高微,上下界的問題
For any x=(x1,x2) in R^2, we denote ││x││=(x1^2+x2^2)^1/2.
Let A be a compact subset in R^2 and x0 不屬於A.
(a)Prove that dist(x0,A)=inf{││x-y││:y屬於A} > 0.
(b)Prove that there is a point y0屬於A so that ││x0-y0││=dist(x0,A).
1 個解答
- LLv 71 0 年前最佳解答
本題只需 A 為 closed 即可。
(a)
Since A is closed
=> there is d>0 so that the intersection of B_d(x_0) and A = empty set
=> ∥y - x_0∥ ≧ d for any y in A (Since the space is |R^2)
=> d(x_0,A) = inf{∥y - x_0∥: y in A} ≧ d
(b)
Put d(x_0,A) = d and define
F_n = the intersection of (B_(d + 1/n))(x_0) and A. Clearly,
1) Each F_n is nonempty and is contained in A
2) F_n↓
Hence the Cantor intersection theorem implies ∩F_n is nonempty. Choose y in ∩F_n (i.e. y in A), then
d ≦∥y - x_0∥< d + 1/n for each n
Hence∥y - x_0∥= d
2007-02-05 02:04:15 補充:
∥ ∥ 是 norm
≧ : 大於等於