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 quasi-compact.

• 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 ${𝔸}_k^n$ (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 ${𝔸}_k^n$, we know that if$Y_1\supseteq Y_2\supseteq \mathrm{\cdots }$ is a descending chain of Zariski-closed 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 one-to-one correspondence between radical ideals of $k[x_1,\mathrm{\dots },x_n]$ and Zariski-closed sets in ${𝔸}_k^n$, we have $V(I(Y_i))=Y_i$ for all $i$. Hence$Y_m=Y_{m+1}=\mathrm{\cdots }$ as required.