[分析] 點集拓墣

看板Math作者 (QQ)時間13年前 (2012/10/03 01:38), 編輯推噓5(5推 0噓 5→)
留言10則, 3人參與, 最新討論串1/1
我已經證下列這件事： ------------------------------------------- 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) -------------------------------------------- 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！ -- ※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 1.171.11.61
→
10/03 02:23, , 1F
不好意思我造到反例了 是造A與M都是不連通集才得到
10/03 02:23, 1F
→
10/03 04:57, , 2F
有說A是開集或閉集嗎?? r可以伸到邊界上不太自然呢
10/03 04:57, 2F
推
10/03 09:00, , 3F
for all x€D(a:d(a,bd(A))) let r'=d(a,x)<d(a,bd(A
10/03 09:00, 3F
→
10/03 09:03, , 4F
then x€D(a:(r+r')/2) which is a subset of A
10/03 09:03, 4F
→
10/03 09:04, , 5F
so x€A, and hence D(a:d(a,bd(A)))€A
10/03 09:04, 5F
推
10/03 09:08, , 6F
確認一下 D(a:r)是指 { x | d(a,x)<r } 沒錯吧
10/03 09:08, 6F
推
10/03 09:14, , 7F
上面這段要在嚴謹一點的話 要證明D(a:r)包含於A
10/03 09:14, 7F
→
10/03 09:14, , 8F
for all r < d(a,bd(A))
10/03 09:14, 8F
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
10/03 16:07, , 9F
婀..你的boundary定義是甚麼? bd(A)應該有0吧
10/03 16:07, 9F
推
10/03 16:12, , 10F
阿 你是對的
10/03 16:12, 10F
‣ 更多 znmkhxrw 的文章
文章代碼(AID): #1GQoPwg_ (Math)