青ㄟ
Lv 6
青ㄟ asked in 科學數學 · 1 decade ago

高等微積分

Find the closure of A={(x,y)€ R^2,x>y^2}

2 Answers

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  • Anonymous
    1 decade ago
    Favorite Answer

    Answer: {(x,y)€ R^2: x>= y^2}.

    proof:

    Let A={(x,y)€ R^2: x>y^2}, B={(x,y)€ R^2: x = y^2 },

    C={(x,y)€ R^2: x>= y^2}.

    We claim that the closure of A is C.

    Let p = (x,y) be a point in C, then every neighborhood of p contains a

    point q different from p, such that q in A. So every point of C is a limit

    point (cluster point ) of A. By definition of clousre, the closure of A is

    the union of A and all its limit points. Therefore, the closure of A is C.

    p.s.

    How do you find the set of all limit points of A?

    At first A is open, so every point of A is of course

    a limit point of A.

    Secondly, the point in the boundary is also a limit point of A

    However, the point in the complement of C forms a open set , so every

    point lying this set is not a limit point of A.

    Hence, the set of all limit points of A is C, i.e., the union of A and B.

    So, you just add the boundary B of A to A, you would get the closure of

    A , the smallest closed set containing A.

    2007-12-14 12:30:39 補充:

    如何判斷limit point?

    要判斷平面上任意一點 P是否為集合A的 limit point, 就以P點為圓心

    劃一個圓, 不管此圓的半徑有多小, 此圓的內部除去圓心的部分至少

    要與A集合有一交點, 如果是這種情形的話, P點就是集合A的 limit

    point 了!

  • 進哥
    Lv 7
    1 decade ago

    應該就是把不等號改成大於等於而已吧,

    高微,好久以前的東西了,現在記憶中剩不到1/10.

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