Re: [分析] 一題點拓
※ 引述《honsan (honsan)》之銘言：
: Let ordered pair (X,d) be a metric space, and Sr(x) is an open ball,
: proof: diam(Sr(x)) <= 2r. diam(S) = sup{d(x,y)|x,y belong to S}
: This is my provement:
: Let y, z belongs to Sr(x), then d(y,x) < r and d(x,z) < r
: -> d(y,z) <= d(y,x)+d(x,z) < 2r (triangle inequality)
: so every element in {d(y,z)|y,z belong to Sr(x)} is smaller than 2r...
: 我證明到這裡就卡住了，請問要怎麼樣才能推到結論呢?
suppose diam>2r
say diam=2r+a a>0
then 2r+a>=d(y,z)>2r+a/2 for some y,z in Sr(x)
contradiction
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※ 編輯: linshihhua 來自: 36.233.106.222 (07/31 02:53)
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