[分析] 點集拓墣

[znmkhxrw (QQ)]

我已經證下列這件事：
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M is metric space, A is a nonempty subset of M , a€A

if

boundary of A (denoted by bd(A)) is nonempty

then

d(a,bd(A)) = inf{d(a,x),x€bd(A)} > 0

iff

a€int(A) (there exists r > 0 , s.t. D(a:r) is a subset of A)
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想證明兩件事(A)、(B) (條件同上)

(A)：if d(a,bd(A)) > 0

     then (1) r <= d(a,bd(A))

          (2) r can be d(a,bd(A))

          (3) r cannot be bigger than d(a,bd(A))

(B)：if a€int(A)

     then (1) r <= d(a,bd(A))

          (2) r can be d(a,bd(A))

          (3) r cannot be bigger than d(a,bd(A))

已經想很久了..不知道怎麼估，在R^n空間畫圖蠻顯然的...

不知道是不是因為有反例才證不出來??

感謝幫忙！

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◆ From: 1.171.11.61
→ znmkhxrw    :不好意思我造到反例了 是造A與M都是不連通集才得到
→ jetzake     :有說A是開集或閉集嗎?? r可以伸到邊界上不太自然呢
推 silvermare  :for all x€D(a:d(a,bd(A))) let r'=d(a,x)<d(a,bd(A
→ silvermare  :then x€D(a:(r+r')/2) which is a subset of A
→ silvermare  :so x€A, and hence D(a:d(a,bd(A)))€A
推 silvermare  :確認一下 D(a:r)是指 { x | d(a,x)<r } 沒錯吧
推 silvermare  :上面這段要在嚴謹一點的話 要證明D(a:r)包含於A
→ silvermare  :for all r < d(a,bd(A))
反例是：M = (-inf,-1) U {0} U (1,+inf)

        A = [-3,-2] U {0} U [2,3]

        bd(A) = {-3,-2,2,3}

        for a = 0 € A , a€int(A)

        and d(0,bd(A)) = 2

        but D(0;d(0,bd(A))) = D(0;2) = (-2,-1) U {0} U (1,2)

            is not a subset of A
※ 編輯: znmkhxrw        來自: 140.114.81.83        (10/03 13:18)
推 silvermare  :婀..你的boundary定義是甚麼? bd(A)應該有0吧
推 silvermare  :阿 你是對的