Re: [分析] 證明子集交集為零 in Metric Space

[FANggot (憨吉)]

※ 引述《weibo (打混鬼)》之銘言：
:   Q: (X,d) be a metric space. Prove that F1 and F2 are two disjoint
:       closed sets in X, then exists two open sets in X, O1)F1, and O2)F2,
:       such taht O1 interaction O2 = empty
:     請大家幫幫忙了！！我想了好久都不知道怎麼下筆阿！！
:     謝謝囉!!



我的想法是直接建構


Given that F1 and F2 are two disjoint closed sets in X,

we know that any p in F1 does not lie in the closure of F2.


Then there exists r>0 such that N (p) does not intersect F2.
                                 r

(here N (p) denotes the open neighborhood of p with radius r)
       r


Not let L be the set of open neighborhoods mentioned above, with radius r/2

i.e. L={ N   (p) | p in F1 and N (p) does not intersect F2}
          r/2                   r


and let M={ N   (p) | p in F2 and N (p) does not intersect F1}
             r/2                   r

Then the Union of L is an open cover of F1, the Union of M is an

open cover of F2, and L and M are disjoint.

Thus we have proved their existence.


以上是我的想法

有錯請指正





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◆ From: 165.124.140.16
推 math1209    :請問由哪知:L and M are disjoint?
→ math1209    :我的顧慮是: L 中的球 與 M 中的球或許有相交.
→ FANggot     :所以才會用r/2 不然球會相交
推 math1209    :還是有可能會交在一起. 畢竟存在 d(F_1,F_2) = 0.

讓我講述一下我的想法&疑問

If p is in f1, and f1 and f2 are disjoint closed sets,

then f1 and f2 are seperated.

Which inplies that d(p,f2) = inf{d(p,y):y in f2}=/=0

http://ppt.cc/982y (wiki連結 該條目第四行說 d(p,f2)=0 if and only if
                    p belongs to the closure of f2).

Because we know p does not lie in the closure of f2 (by the fact that
f1 and f2 are seperated), we know d(p,f2)=/=0


所以就每一點p in f1而言,

都可以有一個0<r<inf{d(p,y):y in f2}=/=0 讓N (p) 與f2不相交
                                           r




以下是證明L,M does not intersect:

If some q lies in the intersection of L and M,

then q is in some N      (p1), p1 in f1,
                   (r1)/2

and q is also in some N      (p2), p2 in f2.
                       (r2)/2

We know d(p1,p2) > (r1) and d(p1,p2) > (r2), (this implies d(p1,p2)>(r1+r2)/2)

However d(q,p1) < (r1)/2, d(q,p2) < (r2)/2,

so d(p1,p2) < d(p1,q) + d(q,p2) < (r1 + r2)/2
            =

which is a contradiction.




但是我的疑問是 d(f1,f2)=0

我想不透為什麼這樣就表示球會相交

有點搞不懂

若是定義
d(f1,f2) = inf{d(x,y) : x in f1 and y in f2}

then by definition of f1 and f2,

0 is NOT a member of {d(x,y): x in f1 and y in f2}

(assume 0 is a member of {d(x,y): x in f1 and y in f2}

then some d(x,y)=0, but that is equivalent to saying x=y, which would

contradict to the fact that f1 and f2 are disjoint closed sets )



所以d(x,y)>0 for all x in f1 and y in f2

就算infimum 是0,

那只表示for and d(x,y)>0, there is a d(x',y') such that

0<d(x',y')<d(x,y)

但是這些x',y' 如果x'=x (或是y'=y)

那上面已經證明 0<d(x,f1)<d(x, y) for all y in f2 了
                        =

若是x', y'=/= x, y   那也與我的論點也不相違背

所以我不知道d(f1,f2)=0會怎麼矛盾

可以指出我的錯誤嗎?

※ 編輯: FANggot         來自: 199.74.85.253        (10/23 07:41)
→ FANggot     :請板友幫忙解惑<(_ _)>
※ 編輯: FANggot         來自: 199.74.85.253        (10/23 08:06)
※ 編輯: FANggot         來自: 199.74.85.253        (10/23 08:30)
※ 編輯: FANggot         來自: 199.74.85.253        (10/23 08:34)
※ 編輯: FANggot         來自: 199.74.85.253        (10/23 08:36)
推 math1209    :你可以不用管我說的. 我想知道你證明中...
→ math1209    :We know d(p1,p2) > (r1) and d(p1,p2) > (r2),why?
→ FANggot     :依照我建構的方法 N  (p1)與f2不相交, p2 is in f2
→ FANggot     :                  r1
→ FANggot     :所以 d(p2,p1)>r1   同裡d(p1,p2)>r2
推 math1209    :ok. It is good.