維正 發問時間: 科學數學 · 1 0 年前

一題度量空間證明 (有關開集合的充要條件)

Prove: A set E is open if and only if E = int E.

(int E 就是 E 的 interior。詳細定義請見下方。)

註:

metric function 和 metric space 的定義:

Suppose that there exists a real-valued function ρ defined for every ordered pair (x, y) of points of the universal set X, having the following properties:

(i) ρ(x, y) ≧ 0, and ρ(x, y) = 0 if and only if x = y.

(ii) ρ(x, y) = ρ(y, x) (symmetry).

(iii) ρ(x, z) ≦ ρ(x, y) + ρ(y, z) (the triangle inequality).

The function ρ is then called a metric function (or a distance function) on X, and the pair (X, ρ) is called a metric space.

open ball 的定義:

For any x∈X, ε>0, the set

B(y, ε) = {y; ρ(x, y) < ε}

is called the open ball with center y and radius ε, or an ε-neighborhood of y.

開集合的定義:

A set E is called an open set if for any y∈E there is a ball B(y, ε) that is contained in E.

interior的定義:

The interior of a set E is the set of all points y in E such that there is an ε-neighborhood of y contained in E. We denote the interior of E by int E.

1 個解答

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  • 快意
    Lv 4
    1 0 年前
    最佳解答

    (→)A set E is open. Claim:E=int E.

    Let y be any point in E. Since E is an open set, there exists a ball B(y, ε) that is contained in E, hence y∈int E. E⊂intE.

    Let y be any point in int E. By definition, there exists some ε-neighborhood B(y, ε) contained in E. y∈B(y, ε)⊂E→y∈E→int E ⊂ E.

    Hence,int E=E.

    (←)E=int E. Claim:E is open.

    Let y ∈E. Since E=int E, hence y ∈int E, and there is an ε-neighborhood B(y, ε) contained in E, it follows that E is open.

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