[分析] 一題點拓

[honsan (honsan)]

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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推 yusd24      :taking sup
→ honsan      :What does you mean?
→ THEJOY      :{d(x,y)}這些東西取了sup會有等於2r的情況出現
→ honsan      :意思是說不等式那行有錯?
→ c76068      :sup{d(x,y)| x,y 屬於 Sr(x)}<= 2r
→ c76068      :這樣不就結束了嗎?
推 physicist512:It is easy to see...
→ Lpspace     :trivial....