[分析] 一題點拓
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?
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