L
Lv 7
L asked in 科學數學 · 9 years ago

[高微] 連通性相關證明

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

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2 Answers

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  • Sam
    Lv 6
    9 years ago
    Favorite Answer

    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:

    This is a Contradiction, so

    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.

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