Copestone 發問時間： 科學數學 · 1 0 年前

# 〔拓璞〕Noetherian 空間之刻劃

### 3 個解答

• 1 0 年前
最佳解答

1. Suppose every open set in X is compact, and {G_k} is a sequence of ascending open sets.

Let G= ⋃ G_k. Then G is open and {G_k} is an open covering of G.

By compactness of G, G_n = G for some n.

Therefore G_k=G_n=G for all k>=n.

Thus X is Neotherian.

2. Suppose X is Neotherian, G is an open in X, and {H_k} is an open covering of G.

Let G_k = ⋃ {H_1, H_2, ... H_k}.

Then {G_k} is an open covering of G, and it is a sequence of ascending open sets. Since X is Neotherian, there exist an integer n such that G_k=G_n for all k>=n, hence G_n=G, and {H_1, ... H_n} is a finite subcovering. Therefore G is compact.

2007-05-17 00:50:34 補充：

修正一下 2. 因為 open covering 不一定 countable, 所以應該用反證法比較好：

2007-05-17 00:51:58 補充：

Suppose X is Neotherian, G is an open but non-compact subset in X. There exits and open covering {H_α} of G that has no finite

subcovering.

2007-05-17 00:52:54 補充：

We *can* choose a sequence {H_k} such that, by letting G_k = ⋃ {H_1, H_2, ... H_k, G_k is an *strictly* ascending sequence of open sets. Contradiction to X being Neotherian.

2007-05-17 11:45:07 補充：

補充不能一次寫太多。你不是說大家都是有素養的人，重點寫一寫就好了嗎？

2007-05-17 11:52:04 補充：

Pick H_{k+1} out of {H_α} in such a way that

(i) H_{k+1} is not a proper subset of ⋃ {G_k}

(ii) G_{k+1} = H_{k+1} ⋃ G_k is not an open covering of G.

This is possible bacause {H_α} has no finite subcovering of G.

• Eric
Lv 6
1 0 年前

Noetherian 空間的例子：

On any set X, we can define the topology

T = {A 包含於 X: X\A is finite or A is empty},

which turns X into a Noetherian space.

• 1 0 年前

你「only if」部份要把它寫完麼？can select ．．．那也太不嚴謹了吧。