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# 高等微積分(點集拓樸)

Prove that :

Every open set on |R can be written as countable disjoint open intervals

### 3 Answers

- ?Lv 71 decade agoFavorite Answer
Let S be open set in |R. For each x in S, there exists r(x)>0 so that (x - r(x) , x + r(x)) is contained in S. Define

a(x) = sup{y : (x,y) is contained in S}

b(x) = inf{y : (y,x) is contained in S}, and

I(x) = (a(x), b(x))

Claim : If I(x)∩I(y)≠empty set, then I(x)＝I(y).

pf of claim :

Since I(x)∪I(y) is contained in S and it contains I(x) and I(y). Hence I(x)∪I(y) = I(x) = I(y). (because I(x) and I(y) is the largest open interval contain x and y, respectively)

since |R is separable, we can write S = ∪_[1,oo] I(x_i) where x_i in S

2007-05-12 14:36:52 補充：

By claim, we can filter some I(x_i), and then the remaining open intervals are the thing we want.

- CopestoneLv 41 decade ago
設 U 為一個 open subset of R。

證明其實很直觀，也有多種方式去寫。例如，可以對任意 U 內之兩點定義一個 equivalence relation, a ~b if and only if a 到 b 整個線段都包含於 U〔自行證明此為 equivalence relation〕。再 argue 每個 equivalence 就是一個 open interval，而 countable 則是顯然的，因為任 open interval 必有一有理點。

2007-05-11 00:11:59 補充：

其他我見過的方法都大同小異，例如 U 內之任一 x，定義兩個數，

a_x = inf{y: (y, x] 包含於 U}

b_x = sup{y: [x, y) 包含於 U}

U 為開集得出 a_x < x < b_x

定義 I_x = (a_x, b_x)，易證 I_x 包含於 U，即 U 為所有 I_x 之聯集。

再證 x,y in U, I_x 與 I_y 的交集非空，則 I_x = I_y 〔也就是前面 equivalence class 的意義一樣〕。至於 countable selection, 其理如上。

2007-05-14 00:58:24 補充：

天猊證明中的 a(x) 及 b(x) 寫反了，不過無傷大雅。他的證明也就是我說的第二個方式。

至於第一個方式，大同小異，那是有版權的，因為那是一篇已發表了的文章：

Labarre, A. E. (1965), Structural Theorem for Open Sets of Real Numbers, American Mathematical Monthly, Vol. 72, No. 10, p. 1114