Lv 6

# 高等微積分

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

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• Anonymous

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

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.