函數 f 皆將 A 映至一區間

連通集定義：若 R^n 中的 A 是連通集，則不存在 R^n 中的開集 U、V 使得

1) U∩A 和 V∩A 都非空集合

2) (U∩A)∪(V∩A) = A

3) (U∩A)∩(V∩A) = empty set[[Pf]]用W表示無窮大00。假設F(A)不是區間，則存在一個實數L，使 (-W，L) ∩F(A) 和(L，W) ∩F(A)都非空。令U=F^(-1)((-W，L))，V= F^(-1)( (L，W))。則 (0) U，V 為連續函數F在開集(-W，L)，(L，W)之逆像，所以為開集，不相交，且U∪V=(R^n-F(-1)(L))包含A。顯然滿足:1) U∩A 和 V∩A 都非空集合

2) (U∩A)∪(V∩A) = A

3) (U∩A)∩(V∩A) = empty set因此 A不是連通集，矛盾。所以F(A)必須為區間。[[DONE]]

2012-03-22 16:17:27 補充：

[[ pf of (only if) ]]

[[i think the proof is tedious, but not so difficult.]]

Preparation :

(*) If A is a subset of Rn(=R^n), and x is an element of Rn, define :

fA(x)=inf{d(x,y):y in A},fA is called the distance from the point x to the set A. And we know that

2012-03-22 16:17:54 補充：

.(1) fA is a continuous function from Rn to R.

.(2) fA(x)=0, if x is in A_bar;and fA(x) > 0 if x is outside of A_bar, where A_bar is the closure of A.

{in the following we use *_bar for closure of *.

{the proof is routine, omitted.}

2012-03-22 16:18:37 補充：

(**)If U, V are open subsets of Rn such that

1) U∩A 和 V∩A 都非空集合

2) (U∩A)∪(V∩A) = A

3) (U∩A)∩(V∩A) = empty set.

Then

If x is in U∩A = > x is not in (V∩A)_bar. …(*0)

2012-03-22 16:19:09 補充：

[pf]

Assume x in (V∩A)_bar,

= > there is a sequence {xn} with element in (V∩A), i.e. xn in V and xn in A, and xn - > x.

Since x is in U∩A,= > x is in U, and since U is open and xn - > x, = > there is xm for so large m such that xm is in U.

= > xm is in U∩A,

= > xm is in (U∩A) ∩(V∩A) = empty set.

2012-03-22 16:19:56 補充：

x is not in (V∩A)_bar.

[[end of pf]]

Similarly, x is in V∩A = > x is not in (U∩A)_bar.

2012-03-22 16:20:30 補充：

Now consider:

f1=f(U∩A)(x), f2=f(V∩A)(x)

and f(x)= f(U∩A)(x)-f(V∩A)(x)=f1(x) - f2(x).

we know that:

.(1)f1, f2, and f are all continuous function from Rn to R,

2012-03-22 16:22:05 補充：

.(2)By Preparation(*) and (*0), if x in (U∩A) then f1(x)=0, and f2(x) > 0, and = > f(x) <0.

Similarly, if x in (V∩A) then f1(x)>0, and f2(x) = 0, and = > f(x)>0.

And for x in A=(U∩A)∪(V∩A), x in (U∩A) or (V∩A), hence f(x) is not 0. …(*1)

2012-03-22 16:22:39 補充：

.(3) since U∩A and V∩A are not empty set, there are u in U∩A and v in V∩A such that

f(u)<0 and f(v)>0. …(*2)

From (*1) and (*2) we have f(A) is not an interval.

2012-03-22 16:23:02 補充：

Until now,we have proved that if A is not connected, we can construct a continuous function f from Rn to R such that f(A) is not an interval, i.e. the (only if ) part.

The proof is complete.

[[DONE]]

• L
Lv 7
9 years ago

3q, (<=) 的方向沒問題. (=>) 的方向感覺比較難, 若 A 不連通, 要找一個 A 上的連續函數使得 f(A) 不是區間.