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

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  • L
    Lv 7
    1 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:

    &par; &par; 是 norm

    &gE; : 大於等於

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