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# 幾何學證明

prove that{ x屬於Rn : |x-a| < r} is open by considering the function f : Rn -> R

with f(x) = | x-a |

hint: if A 屬於Rn,a function f: A -> Rm is continuous iff for every open set U 屬於 Rm

there is some open V 屬於Rn such that f(U)的反函數 = V交集A

沒有利用提示也沒有關係

### 2 Answers

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- ddLv 62 decades agoFavorite Answer
這很trivial了呀因為 f|A: A -> R with f(x) = | x - a|是 continuous且 f(A) = (0,r) is open in R所以 f-1(f(A)) = A is open in Rn

2005-12-04 18:01:46 補充：

忘了寫 A = { x屬於Rn : |x-a| 小於 r}

- Anonymous2 decades ago
這.....這真的是幾何嗎

怎麼看起來像是不等式加上函數

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