Noetherian topological space
A topological space^{} $X$ is called if it satisfies the descending chain condition^{} for closed subsets: for any sequence^{}
$${Y}_{1}\supseteq {Y}_{2}\supseteq \mathrm{\cdots}$$ 
of closed subsets ${Y}_{i}$ of $X$, there is an integer $m$ such that ${Y}_{m}={Y}_{m+1}=\mathrm{\cdots}$.
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 quasicompact.

•
A Hausdorff topological space $X$ is Noetherian if and only if every subspace^{} of $X$ is compact^{}. (i.e. $X$ is hereditarily compact)
Note that if $R$ is a Noetherian ring, then $\text{Spec}(R)$, the prime spectrum of $R$, is a Noetherian topological space.
Example of a Noetherian topological space:
The space ${\mathbb{A}}_{k}^{n}$ (affine $n$space over a field $k$) under the Zariski topology^{} is an example of a Noetherian topological space. By properties of the ideal of a subset of ${\mathbb{A}}_{k}^{n}$, we know that if
${Y}_{1}\supseteq {Y}_{2}\supseteq \mathrm{\cdots}$ is a descending chain of Zariskiclosed subsets, then $I({Y}_{1})\subseteq I({Y}_{2})\subseteq \mathrm{\cdots}$ is an ascending chain of ideals of $k[{x}_{1},\mathrm{\dots},{x}_{n}]$.
Since $k[{x}_{1},\mathrm{\dots},{x}_{n}]$ is a Noetherian ring, there exists an integer $m$ such that $I({Y}_{m})=I({Y}_{m+1})=\mathrm{\cdots}$. But because we have a onetoone correspondence between radical ideals of $k[{x}_{1},\mathrm{\dots},{x}_{n}]$ and Zariskiclosed sets in ${\mathbb{A}}_{k}^{n}$, we have $V(I({Y}_{i}))={Y}_{i}$ for all $i$. Hence ${Y}_{m}={Y}_{m+1}=\mathrm{\cdots}$ as required.
Title  Noetherian topological space 

Canonical name  NoetherianTopologicalSpace 
Date of creation  20130322 13:03:33 
Last modified on  20130322 13:03:33 
Owner  mathcam (2727) 
Last modified by  mathcam (2727) 
Numerical id  18 
Author  mathcam (2727) 
Entry type  Definition 
Classification  msc 14A10 
Related topic  Compact 