# Noetherian topological space

 $Y_{1}\supseteq Y_{2}\supseteq\cdots$

of closed subsets $Y_{i}$ of $X$, there is an integer $m$ such that $Y_{m}=Y_{m+1}=\cdots$.

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

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 $\mathbb{A}^{n}_{k}$ (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 $\mathbb{A}^{n}_{k}$, 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 $m$ 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 $V(I(Y_{i}))=Y_{i}$ for all $i$. Hence $Y_{m}=Y_{m+1}=\cdots$ as required.

Title Noetherian topological space NoetherianTopologicalSpace 2013-03-22 13:03:33 2013-03-22 13:03:33 mathcam (2727) mathcam (2727) 18 mathcam (2727) Definition msc 14A10 Compact