PlanetMath: Noetherian topological space

(more info)

Math for the people, by the people.

Main Menu

sections Encyclopædia
Papers
Books
Expositions

meta Requests (200)
Orphanage
Unclass'd
Unproven (490)
Corrections (6)
Classification

talkback Polls
Forums
Feedback
Bug Reports

downloads Snapshots
PM Book

information News
Docs
Wiki
ChangeLog
TODO List
Legalese
About

Entry average rating: No information on entry rating

Noetherian topological space

(Definition)

A topological space is called Noetherian if it satisfies the descending chain condition for closed subsets: for any sequence

$\displaystyle Y_1 \supseteq Y_2 \supseteq \cdots$

of closed subsets

, there is an integer

such that $Y_m=Y_{m+1}=\cdots$ .

As a first example, note that all finite topological spaces are Noetherian.

There is a lot of interplay between the Noetherian condition and compactness:

Every Noetherian topological space is quasi-compact.
A Hausdorff topological space is Noetherian if and only if every subspace of is compact. (i.e. is hereditarily compact)

Note that if is a Noetherian ring, then Spec, the prime spectrum of , is a Noetherian topological space.

Example of a Noetherian topological space:
The space $\mathbb{A}^n_k$ (affine -space over a field ) under the Zariski topology is an example of a Noetherian topological space. By properties of the ideal of a subset of $\mathbb{A}^n_k$ , we know that if $Y_1 \supseteq Y_2 \supseteq \cdots$ is a descending chain of Zariski-closed subsets, then $I(Y_1) \subseteq I(Y_2) \subseteq \cdots$ is an ascending chain of ideals of $k[x_1,\ldots,x_n]$ .

Since $k[x_1,\ldots,x_n]$ is a Noetherian ring, there exists an integer such that $I(Y_m)=I(Y_{m+1})=\cdots$ . But because we have a one-to-one correspondence between radical ideals of $k[x_1,\ldots,x_n]$ and Zariski-closed sets in $\mathbb{A}^n_k$ , we have for all . Hence $Y_m=Y_{m+1}=\cdots$ as required.